Integrand size = 28, antiderivative size = 210 \[ \int \sec ^8(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=\frac {3 a^3 \text {arctanh}(\sin (c+d x))}{8 d}-\frac {3 a b^2 \text {arctanh}(\sin (c+d x))}{16 d}+\frac {3 a^2 b \sec ^5(c+d x)}{5 d}-\frac {b^3 \sec ^5(c+d x)}{5 d}+\frac {b^3 \sec ^7(c+d x)}{7 d}+\frac {3 a^3 \sec (c+d x) \tan (c+d x)}{8 d}-\frac {3 a b^2 \sec (c+d x) \tan (c+d x)}{16 d}+\frac {a^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {a b^2 \sec ^3(c+d x) \tan (c+d x)}{8 d}+\frac {a b^2 \sec ^5(c+d x) \tan (c+d x)}{2 d} \] Output:
3/8*a^3*arctanh(sin(d*x+c))/d-3/16*a*b^2*arctanh(sin(d*x+c))/d+3/5*a^2*b*s ec(d*x+c)^5/d-1/5*b^3*sec(d*x+c)^5/d+1/7*b^3*sec(d*x+c)^7/d+3/8*a^3*sec(d* x+c)*tan(d*x+c)/d-3/16*a*b^2*sec(d*x+c)*tan(d*x+c)/d+1/4*a^3*sec(d*x+c)^3* tan(d*x+c)/d-1/8*a*b^2*sec(d*x+c)^3*tan(d*x+c)/d+1/2*a*b^2*sec(d*x+c)^5*ta n(d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(637\) vs. \(2(210)=420\).
Time = 2.01 (sec) , antiderivative size = 637, normalized size of antiderivative = 3.03 \[ \int \sec ^8(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=\frac {\sec ^7(c+d x) \left (10752 a^2 b+1536 b^3+3584 \left (3 a^2 b-b^3\right ) \cos (2 (c+d x))-4410 a^3 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2205 a b^2 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-1470 a^3 \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 )+735 a b^2 \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 )-210 a^3 \cos (7 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+105 a b^2 \cos (7 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-3675 a \left (2 a^2-b^2\right ) \cos (c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+4410 a^3 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-2205 a b^2 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+1470 a^3 \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 )-735 a b^2 \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 )+210 a^3 \cos (7 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-105 a b^2 \cos (7 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+4340 a^3 \sin (2 (c+d x))+6790 a b^2 \sin (2 (c+d x))+2800 a^3 \sin (4 (c+d x))-1400 a b^2 \sin (4 (c+d x))+420 a^3 \sin (6 (c+d x))-210 a b^2 \sin (6 (c+d x))\right )}{35840 d} \] Input:
Integrate[Sec[c + d*x]^8*(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]
Output:
(Sec[c + d*x]^7*(10752*a^2*b + 1536*b^3 + 3584*(3*a^2*b - b^3)*Cos[2*(c + d*x)] - 4410*a^3*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2205*a*b^2*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 1470*a^3*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 735*a *b^2*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 210*a^3*C os[7*(c + d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 105*a*b^2*Cos[7 *(c + d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 3675*a*(2*a^2 - b^2 )*Cos[c + d*x]*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d* x)/2] + Sin[(c + d*x)/2]]) + 4410*a^3*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2 ] + Sin[(c + d*x)/2]] - 2205*a*b^2*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 1470*a^3*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] + Sin[ (c + d*x)/2]] - 735*a*b^2*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 210*a^3*Cos[7*(c + d*x)]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/ 2]] - 105*a*b^2*Cos[7*(c + d*x)]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 4340*a^3*Sin[2*(c + d*x)] + 6790*a*b^2*Sin[2*(c + d*x)] + 2800*a^3*Sin[4 *(c + d*x)] - 1400*a*b^2*Sin[4*(c + d*x)] + 420*a^3*Sin[6*(c + d*x)] - 210 *a*b^2*Sin[6*(c + d*x)]))/(35840*d)
Time = 0.45 (sec) , antiderivative size = 210, 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.
