Integrand size = 33, antiderivative size = 251 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=2 a b \left (2 A b^2+a^2 (A+2 C)\right ) x+\frac {b^2 \left (2 A b^2+\left (12 a^2+b^2\right ) C\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {\left (6 A b^2+a^2 (2 A+3 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac {2 A b \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {A \cos ^2(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac {2 a b \left (b^2 (11 A-6 C)+a^2 (2 A+3 C)\right ) \tan (c+d x)}{3 d}-\frac {b^2 \left (3 b^2 (6 A-C)+a^2 (4 A+6 C)\right ) \sec (c+d x) \tan (c+d x)}{6 d} \]
2*a*b*(2*A*b^2+a^2*(A+2*C))*x+1/2*b^2*(2*A*b^2+(12*a^2+b^2)*C)*arctanh(sin (d*x+c))/d+1/3*(6*A*b^2+a^2*(2*A+3*C))*(a+b*sec(d*x+c))^2*sin(d*x+c)/d+2/3 *A*b*cos(d*x+c)*(a+b*sec(d*x+c))^3*sin(d*x+c)/d+1/3*A*cos(d*x+c)^2*(a+b*se c(d*x+c))^4*sin(d*x+c)/d-2/3*a*b*(b^2*(11*A-6*C)+a^2*(2*A+3*C))*tan(d*x+c) /d-1/6*b^2*(3*b^2*(6*A-C)+a^2*(4*A+6*C))*sec(d*x+c)*tan(d*x+c)/d
Time = 4.83 (sec) , antiderivative size = 324, normalized size of antiderivative = 1.29 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {24 a b \left (2 A b^2+a^2 (A+2 C)\right ) (c+d x)-6 b^2 \left (2 A b^2+\left (12 a^2+b^2\right ) C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 b^2 \left (2 A b^2+\left (12 a^2+b^2\right ) C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {3 b^4 C}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {48 a b^3 C \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}-\frac {3 b^4 C}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {48 a b^3 C \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}+3 a^2 \left (24 A b^2+a^2 (3 A+4 C)\right ) \sin (c+d x)+12 a^3 A b \sin (2 (c+d x))+a^4 A \sin (3 (c+d x))}{12 d} \]
(24*a*b*(2*A*b^2 + a^2*(A + 2*C))*(c + d*x) - 6*b^2*(2*A*b^2 + (12*a^2 + b ^2)*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 6*b^2*(2*A*b^2 + (12*a^2 + b^2)*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (3*b^4*C)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (48*a*b^3*C*Sin[(c + d*x)/2])/(Cos[(c + d* x)/2] - Sin[(c + d*x)/2]) - (3*b^4*C)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2] )^2 + (48*a*b^3*C*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 3*a^2*(24*A*b^2 + a^2*(3*A + 4*C))*Sin[c + d*x] + 12*a^3*A*b*Sin[2*(c + d*x)] + a^4*A*Sin[3*(c + d*x)])/(12*d)
Time = 1.27 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {3042, 4583, 3042, 4582, 27, 3042, 4582, 3042, 4536, 2009}
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.
\(\displaystyle \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^4 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
\(\Big \downarrow \) 4583 |
\(\displaystyle \frac {1}{3} \int \cos ^2(c+d x) (a+b \sec (c+d x))^3 \left (-b (2 A-3 C) \sec ^2(c+d x)+a (2 A+3 C) \sec (c+d x)+4 A b\right )dx+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^4}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (-b (2 A-3 C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+a (2 A+3 C) \csc \left (c+d x+\frac {\pi }{2}\right )+4 A b\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^4}{3 d}\) |
\(\Big \downarrow \) 4582 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int 2 \cos (c+d x) (a+b \sec (c+d x))^2 \left (\frac {1}{2} (4 A+6 C) a^2+2 b (A+3 C) \sec (c+d x) a+6 A b^2-3 b^2 (2 A-C) \sec ^2(c+d x)\right )dx+\frac {2 A b \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{d}\right )+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^4}{3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \left (\int \cos (c+d x) (a+b \sec (c+d x))^2 \left ((2 A+3 C) a^2+2 b (A+3 C) \sec (c+d x) a+6 A b^2-3 b^2 (2 A-C) \sec ^2(c+d x)\right )dx+\frac {2 A b \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{d}\right )+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^4}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} \left (\int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left ((2 A+3 C) a^2+2 b (A+3 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a+6 A b^2-3 b^2 (2 A-C) \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 A b \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{d}\right )+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^4}{3 d}\) |
