Integrand size = 31, antiderivative size = 227 \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {1}{2} a b \left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) x+\frac {a^4 A \text {arctanh}(\sin (c+d x))}{d}+\frac {\left (6 a^4 C+2 b^4 (5 A+4 C)+a^2 b^2 (85 A+56 C)\right ) \sin (c+d x)}{15 d}+\frac {a b \left (40 A b^2+6 a^2 C+29 b^2 C\right ) \cos (c+d x) \sin (c+d x)}{30 d}+\frac {\left (3 a^2 C+b^2 (5 A+4 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{15 d}+\frac {a C (a+b \cos (c+d x))^3 \sin (c+d x)}{5 d}+\frac {C (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d} \]
1/2*a*b*(4*a^2*(2*A+C)+b^2*(4*A+3*C))*x+a^4*A*arctanh(sin(d*x+c))/d+1/15*( 6*a^4*C+2*b^4*(5*A+4*C)+a^2*b^2*(85*A+56*C))*sin(d*x+c)/d+1/30*a*b*(40*A*b ^2+6*C*a^2+29*C*b^2)*cos(d*x+c)*sin(d*x+c)/d+1/15*(3*a^2*C+b^2*(5*A+4*C))* (a+b*cos(d*x+c))^2*sin(d*x+c)/d+1/5*a*C*(a+b*cos(d*x+c))^3*sin(d*x+c)/d+1/ 5*C*(a+b*cos(d*x+c))^4*sin(d*x+c)/d
Time = 2.89 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.00 \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {120 a b \left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) (c+d x)-240 a^4 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+240 a^4 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+30 \left (8 a^4 C+12 a^2 b^2 (4 A+3 C)+b^4 (6 A+5 C)\right ) \sin (c+d x)+240 a b \left (A b^2+\left (a^2+b^2\right ) C\right ) \sin (2 (c+d x))+5 b^2 \left (4 A b^2+24 a^2 C+5 b^2 C\right ) \sin (3 (c+d x))+30 a b^3 C \sin (4 (c+d x))+3 b^4 C \sin (5 (c+d x))}{240 d} \]
(120*a*b*(4*a^2*(2*A + C) + b^2*(4*A + 3*C))*(c + d*x) - 240*a^4*A*Log[Cos [(c + d*x)/2] - Sin[(c + d*x)/2]] + 240*a^4*A*Log[Cos[(c + d*x)/2] + Sin[( c + d*x)/2]] + 30*(8*a^4*C + 12*a^2*b^2*(4*A + 3*C) + b^4*(6*A + 5*C))*Sin [c + d*x] + 240*a*b*(A*b^2 + (a^2 + b^2)*C)*Sin[2*(c + d*x)] + 5*b^2*(4*A* b^2 + 24*a^2*C + 5*b^2*C)*Sin[3*(c + d*x)] + 30*a*b^3*C*Sin[4*(c + d*x)] + 3*b^4*C*Sin[5*(c + d*x)])/(240*d)
Time = 1.71 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.05, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.516, Rules used = {3042, 3529, 3042, 3528, 27, 3042, 3528, 3042, 3512, 3042, 3502, 27, 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.