| \(\displaystyle \int \sec ^8(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \cos (c+d x)+b \sin (c+d x))^3}{\cos (c+d x)^8}dx\) |
| \(\Big \downarrow \) 3569 |
| \(\displaystyle \int \left (a^3 \sec ^5(c+d x)+3 a^2 b \tan (c+d x) \sec ^5(c+d x)+3 a b^2 \tan ^2(c+d x) \sec ^5(c+d x)+b^3 \tan ^3(c+d x) \sec ^5(c+d x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {3 a^3 \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^3 \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {3 a^3 \tan (c+d x) \sec (c+d x)}{8 d}+\frac {3 a^2 b \sec ^5(c+d x)}{5 d}-\frac {3 a b^2 \text {arctanh}(\sin (c+d x))}{16 d}+\frac {a b^2 \tan (c+d x) \sec ^5(c+d x)}{2 d}-\frac {a b^2 \tan (c+d x) \sec ^3(c+d x)}{8 d}-\frac {3 a b^2 \tan (c+d x) \sec (c+d x)}{16 d}+\frac {b^3 \sec ^7(c+d x)}{7 d}-\frac {b^3 \sec ^5(c+d x)}{5 d}\) |
Input:
Int[Sec[c + d*x]^8*(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]
Output:
(3*a^3*ArcTanh[Sin[c + d*x]])/(8*d) - (3*a*b^2*ArcTanh[Sin[c + d*x]])/(16* d) + (3*a^2*b*Sec[c + d*x]^5)/(5*d) - (b^3*Sec[c + d*x]^5)/(5*d) + (b^3*Se c[c + d*x]^7)/(7*d) + (3*a^3*Sec[c + d*x]*Tan[c + d*x])/(8*d) - (3*a*b^2*S ec[c + d*x]*Tan[c + d*x])/(16*d) + (a^3*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) - (a*b^2*Sec[c + d*x]^3*Tan[c + d*x])/(8*d) + (a*b^2*Sec[c + d*x]^5*Tan[c + d*x])/(2*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 = 1.16 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.89
| method | result | size |
| parts | \(\frac {a^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {b^{3} \left (\frac {\sec \left (d x +c \right )^{7}}{7}-\frac {\sec \left (d x +c \right )^{5}}{5}\right )}{d}+\frac {3 a^{2} b \sec \left (d x +c \right )^{5}}{5 d}+\frac {3 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )}{d}\) | \(186\) |
| derivativedivides | \(\frac {a^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {3 a^{2} b}{5 \cos \left (d x +c \right )^{5}}+3 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+b^{3} \left (\frac {\sin \left (d x +c \right )^{4}}{7 \cos \left (d x +c \right )^{7}}+\frac {3 \sin \left (d x +c \right )^{4}}{35 \cos \left (d x +c \right )^{5}}+\frac {\sin \left (d x +c \right )^{4}}{35 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{4}}{35 \cos \left (d x +c \right )}-\frac {\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{35}\right )}{d}\) | \(248\) |
| default | \(\frac {a^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {3 a^{2} b}{5 \cos \left (d x +c \right )^{5}}+3 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+b^{3} \left (\frac {\sin \left (d x +c \right )^{4}}{7 \cos \left (d x +c \right )^{7}}+\frac {3 \sin \left (d x +c \right )^{4}}{35 \cos \left (d x +c \right )^{5}}+\frac {\sin \left (d x +c \right )^{4}}{35 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{4}}{35 \cos \left (d x +c \right )}-\frac {\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{35}\right )}{d}\) | \(248\) |
| parallelrisch | \(\frac {-210 \left (a^{2}-\frac {b^{2}}{2}\right ) \left (\cos \left (7 d x +7 c \right )+7 \cos \left (5 d x +5 c \right )+21 \cos \left (3 d x +3 c \right )+35 \cos \left (d x +c \right )\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+210 \left (a^{2}-\frac {b^{2}}{2}\right ) \left (\cos \left (7 d x +7 c \right )+7 \cos \left (5 d x +5 c \right )+21 \cos \left (3 d x +3 c \right )+35 \cos \left (d x +c \right )\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (7056 a^{2} b -672 b^{3}\right ) \cos \left (3 d x +3 c \right )+\left (2352 a^{2} b -224 b^{3}\right ) \cos \left (5 d x +5 c \right )+\left (336 a^{2} b -32 b^{3}\right ) \cos \left (7 d x +7 c \right )+\left (10752 a^{2} b -3584 b^{3}\right ) \cos \left (2 d x +2 c \right )+\left (4340 a^{3}+6790 a \,b^{2}\right ) \sin \left (2 d x +2 c \right )+\left (2800 a^{3}-1400 a \,b^{2}\right ) \sin \left (4 d x +4 c \right )+\left (420 a^{3}-210 a \,b^{2}\right ) \sin \left (6 d x +6 c \right )+\left (11760 a^{2} b -1120 b^{3}\right ) \cos \left (d x +c \right )+10752 a^{2} b +1536 b^{3}}{560 d \left (\cos \left (7 d x +7 c \right )+7 \cos \left (5 d x +5 c \right )+21 \cos \left (3 d x +3 c \right )+35 \cos \left (d x +c \right )\right )}\) | \(361\) |
