Integrand size = 28, antiderivative size = 205 \[ \int \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\frac {5 a b^4 \text {arctanh}(\sin (c+d x))}{d}-\frac {10 a^2 b^3 \cos (c+d x)}{d}+\frac {2 b^5 \cos (c+d x)}{d}-\frac {5 a^4 b \cos ^3(c+d x)}{3 d}+\frac {10 a^2 b^3 \cos ^3(c+d x)}{3 d}-\frac {b^5 \cos ^3(c+d x)}{3 d}+\frac {b^5 \sec (c+d x)}{d}+\frac {a^5 \sin (c+d x)}{d}-\frac {5 a b^4 \sin (c+d x)}{d}-\frac {a^5 \sin ^3(c+d x)}{3 d}+\frac {10 a^3 b^2 \sin ^3(c+d x)}{3 d}-\frac {5 a b^4 \sin ^3(c+d x)}{3 d} \] Output:
5*a*b^4*arctanh(sin(d*x+c))/d-10*a^2*b^3*cos(d*x+c)/d+2*b^5*cos(d*x+c)/d-5 /3*a^4*b*cos(d*x+c)^3/d+10/3*a^2*b^3*cos(d*x+c)^3/d-1/3*b^5*cos(d*x+c)^3/d +b^5*sec(d*x+c)/d+a^5*sin(d*x+c)/d-5*a*b^4*sin(d*x+c)/d-1/3*a^5*sin(d*x+c) ^3/d+10/3*a^3*b^2*sin(d*x+c)^3/d-5/3*a*b^4*sin(d*x+c)^3/d
Leaf count is larger than twice the leaf count of optimal. \(632\) vs. \(2(205)=410\).
Time = 7.89 (sec) , antiderivative size = 632, normalized size of antiderivative = 3.08 \[ \int \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\frac {b^5 \cos ^5(c+d x) (a+b \tan (c+d x))^5}{d (a \cos (c+d x)+b \sin (c+d x))^5}-\frac {b \left (5 a^4+30 a^2 b^2-7 b^4\right ) \cos ^6(c+d x) (a+b \tan (c+d x))^5}{4 d (a \cos (c+d x)+b \sin (c+d x))^5}-\frac {b \left (5 a^4-10 a^2 b^2+b^4\right ) \cos ^5(c+d x) \cos (3 (c+d x)) (a+b \tan (c+d x))^5}{12 d (a \cos (c+d x)+b \sin (c+d x))^5}-\frac {5 a b^4 \cos ^5(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^5}{d (a \cos (c+d x)+b \sin (c+d x))^5}+\frac {5 a b^4 \cos ^5(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^5}{d (a \cos (c+d x)+b \sin (c+d x))^5}+\frac {b^5 \cos ^5(c+d x) \sin \left (\frac {1}{2} (c+d x)\right ) (a+b \tan (c+d x))^5}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^5}-\frac {b^5 \cos ^5(c+d x) \sin \left (\frac {1}{2} (c+d x)\right ) (a+b \tan (c+d x))^5}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^5}+\frac {a \left (3 a^4+10 a^2 b^2-25 b^4\right ) \cos ^5(c+d x) \sin (c+d x) (a+b \tan (c+d x))^5}{4 d (a \cos (c+d x)+b \sin (c+d x))^5}+\frac {a \left (a^4-10 a^2 b^2+5 b^4\right ) \cos ^5(c+d x) \sin (3 (c+d x)) (a+b \tan (c+d x))^5}{12 d (a \cos (c+d x)+b \sin (c+d x))^5} \] Input:
Integrate[Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^5,x]
Output:
(b^5*Cos[c + d*x]^5*(a + b*Tan[c + d*x])^5)/(d*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) - (b*(5*a^4 + 30*a^2*b^2 - 7*b^4)*Cos[c + d*x]^6*(a + b*Tan[c + d*x])^5)/(4*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) - (b*(5*a^4 - 10*a^2*b^ 2 + b^4)*Cos[c + d*x]^5*Cos[3*(c + d*x)]*(a + b*Tan[c + d*x])^5)/(12*d*(a* Cos[c + d*x] + b*Sin[c + d*x])^5) - (5*a*b^4*Cos[c + d*x]^5*Log[Cos[(c + d *x)/2] - Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x])^5)/(d*(a*Cos[c + d*x] + b* Sin[c + d*x])^5) + (5*a*b^4*Cos[c + d*x]^5*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x])^5)/(d*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) + (b^5*Cos[c + d*x]^5*Sin[(c + d*x)/2]*(a + b*Tan[c + d*x])^5)/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) - (b^5* Cos[c + d*x]^5*Sin[(c + d*x)/2]*(a + b*Tan[c + d*x])^5)/(d*(Cos[(c + d*x)/ 2] + Sin[(c + d*x)/2])*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) + (a*(3*a^4 + 10*a^2*b^2 - 25*b^4)*Cos[c + d*x]^5*Sin[c + d*x]*(a + b*Tan[c + d*x])^5)/( 4*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) + (a*(a^4 - 10*a^2*b^2 + 5*b^4)*C os[c + d*x]^5*Sin[3*(c + d*x)]*(a + b*Tan[c + d*x])^5)/(12*d*(a*Cos[c + d* x] + b*Sin[c + d*x])^5)
Time = 0.48 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {3042, 3569, 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.