\(\Big \downarrow \) 4582 |
\(\displaystyle \frac {1}{3} \left (\int (a+b \sec (c+d x)) \left (-a (4 A-9 C) \sec (c+d x) b^2-\left ((4 A+6 C) a^2+3 b^2 (6 A-C)\right ) \sec ^2(c+d x) b+6 \left ((A+2 C) a^2+2 A b^2\right ) b\right )dx+\frac {\left (a^2 (2 A+3 C)+6 A b^2\right ) \sin (c+d x) (a+b \sec (c+d x))^2}{d}+\frac {2 A b \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{d}\right )+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^4}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} \left (\int \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (-a (4 A-9 C) \csc \left (c+d x+\frac {\pi }{2}\right ) b^2-\left ((4 A+6 C) a^2+3 b^2 (6 A-C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2 b+6 \left ((A+2 C) a^2+2 A b^2\right ) b\right )dx+\frac {\left (a^2 (2 A+3 C)+6 A b^2\right ) \sin (c+d x) (a+b \sec (c+d x))^2}{d}+\frac {2 A b \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{d}\right )+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^4}{3 d}\) |
\(\Big \downarrow \) 4536 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \left (3 \left (2 A b^2+\left (12 a^2+b^2\right ) C\right ) \sec (c+d x) b^2-4 a \left ((2 A+3 C) a^2+b^2 (11 A-6 C)\right ) \sec ^2(c+d x) b+12 a \left ((A+2 C) a^2+2 A b^2\right ) b\right )dx+\frac {\left (a^2 (2 A+3 C)+6 A b^2\right ) \sin (c+d x) (a+b \sec (c+d x))^2}{d}-\frac {b^2 \left (a^2 (4 A+6 C)+3 b^2 (6 A-C)\right ) \tan (c+d x) \sec (c+d x)}{2 d}+\frac {2 A b \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{d}\right )+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^4}{3 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\frac {3 b^2 \left (C \left (12 a^2+b^2\right )+2 A b^2\right ) \text {arctanh}(\sin (c+d x))}{d}-\frac {4 a b \left (a^2 (2 A+3 C)+b^2 (11 A-6 C)\right ) \tan (c+d x)}{d}+12 a b x \left (a^2 (A+2 C)+2 A b^2\right )\right )+\frac {\left (a^2 (2 A+3 C)+6 A b^2\right ) \sin (c+d x) (a+b \sec (c+d x))^2}{d}-\frac {b^2 \left (a^2 (4 A+6 C)+3 b^2 (6 A-C)\right ) \tan (c+d x) \sec (c+d x)}{2 d}+\frac {2 A b \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{d}\right )+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^4}{3 d}\) |
(A*Cos[c + d*x]^2*(a + b*Sec[c + d*x])^4*Sin[c + d*x])/(3*d) + (((6*A*b^2 + a^2*(2*A + 3*C))*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/d + (2*A*b*Cos[c + d*x]*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/d - (b^2*(3*b^2*(6*A - C) + a^2 *(4*A + 6*C))*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (12*a*b*(2*A*b^2 + a^2*(A + 2*C))*x + (3*b^2*(2*A*b^2 + (12*a^2 + b^2)*C)*ArcTanh[Sin[c + d*x]])/d - (4*a*b*(b^2*(11*A - 6*C) + a^2*(2*A + 3*C))*Tan[c + d*x])/d)/2)/3
3.7.68.3.1 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_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(-b)*C*Csc[e + f*x]*(Cot[e + f*x]/(2*f)), x] + Simp[1/2 Int[Simp[2*A*a + (2*B*a + b*(2* A + C))*Csc[e + f*x] + 2*(a*C + B*b)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a , b, e, f, A, B, C}, x]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(d*n) Int[(a + b*Csc[e + f*x])^(m - 1)*(d* Csc[e + f*x])^(n + 1)*Simp[A*b*m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Cs c[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a , b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. ))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[A*Co t[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/( d*n) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*m - a*(C*n + A*(n + 1))*Csc[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^ 2, x], x], x] /; FreeQ[{a, b, d, e, f, A, C}, x] && NeQ[a^2 - b^2, 0] && Gt Q[m, 0] && LeQ[n, -1]