| \(\displaystyle \int \sec (c+d x) (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^4 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3529 |
| \(\displaystyle \frac {1}{5} \int (a+b \cos (c+d x))^3 \left (4 a C \cos ^2(c+d x)+b (5 A+4 C) \cos (c+d x)+5 a A\right ) \sec (c+d x)dx+\frac {C \sin (c+d x) (a+b \cos (c+d x))^4}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (4 a C \sin \left (c+d x+\frac {\pi }{2}\right )^2+b (5 A+4 C) \sin \left (c+d x+\frac {\pi }{2}\right )+5 a A\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {C \sin (c+d x) (a+b \cos (c+d x))^4}{5 d}\) |
| \(\Big \downarrow \) 3528 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int 4 (a+b \cos (c+d x))^2 \left (5 A a^2+b (10 A+7 C) \cos (c+d x) a+\left (3 C a^2+b^2 (5 A+4 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x)dx+\frac {a C \sin (c+d x) (a+b \cos (c+d x))^3}{d}\right )+\frac {C \sin (c+d x) (a+b \cos (c+d x))^4}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\int (a+b \cos (c+d x))^2 \left (5 A a^2+b (10 A+7 C) \cos (c+d x) a+\left (3 C a^2+b^2 (5 A+4 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x)dx+\frac {a C \sin (c+d x) (a+b \cos (c+d x))^3}{d}\right )+\frac {C \sin (c+d x) (a+b \cos (c+d x))^4}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (5 A a^2+b (10 A+7 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a+\left (3 C a^2+b^2 (5 A+4 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a C \sin (c+d x) (a+b \cos (c+d x))^3}{d}\right )+\frac {C \sin (c+d x) (a+b \cos (c+d x))^4}{5 d}\) |
| \(\Big \downarrow \) 3528 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int (a+b \cos (c+d x)) \left (15 A a^3+\left (6 C a^2+40 A b^2+29 b^2 C\right ) \cos ^2(c+d x) a+b \left (9 (5 A+3 C) a^2+2 b^2 (5 A+4 C)\right ) \cos (c+d x)\right ) \sec (c+d x)dx+\frac {\left (3 a^2 C+5 A b^2+4 b^2 C\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {a C \sin (c+d x) (a+b \cos (c+d x))^3}{d}\right )+\frac {C \sin (c+d x) (a+b \cos (c+d x))^4}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (15 A a^3+\left (6 C a^2+40 A b^2+29 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a+b \left (9 (5 A+3 C) a^2+2 b^2 (5 A+4 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {\left (3 a^2 C+5 A b^2+4 b^2 C\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {a C \sin (c+d x) (a+b \cos (c+d x))^3}{d}\right )+\frac {C \sin (c+d x) (a+b \cos (c+d x))^4}{5 d}\) |
| \(\Big \downarrow \) 3512 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {1}{2} \int \left (30 A a^4+15 b \left (4 (2 A+C) a^2+b^2 (4 A+3 C)\right ) \cos (c+d x) a+2 \left (6 C a^4+b^2 (85 A+56 C) a^2+2 b^4 (5 A+4 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x)dx+\frac {a b \left (6 a^2 C+40 A b^2+29 b^2 C\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\left (3 a^2 C+5 A b^2+4 b^2 C\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {a C \sin (c+d x) (a+b \cos (c+d x))^3}{d}\right )+\frac {C \sin (c+d x) (a+b \cos (c+d x))^4}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {1}{2} \int \frac {30 A a^4+15 b \left (4 (2 A+C) a^2+b^2 (4 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+2 \left (6 C a^4+b^2 (85 A+56 C) a^2+2 b^4 (5 A+4 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a b \left (6 a^2 C+40 A b^2+29 b^2 C\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\left (3 a^2 C+5 A b^2+4 b^2 C\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {a C \sin (c+d x) (a+b \cos (c+d x))^3}{d}\right )+\frac {C \sin (c+d x) (a+b \cos (c+d x))^4}{5 d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {1}{2} \left (\int 15 \left (2 A a^4+b \left (4 (2 A+C) a^2+b^2 (4 A+3 C)\right ) \cos (c+d x) a\right ) \sec (c+d x)dx+\frac {2 \left (6 a^4 C+a^2 b^2 (85 A+56 C)+2 b^4 (5 A+4 C)\right ) \sin (c+d x)}{d}\right )+\frac {a b \left (6 a^2 C+40 A b^2+29 b^2 C\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\left (3 a^2 C+5 A b^2+4 b^2 C\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {a C \sin (c+d x) (a+b \cos (c+d x))^3}{d}\right )+\frac {C \sin (c+d x) (a+b \cos (c+d x))^4}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 \int \left (2 A a^4+b \left (4 (2 A+C) a^2+b^2 (4 A+3 C)\right ) \cos (c+d x) a\right ) \sec (c+d x)dx+\frac {2 \left (6 a^4 C+a^2 b^2 (85 A+56 C)+2 b^4 (5 A+4 C)\right ) \sin (c+d x)}{d}\right )+\frac {a b \left (6 a^2 C+40 A b^2+29 b^2 C\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\left (3 a^2 C+5 A b^2+4 b^2 C\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {a C \sin (c+d x) (a+b \cos (c+d x))^3}{d}\right )+\frac {C \sin (c+d x) (a+b \cos (c+d x))^4}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 \int \frac {2 A a^4+b \left (4 (2 A+C) a^2+b^2 (4 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \left (6 a^4 C+a^2 b^2 (85 A+56 C)+2 b^4 (5 A+4 C)\right ) \sin (c+d x)}{d}\right )+\frac {a b \left (6 a^2 C+40 A b^2+29 b^2 C\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\left (3 a^2 C+5 A b^2+4 b^2 C\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {a C \sin (c+d x) (a+b \cos (c+d x))^3}{d}\right )+\frac {C \sin (c+d x) (a+b \cos (c+d x))^4}{5 d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 \left (2 a^4 A \int \sec (c+d x)dx+a b x \left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right )\right )+\frac {2 \left (6 a^4 C+a^2 b^2 (85 A+56 C)+2 b^4 (5 A+4 C)\right ) \sin (c+d x)}{d}\right )+\frac {a b \left (6 a^2 C+40 A b^2+29 b^2 C\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\left (3 a^2 C+5 A b^2+4 b^2 C\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {a C \sin (c+d x) (a+b \cos (c+d x))^3}{d}\right )+\frac {C \sin (c+d x) (a+b \cos (c+d x))^4}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 \left (2 a^4 A \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+a b x \left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right )\right )+\frac {2 \left (6 a^4 C+a^2 b^2 (85 A+56 C)+2 b^4 (5 A+4 C)\right ) \sin (c+d x)}{d}\right )+\frac {a b \left (6 a^2 C+40 A b^2+29 b^2 C\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\left (3 a^2 C+5 A b^2+4 b^2 C\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {a C \sin (c+d x) (a+b \cos (c+d x))^3}{d}\right )+\frac {C \sin (c+d x) (a+b \cos (c+d x))^4}{5 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{5} \left (\frac {\left (3 a^2 C+5 A b^2+4 b^2 C\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {1}{3} \left (\frac {a b \left (6 a^2 C+40 A b^2+29 b^2 C\right ) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {1}{2} \left (15 \left (\frac {2 a^4 A \text {arctanh}(\sin (c+d x))}{d}+a b x \left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right )\right )+\frac {2 \left (6 a^4 C+a^2 b^2 (85 A+56 C)+2 b^4 (5 A+4 C)\right ) \sin (c+d x)}{d}\right )\right )+\frac {a C \sin (c+d x) (a+b \cos (c+d x))^3}{d}\right )+\frac {C \sin (c+d x) (a+b \cos (c+d x))^4}{5 d}\) |
(C*(a + b*Cos[c + d*x])^4*Sin[c + d*x])/(5*d) + (((5*A*b^2 + 3*a^2*C + 4*b ^2*C)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(3*d) + (a*C*(a + b*Cos[c + d*x ])^3*Sin[c + d*x])/d + ((a*b*(40*A*b^2 + 6*a^2*C + 29*b^2*C)*Cos[c + d*x]* Sin[c + d*x])/(2*d) + (15*(a*b*(4*a^2*(2*A + C) + b^2*(4*A + 3*C))*x + (2* a^4*A*ArcTanh[Sin[c + d*x]])/d) + (2*(6*a^4*C + 2*b^4*(5*A + 4*C) + a^2*b^ 2*(85*A + 56*C))*Sin[c + d*x])/d)/2)/3)/5
3.6.51.3.1 Defintions of rubi rules used
