Integrand size = 33, antiderivative size = 198 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {1}{2} a^4 (2 A+13 C) x+\frac {2 a^4 (3 A+2 C) \text {arctanh}(\sin (c+d x))}{d}-\frac {5 a^4 (2 A-C) \sin (c+d x)}{2 d}-\frac {(22 A+3 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {(8 A+3 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \tan (c+d x)}{3 d}+\frac {2 a A (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{3 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d} \] Output:
1/2*a^4*(2*A+13*C)*x+2*a^4*(3*A+2*C)*arctanh(sin(d*x+c))/d-5/2*a^4*(2*A-C) *sin(d*x+c)/d-1/6*(22*A+3*C)*(a^4+a^4*cos(d*x+c))*sin(d*x+c)/d+1/3*(8*A+3* C)*(a^2+a^2*cos(d*x+c))^2*tan(d*x+c)/d+2/3*a*A*(a+a*cos(d*x+c))^3*sec(d*x+ c)*tan(d*x+c)/d+1/3*A*(a+a*cos(d*x+c))^4*sec(d*x+c)^2*tan(d*x+c)/d
Time = 12.09 (sec) , antiderivative size = 386, normalized size of antiderivative = 1.95 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=a^4 \left (\frac {(2 A+13 C) (c+d x)}{2 d}-\frac {2 (3 A+2 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {2 (3 A+2 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {13 A}{12 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {A \sin \left (\frac {1}{2} (c+d x)\right )}{6 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {A \sin \left (\frac {1}{2} (c+d x)\right )}{6 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}-\frac {13 A}{12 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {20 A \sin \left (\frac {1}{2} (c+d x)\right )+3 C \sin \left (\frac {1}{2} (c+d x)\right )}{3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {20 A \sin \left (\frac {1}{2} (c+d x)\right )+3 C \sin \left (\frac {1}{2} (c+d x)\right )}{3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {4 C \sin (c+d x)}{d}+\frac {C \sin (2 (c+d x))}{4 d}\right ) \] Input:
Integrate[(a + a*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]
Output:
a^4*(((2*A + 13*C)*(c + d*x))/(2*d) - (2*(3*A + 2*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (2*(3*A + 2*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d* x)/2]])/d + (13*A)/(12*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (A*Sin [(c + d*x)/2])/(6*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3) + (A*Sin[(c + d*x)/2])/(6*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3) - (13*A)/(12*d*(Co s[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (20*A*Sin[(c + d*x)/2] + 3*C*Sin[( c + d*x)/2])/(3*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (20*A*Sin[(c + d*x)/2] + 3*C*Sin[(c + d*x)/2])/(3*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) ) + (4*C*Sin[c + d*x])/d + (C*Sin[2*(c + d*x)])/(4*d))
Time = 1.61 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.02, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {3042, 3523, 3042, 3454, 27, 3042, 3454, 3042, 3455, 27, 3042, 3447, 3042, 3502, 3042, 3214, 3042, 4257}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \sec ^4(c+d x) (a \cos (c+d x)+a)^4 \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^4 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 3523 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^4 (4 a A-a (2 A-3 C) \cos (c+d x)) \sec ^3(c+d x)dx}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^4 \left (4 a A-a (2 A-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {\frac {1}{2} \int 2 (\cos (c+d x) a+a)^3 \left (a^2 (8 A+3 C)-3 a^2 (2 A-C) \cos (c+d x)\right ) \sec ^2(c+d x)dx+\frac {2 a^2 A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^3 \left (a^2 (8 A+3 C)-3 a^2 (2 A-C) \cos (c+d x)\right ) \sec ^2(c+d x)dx+\frac {2 a^2 A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (a^2 (8 A+3 C)-3 a^2 (2 A-C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {2 a^2 A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^2 \left (6 a^3 (3 A+2 C)-a^3 (22 A+3 C) \cos (c+d x)\right ) \sec (c+d x)dx+\frac {a^3 (8 A+3 C) \tan (c+d x) (a \cos (c+d x)+a)^2}{d}+\frac {2 a^2 A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (6 a^3 (3 A+2 C)-a^3 (22 A+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a^3 (8 A+3 C) \tan (c+d x) (a \cos (c+d x)+a)^2}{d}+\frac {2 a^2 A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {\frac {1}{2} \int 3 (\cos (c+d x) a+a) \left (4 a^4 (3 A+2 C)-5 a^4 (2 A-C) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {(22 A+3 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{2 d}+\frac {a^3 (8 A+3 C) \tan (c+d x) (a \cos (c+d x)+a)^2}{d}+\frac {2 a^2 A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {3}{2} \int (\cos (c+d x) a+a) \left (4 a^4 (3 A+2 C)-5 a^4 (2 A-C) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {(22 A+3 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{2 d}+\frac {a^3 (8 A+3 C) \tan (c+d x) (a \cos (c+d x)+a)^2}{d}+\frac {2 a^2 A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{2} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (4 a^4 (3 A+2 