Time = 1.00 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.79
method | result | size |
derivativedivides | \(\frac {\frac {a^{4} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a^{4} C \sin \left (d x +c \right )+4 A \,a^{3} b \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{3} b C \left (d x +c \right )+6 A \,a^{2} b^{2} \sin \left (d x +c \right )+6 C \,a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a A \,b^{3} \left (d x +c \right )+4 C \tan \left (d x +c \right ) a \,b^{3}+A \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,b^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(198\) |
default | \(\frac {\frac {a^{4} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a^{4} C \sin \left (d x +c \right )+4 A \,a^{3} b \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{3} b C \left (d x +c \right )+6 A \,a^{2} b^{2} \sin \left (d x +c \right )+6 C \,a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a A \,b^{3} \left (d x +c \right )+4 C \tan \left (d x +c \right ) a \,b^{3}+A \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,b^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(198\) |
parallelrisch | \(\frac {-24 b^{2} \left (1+\cos \left (2 d x +2 c \right )\right ) \left (\left (A +\frac {C}{2}\right ) b^{2}+6 C \,a^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+24 b^{2} \left (1+\cos \left (2 d x +2 c \right )\right ) \left (\left (A +\frac {C}{2}\right ) b^{2}+6 C \,a^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+48 \left (2 A \,b^{2}+a^{2} \left (A +2 C \right )\right ) a x b d \cos \left (2 d x +2 c \right )+\left (72 A \,a^{2} b^{2}+11 \left (A +\frac {12 C}{11}\right ) a^{4}\right ) \sin \left (3 d x +3 c \right )+\left (24 A \,a^{3} b +96 C a \,b^{3}\right ) \sin \left (2 d x +2 c \right )+12 A \,a^{3} b \sin \left (4 d x +4 c \right )+a^{4} A \sin \left (5 d x +5 c \right )+\left (24 C \,b^{4}+72 A \,a^{2} b^{2}+10 \left (A +\frac {6 C}{5}\right ) a^{4}\right ) \sin \left (d x +c \right )+48 \left (2 A \,b^{2}+a^{2} \left (A +2 C \right )\right ) a x b d}{24 d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(279\) |
risch | \(2 a^{3} A b x +4 A a \,b^{3} x +4 C \,a^{3} b x -\frac {i a^{4} A \,{\mathrm e}^{3 i \left (d x +c \right )}}{24 d}+\frac {3 i a^{4} A \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} A \,a^{2} b^{2}}{d}-\frac {i A \,a^{3} b \,{\mathrm e}^{2 i \left (d x +c \right )}}{2 d}-\frac {3 i a^{4} A \,{\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {i C \,b^{3} \left (b \,{\mathrm e}^{3 i \left (d x +c \right )}-8 a \,{\mathrm e}^{2 i \left (d x +c \right )}-b \,{\mathrm e}^{i \left (d x +c \right )}-8 a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {i A \,a^{3} b \,{\mathrm e}^{-2 i \left (d x +c \right )}}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{4} C}{2 d}+\frac {i a^{4} A \,{\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{4} C}{2 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} A \,a^{2} b^{2}}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{4}}{d}+\frac {6 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C \,a^{2} b^{2}}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C \,b^{4}}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{4}}{d}-\frac {6 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C \,a^{2} b^{2}}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C \,b^{4}}{2 d}\) | \(423\) |