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_.)*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_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f _.)*(x_)]^2), x_Symbol] :> Simp[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si n[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3)) Int[(a + b*Si n[e + f*x])^m*Simp[a*C*d + A*b*c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e + f*x]^2 , x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && !LtQ[m, -1]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ .) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x ])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + n + 2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a *c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n} , x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[ m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
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)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + n + 2)) Int[(a + b*Sin[e + f*x ])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*( n + 1)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + C* (a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f , A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 6.20 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.86
| method | result | size |
| parallelrisch | \(\frac {-240 a^{4} A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+240 a^{4} A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+240 b \left (b^{2} \left (A +C \right )+a^{2} C \right ) a \sin \left (2 d x +2 c \right )+20 b^{2} \left (\left (A +\frac {5 C}{4}\right ) b^{2}+6 a^{2} C \right ) \sin \left (3 d x +3 c \right )+30 C \sin \left (4 d x +4 c \right ) a \,b^{3}+3 C \sin \left (5 d x +5 c \right ) b^{4}+30 \left (b^{4} \left (6 A +5 C \right )+48 \left (A +\frac {3 C}{4}\right ) a^{2} b^{2}+8 C \,a^{4}\right ) \sin \left (d x +c \right )+960 x b d a \left (\frac {\left (A +\frac {3 C}{4}\right ) b^{2}}{2}+a^{2} \left (A +\frac {C}{2}\right )\right )}{240 d}\) | \(195\) |
| parts | \(\frac {a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (A \,b^{4}+6 C \,a^{2} b^{2}\right ) \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {\left (4 A a \,b^{3}+4 C \,a^{3} b \right ) \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (6 A \,a^{2} b^{2}+C \,a^{4}\right ) \sin \left (d x +c \right )}{d}+\frac {C \,b^{4} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5 d}+\frac {4 A \,a^{3} b \left (d x +c \right )}{d}+\frac {4 C a \,b^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(219\) |
| derivativedivides | \(\frac {a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,a^{4} \sin \left (d x +c \right )+4 A \,a^{3} b \left (d x +c \right )+4 C \,a^{3} b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+6 A \sin \left (d x +c \right ) a^{2} b^{2}+2 C \,a^{2} b^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+4 A a \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 C a \,b^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {A \,b^{4} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {C \,b^{4} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\) | \(240\) |
| default | \(\frac {a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,a^{4} \sin \left (d x +c \right )+4 A \,a^{3} b \left (d x +c \right )+4 C \,a^{3} b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+6 A \sin \left (d x +c \right ) a^{2} b^{2}+2 C \,a^{2} b^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+4 A a \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 C a \,b^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {A \,b^{4} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {C \,b^{4} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\) | \(240\) |
