\(\int \sec ^6(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx\) [103]

Optimal result
Mathematica [B] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 28, antiderivative size = 224 \[ \int \sec ^6(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\frac {a^5 \text {arctanh}(\sin (c+d x))}{d}-\frac {5 a^3 b^2 \text {arctanh}(\sin (c+d x))}{d}+\frac {15 a b^4 \text {arctanh}(\sin (c+d x))}{8 d}+\frac {5 a^4 b \sec (c+d x)}{d}-\frac {10 a^2 b^3 \sec (c+d x)}{d}+\frac {b^5 \sec (c+d x)}{d}+\frac {10 a^2 b^3 \sec ^3(c+d x)}{3 d}-\frac {2 b^5 \sec ^3(c+d x)}{3 d}+\frac {b^5 \sec ^5(c+d x)}{5 d}+\frac {5 a^3 b^2 \sec (c+d x) \tan (c+d x)}{d}-\frac {15 a b^4 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {5 a b^4 \sec (c+d x) \tan ^3(c+d x)}{4 d} \] Output:

a^5*arctanh(sin(d*x+c))/d-5*a^3*b^2*arctanh(sin(d*x+c))/d+15/8*a*b^4*arcta 
nh(sin(d*x+c))/d+5*a^4*b*sec(d*x+c)/d-10*a^2*b^3*sec(d*x+c)/d+b^5*sec(d*x+ 
c)/d+10/3*a^2*b^3*sec(d*x+c)^3/d-2/3*b^5*sec(d*x+c)^3/d+1/5*b^5*sec(d*x+c) 
^5/d+5*a^3*b^2*sec(d*x+c)*tan(d*x+c)/d-15/8*a*b^4*sec(d*x+c)*tan(d*x+c)/d+ 
5/4*a*b^4*sec(d*x+c)*tan(d*x+c)^3/d
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(1219\) vs. \(2(224)=448\).

Time = 7.74 (sec) , antiderivative size = 1219, normalized size of antiderivative = 5.44 \[ \int \sec ^6(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx =\text {Too large to display} \] Input:

Integrate[Sec[c + d*x]^6*(a*Cos[c + d*x] + b*Sin[c + d*x])^5,x]
 

Output:

(b*(600*a^4 - 1000*a^2*b^2 + 89*b^4)*Cos[c + d*x]^5*(a + b*Tan[c + d*x])^5 
)/(120*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) + ((-8*a^5 + 40*a^3*b^2 - 15 
*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)/(8*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) + ((8*a^5 - 40*a^3 
*b^2 + 15*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)/(8*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) + ((25*a* 
b^4 + 2*b^5)*Cos[c + d*x]^5*(a + b*Tan[c + d*x])^5)/(80*d*(Cos[(c + d*x)/2 
] - Sin[(c + d*x)/2])^4*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) + ((600*a^3*b 
^2 + 200*a^2*b^3 - 375*a*b^4 - 31*b^5)*Cos[c + d*x]^5*(a + b*Tan[c + d*x]) 
^5)/(240*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^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)/(20*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5*(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) 
/(20*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5*(a*Cos[c + d*x] + b*Sin[c + 
 d*x])^5) + ((-25*a*b^4 + 2*b^5)*Cos[c + d*x]^5*(a + b*Tan[c + d*x])^5)/(8 
0*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4*(a*Cos[c + d*x] + b*Sin[c + d* 
x])^5) + ((-600*a^3*b^2 + 200*a^2*b^3 + 375*a*b^4 - 31*b^5)*Cos[c + d*x]^5 
*(a + b*Tan[c + d*x])^5)/(240*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2*(a 
*Cos[c + d*x] + b*Sin[c + d*x])^5) + (Cos[c + d*x]^5*(-600*a^4*b*Sin[(c + 
d*x)/2] + 1000*a^2*b^3*Sin[(c + d*x)/2] - 89*b^5*Sin[(c + d*x)/2])*(a +...
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 224, 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.