C)-5 a^4 (2 A-C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {(22 A+3 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{2 d}+\frac {a^3 (8 A+3 C) \tan (c+d x) (a \cos (c+d x)+a)^2}{d}+\frac {2 a^2 A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\frac {3}{2} \int \left (-5 (2 A-C) \cos ^2(c+d x) a^5+4 (3 A+2 C) a^5+\left (4 a^5 (3 A+2 C)-5 a^5 (2 A-C)\right ) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {(22 A+3 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{2 d}+\frac {a^3 (8 A+3 C) \tan (c+d x) (a \cos (c+d x)+a)^2}{d}+\frac {2 a^2 A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{2} \int \frac {-5 (2 A-C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^5+4 (3 A+2 C) a^5+\left (4 a^5 (3 A+2 C)-5 a^5 (2 A-C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {(22 A+3 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{2 d}+\frac {a^3 (8 A+3 C) \tan (c+d x) (a \cos (c+d x)+a)^2}{d}+\frac {2 a^2 A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\frac {3}{2} \left (\int \left (4 (3 A+2 C) a^5+(2 A+13 C) \cos (c+d x) a^5\right ) \sec (c+d x)dx-\frac {5 a^5 (2 A-C) \sin (c+d x)}{d}\right )-\frac {(22 A+3 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{2 d}+\frac {a^3 (8 A+3 C) \tan (c+d x) (a \cos (c+d x)+a)^2}{d}+\frac {2 a^2 A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{2} \left (\int \frac {4 (3 A+2 C) a^5+(2 A+13 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^5}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {5 a^5 (2 A-C) \sin (c+d x)}{d}\right )-\frac {(22 A+3 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{2 d}+\frac {a^3 (8 A+3 C) \tan (c+d x) (a \cos (c+d x)+a)^2}{d}+\frac {2 a^2 A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {\frac {3}{2} \left (4 a^5 (3 A+2 C) \int \sec (c+d x)dx-\frac {5 a^5 (2 A-C) \sin (c+d x)}{d}+a^5 x (2 A+13 C)\right )-\frac {(22 A+3 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{2 d}+\frac {a^3 (8 A+3 C) \tan (c+d x) (a \cos (c+d x)+a)^2}{d}+\frac {2 a^2 A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{2} \left (4 a^5 (3 A+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {5 a^5 (2 A-C) \sin (c+d x)}{d}+a^5 x (2 A+13 C)\right )-\frac {(22 A+3 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{2 d}+\frac {a^3 (8 A+3 C) \tan (c+d x) (a \cos (c+d x)+a)^2}{d}+\frac {2 a^2 A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {3}{2} \left (\frac {4 a^5 (3 A+2 C) \text {arctanh}(\sin (c+d x))}{d}-\frac {5 a^5 (2 A-C) \sin (c+d x)}{d}+a^5 x (2 A+13 C)\right )-\frac {(22 A+3 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{2 d}+\frac {a^3 (8 A+3 C) \tan (c+d x) (a \cos (c+d x)+a)^2}{d}+\frac {2 a^2 A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
Input:
Int[(a + a*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]
Output:
(A*(a + a*Cos[c + d*x])^4*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + (-1/2*((22* A + 3*C)*(a^5 + a^5*Cos[c + d*x])*Sin[c + d*x])/d + (3*(a^5*(2*A + 13*C)*x + (4*a^5*(3*A + 2*C)*ArcTanh[Sin[c + d*x]])/d - (5*a^5*(2*A - C)*Sin[c + d*x])/d))/2 + (a^3*(8*A + 3*C)*(a + a*Cos[c + d*x])^2*Tan[c + d*x])/d + (2 *a^2*A*(a + a*Cos[c + d*x])^3*Sec[c + d*x]*Tan[c + d*x])/d)/(3*a)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-b^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + a*d)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp [a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B *(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0 ])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 ) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1 ] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C + A*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a *d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n + 2) + C* (c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 12.80 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.96
| method | result | size |
| derivativedivides | \(\frac {a^{4} A \left (d x +c \right )+a^{4} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{4} C \sin \left (d x +c \right )+6 a^{4} A \tan \left (d x +c \right )+6 a^{4} C \left (d x +c \right )+4 a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-a^{4} A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+a^{4} C \tan \left (d x +c \right )}{d}\) | \(190\) |
| default | \(\frac {a^{4} A \left (d x +c \right )+a^{4} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{4} C \sin \left (d x +c \right )+6 a^{4} A \tan \left (d x +c \right )+6 a^{4} C \left (d x +c \right )+4 a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-a^{4} A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+a^{4} C \tan \left (d x +c \right )}{d}\) | \(190\) |