| risch | \(4 A \,a^{3} b x +2 x A a \,b^{3}+2 C \,a^{3} b x +\frac {3 a \,b^{3} C x}{2}-\frac {9 i {\mathrm e}^{i \left (d x +c \right )} C \,a^{2} b^{2}}{4 d}+\frac {9 i {\mathrm e}^{-i \left (d x +c \right )} C \,a^{2} b^{2}}{4 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} A \,a^{2} b^{2}}{d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} A \,a^{2} b^{2}}{d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} C \,a^{4}}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} C \,a^{4}}{2 d}-\frac {5 i {\mathrm e}^{i \left (d x +c \right )} C \,b^{4}}{16 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} A \,b^{4}}{8 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} A \,b^{4}}{8 d}+\frac {5 i {\mathrm e}^{-i \left (d x +c \right )} C \,b^{4}}{16 d}+\frac {a^{4} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a^{4} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {\sin \left (5 d x +5 c \right ) C \,b^{4}}{80 d}+\frac {\sin \left (4 d x +4 c \right ) C a \,b^{3}}{8 d}+\frac {\sin \left (3 d x +3 c \right ) A \,b^{4}}{12 d}+\frac {\sin \left (3 d x +3 c \right ) C \,a^{2} b^{2}}{2 d}+\frac {5 \sin \left (3 d x +3 c \right ) C \,b^{4}}{48 d}+\frac {\sin \left (2 d x +2 c \right ) A a \,b^{3}}{d}+\frac {\sin \left (2 d x +2 c \right ) C \,a^{3} b}{d}+\frac {\sin \left (2 d x +2 c \right ) C a \,b^{3}}{d}\) | \(427\) |
| norman | \(\frac {\left (4 A \,a^{3} b +2 A a \,b^{3}+2 C \,a^{3} b +\frac {3}{2} C a \,b^{3}\right ) x +\left (4 A \,a^{3} b +2 A a \,b^{3}+2 C \,a^{3} b +\frac {3}{2} C a \,b^{3}\right ) x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (60 A \,a^{3} b +30 A a \,b^{3}+30 C \,a^{3} b +\frac {45}{2} C a \,b^{3}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (60 A \,a^{3} b +30 A a \,b^{3}+30 C \,a^{3} b +\frac {45}{2} C a \,b^{3}\right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (24 A \,a^{3} b +12 A a \,b^{3}+12 C \,a^{3} b +9 C a \,b^{3}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (24 A \,a^{3} b +12 A a \,b^{3}+12 C \,a^{3} b +9 C a \,b^{3}\right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (80 A \,a^{3} b +40 A a \,b^{3}+40 C \,a^{3} b +30 C a \,b^{3}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {\left (12 A \,a^{2} b^{2}-4 A a \,b^{3}+2 A \,b^{4}+2 C \,a^{4}-4 C \,a^{3} b +12 C \,a^{2} b^{2}-5 C a \,b^{3}+2 C \,b^{4}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (12 A \,a^{2} b^{2}+4 A a \,b^{3}+2 A \,b^{4}+2 C \,a^{4}+4 C \,a^{3} b +12 C \,a^{2} b^{2}+5 C a \,b^{3}+2 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (180 A \,a^{2} b^{2}-36 A a \,b^{3}+22 A \,b^{4}+30 C \,a^{4}-36 C \,a^{3} b +132 C \,a^{2} b^{2}-21 C a \,b^{3}+14 C \,b^{4}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {\left (180 A \,a^{2} b^{2}+36 A a \,b^{3}+22 A \,b^{4}+30 C \,a^{4}+36 C \,a^{3} b +132 C \,a^{2} b^{2}+21 C a \,b^{3}+14 C \,b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 \left (300 A \,a^{2} b^{2}-20 A a \,b^{3}+30 A \,b^{4}+50 C \,a^{4}-20 C \,a^{3} b +180 C \,a^{2} b^{2}-5 C a \,b^{3}+26 C \,b^{4}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {2 \left (300 A \,a^{2} b^{2}+20 A a \,b^{3}+30 A \,b^{4}+50 C \,a^{4}+20 C \,a^{3} b +180 C \,a^{2} b^{2}+5 C a \,b^{3}+26 C \,b^{4}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {a^{4} A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a^{4} A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(783\) |
1/240*(-240*a^4*A*ln(tan(1/2*d*x+1/2*c)-1)+240*a^4*A*ln(tan(1/2*d*x+1/2*c) +1)+240*b*(b^2*(A+C)+a^2*C)*a*sin(2*d*x+2*c)+20*b^2*((A+5/4*C)*b^2+6*a^2*C )*sin(3*d*x+3*c)+30*C*sin(4*d*x+4*c)*a*b^3+3*C*sin(5*d*x+5*c)*b^4+30*(b^4* (6*A+5*C)+48*(A+3/4*C)*a^2*b^2+8*C*a^4)*sin(d*x+c)+960*x*b*d*a*(1/2*(A+3/4 *C)*b^2+a^2*(A+1/2*C)))/d
Time = 0.29 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.86 \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {15 \, A a^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, A a^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 15 \, {\left (4 \, {\left (2 \, A + C\right )} a^{3} b + {\left (4 \, A + 3 \, C\right )} a b^{3}\right )} d x + {\left (6 \, C b^{4} \cos \left (d x + c\right )^{4} + 30 \, C a b^{3} \cos \left (d x + c\right )^{3} + 30 \, C a^{4} + 60 \, {\left (3 \, A + 2 \, C\right )} a^{2} b^{2} + 4 \, {\left (5 \, A + 4 \, C\right )} b^{4} + 2 \, {\left (30 \, C a^{2} b^{2} + {\left (5 \, A + 4 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (4 \, C a^{3} b + {\left (4 \, A + 3 \, C\right )} a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{30 \, d} \]