\(\displaystyle \int \sec ^6(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \cos (c+d x)+b \sin (c+d x))^5}{\cos (c+d x)^6}dx\)

\(\Big \downarrow \) 3569

\(\displaystyle \int \left (a^5 \sec (c+d x)+5 a^4 b \tan (c+d x) \sec (c+d x)+10 a^3 b^2 \tan ^2(c+d x) \sec (c+d x)+10 a^2 b^3 \tan ^3(c+d x) \sec (c+d x)+5 a b^4 \tan ^4(c+d x) \sec (c+d x)+b^5 \tan ^5(c+d x) \sec (c+d x)\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a^5 \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^4 b \sec (c+d x)}{d}-\frac {5 a^3 b^2 \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^3 b^2 \tan (c+d x) \sec (c+d x)}{d}+\frac {10 a^2 b^3 \sec ^3(c+d x)}{3 d}-\frac {10 a^2 b^3 \sec (c+d x)}{d}+\frac {15 a b^4 \text {arctanh}(\sin (c+d x))}{8 d}+\frac {5 a b^4 \tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {15 a b^4 \tan (c+d x) \sec (c+d x)}{8 d}+\frac {b^5 \sec ^5(c+d x)}{5 d}-\frac {2 b^5 \sec ^3(c+d x)}{3 d}+\frac {b^5 \sec (c+d x)}{d}\)

Input:

Int[Sec[c + d*x]^6*(a*Cos[c + d*x] + b*Sin[c + d*x])^5,x]
 

Output:

(a^5*ArcTanh[Sin[c + d*x]])/d - (5*a^3*b^2*ArcTanh[Sin[c + d*x]])/d + (15* 
a*b^4*ArcTanh[Sin[c + d*x]])/(8*d) + (5*a^4*b*Sec[c + d*x])/d - (10*a^2*b^ 
3*Sec[c + d*x])/d + (b^5*Sec[c + d*x])/d + (10*a^2*b^3*Sec[c + d*x]^3)/(3* 
d) - (2*b^5*Sec[c + d*x]^3)/(3*d) + (b^5*Sec[c + d*x]^5)/(5*d) + (5*a^3*b^ 
2*Sec[c + d*x]*Tan[c + d*x])/d - (15*a*b^4*Sec[c + d*x]*Tan[c + d*x])/(8*d 
) + (5*a*b^4*Sec[c + d*x]*Tan[c + d*x]^3)/(4*d)
 

Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3569
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]
 
Maple [A] (verified)

