Integrand size = 28, antiderivative size = 67 \[ \int \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {a^2 \text {arctanh}(\sin (c+d x))}{d}-\frac {b^2 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {2 a b \sec (c+d x)}{d}+\frac {b^2 \sec (c+d x) \tan (c+d x)}{2 d} \] Output:
a^2*arctanh(sin(d*x+c))/d-1/2*b^2*arctanh(sin(d*x+c))/d+2*a*b*sec(d*x+c)/d +1/2*b^2*sec(d*x+c)*tan(d*x+c)/d
Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00 \[ \int \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {a^2 \coth ^{-1}(\sin (c+d x))}{d}-\frac {b^2 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {2 a b \sec (c+d x)}{d}+\frac {b^2 \sec (c+d x) \tan (c+d x)}{2 d} \] Input:
Integrate[Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^2,x]
Output:
(a^2*ArcCoth[Sin[c + d*x]])/d - (b^2*ArcTanh[Sin[c + d*x]])/(2*d) + (2*a*b *Sec[c + d*x])/d + (b^2*Sec[c + d*x]*Tan[c + d*x])/(2*d)
Time = 0.29 (sec) , antiderivative size = 67, 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.
| \(\displaystyle \int \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \cos (c+d x)+b \sin (c+d x))^2}{\cos (c+d x)^3}dx\) |
| \(\Big \downarrow \) 3569 |
| \(\displaystyle \int \left (a^2 \sec (c+d x)+2 a b \tan (c+d x) \sec (c+d x)+b^2 \tan ^2(c+d x) \sec (c+d x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^2 \text {arctanh}(\sin (c+d x))}{d}+\frac {2 a b \sec (c+d x)}{d}-\frac {b^2 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {b^2 \tan (c+d x) \sec (c+d x)}{2 d}\) |
Input:
Int[Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^2,x]
Output:
(a^2*ArcTanh[Sin[c + d*x]])/d - (b^2*ArcTanh[Sin[c + d*x]])/(2*d) + (2*a*b *Sec[c + d*x])/d + (b^2*Sec[c + d*x]*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 = 0.38 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.24
| method | result | size |
| derivativedivides | \(\frac {a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+\frac {2 a b}{\cos \left (d x +c \right )}+b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(83\) |
| default | \(\frac {a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+\frac {2 a b}{\cos \left (d x +c \right )}+b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(83\) |
| parts | \(\frac {a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {2 a b \sec \left (d x +c \right )}{d}\) | \(86\) |
| parallelrisch | \(\frac {-\left (a^{2}-\frac {b^{2}}{2}\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (a^{2}-\frac {b^{2}}{2}\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+4 \left (a \cos \left (d x +c \right )-\frac {a \cos \left (2 d x +2 c \right )}{2}+\frac {b \sin \left (d x +c \right )}{4}-\frac {a}{2}\right ) b}{d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(122\) |
| risch | \(\frac {b \,{\mathrm e}^{i \left (d x +c \right )} \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+4 a \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +4 a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2}}{d}+\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2}}{d}-\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}\) | \(151\) |
| norman | \(\frac {\frac {4 a b}{d}+\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {3 b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}+\frac {3 b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}+\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}-\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-\frac {\left (2 a^{2}-b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (2 a^{2}-b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(223\) |
Input:
int(sec(d*x+c)^3*(a*cos(d*x+c)+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(a^2*ln(sec(d*x+c)+tan(d*x+c))+2*a*b/cos(d*x+c)+b^2*(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))))
Time = 0.08 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.43 \[ \int \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 8 \, a b \cos \left (d x + c\right ) + 2 \, b^{2} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate(sec(d*x+c)^3*(a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="fricas" )
Output:
1/4*((2*a^2 - b^2)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (2*a^2 - b^2)*co s(d*x + c)^2*log(-sin(d*x + c) + 1) + 8*a*b*cos(d*x + c) + 2*b^2*sin(d*x + c))/(d*cos(d*x + c)^2)
\[ \int \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\int \left (a \cos {\left (c + d x \right )} + b \sin {\left (c + d x \right )}\right )^{2} \sec ^{3}{\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)**3*(a*cos(d*x+c)+b*sin(d*x+c))**2,x)
Output:
Integral((a*cos(c + d*x) + b*sin(c + d*x))**2*sec(c + d*x)**3, x)
Time = 0.03 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.33 \[ \int \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=-\frac {b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 2 \, a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - \frac {8 \, a b}{\cos \left (d x + c\right )}}{4 \, d} \] Input:
integrate(sec(d*x+c)^3*(a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="maxima" )
Output:
-1/4*(b^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + log(sin(d*x + c) + 1) - l og(sin(d*x + c) - 1)) - 2*a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 8*a*b/cos(d*x + c))/d
Time = 0.17 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.82 \[ \int \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {{\left (2 \, a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (2 \, a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, a b\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \] Input:
integrate(sec(d*x+c)^3*(a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="giac")
Output:
1/2*((2*a^2 - b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (2*a^2 - b^2)*log( abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(b^2*tan(1/2*d*x + 1/2*c)^3 - 4*a*b*tan (1/2*d*x + 1/2*c)^2 + b^2*tan(1/2*d*x + 1/2*c) + 4*a*b)/(tan(1/2*d*x + 1/2 *c)^2 - 1)^2)/d
Time = 17.05 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.58 \[ \int \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+4\,a\,b}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,a^2-b^2\right )}{d} \] Input:
int((a*cos(c + d*x) + b*sin(c + d*x))^2/cos(c + d*x)^3,x)
Output:
(4*a*b + b^2*tan(c/2 + (d*x)/2)^3 + b^2*tan(c/2 + (d*x)/2) - 4*a*b*tan(c/2 + (d*x)/2)^2)/(d*(tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^2 + 1)) + ( atanh(tan(c/2 + (d*x)/2))*(2*a^2 - b^2))/d
Time = 0.17 (sec) , antiderivative size = 221, normalized size of antiderivative = 3.30 \[ \int \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {-4 \cos \left (d x +c \right ) a b -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b^{2}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{2}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{2}-4 \sin \left (d x +c \right )^{2} a b -\sin \left (d x +c \right ) b^{2}+4 a b}{2 d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(sec(d*x+c)^3*(a*cos(d*x+c)+b*sin(d*x+c))^2,x)
Output:
( - 4*cos(c + d*x)*a*b - 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2 + log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**2 + 2*log(tan((c + d*x)/2) - 1)*a**2 - log(tan((c + d*x)/2) - 1)*b**2 + 2*log(tan((c + d*x)/2) + 1)*s in(c + d*x)**2*a**2 - log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b**2 - 2*l og(tan((c + d*x)/2) + 1)*a**2 + log(tan((c + d*x)/2) + 1)*b**2 - 4*sin(c + d*x)**2*a*b - sin(c + d*x)*b**2 + 4*a*b)/(2*d*(sin(c + d*x)**2 - 1))