| \(\displaystyle \int \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \cos (c+d x)+b \sin (c+d x))^5}{\cos (c+d x)^2}dx\) |
| \(\Big \downarrow \) 3569 |
| \(\displaystyle \int \left (a^5 \cos ^3(c+d x)+5 a^4 b \sin (c+d x) \cos ^2(c+d x)+10 a^3 b^2 \sin ^2(c+d x) \cos (c+d x)+10 a^2 b^3 \sin ^3(c+d x)+5 a b^4 \sin ^3(c+d x) \tan (c+d x)+b^5 \sin ^3(c+d x) \tan ^2(c+d x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^5 \sin ^3(c+d x)}{3 d}+\frac {a^5 \sin (c+d x)}{d}-\frac {5 a^4 b \cos ^3(c+d x)}{3 d}+\frac {10 a^3 b^2 \sin ^3(c+d x)}{3 d}+\frac {10 a^2 b^3 \cos ^3(c+d x)}{3 d}-\frac {10 a^2 b^3 \cos (c+d x)}{d}+\frac {5 a b^4 \text {arctanh}(\sin (c+d x))}{d}-\frac {5 a b^4 \sin ^3(c+d x)}{3 d}-\frac {5 a b^4 \sin (c+d x)}{d}-\frac {b^5 \cos ^3(c+d x)}{3 d}+\frac {2 b^5 \cos (c+d x)}{d}+\frac {b^5 \sec (c+d x)}{d}\) |
Input:
Int[Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^5,x]
Output:
(5*a*b^4*ArcTanh[Sin[c + d*x]])/d - (10*a^2*b^3*Cos[c + d*x])/d + (2*b^5*C os[c + d*x])/d - (5*a^4*b*Cos[c + d*x]^3)/(3*d) + (10*a^2*b^3*Cos[c + d*x] ^3)/(3*d) - (b^5*Cos[c + d*x]^3)/(3*d) + (b^5*Sec[c + d*x])/d + (a^5*Sin[c + d*x])/d - (5*a*b^4*Sin[c + d*x])/d - (a^5*Sin[c + d*x]^3)/(3*d) + (10*a ^3*b^2*Sin[c + d*x]^3)/(3*d) - (5*a*b^4*Sin[c + d*x]^3)/(3*d)
Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*si n[(c_.) + (d_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandTrig[cos[c + d*x]^m*(a *cos[c + d*x] + b*sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d}, x] && Inte gerQ[m] && IGtQ[n, 0]
Time = 0.89 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.82
| method | result | size |
| derivativedivides | \(\frac {\frac {a^{5} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}-\frac {5 a^{4} b \cos \left (d x +c \right )^{3}}{3}+\frac {10 a^{3} b^{2} \sin \left (d x +c \right )^{3}}{3}-\frac {10 a^{2} b^{3} \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}+5 b^{4} a \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+b^{5} \left (\frac {\sin \left (d x +c \right )^{6}}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )\right )}{d}\) | \(169\) |
| default | \(\frac {\frac {a^{5} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}-\frac {5 a^{4} b \cos \left (d x +c \right )^{3}}{3}+\frac {10 a^{3} b^{2} \sin \left (d x +c \right )^{3}}{3}-\frac {10 a^{2} b^{3} \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}+5 b^{4} a \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+b^{5} \left (\frac {\sin \left (d x +c \right )^{6}}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )\right )}{d}\) | \(169\) |
| parts | \(\frac {a^{5} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3 d}+\frac {b^{5} \left (\frac {\sin \left (d x +c \right )^{6}}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )\right )}{d}+\frac {10 a^{3} b^{2} \sin \left (d x +c \right )^{3}}{3 d}+\frac {5 b^{4} a \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}-\frac {5 a^{4} b}{3 \sec \left (d x +c \right )^{3} d}+\frac {10 a^{2} b^{3} \left (\frac {\cos \left (d x +c \right )^{3}}{3}-\cos \left (d x +c \right )\right )}{d}\) | \(186\) |