Time = 1.18 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.05

method result size
parts \(\frac {a^{5} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {b^{5} \left (\frac {\sec \left (d x +c \right )^{5}}{5}-\frac {2 \sec \left (d x +c \right )^{3}}{3}+\sec \left (d x +c \right )\right )}{d}+\frac {5 a^{4} b \sec \left (d x +c \right )}{d}+\frac {10 a^{3} b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {10 a^{2} b^{3} \left (\frac {\sec \left (d x +c \right )^{3}}{3}-\sec \left (d x +c \right )\right )}{d}+\frac {5 b^{4} a \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) \(236\)
derivativedivides \(\frac {a^{5} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+\frac {5 a^{4} b}{\cos \left (d x +c \right )}+10 a^{3} b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+10 a^{2} b^{3} \left (\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}\right )+5 b^{4} a \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b^{5} \left (\frac {\sin \left (d x +c \right )^{6}}{5 \cos \left (d x +c \right )^{5}}-\frac {\sin \left (d x +c \right )^{6}}{15 \cos \left (d x +c \right )^{3}}+\frac {\sin \left (d x +c \right )^{6}}{5 \cos \left (d x +c \right )}+\frac {\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )}{5}\right )}{d}\) \(316\)
default \(\frac {a^{5} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+\frac {5 a^{4} b}{\cos \left (d x +c \right )}+10 a^{3} b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+10 a^{2} b^{3} \left (\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}\right )+5 b^{4} a \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b^{5} \left (\frac {\sin \left (d x +c \right )^{6}}{5 \cos \left (d x +c \right )^{5}}-\frac {\sin \left (d x +c \right )^{6}}{15 \cos \left (d x +c \right )^{3}}+\frac {\sin \left (d x +c \right )^{6}}{5 \cos \left (d x +c \right )}+\frac {\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )}{5}\right )}{d}\) \(316\)
parallelrisch \(\frac {-\left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) \left (a^{4}-5 a^{2} b^{2}+\frac {15}{8} b^{4}\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) \left (a^{4}-5 a^{2} b^{2}+\frac {15}{8} b^{4}\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+5 \left (\left (5 a^{4}-\frac {20}{3} a^{2} b^{2}+\frac {8}{15} b^{4}\right ) \cos \left (3 d x +3 c \right )+\left (a^{4}-\frac {4}{3} a^{2} b^{2}+\frac {8}{75} b^{4}\right ) \cos \left (5 d x +5 c \right )+8 \left (a^{4}-\frac {4}{3} a^{2} b^{2}+\frac {1}{15} b^{4}\right ) \cos \left (2 d x +2 c \right )+2 \left (a^{4}-2 a^{2} b^{2}+\frac {1}{5} b^{4}\right ) \cos \left (4 d x +4 c \right )+\left (4 a^{3} b -\frac {1}{2} b^{3} a \right ) \sin \left (2 d x +2 c \right )+\left (2 a^{3} b -\frac {5}{4} b^{3} a \right ) \sin \left (4 d x +4 c \right )+2 \left (5 a^{4}-\frac {20}{3} a^{2} b^{2}+\frac {8}{15} b^{4}\right ) \cos \left (d x +c \right )+6 a^{4}-\frac {20 a^{2} b^{2}}{3}+\frac {58 b^{4}}{75}\right ) b}{d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\) \(365\)
risch \(\frac {b \,{\mathrm e}^{i \left (d x +c \right )} \left (150 i a \,b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+600 i a^{3} b +600 a^{4} {\mathrm e}^{8 i \left (d x +c \right )}-1200 a^{2} b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+120 b^{4} {\mathrm e}^{8 i \left (d x +c \right )}+375 i a \,b^{3} {\mathrm e}^{8 i \left (d x +c \right )}-600 i a^{3} b \,{\mathrm e}^{8 i \left (d x +c \right )}+2400 a^{4} {\mathrm e}^{6 i \left (d x +c \right )}-3200 a^{2} b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+160 b^{4} {\mathrm e}^{6 i \left (d x +c \right )}+3600 a^{4} {\mathrm e}^{4 i \left (d x +c \right )}-4000 a^{2} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+464 b^{4} {\mathrm e}^{4 i \left (d x +c \right )}-375 i b^{3} a -150 i a \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+2400 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}-3200 a^{2} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+160 b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+1200 i a^{3} b \,{\mathrm e}^{2 i \left (d x +c \right )}-1200 i a^{3} b \,{\mathrm e}^{6 i \left (d x +c \right )}+600 a^{4}-1200 a^{2} b^{2}+120 b^{4}\right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {5 a^{3} b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {15 b^{4} a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {5 a^{3} b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {15 b^{4} a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}\) \(510\)

Input:

int(sec(d*x+c)^6*(a*cos(d*x+c)+b*sin(d*x+c))^5,x,method=_RETURNVERBOSE)
 

Output:

a^5/d*ln(sec(d*x+c)+tan(d*x+c))+b^5/d*(1/5*sec(d*x+c)^5-2/3*sec(d*x+c)^3+s 
ec(d*x+c))+5*a^4*b*sec(d*x+c)/d+10*a^3*b^2/d*(1/2*sin(d*x+c)^3/cos(d*x+c)^ 
2+1/2*sin(d*x+c)-1/2*ln(sec(d*x+c)+tan(d*x+c)))+10*a^2*b^3/d*(1/3*sec(d*x+ 
c)^3-sec(d*x+c))+5*b^4*a/d*(1/4*sin(d*x+c)^5/cos(d*x+c)^4-1/8*sin(d*x+c)^5 
/cos(d*x+c)^2-1/8*sin(d*x+c)^3-3/8*sin(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c) 
))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.88 \[ \int \sec ^6(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\frac {15 \, {\left (8 \, a^{5} - 40 \, a^{3} b^{2} + 15 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (8 \, a^{5} - 40 \, a^{3} b^{2} + 15 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 48 \, b^{5} + 240 \, {\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{4} + 160 \, {\left (5 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} + 150 \, {\left (2 \, a b^{4} \cos \left (d x + c\right ) + {\left (8 \, a^{3} b^{2} - 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \] Input:

integrate(sec(d*x+c)^6*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="fricas" 
)
 

Output:

