Integrand size = 33, antiderivative size = 181 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {1}{8} a^4 (52 A+35 C) x+\frac {4 a^4 A \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^4 (4 A+7 C) \sin (c+d x)}{8 d}-\frac {a (4 A-C) (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}-\frac {(12 A-7 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{12 d}-\frac {(12 A-35 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{24 d}+\frac {A (a+a \cos (c+d x))^4 \tan (c+d x)}{d} \] Output:
1/8*a^4*(52*A+35*C)*x+4*a^4*A*arctanh(sin(d*x+c))/d+5/8*a^4*(4*A+7*C)*sin( d*x+c)/d-1/4*a*(4*A-C)*(a+a*cos(d*x+c))^3*sin(d*x+c)/d-1/12*(12*A-7*C)*(a^ 2+a^2*cos(d*x+c))^2*sin(d*x+c)/d-1/24*(12*A-35*C)*(a^4+a^4*cos(d*x+c))*sin (d*x+c)/d+A*(a+a*cos(d*x+c))^4*tan(d*x+c)/d
Time = 8.61 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.87 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {a^4 (1+\cos (c+d x))^4 \sec ^8\left (\frac {1}{2} (c+d x)\right ) \left (12 (52 A+35 C) x-\frac {384 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {384 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {96 (4 A+7 C) \cos (d x) \sin (c)}{d}+\frac {24 (A+7 C) \cos (2 d x) \sin (2 c)}{d}+\frac {32 C \cos (3 d x) \sin (3 c)}{d}+\frac {3 C \cos (4 d x) \sin (4 c)}{d}+\frac {96 (4 A+7 C) \cos (c) \sin (d x)}{d}+\frac {24 (A+7 C) \cos (2 c) \sin (2 d x)}{d}+\frac {32 C \cos (3 c) \sin (3 d x)}{d}+\frac {3 C \cos (4 c) \sin (4 d x)}{d}+\frac {96 A \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {96 A \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )}{1536} \] Input:
Integrate[(a + a*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^2,x]
Output:
(a^4*(1 + Cos[c + d*x])^4*Sec[(c + d*x)/2]^8*(12*(52*A + 35*C)*x - (384*A* Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (384*A*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/d + (96*(4*A + 7*C)*Cos[d*x]*Sin[c])/d + (24*(A + 7*C )*Cos[2*d*x]*Sin[2*c])/d + (32*C*Cos[3*d*x]*Sin[3*c])/d + (3*C*Cos[4*d*x]* Sin[4*c])/d + (96*(4*A + 7*C)*Cos[c]*Sin[d*x])/d + (24*(A + 7*C)*Cos[2*c]* Sin[2*d*x])/d + (32*C*Cos[3*c]*Sin[3*d*x])/d + (3*C*Cos[4*c]*Sin[4*d*x])/d + (96*A*Sin[(d*x)/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (96*A*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x) /2] + Sin[(c + d*x)/2]))))/1536
Time = 1.54 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.09, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.515, Rules used = {3042, 3523, 3042, 3455, 3042, 3455, 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.
| \(\displaystyle \int \sec ^2(c+d x) (a \cos (c+d x)+a)^4 \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^4 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 3523 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^4 (4 a A-a (4 A-C) \cos (c+d x)) \sec (c+d x)dx}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^4}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^4 \left (4 a A-a (4 A-C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^4}{d}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {\frac {1}{4} \int (\cos (c+d x) a+a)^3 \left (16 a^2 A-a^2 (12 A-7 C) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {a^2 (4 A-C) \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^4}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{4} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (16 a^2 A-a^2 (12 A-7 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {a^2 (4 A-C) \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^4}{d}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \int (\cos (c+d x) a+a)^2 \left (48 a^3 A-a^3 (12 A-35 C) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {a^3 (12 A-7 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}\right )-\frac {a^2 (4 A-C) \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^4}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (48 a^3 A-a^3 (12 A-35 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {a^3 (12 A-7 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}\right )-\frac {a^2 (4 A-C) \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^4}{d}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int 3 (\cos (c+d x) a+a) \left (32 A a^4+5 (4 A+7 C) \cos (c+d x) a^4\right ) \sec (c+d x)dx-\frac {(12 A-35 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{2 d}\right )-\frac {a^3 (12 A-7 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}\right )-\frac {a^2 (4 A-C) \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^4}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {3}{2} \int (\cos (c+d x) a+a) \left (32 A a^4+5 (4 A+7 C) \cos (c+d x) a^4\right ) \sec (c+d x)dx-\frac {(12 A-35 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{2 d}\right )-\frac {a^3 (12 A-7 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}\right )-\frac {a^2 (4 A-C) \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^4}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {3}{2} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (32 A a^4+5 (4 A+7 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^4\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {(12 