| parallelrisch | \(\frac {-120 \cos \left (d x +c \right ) b^{4} a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+120 \cos \left (d x +c \right ) b^{4} a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+20 \left (-a^{4} b -4 a^{2} b^{3}+b^{5}\right ) \cos \left (2 d x +2 c \right )+\left (-5 a^{4} b +10 a^{2} b^{3}-b^{5}\right ) \cos \left (4 d x +4 c \right )+10 \left (a^{5}+2 a^{3} b^{2}-7 b^{4} a \right ) \sin \left (2 d x +2 c \right )+a \left (a^{4}-10 a^{2} b^{2}+5 b^{4}\right ) \sin \left (4 d x +4 c \right )-40 \left (\left (a^{4}+4 a^{2} b^{2}-\frac {8}{5} b^{4}\right ) \cos \left (d x +c \right )+\frac {3 a^{4}}{8}+\frac {9 a^{2} b^{2}}{4}-\frac {9 b^{4}}{8}\right ) b}{24 d \cos \left (d x +c \right )}\) | \(225\) |
| risch | \(-\frac {5 \,{\mathrm e}^{i \left (d x +c \right )} a^{4} b}{8 d}-\frac {15 \,{\mathrm e}^{i \left (d x +c \right )} a^{2} b^{3}}{4 d}+\frac {7 \,{\mathrm e}^{i \left (d x +c \right )} b^{5}}{8 d}+\frac {25 i {\mathrm e}^{i \left (d x +c \right )} b^{4} a}{8 d}-\frac {5 i {\mathrm e}^{i \left (d x +c \right )} a^{3} b^{2}}{4 d}+\frac {5 i {\mathrm e}^{-i \left (d x +c \right )} a^{3} b^{2}}{4 d}-\frac {5 \,{\mathrm e}^{-i \left (d x +c \right )} a^{4} b}{8 d}-\frac {15 \,{\mathrm e}^{-i \left (d x +c \right )} a^{2} b^{3}}{4 d}+\frac {7 \,{\mathrm e}^{-i \left (d x +c \right )} b^{5}}{8 d}-\frac {25 i {\mathrm e}^{-i \left (d x +c \right )} b^{4} a}{8 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} a^{5}}{8 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} a^{5}}{8 d}+\frac {2 b^{5} {\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {5 b^{4} a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {5 b^{4} a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {5 b \cos \left (3 d x +3 c \right ) a^{4}}{12 d}+\frac {5 b^{3} \cos \left (3 d x +3 c \right ) a^{2}}{6 d}-\frac {b^{5} \cos \left (3 d x +3 c \right )}{12 d}+\frac {a^{5} \sin \left (3 d x +3 c \right )}{12 d}-\frac {5 a^{3} \sin \left (3 d x +3 c \right ) b^{2}}{6 d}+\frac {5 a \sin \left (3 d x +3 c \right ) b^{4}}{12 d}\) | \(412\) |
| norman | \(\frac {\frac {10 a^{4} b +40 a^{2} b^{3}-16 b^{5}}{3 d}-\frac {10 a^{4} b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}-\frac {5 \left (2 a^{4} b +8 a^{2} b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}+\frac {5 \left (4 a^{4} b +16 a^{2} b^{3}-16 b^{5}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 d}+\frac {2 \left (5 a^{4} b +80 a^{2} b^{3}-32 b^{5}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3 d}-\frac {2 a \left (a^{4}-5 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a \left (a^{4}-5 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}-\frac {10 a \left (a^{4}+8 a^{2} b^{2}-13 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {10 a \left (a^{4}+8 a^{2} b^{2}-13 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}-\frac {4 a \left (a^{4}+20 a^{2} b^{2}-25 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}+\frac {4 a \left (a^{4}+20 a^{2} b^{2}-25 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}+\frac {4 b \left (5 a^{4}-40 a^{2} b^{2}-8 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}-\frac {5 b^{4} a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {5 b^{4} a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\) | \(446\) |
Input:
int(sec(d*x+c)^2*(a*cos(d*x+c)+b*sin(d*x+c))^5,x,method=_RETURNVERBOSE)
Output:
