Integrand size = 28, antiderivative size = 168 \[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=\frac {a^4 \text {arctanh}(\sin (c+d x))}{d}-\frac {3 a^2 b^2 \text {arctanh}(\sin (c+d x))}{d}+\frac {3 b^4 \text {arctanh}(\sin (c+d x))}{8 d}+\frac {4 a^3 b \sec (c+d x)}{d}-\frac {4 a b^3 \sec (c+d x)}{d}+\frac {4 a b^3 \sec ^3(c+d x)}{3 d}+\frac {3 a^2 b^2 \sec (c+d x) \tan (c+d x)}{d}-\frac {3 b^4 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b^4 \sec (c+d x) \tan ^3(c+d x)}{4 d} \] Output:
a^4*arctanh(sin(d*x+c))/d-3*a^2*b^2*arctanh(sin(d*x+c))/d+3/8*b^4*arctanh( sin(d*x+c))/d+4*a^3*b*sec(d*x+c)/d-4*a*b^3*sec(d*x+c)/d+4/3*a*b^3*sec(d*x+ c)^3/d+3*a^2*b^2*sec(d*x+c)*tan(d*x+c)/d-3/8*b^4*sec(d*x+c)*tan(d*x+c)/d+1 /4*b^4*sec(d*x+c)*tan(d*x+c)^3/d
Leaf count is larger than twice the leaf count of optimal. \(936\) vs. \(2(168)=336\).
Time = 6.99 (sec) , antiderivative size = 936, normalized size of antiderivative = 5.57 \[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx =\text {Too large to display} \] Input:
Integrate[Sec[c + d*x]^5*(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]
Output:
(2*a*b*(6*a^2 - 5*b^2)*Cos[c + d*x]^4*(a + b*Tan[c + d*x])^4)/(3*d*(a*Cos[ c + d*x] + b*Sin[c + d*x])^4) + ((-8*a^4 + 24*a^2*b^2 - 3*b^4)*Cos[c + d*x ]^4*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x])^4)/(8*d* (a*Cos[c + d*x] + b*Sin[c + d*x])^4) + ((8*a^4 - 24*a^2*b^2 + 3*b^4)*Cos[c + d*x]^4*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x])^4) /(8*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + (b^4*Cos[c + d*x]^4*(a + b*Ta n[c + d*x])^4)/(16*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^4*(a*Cos[c + d* x] + b*Sin[c + d*x])^4) + ((72*a^2*b^2 + 16*a*b^3 - 15*b^4)*Cos[c + d*x]^4 *(a + b*Tan[c + d*x])^4)/(48*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2*(a* Cos[c + d*x] + b*Sin[c + d*x])^4) + (2*a*b^3*Cos[c + d*x]^4*Sin[(c + d*x)/ 2]*(a + b*Tan[c + d*x])^4)/(3*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*(a *Cos[c + d*x] + b*Sin[c + d*x])^4) - (b^4*Cos[c + d*x]^4*(a + b*Tan[c + d* x])^4)/(16*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4*(a*Cos[c + d*x] + b*S in[c + d*x])^4) - (2*a*b^3*Cos[c + d*x]^4*Sin[(c + d*x)/2]*(a + b*Tan[c + d*x])^4)/(3*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3*(a*Cos[c + d*x] + b* Sin[c + d*x])^4) + ((-72*a^2*b^2 + 16*a*b^3 + 15*b^4)*Cos[c + d*x]^4*(a + b*Tan[c + d*x])^4)/(48*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + (2*Cos[c + d*x]^4*(6*a^3*b*Sin[(c + d*x)/2] - 5*a*b^3*Sin[(c + d*x)/2])*(a + b*Tan[c + d*x])^4)/(3*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) - (2*Cos[c + ...