A-35 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{2 d}\right )-\frac {a^3 (12 A-7 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}\right )-\frac {a^2 (4 A-C) \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^4}{d}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {3}{2} \int \left (5 (4 A+7 C) \cos ^2(c+d x) a^5+32 A a^5+\left (32 A a^5+5 (4 A+7 C) a^5\right ) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {(12 A-35 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{2 d}\right )-\frac {a^3 (12 A-7 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}\right )-\frac {a^2 (4 A-C) \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^4}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {3}{2} \int \frac {5 (4 A+7 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^5+32 A a^5+\left (32 A a^5+5 (4 A+7 C) a^5\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {(12 A-35 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{2 d}\right )-\frac {a^3 (12 A-7 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}\right )-\frac {a^2 (4 A-C) \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^4}{d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {3}{2} \left (\int \left (32 A a^5+(52 A+35 C) \cos (c+d x) a^5\right ) \sec (c+d x)dx+\frac {5 a^5 (4 A+7 C) \sin (c+d x)}{d}\right )-\frac {(12 A-35 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{2 d}\right )-\frac {a^3 (12 A-7 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}\right )-\frac {a^2 (4 A-C) \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^4}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {3}{2} \left (\int \frac {32 A a^5+(52 A+35 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^5}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {5 a^5 (4 A+7 C) \sin (c+d x)}{d}\right )-\frac {(12 A-35 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{2 d}\right )-\frac {a^3 (12 A-7 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}\right )-\frac {a^2 (4 A-C) \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^4}{d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {3}{2} \left (32 a^5 A \int \sec (c+d x)dx+\frac {5 a^5 (4 A+7 C) \sin (c+d x)}{d}+a^5 x (52 A+35 C)\right )-\frac {(12 A-35 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{2 d}\right )-\frac {a^3 (12 A-7 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}\right )-\frac {a^2 (4 A-C) \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^4}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {3}{2} \left (32 a^5 A \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {5 a^5 (4 A+7 C) \sin (c+d x)}{d}+a^5 x (52 A+35 C)\right )-\frac {(12 A-35 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{2 d}\right )-\frac {a^3 (12 A-7 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}\right )-\frac {a^2 (4 A-C) \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^4}{d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {3}{2} \left (\frac {32 a^5 A \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^5 (4 A+7 C) \sin (c+d x)}{d}+a^5 x (52 A+35 C)\right )-\frac {(12 A-35 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{2 d}\right )-\frac {a^3 (12 A-7 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}\right )-\frac {a^2 (4 A-C) \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^4}{d}\) |
Input:
Int[(a + a*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^2,x]
Output:
(-1/4*(a^2*(4*A - C)*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/d + (-1/3*(a^3*( 12*A - 7*C)*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/d + (-1/2*((12*A - 35*C)* (a^5 + a^5*Cos[c + d*x])*Sin[c + d*x])/d + (3*(a^5*(52*A + 35*C)*x + (32*a ^5*A*ArcTanh[Sin[c + d*x]])/d + (5*a^5*(4*A + 7*C)*Sin[c + d*x])/d))/2)/3) /4)/a + (A*(a + a*Cos[c + d*x])^4*Tan[c + d*x])/d
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)*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 = 12.33 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.80
| method | result | size |
| parallelrisch | \(\frac {a^{4} \left (-32 \cos \left (d x +c \right ) A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+32 \cos \left (d x +c \right ) A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (16 A +\frac {88 C}{3}\right ) \sin \left (2 d x +2 c \right )+\left (A +\frac {57 C}{8}\right ) \sin \left (3 d x +3 c \right )+\frac {4 \sin \left (4 d x +4 c \right ) C}{3}+\frac {C \sin \left (5 d x +5 c \right )}{8}+52 x \left (A +\frac {35 C}{52}\right ) d \cos \left (d x +c \right )+9 \sin \left (d x +c \right ) \left (A +\frac {7 C}{9}\right )\right )}{8 d \cos \left (d x +c \right )}\) | \(144\) |
| parts | \(\frac {a^{4} A \tan \left (d x +c \right )}{d}+\frac {\left (a^{4} A +6 a^{4} C \right ) \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (4 a^{4} A +4 a^{4} C \right ) \sin \left (d x +c \right )}{d}+\frac {\left (6 a^{4} A +a^{4} C \right ) \left (d x +c \right )}{d}+\frac {a^{4} C \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}+\frac {4 a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {4 a^{4} C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3 d}\) | \(187\) |