1/240*(15*(8*a^5 - 40*a^3*b^2 + 15*a*b^4)*cos(d*x + c)^5*log(sin(d*x + c) 
+ 1) - 15*(8*a^5 - 40*a^3*b^2 + 15*a*b^4)*cos(d*x + c)^5*log(-sin(d*x + c) 
 + 1) + 48*b^5 + 240*(5*a^4*b - 10*a^2*b^3 + b^5)*cos(d*x + c)^4 + 160*(5* 
a^2*b^3 - b^5)*cos(d*x + c)^2 + 150*(2*a*b^4*cos(d*x + c) + (8*a^3*b^2 - 5 
*a*b^4)*cos(d*x + c)^3)*sin(d*x + c))/(d*cos(d*x + c)^5)
 

Sympy [F(-1)]

Timed out. \[ \int \sec ^6(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\text {Timed out} \] Input:

integrate(sec(d*x+c)**6*(a*cos(d*x+c)+b*sin(d*x+c))**5,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.03 \[ \int \sec ^6(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\frac {75 \, a b^{4} {\left (\frac {2 \, {\left (5 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} + 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 600 \, a^{3} b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 120 \, a^{5} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + \frac {1200 \, a^{4} b}{\cos \left (d x + c\right )} - \frac {800 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a^{2} b^{3}}{\cos \left (d x + c\right )^{3}} + \frac {16 \, {\left (15 \, \cos \left (d x + c\right )^{4} - 10 \, \cos \left (d x + c\right )^{2} + 3\right )} b^{5}}{\cos \left (d x + c\right )^{5}}}{240 \, d} \] Input:

integrate(sec(d*x+c)^6*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="maxima" 
)
 

Output:

1/240*(75*a*b^4*(2*(5*sin(d*x + c)^3 - 3*sin(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)) 
- 600*a^3*b^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + log(sin(d*x + c) + 1) 
 - log(sin(d*x + c) - 1)) + 120*a^5*(log(sin(d*x + c) + 1) - log(sin(d*x + 
 c) - 1)) + 1200*a^4*b/cos(d*x + c) - 800*(3*cos(d*x + c)^2 - 1)*a^2*b^3/c 
os(d*x + c)^3 + 16*(15*cos(d*x + c)^4 - 10*cos(d*x + c)^2 + 3)*b^5/cos(d*x 
 + c)^5)/d
 

Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 410, normalized size of antiderivative = 1.83 \[ \int \sec ^6(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\frac {15 \, {\left (8 \, a^{5} - 40 \, a^{3} b^{2} + 15 \, a b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (8 \, a^{5} - 40 \, a^{3} b^{2} + 15 \, a b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (600 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 225 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 600 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 1200 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1050 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 2400 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 2400 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 3600 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 5600 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 640 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 1200 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1050 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2400 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4000 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 320 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 600 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 225 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 600 \, a^{4} b + 800 \, a^{2} b^{3} - 64 \, b^{5}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \] Input:

integrate(sec(d*x+c)^6*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="giac")
 

Output:

1/120*(15*(8*a^5 - 40*a^3*b^2 + 15*a*b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1 
)) - 15*(8*a^5 - 40*a^3*b^2 + 15*a*b^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) 
 + 2*(600*a^3*b^2*tan(1/2*d*x + 1/2*c)^9 - 225*a*b^4*tan(1/2*d*x + 1/2*c)^ 
9 - 600*a^4*b*tan(1/2*d*x + 1/2*c)^8 - 1200*a^3*b^2*tan(1/2*d*x + 1/2*c)^7 
 + 1050*a*b^4*tan(1/2*d*x + 1/2*c)^7 + 2400*a^4*b*tan(1/2*d*x + 1/2*c)^6 - 
 2400*a^2*b^3*tan(1/2*d*x + 1/2*c)^6 - 3600*a^4*b*tan(1/2*d*x + 1/2*c)^4 + 
 5600*a^2*b^3*tan(1/2*d*x + 1/2*c)^4 - 640*b^5*tan(1/2*d*x + 1/2*c)^4 + 12 
00*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 - 1050*a*b^4*tan(1/2*d*x + 1/2*c)^3 + 24 
00*a^4*b*tan(1/2*d*x + 1/2*c)^2 - 4000*a^2*b^3*tan(1/2*d*x + 1/2*c)^2 + 32 
0*b^5*tan(1/2*d*x + 1/2*c)^2 - 600*a^3*b^2*tan(1/2*d*x + 1/2*c) + 225*a*b^ 
4*tan(1/2*d*x + 1/2*c) - 600*a^4*b + 800*a^2*b^3 - 64*b^5)/(tan(1/2*d*x + 
1/2*c)^2 - 1)^5)/d
 