Time = 0.41 (sec) , antiderivative size = 168, 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.
| \(\displaystyle \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \cos (c+d x)+b \sin (c+d x))^4}{\cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3569 |
| \(\displaystyle \int \left (a^4 \sec (c+d x)+4 a^3 b \tan (c+d x) \sec (c+d x)+6 a^2 b^2 \tan ^2(c+d x) \sec (c+d x)+4 a b^3 \tan ^3(c+d x) \sec (c+d x)+b^4 \tan ^4(c+d x) \sec (c+d x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^4 \text {arctanh}(\sin (c+d x))}{d}+\frac {4 a^3 b \sec (c+d x)}{d}-\frac {3 a^2 b^2 \text {arctanh}(\sin (c+d x))}{d}+\frac {3 a^2 b^2 \tan (c+d x) \sec (c+d x)}{d}+\frac {4 a b^3 \sec ^3(c+d x)}{3 d}-\frac {4 a b^3 \sec (c+d x)}{d}+\frac {3 b^4 \text {arctanh}(\sin (c+d x))}{8 d}+\frac {b^4 \tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3 b^4 \tan (c+d x) \sec (c+d x)}{8 d}\) |
Input:
Int[Sec[c + d*x]^5*(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]
Output:
(a^4*ArcTanh[Sin[c + d*x]])/d - (3*a^2*b^2*ArcTanh[Sin[c + d*x]])/d + (3*b ^4*ArcTanh[Sin[c + d*x]])/(8*d) + (4*a^3*b*Sec[c + d*x])/d - (4*a*b^3*Sec[ c + d*x])/d + (4*a*b^3*Sec[c + d*x]^3)/(3*d) + (3*a^2*b^2*Sec[c + d*x]*Tan [c + d*x])/d - (3*b^4*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (b^4*Sec[c + d*x] *Tan[c + d*x]^3)/(4*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 = 0.91 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.18
| method | result | size |
| parts | \(\frac {a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {b^{4} \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {4 a^{3} b \sec \left (d x +c \right )}{d}+\frac {6 a^{2} b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {4 b^{3} a \left (\frac {\sec \left (d x +c \right )^{3}}{3}-\sec \left (d x +c \right )\right )}{d}\) | \(198\) |
| derivativedivides | \(\frac {a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+\frac {4 a^{3} b}{\cos \left (d x +c \right )}+6 a^{2} b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 b^{3} a \left (\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}\right )+b^{4} \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(225\) |
| default | \(\frac {a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+\frac {4 a^{3} b}{\cos \left (d x +c \right )}+6 a^{2} b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 b^{3} a \left (\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}\right )+b^{4} \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(225\) |
| parallelrisch | \(\frac {-4 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a^{4}-3 a^{2} b^{2}+\frac {3}{8} b^{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+4 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a^{4}-3 a^{2} b^{2}+\frac {3}{8} b^{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+24 \left (\frac {2 \left (a^{3}-\frac {2}{3} a \,b^{2}\right ) \cos \left (2 d x +2 c \right )}{3}+\frac {\left (\frac {1}{2} a^{3}-\frac {1}{3} a \,b^{2}\right ) \cos \left (4 d x +4 c \right )}{3}+\frac {\left (a^{3}-a \,b^{2}\right ) \cos \left (3 d x +3 c \right )}{3}+\frac {\left (a^{2} b -\frac {5}{24} b^{3}\right ) \sin \left (3 d x +3 c \right )}{4}+\left (a^{3}-\frac {5}{9} a \,b^{2}\right ) \cos \left (d x +c \right )+\frac {\left (a^{2} b +\frac {1}{8} b^{3}\right ) \sin \left (d x +c \right )}{4}+\frac {a^{3}}{2}-\frac {a \,b^{2}}{3}\right ) b}{d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(274\) |
| risch | \(\frac {b \,{\mathrm e}^{i \left (d x +c \right )} \left (72 i a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}+9 i b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+96 a^{3} {\mathrm e}^{6 i \left (d x +c \right )}-96 a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+72 i a^{2} b +15 i b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+288 a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-160 a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-72 i a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}-9 i b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+288 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-160 a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-15 i b^{3}-72 i a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+96 a^{3}-96 a \,b^{2}\right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{4}}{d}+\frac {3 b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2}}{d}-\frac {3 b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{4}}{d}-\frac {3 b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2}}{d}+\frac {3 b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}\) | \(392\) |