| derivativedivides | \(\frac {a^{4} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{4} C \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+4 a^{4} A \sin \left (d x +c \right )+\frac {4 a^{4} C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+6 a^{4} A \left (d x +c \right )+6 a^{4} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{4} C \sin \left (d x +c \right )+a^{4} A \tan \left (d x +c \right )+a^{4} C \left (d x +c \right )}{d}\) | \(197\) |
| default | \(\frac {a^{4} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{4} C \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+4 a^{4} A \sin \left (d x +c \right )+\frac {4 a^{4} C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+6 a^{4} A \left (d x +c \right )+6 a^{4} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{4} C \sin \left (d x +c \right )+a^{4} A \tan \left (d x +c \right )+a^{4} C \left (d x +c \right )}{d}\) | \(197\) |
| risch | \(\frac {13 a^{4} x A}{2}+\frac {35 a^{4} C x}{8}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} a^{4} A}{8 d}-\frac {7 i {\mathrm e}^{2 i \left (d x +c \right )} a^{4} C}{8 d}-\frac {2 i {\mathrm e}^{i \left (d x +c \right )} a^{4} A}{d}-\frac {7 i {\mathrm e}^{i \left (d x +c \right )} a^{4} C}{2 d}+\frac {2 i {\mathrm e}^{-i \left (d x +c \right )} a^{4} A}{d}+\frac {7 i {\mathrm e}^{-i \left (d x +c \right )} a^{4} C}{2 d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} a^{4} A}{8 d}+\frac {7 i {\mathrm e}^{-2 i \left (d x +c \right )} a^{4} C}{8 d}+\frac {2 i a^{4} A}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {4 a^{4} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {4 a^{4} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {\sin \left (4 d x +4 c \right ) a^{4} C}{32 d}+\frac {\sin \left (3 d x +3 c \right ) a^{4} C}{3 d}\) | \(271\) |
| norman | \(\frac {\left (-\frac {13}{2} a^{4} A -\frac {35}{8} a^{4} C \right ) x +\left (-\frac {117}{2} a^{4} A -\frac {315}{8} a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-\frac {65}{2} a^{4} A -\frac {175}{8} a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-\frac {65}{2} a^{4} A -\frac {175}{8} a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (\frac {13}{2} a^{4} A +\frac {35}{8} a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}+\left (\frac {65}{2} a^{4} A +\frac {175}{8} a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (\frac {65}{2} a^{4} A +\frac {175}{8} a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (\frac {117}{2} a^{4} A +\frac {315}{8} a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\frac {5 a^{4} \left (4 A +7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{4 d}-\frac {a^{4} \left (36 A -25 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {a^{4} \left (44 A +93 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a^{4} \left (108 A +245 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{6 d}+\frac {a^{4} \left (132 A +791 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 d}-\frac {a^{4} \left (276 A +395 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}-\frac {a^{4} \left (828 A +617 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}-\frac {4 a^{4} A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {4 a^{4} A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\) | \(453\) |
Input:
int((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^2,x,method=_RETURNVER BOSE)
Output:
1/8*a^4*(-32*cos(d*x+c)*A*ln(tan(1/2*d*x+1/2*c)-1)+32*cos(d*x+c)*A*ln(tan( 1/2*d*x+1/2*c)+1)+(16*A+88/3*C)*sin(2*d*x+2*c)+(A+57/8*C)*sin(3*d*x+3*c)+4 /3*sin(4*d*x+4*c)*C+1/8*C*sin(5*d*x+5*c)+52*x*(A+35/52*C)*d*cos(d*x+c)+9*s in(d*x+c)*(A+7/9*C))/d/cos(d*x+c)
Time = 0.12 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.87 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {3 \, {\left (52 \, A + 35 \, C\right )} a^{4} d x \cos \left (d x + c\right ) + 48 \, A a^{4} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 48 \, A a^{4} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (6 \, C a^{4} \cos \left (d x + c\right )^{4} + 32 \, C a^{4} \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, A + 27 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 32 \, {\left (3 \, A + 5 \, C\right )} a^{4} \cos \left (d x + c\right ) + 24 \, A a^{4}\right )} \sin \left (d x + c\right )}{24 \, d \cos \left (d x + c\right )} \] Input:
integrate((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, algorithm= "fricas")
Output:
1/24*(3*(52*A + 35*C)*a^4*d*x*cos(d*x + c) + 48*A*a^4*cos(d*x + c)*log(sin (d*x + c) + 1) - 48*A*a^4*cos(d*x + c)*log(-sin(d*x + c) + 1) + (6*C*a^4*c os(d*x + c)^4 + 32*C*a^4*cos(d*x + c)^3 + 3*(4*A + 27*C)*a^4*cos(d*x + c)^ 2 + 32*(3*A + 5*C)*a^4*cos(d*x + c) + 24*A*a^4)*sin(d*x + c))/(d*cos(d*x + c))