norman | \(\frac {\left (-2 A \,a^{3} b -4 a A \,b^{3}-4 a^{3} b C \right ) x +\left (-12 A \,a^{3} b -24 a A \,b^{3}-24 a^{3} b C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (-4 A \,a^{3} b -8 a A \,b^{3}-8 a^{3} b C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (-4 A \,a^{3} b -8 a A \,b^{3}-8 a^{3} b C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}+\left (2 A \,a^{3} b +4 a A \,b^{3}+4 a^{3} b C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{16}+\left (4 A \,a^{3} b +8 a A \,b^{3}+8 a^{3} b C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (4 A \,a^{3} b +8 a A \,b^{3}+8 a^{3} b C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (12 A \,a^{3} b +24 a A \,b^{3}+24 a^{3} b C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\frac {\left (2 a^{4} A -4 A \,a^{3} b +12 A \,a^{2} b^{2}+2 a^{4} C -8 C a \,b^{3}+C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{d}-\frac {\left (2 a^{4} A +4 A \,a^{3} b +12 A \,a^{2} b^{2}+2 a^{4} C +8 C a \,b^{3}+C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {\left (26 a^{4} A -60 A \,a^{3} b +108 A \,a^{2} b^{2}+18 a^{4} C -24 C a \,b^{3}-3 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{3 d}+\frac {\left (26 a^{4} A +60 A \,a^{3} b +108 A \,a^{2} b^{2}+18 a^{4} C +24 C a \,b^{3}-3 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {\left (46 a^{4} A -108 A \,a^{3} b +36 A \,a^{2} b^{2}+6 a^{4} C +72 C a \,b^{3}-9 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}-\frac {\left (46 a^{4} A +108 A \,a^{3} b +36 A \,a^{2} b^{2}+6 a^{4} C -72 C a \,b^{3}-9 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}-\frac {\left (50 a^{4} A -60 A \,a^{3} b -180 A \,a^{2} b^{2}-30 a^{4} C +72 C a \,b^{3}+9 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}+\frac {\left (50 a^{4} A +60 A \,a^{3} b -180 A \,a^{2} b^{2}-30 a^{4} C -72 C a \,b^{3}+9 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5}}-\frac {b^{2} \left (2 A \,b^{2}+12 C \,a^{2}+C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {b^{2} \left (2 A \,b^{2}+12 C \,a^{2}+C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(838\) |
1/d*(1/3*a^4*A*(2+cos(d*x+c)^2)*sin(d*x+c)+a^4*C*sin(d*x+c)+4*A*a^3*b*(1/2 *sin(d*x+c)*cos(d*x+c)+1/2*d*x+1/2*c)+4*a^3*b*C*(d*x+c)+6*A*a^2*b^2*sin(d* x+c)+6*C*a^2*b^2*ln(sec(d*x+c)+tan(d*x+c))+4*a*A*b^3*(d*x+c)+4*C*tan(d*x+c )*a*b^3+A*b^4*ln(sec(d*x+c)+tan(d*x+c))+C*b^4*(1/2*sec(d*x+c)*tan(d*x+c)+1 /2*ln(sec(d*x+c)+tan(d*x+c))))
Time = 0.30 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.84 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {24 \, {\left ({\left (A + 2 \, C\right )} a^{3} b + 2 \, A a b^{3}\right )} d x \cos \left (d x + c\right )^{2} + 3 \, {\left (12 \, C a^{2} b^{2} + {\left (2 \, A + C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (12 \, C a^{2} b^{2} + {\left (2 \, A + C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, A a^{4} \cos \left (d x + c\right )^{4} + 12 \, A a^{3} b \cos \left (d x + c\right )^{3} + 24 \, C a b^{3} \cos \left (d x + c\right ) + 3 \, C b^{4} + 2 \, {\left ({\left (2 \, A + 3 \, C\right )} a^{4} + 18 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{2}} \]
1/12*(24*((A + 2*C)*a^3*b + 2*A*a*b^3)*d*x*cos(d*x + c)^2 + 3*(12*C*a^2*b^ 2 + (2*A + C)*b^4)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - 3*(12*C*a^2*b^2 + (2*A + C)*b^4)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(2*A*a^4*cos(d* x + c)^4 + 12*A*a^3*b*cos(d*x + c)^3 + 24*C*a*b^3*cos(d*x + c) + 3*C*b^4 + 2*((2*A + 3*C)*a^4 + 18*A*a^2*b^2)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d *x + c)^2)
Timed out. \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Time = 0.20 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.88 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} - 12 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} b - 48 \, {\left (d x + c\right )} C a^{3} b - 48 \, {\left (d x + c\right )} A a b^{3} + 3 \, C b^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, C a^{2} b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 6 \, A b^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, C a^{4} \sin \left (d x + c\right ) - 72 \, A a^{2} b^{2} \sin \left (d x + c\right ) - 48 \, C a b^{3} \tan \left (d x + c\right )}{12 \, d} \]