1/30*(15*A*a^4*log(sin(d*x + c) + 1) - 15*A*a^4*log(-sin(d*x + c) + 1) + 1 5*(4*(2*A + C)*a^3*b + (4*A + 3*C)*a*b^3)*d*x + (6*C*b^4*cos(d*x + c)^4 + 30*C*a*b^3*cos(d*x + c)^3 + 30*C*a^4 + 60*(3*A + 2*C)*a^2*b^2 + 4*(5*A + 4 *C)*b^4 + 2*(30*C*a^2*b^2 + (5*A + 4*C)*b^4)*cos(d*x + c)^2 + 15*(4*C*a^3* b + (4*A + 3*C)*a*b^3)*cos(d*x + c))*sin(d*x + c))/d
\[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\int \left (A + C \cos ^{2}{\left (c + d x \right )}\right ) \left (a + b \cos {\left (c + d x \right )}\right )^{4} \sec {\left (c + d x \right )}\, dx \]
Time = 0.21 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.02 \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {480 \, {\left (d x + c\right )} A a^{3} b + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} b - 240 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} b^{2} + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a b^{3} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a b^{3} - 40 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A b^{4} + 8 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C b^{4} + 120 \, A a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 120 \, C a^{4} \sin \left (d x + c\right ) + 720 \, A a^{2} b^{2} \sin \left (d x + c\right )}{120 \, d} \]
1/120*(480*(d*x + c)*A*a^3*b + 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^3* b - 240*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^2*b^2 + 120*(2*d*x + 2*c + s in(2*d*x + 2*c))*A*a*b^3 + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2* d*x + 2*c))*C*a*b^3 - 40*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*b^4 + 8*(3*si n(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*C*b^4 + 120*A*a^4*log( sec(d*x + c) + tan(d*x + c)) + 120*C*a^4*sin(d*x + c) + 720*A*a^2*b^2*sin( d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 753 vs. \(2 (215) = 430\).
Time = 0.35 (sec) , antiderivative size = 753, normalized size of antiderivative = 3.32 \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\text {Too large to display} \]
1/30*(30*A*a^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 30*A*a^4*log(abs(tan(1 /2*d*x + 1/2*c) - 1)) + 15*(8*A*a^3*b + 4*C*a^3*b + 4*A*a*b^3 + 3*C*a*b^3) *(d*x + c) + 2*(30*C*a^4*tan(1/2*d*x + 1/2*c)^9 - 60*C*a^3*b*tan(1/2*d*x + 1/2*c)^9 + 180*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 + 180*C*a^2*b^2*tan(1/2*d *x + 1/2*c)^9 - 60*A*a*b^3*tan(1/2*d*x + 1/2*c)^9 - 75*C*a*b^3*tan(1/2*d*x + 1/2*c)^9 + 30*A*b^4*tan(1/2*d*x + 1/2*c)^9 + 30*C*b^4*tan(1/2*d*x + 1/2 *c)^9 + 120*C*a^4*tan(1/2*d*x + 1/2*c)^7 - 120*C*a^3*b*tan(1/2*d*x + 1/2*c )^7 + 720*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 + 480*C*a^2*b^2*tan(1/2*d*x + 1 /2*c)^7 - 120*A*a*b^3*tan(1/2*d*x + 1/2*c)^7 - 30*C*a*b^3*tan(1/2*d*x + 1/ 2*c)^7 + 80*A*b^4*tan(1/2*d*x + 1/2*c)^7 + 40*C*b^4*tan(1/2*d*x + 1/2*c)^7 + 180*C*a^4*tan(1/2*d*x + 1/2*c)^5 + 1080*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^ 5 + 600*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 100*A*b^4*tan(1/2*d*x + 1/2*c)^ 5 + 116*C*b^4*tan(1/2*d*x + 1/2*c)^5 + 120*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 120*C*a^3*b*tan(1/2*d*x + 1/2*c)^3 + 720*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 480*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 120*A*a*b^3*tan(1/2*d*x + 1/2*c)^ 3 + 30*C*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 80*A*b^4*tan(1/2*d*x + 1/2*c)^3 + 40*C*b^4*tan(1/2*d*x + 1/2*c)^3 + 30*C*a^4*tan(1/2*d*x + 1/2*c) + 60*C*a^3 *b*tan(1/2*d*x + 1/2*c) + 180*A*a^2*b^2*tan(1/2*d*x + 1/2*c) + 180*C*a^2*b ^2*tan(1/2*d*x + 1/2*c) + 60*A*a*b^3*tan(1/2*d*x + 1/2*c) + 75*C*a*b^3*tan (1/2*d*x + 1/2*c) + 30*A*b^4*tan(1/2*d*x + 1/2*c) + 30*C*b^4*tan(1/2*d*...