| parts | \(-\frac {a^{4} A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (a^{4} A +6 a^{4} C \right ) \left (d x +c \right )}{d}+\frac {\left (4 a^{4} A +4 a^{4} C \right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (6 a^{4} A +a^{4} C \right ) \tan \left (d x +c \right )}{d}+\frac {a^{4} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a^{4} A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {2 a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {4 \sin \left (d x +c \right ) a^{4} C}{d}\) | \(192\) |
| parallelrisch | \(\frac {4 \left (-\frac {9 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (A +\frac {2 C}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}+\frac {9 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (A +\frac {2 C}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}+\frac {\left (A +\frac {13 C}{2}\right ) x d \cos \left (3 d x +3 c \right )}{4}+\left (A +C \right ) \sin \left (2 d x +2 c \right )+\left (\frac {5 A}{3}+\frac {11 C}{32}\right ) \sin \left (3 d x +3 c \right )+\frac {\sin \left (4 d x +4 c \right ) C}{2}+\frac {C \sin \left (5 d x +5 c \right )}{32}+\frac {3 \left (A +\frac {13 C}{2}\right ) x d \cos \left (d x +c \right )}{4}+2 \sin \left (d x +c \right ) \left (A +\frac {5 C}{32}\right )\right ) a^{4}}{d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\) | \(204\) |
| risch | \(a^{4} x A +\frac {13 a^{4} C x}{2}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} a^{4} C}{8 d}-\frac {2 i {\mathrm e}^{i \left (d x +c \right )} a^{4} C}{d}+\frac {2 i {\mathrm e}^{-i \left (d x +c \right )} a^{4} C}{d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} a^{4} C}{8 d}-\frac {2 i a^{4} \left (6 A \,{\mathrm e}^{5 i \left (d x +c \right )}-18 A \,{\mathrm e}^{4 i \left (d x +c \right )}-3 C \,{\mathrm e}^{4 i \left (d x +c \right )}-42 A \,{\mathrm e}^{2 i \left (d x +c \right )}-6 C \,{\mathrm e}^{2 i \left (d x +c \right )}-6 A \,{\mathrm e}^{i \left (d x +c \right )}-20 A -3 C \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {6 a^{4} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {4 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {6 a^{4} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {4 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}\) | \(280\) |
Input:
int((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x,method=_RETURNVER BOSE)
Output:
1/d*(a^4*A*(d*x+c)+a^4*C*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+4*a^4*A *ln(sec(d*x+c)+tan(d*x+c))+4*a^4*C*sin(d*x+c)+6*a^4*A*tan(d*x+c)+6*a^4*C*( d*x+c)+4*a^4*A*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+4 *a^4*C*ln(sec(d*x+c)+tan(d*x+c))-a^4*A*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+ a^4*C*tan(d*x+c))
Time = 0.11 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.86 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {3 \, {\left (2 \, A + 13 \, C\right )} a^{4} d x \cos \left (d x + c\right )^{3} + 6 \, {\left (3 \, A + 2 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 6 \, {\left (3 \, A + 2 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (3 \, C a^{4} \cos \left (d x + c\right )^{4} + 24 \, C a^{4} \cos \left (d x + c\right )^{3} + 2 \, {\left (20 \, A + 3 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 12 \, A a^{4} \cos \left (d x + c\right ) + 2 \, A a^{4}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{3}} \] Input:
integrate((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm= "fricas")
Output:
1/6*(3*(2*A + 13*C)*a^4*d*x*cos(d*x + c)^3 + 6*(3*A + 2*C)*a^4*cos(d*x + c )^3*log(sin(d*x + c) + 1) - 6*(3*A + 2*C)*a^4*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + (3*C*a^4*cos(d*x + c)^4 + 24*C*a^4*cos(d*x + c)^3 + 2*(20*A + 3*C)*a^4*cos(d*x + c)^2 + 12*A*a^4*cos(d*x + c) + 2*A*a^4)*sin(d*x + c))/( d*cos(d*x + c)^3)
Timed out. \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+a*cos(d*x+c))**4*(A+C*cos(d*x+c)**2)*sec(d*x+c)**4,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.07 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 12 \, {\left (d x + c\right )} A a^{4} + 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} + 72 \, {\left (d x + c\right )} C a^{4} - 12 \, A a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, A a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, C a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, C a^{4} \sin \left (d x + c\right ) + 72 \, A a^{4} \tan \left (d x + c\right ) + 12 \, C a^{4} \tan \left (d x + c\right )}{12 \, d} \] Input:
integrate((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm= "maxima")
Output:
1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^4 + 12*(d*x + c)*A*a^4 + 3*( 2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^4 + 72*(d*x + c)*C*a^4 - 12*A*a^4*(2*s in(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c ) - 1)) + 24*A*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 24*C* a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 48*C*a^4*sin(d*x + c ) + 72*A*a^4*tan(d*x + c) + 12*C*a^4*tan(d*x + c))/d