| risch | \(-\frac {{\mathrm e}^{i \left (d x +c \right )} \left (2170 i a^{3} {\mathrm e}^{8 i \left (d x +c \right )}+105 i a \,b^{2}+1400 i a^{3} {\mathrm e}^{10 i \left (d x +c \right )}+3395 i a \,b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+700 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-210 i a^{3}-5376 a^{2} b \,{\mathrm e}^{8 i \left (d x +c \right )}+1792 b^{3} {\mathrm e}^{8 i \left (d x +c \right )}-10752 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}-1536 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+210 i a^{3} {\mathrm e}^{12 i \left (d x +c \right )}-700 i a \,b^{2} {\mathrm e}^{10 i \left (d x +c \right )}-5376 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+1792 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-105 i a \,b^{2} {\mathrm e}^{12 i \left (d x +c \right )}-3395 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-1400 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-2170 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}\right )}{280 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{7}}+\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {3 a \,b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{16 d}-\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}+\frac {3 a \,b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{16 d}\) | \(405\) |
Input:
int(sec(d*x+c)^8*(a*cos(d*x+c)+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
a^3/d*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+ta n(d*x+c)))+b^3/d*(1/7*sec(d*x+c)^7-1/5*sec(d*x+c)^5)+3/5*a^2*b*sec(d*x+c)^ 5/d+3*a*b^2/d*(1/6*sin(d*x+c)^3/cos(d*x+c)^6+1/8*sin(d*x+c)^3/cos(d*x+c)^4 +1/16*sin(d*x+c)^3/cos(d*x+c)^2+1/16*sin(d*x+c)-1/16*ln(sec(d*x+c)+tan(d*x +c)))
Time = 0.09 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.81 \[ \int \sec ^8(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=\frac {105 \, {\left (2 \, a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{7} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left (2 \, a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{7} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 160 \, b^{3} + 224 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} + 70 \, {\left (3 \, {\left (2 \, a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{5} + 8 \, a b^{2} \cos \left (d x + c\right ) + 2 \, {\left (2 \, a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{1120 \, d \cos \left (d x + c\right )^{7}} \] Input:
integrate(sec(d*x+c)^8*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="fricas" )
Output:
1/1120*(105*(2*a^3 - a*b^2)*cos(d*x + c)^7*log(sin(d*x + c) + 1) - 105*(2* a^3 - a*b^2)*cos(d*x + c)^7*log(-sin(d*x + c) + 1) + 160*b^3 + 224*(3*a^2* b - b^3)*cos(d*x + c)^2 + 70*(3*(2*a^3 - a*b^2)*cos(d*x + c)^5 + 8*a*b^2*c os(d*x + c) + 2*(2*a^3 - a*b^2)*cos(d*x + c)^3)*sin(d*x + c))/(d*cos(d*x + c)^7)
Timed out. \[ \int \sec ^8(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**8*(a*cos(d*x+c)+b*sin(d*x+c))**3,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.99 \[ \int \sec ^8(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=\frac {35 \, a b^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 8 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 70 \, a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + \frac {672 \, a^{2} b}{\cos \left (d x + c\right )^{5}} - \frac {32 \, {\left (7 \, \cos \left (d x + c\right )^{2} - 5\right )} b^{3}}{\cos \left (d x + c\right )^{7}}}{1120 \, d} \] Input:
integrate(sec(d*x+c)^8*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="maxima" )
Output:
1/1120*(35*a*b^2*(2*(3*sin(d*x + c)^5 - 8*sin(d*x + c)^3 - 3*sin(d*x + c)) /(sin(d*x + c)^6 - 3*sin(d*x + c)^4 + 3*sin(d*x + c)^2 - 1) - 3*log(sin(d* x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 70*a^3*(2*(3*sin(d*x + c)^3 - 5*s in(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) + 672*a^2*b/cos(d*x + c)^5 - 32*(7*cos(d*x + c)^2 - 5)*b^3/cos(d*x + c)^7)/d
Leaf count of result is larger than twice the leaf count of optimal. 465 vs. \(2 (190) = 380\).