Timed out. \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+a*cos(d*x+c))**4*(A+C*cos(d*x+c)**2)*sec(d*x+c)**2,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.07 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} + 576 \, {\left (d x + c\right )} A a^{4} - 128 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{4} + 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} + 144 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} + 96 \, {\left (d x + c\right )} C a^{4} + 192 \, A a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 384 \, A a^{4} \sin \left (d x + c\right ) + 384 \, C a^{4} \sin \left (d x + c\right ) + 96 \, A a^{4} \tan \left (d x + c\right )}{96 \, d} \] Input:
integrate((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, algorithm= "maxima")
Output:
1/96*(24*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^4 + 576*(d*x + c)*A*a^4 - 12 8*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^4 + 3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C*a^4 + 144*(2*d*x + 2*c + sin(2*d*x + 2*c))*C *a^4 + 96*(d*x + c)*C*a^4 + 192*A*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 384*A*a^4*sin(d*x + c) + 384*C*a^4*sin(d*x + c) + 96*A*a^4*t an(d*x + c))/d
Time = 0.38 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.35 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {96 \, A a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 96 \, A a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {48 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + 3 \, {\left (52 \, A a^{4} + 35 \, C a^{4}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (84 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 105 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 276 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 385 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 300 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 511 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 108 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 279 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \] Input:
integrate((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, algorithm= "giac")
Output:
1/24*(96*A*a^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 96*A*a^4*log(abs(tan(1 /2*d*x + 1/2*c) - 1)) - 48*A*a^4*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c )^2 - 1) + 3*(52*A*a^4 + 35*C*a^4)*(d*x + c) + 2*(84*A*a^4*tan(1/2*d*x + 1 /2*c)^7 + 105*C*a^4*tan(1/2*d*x + 1/2*c)^7 + 276*A*a^4*tan(1/2*d*x + 1/2*c )^5 + 385*C*a^4*tan(1/2*d*x + 1/2*c)^5 + 300*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 511*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 108*A*a^4*tan(1/2*d*x + 1/2*c) + 279* C*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d
Time = 0.24 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.29 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {4\,A\,a^4\,\sin \left (c+d\,x\right )}{d}+\frac {20\,C\,a^4\,\sin \left (c+d\,x\right )}{3\,d}+\frac {13\,A\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {8\,A\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {35\,C\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4\,d}+\frac {A\,a^4\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {4\,C\,a^4\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d}+\frac {C\,a^4\,{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {A\,a^4\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d}+\frac {27\,C\,a^4\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{8\,d} \] Input:
int(((A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^4)/cos(c + d*x)^2,x)
Output:
(4*A*a^4*sin(c + d*x))/d + (20*C*a^4*sin(c + d*x))/(3*d) + (13*A*a^4*atan( sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (8*A*a^4*atanh(sin(c/2 + (d*x) /2)/cos(c/2 + (d*x)/2)))/d + (35*C*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + ( d*x)/2)))/(4*d) + (A*a^4*sin(c + d*x))/(d*cos(c + d*x)) + (4*C*a^4*cos(c + d*x)^2*sin(c + d*x))/(3*d) + (C*a^4*cos(c + d*x)^3*sin(c + d*x))/(4*d) + (A*a^4*cos(c + d*x)*sin(c + d*x))/(2*d) + (27*C*a^4*cos(c + d*x)*sin(c + d *x))/(8*d)
Time = 0.17 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.05 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {a^{4} \left (-6 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{3} c +12 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right ) a +87 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right ) c -96 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +96 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -32 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} c +96 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a +192 \cos \left (d x +c \right ) \sin \left (d x +c \right ) c +156 \cos \left (d x +c \right ) a d x +105 \cos \left (d x +c \right ) c d x +24 \sin \left (d x +c \right ) a \right )}{24 \cos \left (d x +c \right ) d} \] Input:
int((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^2,x)
Output:
(a**4*( - 6*cos(c + d*x)**2*sin(c + d*x)**3*c + 12*cos(c + d*x)**2*sin(c + d*x)*a + 87*cos(c + d*x)**2*sin(c + d*x)*c - 96*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a + 96*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a - 32*cos(c + d*x)*sin(c + d*x)**3*c + 96*cos(c + d*x)*sin(c + d*x)*a + 192*cos(c + d*x )*sin(c + d*x)*c + 156*cos(c + d*x)*a*d*x + 105*cos(c + d*x)*c*d*x + 24*si n(c + d*x)*a))/(24*cos(c + d*x)*d)