| norman | \(\frac {\frac {24 a^{3} b -16 b^{3} a}{3 d}-\frac {4 \left (2 a^{3} b +4 b^{3} a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{d}+\frac {2 \left (12 a^{3} b -8 b^{3} a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}-\frac {4 \left (18 a^{3} b -28 b^{3} a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 d}+\frac {8 a^{3} b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}-\frac {8 a^{3} b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{d}+\frac {3 b^{2} \left (8 a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {3 b^{2} \left (8 a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{4 d}-\frac {5 b^{2} \left (24 a^{2}-19 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}-\frac {5 b^{2} \left (24 a^{2}-19 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}+\frac {b^{2} \left (24 a^{2}+37 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 d}+\frac {b^{2} \left (24 a^{2}+37 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{4 d}+\frac {b^{2} \left (72 a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 d}+\frac {b^{2} \left (72 a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{4 d}-\frac {8 a b \left (9 a^{2}-16 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d}+\frac {8 a b \left (9 a^{2}-16 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{3 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}-\frac {\left (8 a^{4}-24 a^{2} b^{2}+3 b^{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (8 a^{4}-24 a^{2} b^{2}+3 b^{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(547\) |
Input:
int(sec(d*x+c)^5*(a*cos(d*x+c)+b*sin(d*x+c))^4,x,method=_RETURNVERBOSE)
Output:
a^4/d*ln(sec(d*x+c)+tan(d*x+c))+b^4/d*(1/4*sin(d*x+c)^5/cos(d*x+c)^4-1/8*s in(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)))+4*a^3*b*sec(d*x+c)/d+6*a^2*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)))+4*b^3*a/d*(1/3*sec(d*x+c )^3-sec(d*x+c))
Time = 0.09 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.97 \[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=\frac {3 \, {\left (8 \, a^{4} - 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (8 \, a^{4} - 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 64 \, a b^{3} \cos \left (d x + c\right ) + 192 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3} + 6 \, {\left (2 \, b^{4} + {\left (24 \, a^{2} b^{2} - 5 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \] Input:
integrate(sec(d*x+c)^5*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="fricas" )
Output:
1/48*(3*(8*a^4 - 24*a^2*b^2 + 3*b^4)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(8*a^4 - 24*a^2*b^2 + 3*b^4)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 6 4*a*b^3*cos(d*x + c) + 192*(a^3*b - a*b^3)*cos(d*x + c)^3 + 6*(2*b^4 + (24 *a^2*b^2 - 5*b^4)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^4)
Timed out. \[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**5*(a*cos(d*x+c)+b*sin(d*x+c))**4,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.14 \[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=\frac {3 \, b^{4} {\left (\frac {2 \, {\left (5 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} + 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 72 \, a^{2} b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + \frac {192 \, a^{3} b}{\cos \left (d x + c\right )} - \frac {64 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a b^{3}}{\cos \left (d x + c\right )^{3}}}{48 \, d} \] Input:
integrate(sec(d*x+c)^5*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="maxima" )
Output:
1/48*(3*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)) - 72 *a^2*b^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + log(sin(d*x + c) + 1) - lo g(sin(d*x + c) - 1)) + 24*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 192*a^3*b/cos(d*x + c) - 64*(3*cos(d*x + c)^2 - 1)*a*b^3/cos(d*x + c )^3)/d
Leaf count of result is larger than twice the leaf count of optimal. 325 vs. \(2 (160) = 320\).