Mupad [B] (verification not implemented)

Time = 19.16 (sec) , antiderivative size = 345, normalized size of antiderivative = 1.54 \[ \int \sec ^6(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,a^5-10\,a^3\,b^2+\frac {15\,a\,b^4}{4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {15\,a\,b^4}{4}-10\,a^3\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {35\,a\,b^4}{2}-20\,a^3\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {35\,a\,b^4}{2}-20\,a^3\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (40\,a^4\,b-40\,a^2\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (40\,a^4\,b-\frac {200\,a^2\,b^3}{3}+\frac {16\,b^5}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (60\,a^4\,b-\frac {280\,a^2\,b^3}{3}+\frac {32\,b^5}{3}\right )+10\,a^4\,b+\frac {16\,b^5}{15}-\frac {40\,a^2\,b^3}{3}-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {15\,a\,b^4}{4}-10\,a^3\,b^2\right )+10\,a^4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:

int((a*cos(c + d*x) + b*sin(c + d*x))^5/cos(c + d*x)^6,x)
 

Output:

(atanh(tan(c/2 + (d*x)/2))*((15*a*b^4)/4 + 2*a^5 - 10*a^3*b^2))/d - (tan(c 
/2 + (d*x)/2)^9*((15*a*b^4)/4 - 10*a^3*b^2) + tan(c/2 + (d*x)/2)^3*((35*a* 
b^4)/2 - 20*a^3*b^2) - tan(c/2 + (d*x)/2)^7*((35*a*b^4)/2 - 20*a^3*b^2) - 
tan(c/2 + (d*x)/2)^6*(40*a^4*b - 40*a^2*b^3) - tan(c/2 + (d*x)/2)^2*(40*a^ 
4*b + (16*b^5)/3 - (200*a^2*b^3)/3) + tan(c/2 + (d*x)/2)^4*(60*a^4*b + (32 
*b^5)/3 - (280*a^2*b^3)/3) + 10*a^4*b + (16*b^5)/15 - (40*a^2*b^3)/3 - tan 
(c/2 + (d*x)/2)*((15*a*b^4)/4 - 10*a^3*b^2) + 10*a^4*b*tan(c/2 + (d*x)/2)^ 
8)/(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))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 914, normalized size of antiderivative = 4.08 \[ \int \sec ^6(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx =\text {Too large to display} \] Input:

int(sec(d*x+c)^6*(a*cos(d*x+c)+b*sin(d*x+c))^5,x)
 

Output:

( - 120*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**5 + 600* 
cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**3*b**2 - 225*cos 
(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a*b**4 + 240*cos(c + d 
*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**5 - 1200*cos(c + d*x)*log 
(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**3*b**2 + 450*cos(c + d*x)*log(ta 
n((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b**4 - 120*cos(c + d*x)*log(tan((c + 
 d*x)/2) - 1)*a**5 + 600*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**3*b**2 
- 225*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a*b**4 + 120*cos(c + d*x)*log 
(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a**5 - 600*cos(c + d*x)*log(tan((c 
+ d*x)/2) + 1)*sin(c + d*x)**4*a**3*b**2 + 225*cos(c + d*x)*log(tan((c + d 
*x)/2) + 1)*sin(c + d*x)**4*a*b**4 - 240*cos(c + d*x)*log(tan((c + d*x)/2) 
 + 1)*sin(c + d*x)**2*a**5 + 1200*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*s 
in(c + d*x)**2*a**3*b**2 - 450*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin( 
c + d*x)**2*a*b**4 + 120*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**5 - 600 
*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**3*b**2 + 225*cos(c + d*x)*log(t 
an((c + d*x)/2) + 1)*a*b**4 - 600*cos(c + d*x)*sin(c + d*x)**4*a**4*b + 80 
0*cos(c + d*x)*sin(c + d*x)**4*a**2*b**3 - 64*cos(c + d*x)*sin(c + d*x)**4 
*b**5 - 600*cos(c + d*x)*sin(c + d*x)**3*a**3*b**2 + 375*cos(c + d*x)*sin( 
c + d*x)**3*a*b**4 + 1200*cos(c + d*x)*sin(c + d*x)**2*a**4*b - 1600*cos(c 
 + d*x)*sin(c + d*x)**2*a**2*b**3 + 128*cos(c + d*x)*sin(c + d*x)**2*b*...