Time = 0.36 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.25 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {3 \, {\left (2 \, A a^{4} + 13 \, C a^{4}\right )} {\left (d x + c\right )} + 12 \, {\left (3 \, A a^{4} + 2 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 12 \, {\left (3 \, A a^{4} + 2 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {6 \, {\left (7 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}} - \frac {4 \, {\left (15 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 38 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 27 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \] Input:
integrate((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm= "giac")
Output:
1/6*(3*(2*A*a^4 + 13*C*a^4)*(d*x + c) + 12*(3*A*a^4 + 2*C*a^4)*log(abs(tan (1/2*d*x + 1/2*c) + 1)) - 12*(3*A*a^4 + 2*C*a^4)*log(abs(tan(1/2*d*x + 1/2 *c) - 1)) + 6*(7*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 9*C*a^4*tan(1/2*d*x + 1/2* c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^2 - 4*(15*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^4*tan(1/2*d*x + 1/2*c)^5 - 38*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 6*C*a^4 *tan(1/2*d*x + 1/2*c)^3 + 27*A*a^4*tan(1/2*d*x + 1/2*c) + 3*C*a^4*tan(1/2* d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d
Time = 0.27 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.27 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {4\,C\,a^4\,\sin \left (c+d\,x\right )}{d}+\frac {2\,A\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {12\,A\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {13\,C\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {8\,C\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {20\,A\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {2\,A\,a^4\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^2}+\frac {A\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3}+\frac {C\,a^4\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {C\,a^4\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d} \] Input:
int(((A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^4)/cos(c + d*x)^4,x)
Output:
(4*C*a^4*sin(c + d*x))/d + (2*A*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x )/2)))/d + (12*A*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (13 *C*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (8*C*a^4*atanh(sin (c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (20*A*a^4*sin(c + d*x))/(3*d*cos( c + d*x)) + (2*A*a^4*sin(c + d*x))/(d*cos(c + d*x)^2) + (A*a^4*sin(c + d*x ))/(3*d*cos(c + d*x)^3) + (C*a^4*sin(c + d*x))/(d*cos(c + d*x)) + (C*a^4*c os(c + d*x)*sin(c + d*x))/(2*d)
Time = 0.24 (sec) , antiderivative size = 412, normalized size of antiderivative = 2.08 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {a^{4} \left (3 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{3} c -3 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right ) c -36 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a -24 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} c +36 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +24 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) c +36 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a +24 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} c -36 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -24 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) c +24 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} c +6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a d x +39 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} c d x -12 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a -24 \cos \left (d x +c \right ) \sin \left (d x +c \right ) c -6 \cos \left (d x +c \right ) a d x -39 \cos \left (d x +c \right ) c d x +40 \sin \left (d x +c \right )^{3} a +6 \sin \left (d x +c \right )^{3} c -42 \sin \left (d x +c \right ) a -6 \sin \left (d x +c \right ) c \right )}{6 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x)
Output:
(a**4*(3*cos(c + d*x)**2*sin(c + d*x)**3*c - 3*cos(c + d*x)**2*sin(c + d*x )*c - 36*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a - 24*cos (c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*c + 36*cos(c + d*x)*lo g(tan((c + d*x)/2) - 1)*a + 24*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*c + 36*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a + 24*cos(c + d *x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*c - 36*cos(c + d*x)*log(tan( (c + d*x)/2) + 1)*a - 24*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*c + 24*cos (c + d*x)*sin(c + d*x)**3*c + 6*cos(c + d*x)*sin(c + d*x)**2*a*d*x + 39*co s(c + d*x)*sin(c + d*x)**2*c*d*x - 12*cos(c + d*x)*sin(c + d*x)*a - 24*cos (c + d*x)*sin(c + d*x)*c - 6*cos(c + d*x)*a*d*x - 39*cos(c + d*x)*c*d*x + 40*sin(c + d*x)**3*a + 6*sin(c + d*x)**3*c - 42*sin(c + d*x)*a - 6*sin(c + d*x)*c))/(6*cos(c + d*x)*d*(sin(c + d*x)**2 - 1))