Time = 0.22 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.93 \[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=\frac {3 \, {\left (8 \, a^{4} - 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (8 \, a^{4} - 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 9 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 96 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 33 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 288 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 192 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 33 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 288 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 256 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 96 \, a^{3} b - 64 \, a b^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \] Input:
integrate(sec(d*x+c)^5*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="giac")
Output:
1/24*(3*(8*a^4 - 24*a^2*b^2 + 3*b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(8*a^4 - 24*a^2*b^2 + 3*b^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(72* a^2*b^2*tan(1/2*d*x + 1/2*c)^7 - 9*b^4*tan(1/2*d*x + 1/2*c)^7 - 96*a^3*b*t an(1/2*d*x + 1/2*c)^6 - 72*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 33*b^4*tan(1/2 *d*x + 1/2*c)^5 + 288*a^3*b*tan(1/2*d*x + 1/2*c)^4 - 192*a*b^3*tan(1/2*d*x + 1/2*c)^4 - 72*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 33*b^4*tan(1/2*d*x + 1/2 *c)^3 - 288*a^3*b*tan(1/2*d*x + 1/2*c)^2 + 256*a*b^3*tan(1/2*d*x + 1/2*c)^ 2 + 72*a^2*b^2*tan(1/2*d*x + 1/2*c) - 9*b^4*tan(1/2*d*x + 1/2*c) + 96*a^3* b - 64*a*b^3)/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d
Time = 20.09 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.65 \[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,a^4-6\,a^2\,b^2+\frac {3\,b^4}{4}\right )}{d}-\frac {\frac {16\,a\,b^3}{3}-8\,a^3\,b+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,b^4}{4}-6\,a^2\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {3\,b^4}{4}-6\,a^2\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {11\,b^4}{4}-6\,a^2\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {11\,b^4}{4}-6\,a^2\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (16\,a\,b^3-24\,a^3\,b\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {64\,a\,b^3}{3}-24\,a^3\,b\right )+8\,a^3\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \] Input:
int((a*cos(c + d*x) + b*sin(c + d*x))^4/cos(c + d*x)^5,x)
Output:
(atanh(tan(c/2 + (d*x)/2))*(2*a^4 + (3*b^4)/4 - 6*a^2*b^2))/d - ((16*a*b^3 )/3 - 8*a^3*b + tan(c/2 + (d*x)/2)*((3*b^4)/4 - 6*a^2*b^2) + tan(c/2 + (d* x)/2)^7*((3*b^4)/4 - 6*a^2*b^2) - tan(c/2 + (d*x)/2)^3*((11*b^4)/4 - 6*a^2 *b^2) - tan(c/2 + (d*x)/2)^5*((11*b^4)/4 - 6*a^2*b^2) + tan(c/2 + (d*x)/2) ^4*(16*a*b^3 - 24*a^3*b) - tan(c/2 + (d*x)/2)^2*((64*a*b^3)/3 - 24*a^3*b) + 8*a^3*b*tan(c/2 + (d*x)/2)^6)/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + ( d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1))
Time = 0.16 (sec) , antiderivative size = 634, normalized size of antiderivative = 3.77 \[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^5*(a*cos(d*x+c)+b*sin(d*x+c))^4,x)
Output:
( - 96*cos(c + d*x)*sin(c + d*x)**2*a**3*b + 96*cos(c + d*x)*sin(c + d*x)* *2*a*b**3 + 96*cos(c + d*x)*a**3*b - 64*cos(c + d*x)*a*b**3 - 24*log(tan(( c + d*x)/2) - 1)*sin(c + d*x)**4*a**4 + 72*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**2*b**2 - 9*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b**4 + 48*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**4 - 144*log(tan((c + d*x)/ 2) - 1)*sin(c + d*x)**2*a**2*b**2 + 18*log(tan((c + d*x)/2) - 1)*sin(c + d *x)**2*b**4 - 24*log(tan((c + d*x)/2) - 1)*a**4 + 72*log(tan((c + d*x)/2) - 1)*a**2*b**2 - 9*log(tan((c + d*x)/2) - 1)*b**4 + 24*log(tan((c + d*x)/2 ) + 1)*sin(c + d*x)**4*a**4 - 72*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4 *a**2*b**2 + 9*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*b**4 - 48*log(tan ((c + d*x)/2) + 1)*sin(c + d*x)**2*a**4 + 144*log(tan((c + d*x)/2) + 1)*si n(c + d*x)**2*a**2*b**2 - 18*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b** 4 + 24*log(tan((c + d*x)/2) + 1)*a**4 - 72*log(tan((c + d*x)/2) + 1)*a**2* b**2 + 9*log(tan((c + d*x)/2) + 1)*b**4 - 96*sin(c + d*x)**4*a**3*b + 64*s in(c + d*x)**4*a*b**3 - 72*sin(c + d*x)**3*a**2*b**2 + 15*sin(c + d*x)**3* b**4 + 192*sin(c + d*x)**2*a**3*b - 128*sin(c + d*x)**2*a*b**3 + 72*sin(c + d*x)*a**2*b**2 - 9*sin(c + d*x)*b**4 - 96*a**3*b + 64*a*b**3)/(24*d*(sin (c + d*x)**4 - 2*sin(c + d*x)**2 + 1))