Time = 0.19 (sec) , antiderivative size = 465, normalized size of antiderivative = 2.21 \[ \int \sec ^8(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx =\text {Too large to display} \] Input:
integrate(sec(d*x+c)^8*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="giac")
Output:
1/560*(105*(2*a^3 - a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 105*(2*a^3 - a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(350*a^3*tan(1/2*d*x + 1/ 2*c)^13 + 105*a*b^2*tan(1/2*d*x + 1/2*c)^13 - 1680*a^2*b*tan(1/2*d*x + 1/2 *c)^12 - 840*a^3*tan(1/2*d*x + 1/2*c)^11 + 1540*a*b^2*tan(1/2*d*x + 1/2*c) ^11 + 3360*a^2*b*tan(1/2*d*x + 1/2*c)^10 - 1120*b^3*tan(1/2*d*x + 1/2*c)^1 0 + 630*a^3*tan(1/2*d*x + 1/2*c)^9 + 1085*a*b^2*tan(1/2*d*x + 1/2*c)^9 - 5 040*a^2*b*tan(1/2*d*x + 1/2*c)^8 - 1120*b^3*tan(1/2*d*x + 1/2*c)^8 + 6720* a^2*b*tan(1/2*d*x + 1/2*c)^6 - 2240*b^3*tan(1/2*d*x + 1/2*c)^6 - 630*a^3*t an(1/2*d*x + 1/2*c)^5 - 1085*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 3696*a^2*b*tan (1/2*d*x + 1/2*c)^4 - 448*b^3*tan(1/2*d*x + 1/2*c)^4 + 840*a^3*tan(1/2*d*x + 1/2*c)^3 - 1540*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 672*a^2*b*tan(1/2*d*x + 1/2*c)^2 - 224*b^3*tan(1/2*d*x + 1/2*c)^2 - 350*a^3*tan(1/2*d*x + 1/2*c) - 105*a*b^2*tan(1/2*d*x + 1/2*c) - 336*a^2*b + 32*b^3)/(tan(1/2*d*x + 1/2*c )^2 - 1)^7)/d
Time = 19.61 (sec) , antiderivative size = 423, normalized size of antiderivative = 2.01 \[ \int \sec ^8(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=\frac {3\,a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,a^2-b^2\right )}{8\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {5\,a^3}{4}+\frac {3\,a\,b^2}{8}\right )+\frac {6\,a^2\,b}{5}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {11\,a\,b^2}{2}-3\,a^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\left (\frac {11\,a\,b^2}{2}-3\,a^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}\,\left (\frac {5\,a^3}{4}+\frac {3\,a\,b^2}{8}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {9\,a^3}{4}+\frac {31\,a\,b^2}{8}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {9\,a^3}{4}+\frac {31\,a\,b^2}{8}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (12\,a^2\,b-4\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {12\,a^2\,b}{5}-\frac {4\,b^3}{5}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (18\,a^2\,b+4\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (24\,a^2\,b-8\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {66\,a^2\,b}{5}+\frac {8\,b^3}{5}\right )-\frac {4\,b^3}{35}+6\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}-7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+7\,{\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))^3/cos(c + d*x)^8,x)