Time = 3.62 (sec) , antiderivative size = 2241, normalized size of antiderivative = 9.87 \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\text {Too large to display} \]
(tan(c/2 + (d*x)/2)*(2*A*b^4 + 2*C*a^4 + 2*C*b^4 + 12*A*a^2*b^2 + 12*C*a^2 *b^2 + 4*A*a*b^3 + 5*C*a*b^3 + 4*C*a^3*b) + tan(c/2 + (d*x)/2)^5*((20*A*b^ 4)/3 + 12*C*a^4 + (116*C*b^4)/15 + 72*A*a^2*b^2 + 40*C*a^2*b^2) + tan(c/2 + (d*x)/2)^9*(2*A*b^4 + 2*C*a^4 + 2*C*b^4 + 12*A*a^2*b^2 + 12*C*a^2*b^2 - 4*A*a*b^3 - 5*C*a*b^3 - 4*C*a^3*b) + tan(c/2 + (d*x)/2)^3*((16*A*b^4)/3 + 8*C*a^4 + (8*C*b^4)/3 + 48*A*a^2*b^2 + 32*C*a^2*b^2 + 8*A*a*b^3 + 2*C*a*b^ 3 + 8*C*a^3*b) + tan(c/2 + (d*x)/2)^7*((16*A*b^4)/3 + 8*C*a^4 + (8*C*b^4)/ 3 + 48*A*a^2*b^2 + 32*C*a^2*b^2 - 8*A*a*b^3 - 2*C*a*b^3 - 8*C*a^3*b))/(d*( 5*tan(c/2 + (d*x)/2)^2 + 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 + 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1)) - (A*a^4*atan((A*a ^4*(tan(c/2 + (d*x)/2)*(32*A^2*a^8 + 128*A^2*a^2*b^6 + 512*A^2*a^4*b^4 + 5 12*A^2*a^6*b^2 + 72*C^2*a^2*b^6 + 192*C^2*a^4*b^4 + 128*C^2*a^6*b^2 + 192* A*C*a^2*b^6 + 640*A*C*a^4*b^4 + 512*A*C*a^6*b^2) + A*a^4*(32*A*a^4 + 64*A* a*b^3 + 128*A*a^3*b + 48*C*a*b^3 + 64*C*a^3*b))*1i + A*a^4*(tan(c/2 + (d*x )/2)*(32*A^2*a^8 + 128*A^2*a^2*b^6 + 512*A^2*a^4*b^4 + 512*A^2*a^6*b^2 + 7 2*C^2*a^2*b^6 + 192*C^2*a^4*b^4 + 128*C^2*a^6*b^2 + 192*A*C*a^2*b^6 + 640* A*C*a^4*b^4 + 512*A*C*a^6*b^2) - A*a^4*(32*A*a^4 + 64*A*a*b^3 + 128*A*a^3* b + 48*C*a*b^3 + 64*C*a^3*b))*1i)/(256*A^3*a^6*b^6 - 256*A^3*a^11*b + 1024 *A^3*a^8*b^4 - 128*A^3*a^9*b^3 + 1024*A^3*a^10*b^2 + A*a^4*(tan(c/2 + (d*x )/2)*(32*A^2*a^8 + 128*A^2*a^2*b^6 + 512*A^2*a^4*b^4 + 512*A^2*a^6*b^2 ...