1/d*(1/3*a^5*(2+cos(d*x+c)^2)*sin(d*x+c)-5/3*a^4*b*cos(d*x+c)^3+10/3*a^3*b ^2*sin(d*x+c)^3-10/3*a^2*b^3*(2+sin(d*x+c)^2)*cos(d*x+c)+5*b^4*a*(-1/3*sin (d*x+c)^3-sin(d*x+c)+ln(sec(d*x+c)+tan(d*x+c)))+b^5*(sin(d*x+c)^6/cos(d*x+ c)+(8/3+sin(d*x+c)^4+4/3*sin(d*x+c)^2)*cos(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.86 \[ \int \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\frac {15 \, a b^{4} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, a b^{4} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 6 \, b^{5} - 2 \, {\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{4} - 12 \, {\left (5 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left ({\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (a^{5} + 5 \, a^{3} b^{2} - 10 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )} \] Input:
integrate(sec(d*x+c)^2*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="fricas" )
Output:
1/6*(15*a*b^4*cos(d*x + c)*log(sin(d*x + c) + 1) - 15*a*b^4*cos(d*x + c)*l og(-sin(d*x + c) + 1) + 6*b^5 - 2*(5*a^4*b - 10*a^2*b^3 + b^5)*cos(d*x + c )^4 - 12*(5*a^2*b^3 - b^5)*cos(d*x + c)^2 + 2*((a^5 - 10*a^3*b^2 + 5*a*b^4 )*cos(d*x + c)^3 + 2*(a^5 + 5*a^3*b^2 - 10*a*b^4)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c))
Timed out. \[ \int \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**2*(a*cos(d*x+c)+b*sin(d*x+c))**5,x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.79 \[ \int \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=-\frac {10 \, a^{4} b \cos \left (d x + c\right )^{3} - 20 \, a^{3} b^{2} \sin \left (d x + c\right )^{3} + 2 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{5} - 20 \, {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a^{2} b^{3} + 5 \, {\left (2 \, \sin \left (d x + c\right )^{3} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, \sin \left (d x + c\right )\right )} a b^{4} + 2 \, {\left (\cos \left (d x + c\right )^{3} - \frac {3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} b^{5}}{6 \, d} \] Input:
integrate(sec(d*x+c)^2*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="maxima" )
Output:
-1/6*(10*a^4*b*cos(d*x + c)^3 - 20*a^3*b^2*sin(d*x + c)^3 + 2*(sin(d*x + c )^3 - 3*sin(d*x + c))*a^5 - 20*(cos(d*x + c)^3 - 3*cos(d*x + c))*a^2*b^3 + 5*(2*sin(d*x + c)^3 - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1) + 6*sin(d*x + c))*a*b^4 + 2*(cos(d*x + c)^3 - 3/cos(d*x + c) - 6*cos(d*x + c))*b^5)/d
Time = 0.25 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.38 \[ \int \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\frac {15 \, a b^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, a b^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {6 \, b^{5}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + \frac {2 \, {\left (3 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 3 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 50 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 60 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, a^{4} b - 20 \, a^{2} b^{3} + 5 \, b^{5}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{3 \, d} \] Input:
integrate(sec(d*x+c)^2*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="giac")
Output:
1/3*(15*a*b^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*a*b^4*log(abs(tan(1/ 2*d*x + 1/2*c) - 1)) - 6*b^5/(tan(1/2*d*x + 1/2*c)^2 - 1) + 2*(3*a^5*tan(1 /2*d*x + 1/2*c)^5 - 15*a*b^4*tan(1/2*d*x + 1/2*c)^5 - 15*a^4*b*tan(1/2*d*x + 1/2*c)^4 + 3*b^5*tan(1/2*d*x + 1/2*c)^4 + 2*a^5*tan(1/2*d*x + 1/2*c)^3 + 40*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 - 50*a*b^4*tan(1/2*d*x + 1/2*c)^3 - 60 *a^2*b^3*tan(1/2*d*x + 1/2*c)^2 + 12*b^5*tan(1/2*d*x + 1/2*c)^2 + 3*a^5*ta n(1/2*d*x + 1/2*c) - 15*a*b^4*tan(1/2*d*x + 1/2*c) - 5*a^4*b - 20*a^2*b^3 + 5*b^5)/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d