Output:
(3*a*atanh(tan(c/2 + (d*x)/2))*(2*a^2 - b^2))/(8*d) - (tan(c/2 + (d*x)/2)* ((3*a*b^2)/8 + (5*a^3)/4) + (6*a^2*b)/5 + tan(c/2 + (d*x)/2)^3*((11*a*b^2) /2 - 3*a^3) - tan(c/2 + (d*x)/2)^11*((11*a*b^2)/2 - 3*a^3) - tan(c/2 + (d* x)/2)^13*((3*a*b^2)/8 + (5*a^3)/4) + tan(c/2 + (d*x)/2)^5*((31*a*b^2)/8 + (9*a^3)/4) - tan(c/2 + (d*x)/2)^9*((31*a*b^2)/8 + (9*a^3)/4) - tan(c/2 + ( d*x)/2)^10*(12*a^2*b - 4*b^3) - tan(c/2 + (d*x)/2)^2*((12*a^2*b)/5 - (4*b^ 3)/5) + tan(c/2 + (d*x)/2)^8*(18*a^2*b + 4*b^3) - tan(c/2 + (d*x)/2)^6*(24 *a^2*b - 8*b^3) + tan(c/2 + (d*x)/2)^4*((66*a^2*b)/5 + (8*b^3)/5) - (4*b^3 )/35 + 6*a^2*b*tan(c/2 + (d*x)/2)^12)/(d*(7*tan(c/2 + (d*x)/2)^2 - 21*tan( c/2 + (d*x)/2)^4 + 35*tan(c/2 + (d*x)/2)^6 - 35*tan(c/2 + (d*x)/2)^8 + 21* tan(c/2 + (d*x)/2)^10 - 7*tan(c/2 + (d*x)/2)^12 + tan(c/2 + (d*x)/2)^14 - 1))
Time = 0.16 (sec) , antiderivative size = 809, normalized size of antiderivative = 3.85 \[ \int \sec ^8(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^8*(a*cos(d*x+c)+b*sin(d*x+c))^3,x)
Output:
( - 210*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**6*a**3 + 105* cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**6*a*b**2 + 630*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**3 - 315*cos(c + d*x)*l og(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a*b**2 - 630*cos(c + d*x)*log(tan ((c + d*x)/2) - 1)*sin(c + d*x)**2*a**3 + 315*cos(c + d*x)*log(tan((c + d* x)/2) - 1)*sin(c + d*x)**2*a*b**2 + 210*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**3 - 105*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a*b**2 + 210*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**6*a**3 - 105*cos(c + d*x)*l og(tan((c + d*x)/2) + 1)*sin(c + d*x)**6*a*b**2 - 630*cos(c + d*x)*log(tan ((c + d*x)/2) + 1)*sin(c + d*x)**4*a**3 + 315*cos(c + d*x)*log(tan((c + d* x)/2) + 1)*sin(c + d*x)**4*a*b**2 + 630*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**3 - 315*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin (c + d*x)**2*a*b**2 - 210*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**3 + 10 5*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a*b**2 - 336*cos(c + d*x)*sin(c + d*x)**6*a**2*b + 32*cos(c + d*x)*sin(c + d*x)**6*b**3 - 210*cos(c + d*x)* sin(c + d*x)**5*a**3 + 105*cos(c + d*x)*sin(c + d*x)**5*a*b**2 + 1008*cos( c + d*x)*sin(c + d*x)**4*a**2*b - 96*cos(c + d*x)*sin(c + d*x)**4*b**3 + 5 60*cos(c + d*x)*sin(c + d*x)**3*a**3 - 280*cos(c + d*x)*sin(c + d*x)**3*a* b**2 - 1008*cos(c + d*x)*sin(c + d*x)**2*a**2*b + 96*cos(c + d*x)*sin(c + d*x)**2*b**3 - 350*cos(c + d*x)*sin(c + d*x)*a**3 - 105*cos(c + d*x)*si...