-1/12*(4*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^4 - 12*(2*d*x + 2*c + sin(2 *d*x + 2*c))*A*a^3*b - 48*(d*x + c)*C*a^3*b - 48*(d*x + c)*A*a*b^3 + 3*C*b ^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin( d*x + c) - 1)) - 36*C*a^2*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 6*A*b^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 12*C*a^4*s in(d*x + c) - 72*A*a^2*b^2*sin(d*x + c) - 48*C*a*b^3*tan(d*x + c))/d
Time = 0.38 (sec) , antiderivative size = 398, normalized size of antiderivative = 1.59 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {12 \, {\left (A a^{3} b + 2 \, C a^{3} b + 2 \, A a b^{3}\right )} {\left (d x + c\right )} + 3 \, {\left (12 \, C a^{2} b^{2} + 2 \, A b^{4} + C b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (12 \, C a^{2} b^{2} + 2 \, A b^{4} + C b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {6 \, {\left (8 \, C a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, C a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}} + \frac {4 \, {\left (3 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]
1/6*(12*(A*a^3*b + 2*C*a^3*b + 2*A*a*b^3)*(d*x + c) + 3*(12*C*a^2*b^2 + 2* A*b^4 + C*b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(12*C*a^2*b^2 + 2*A* b^4 + C*b^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 6*(8*C*a*b^3*tan(1/2*d*x + 1/2*c)^3 - C*b^4*tan(1/2*d*x + 1/2*c)^3 - 8*C*a*b^3*tan(1/2*d*x + 1/2*c ) - C*b^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2 + 4*(3*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 - 6*A*a^3*b*tan( 1/2*d*x + 1/2*c)^5 + 18*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 2*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 + 36*A*a^2*b^2*tan(1/2*d* x + 1/2*c)^3 + 3*A*a^4*tan(1/2*d*x + 1/2*c) + 3*C*a^4*tan(1/2*d*x + 1/2*c) + 6*A*a^3*b*tan(1/2*d*x + 1/2*c) + 18*A*a^2*b^2*tan(1/2*d*x + 1/2*c))/(ta n(1/2*d*x + 1/2*c)^2 + 1)^3)/d
Time = 18.05 (sec) , antiderivative size = 2660, normalized size of antiderivative = 10.60 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]
(tan(c/2 + (d*x)/2)^5*((4*A*a^4)/3 - 4*C*a^4 + 6*C*b^4 - 24*A*a^2*b^2) - t an(c/2 + (d*x)/2)^3*((8*A*a^4)/3 - 4*C*b^4 + 8*A*a^3*b - 16*C*a*b^3) - tan (c/2 + (d*x)/2)^7*((8*A*a^4)/3 - 4*C*b^4 - 8*A*a^3*b + 16*C*a*b^3) + tan(c /2 + (d*x)/2)*(2*A*a^4 + 2*C*a^4 + C*b^4 + 12*A*a^2*b^2 + 4*A*a^3*b + 8*C* a*b^3) + tan(c/2 + (d*x)/2)^9*(2*A*a^4 + 2*C*a^4 + C*b^4 + 12*A*a^2*b^2 - 4*A*a^3*b - 8*C*a*b^3))/(d*(tan(c/2 + (d*x)/2)^2 - 2*tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1)) - (atan((((A*b^4 + (C*b^4)/2 + 6*C*a^2*b^2)*(32*A*b^4 + 16*C*b^4 + 192 *C*a^2*b^2 + 128*A*a*b^3 + 64*A*a^3*b + 128*C*a^3*b) + tan(c/2 + (d*x)/2)* (32*A^2*b^8 + 8*C^2*b^8 + 512*A^2*a^2*b^6 + 512*A^2*a^4*b^4 + 128*A^2*a^6* b^2 + 192*C^2*a^2*b^6 + 1152*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*b^8 + 384*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 512*A*C*a^6*b^2))*(A*b^4 + (C*b^4)/2 + 6*C*a^2*b^2)*1i - ((A*b^4 + (C*b^4)/2 + 6*C*a^2*b^2)*(32*A*b^4 + 16*C*b^ 4 + 192*C*a^2*b^2 + 128*A*a*b^3 + 64*A*a^3*b + 128*C*a^3*b) - tan(c/2 + (d *x)/2)*(32*A^2*b^8 + 8*C^2*b^8 + 512*A^2*a^2*b^6 + 512*A^2*a^4*b^4 + 128*A ^2*a^6*b^2 + 192*C^2*a^2*b^6 + 1152*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C *b^8 + 384*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 512*A*C*a^6*b^2))*(A*b^4 + (C* b^4)/2 + 6*C*a^2*b^2)*1i)/(((A*b^4 + (C*b^4)/2 + 6*C*a^2*b^2)*(32*A*b^4 + 16*C*b^4 + 192*C*a^2*b^2 + 128*A*a*b^3 + 64*A*a^3*b + 128*C*a^3*b) + tan(c /2 + (d*x)/2)*(32*A^2*b^8 + 8*C^2*b^8 + 512*A^2*a^2*b^6 + 512*A^2*a^4*b...