Time = 20.23 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.35 \[ \int \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\frac {10\,a\,b^4\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (10\,a\,b^4-2\,a^5\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (10\,a^4\,b-40\,a^2\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {2\,a^5}{3}-\frac {80\,a^3\,b^2}{3}+\frac {70\,a\,b^4}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {2\,a^5}{3}-\frac {80\,a^3\,b^2}{3}+\frac {70\,a\,b^4}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {10\,a^4\,b}{3}-\frac {80\,a^2\,b^3}{3}+\frac {32\,b^5}{3}\right )+\frac {10\,a^4\,b}{3}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (10\,a\,b^4-2\,a^5\right )-\frac {16\,b^5}{3}+\frac {40\,a^2\,b^3}{3}-10\,a^4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \] Input:
int((a*cos(c + d*x) + b*sin(c + d*x))^5/cos(c + d*x)^2,x)
Output:
(10*a*b^4*atanh(tan(c/2 + (d*x)/2)))/d - (tan(c/2 + (d*x)/2)*(10*a*b^4 - 2 *a^5) + tan(c/2 + (d*x)/2)^4*(10*a^4*b - 40*a^2*b^3) + tan(c/2 + (d*x)/2)^ 3*((70*a*b^4)/3 + (2*a^5)/3 - (80*a^3*b^2)/3) - tan(c/2 + (d*x)/2)^5*((70* a*b^4)/3 + (2*a^5)/3 - (80*a^3*b^2)/3) - tan(c/2 + (d*x)/2)^2*((10*a^4*b)/ 3 + (32*b^5)/3 - (80*a^2*b^3)/3) + (10*a^4*b)/3 - tan(c/2 + (d*x)/2)^7*(10 *a*b^4 - 2*a^5) - (16*b^5)/3 + (40*a^2*b^3)/3 - 10*a^4*b*tan(c/2 + (d*x)/2 )^6)/(d*(2*tan(c/2 + (d*x)/2)^2 - 2*tan(c/2 + (d*x)/2)^6 - tan(c/2 + (d*x) /2)^8 + 1))
Time = 0.15 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.46 \[ \int \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\frac {-15 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a \,b^{4}+15 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a \,b^{4}-\cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{5}+10 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{3} b^{2}-5 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a \,b^{4}+3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{5}-15 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{4}+5 \cos \left (d x +c \right ) a^{4} b +20 \cos \left (d x +c \right ) a^{2} b^{3}-8 \cos \left (d x +c \right ) b^{5}-5 \sin \left (d x +c \right )^{4} a^{4} b +10 \sin \left (d x +c \right )^{4} a^{2} b^{3}-\sin \left (d x +c \right )^{4} b^{5}+10 \sin \left (d x +c \right )^{2} a^{4} b +10 \sin \left (d x +c \right )^{2} a^{2} b^{3}-4 \sin \left (d x +c \right )^{2} b^{5}-5 a^{4} b -20 a^{2} b^{3}+8 b^{5}}{3 \cos \left (d x +c \right ) d} \] Input:
int(sec(d*x+c)^2*(a*cos(d*x+c)+b*sin(d*x+c))^5,x)
Output:
( - 15*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a*b**4 + 15*cos(c + d*x)*log (tan((c + d*x)/2) + 1)*a*b**4 - cos(c + d*x)*sin(c + d*x)**3*a**5 + 10*cos (c + d*x)*sin(c + d*x)**3*a**3*b**2 - 5*cos(c + d*x)*sin(c + d*x)**3*a*b** 4 + 3*cos(c + d*x)*sin(c + d*x)*a**5 - 15*cos(c + d*x)*sin(c + d*x)*a*b**4 + 5*cos(c + d*x)*a**4*b + 20*cos(c + d*x)*a**2*b**3 - 8*cos(c + d*x)*b**5 - 5*sin(c + d*x)**4*a**4*b + 10*sin(c + d*x)**4*a**2*b**3 - sin(c + d*x)* *4*b**5 + 10*sin(c + d*x)**2*a**4*b + 10*sin(c + d*x)**2*a**2*b**3 - 4*sin (c + d*x)**2*b**5 - 5*a**4*b - 20*a**2*b**3 + 8*b**5)/(3*cos(c + d*x)*d)