Integrand size = 21, antiderivative size = 77 \[ \int (a+b \sec (c+d x))^2 \sin ^2(c+d x) \, dx=\frac {a^2 x}{2}-b^2 x+\frac {2 a b \text {arctanh}(\sin (c+d x))}{d}-\frac {2 a b \sin (c+d x)}{d}-\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {b^2 \tan (c+d x)}{d} \]
1/2*a^2*x-b^2*x+2*a*b*arctanh(sin(d*x+c))/d-2*a*b*sin(d*x+c)/d-1/2*a^2*cos (d*x+c)*sin(d*x+c)/d+b^2*tan(d*x+c)/d
Time = 0.97 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.57 \[ \int (a+b \sec (c+d x))^2 \sin ^2(c+d x) \, dx=-\frac {-2 a^2 c+4 b^2 c-2 a^2 d x+4 b^2 d x+8 a b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-8 a b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+8 a b \sin (c+d x)+a^2 \sin (2 (c+d x))-4 b^2 \tan (c+d x)}{4 d} \]
-1/4*(-2*a^2*c + 4*b^2*c - 2*a^2*d*x + 4*b^2*d*x + 8*a*b*Log[Cos[(c + d*x) /2] - Sin[(c + d*x)/2]] - 8*a*b*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 8*a*b*Sin[c + d*x] + a^2*Sin[2*(c + d*x)] - 4*b^2*Tan[c + d*x])/d
Time = 0.38 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 4359, 3042, 3201, 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.
\(\displaystyle \int \sin ^2(c+d x) (a+b \sec (c+d x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos \left (c+d x-\frac {\pi }{2}\right )^2 \left (a-b \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2dx\) |
\(\Big \downarrow \) 4359 |
\(\displaystyle \int \tan ^2(c+d x) (-a \cos (c+d x)-b)^2dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \sin \left (c+d x-\frac {\pi }{2}\right )-b\right )^2}{\tan \left (c+d x-\frac {\pi }{2}\right )^2}dx\) |
\(\Big \downarrow \) 3201 |
\(\displaystyle \int \left (a^2 \sin ^2(c+d x)+2 a b \sin (c+d x) \tan (c+d x)+b^2 \tan ^2(c+d x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {a^2 x}{2}+\frac {2 a b \text {arctanh}(\sin (c+d x))}{d}-\frac {2 a b \sin (c+d x)}{d}+\frac {b^2 \tan (c+d x)}{d}-b^2 x\) |
(a^2*x)/2 - b^2*x + (2*a*b*ArcTanh[Sin[c + d*x]])/d - (2*a*b*Sin[c + d*x]) /d - (a^2*Cos[c + d*x]*Sin[c + d*x])/(2*d) + (b^2*Tan[c + d*x])/d
3.2.81.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((g_.)*tan[(e_.) + (f_.)*( x_)])^(p_.), x_Symbol] :> Int[ExpandIntegrand[(g*Tan[e + f*x])^p, (a + b*Si n[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m _.), x_Symbol] :> Int[Cot[e + f*x]^p*(b + a*Sin[e + f*x])^m, x] /; FreeQ[{a , b, e, f, p}, x] && IntegerQ[m] && EqQ[m, p]
Time = 1.38 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00
method | result | size |
derivativedivides | \(\frac {a^{2} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 a b \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+b^{2} \left (\tan \left (d x +c \right )-d x -c \right )}{d}\) | \(77\) |
default | \(\frac {a^{2} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 a b \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+b^{2} \left (\tan \left (d x +c \right )-d x -c \right )}{d}\) | \(77\) |
parts | \(\frac {a^{2} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {b^{2} \left (\tan \left (d x +c \right )-d x -c \right )}{d}+\frac {2 a b \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(82\) |
parallelrisch | \(\frac {-16 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )+16 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )-8 a b \sin \left (2 d x +2 c \right )-a^{2} \sin \left (3 d x +3 c \right )+4 d x \left (a^{2}-2 b^{2}\right ) \cos \left (d x +c \right )-\sin \left (d x +c \right ) \left (a^{2}-8 b^{2}\right )}{8 d \cos \left (d x +c \right )}\) | \(122\) |
risch | \(\frac {a^{2} x}{2}-b^{2} x +\frac {i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {i a b \,{\mathrm e}^{i \left (d x +c \right )}}{d}-\frac {i a b \,{\mathrm e}^{-i \left (d x +c \right )}}{d}-\frac {i a^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i b^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {2 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {2 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\) | \(146\) |
norman | \(\frac {\left (-\frac {a^{2}}{2}+b^{2}\right ) x +\left (-\frac {a^{2}}{2}+b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\frac {a^{2}}{2}-b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (\frac {a^{2}}{2}-b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {\left (a^{2}-4 a b -2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {\left (a^{2}+4 a b -2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 \left (a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-\frac {2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\) | \(232\) |
1/d*(a^2*(-1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+2*a*b*(-sin(d*x+c)+ln( sec(d*x+c)+tan(d*x+c)))+b^2*(tan(d*x+c)-d*x-c))
Time = 0.28 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.40 \[ \int (a+b \sec (c+d x))^2 \sin ^2(c+d x) \, dx=\frac {{\left (a^{2} - 2 \, b^{2}\right )} d x \cos \left (d x + c\right ) + 2 \, a b \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 2 \, a b \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (a^{2} \cos \left (d x + c\right )^{2} + 4 \, a b \cos \left (d x + c\right ) - 2 \, b^{2}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]
1/2*((a^2 - 2*b^2)*d*x*cos(d*x + c) + 2*a*b*cos(d*x + c)*log(sin(d*x + c) + 1) - 2*a*b*cos(d*x + c)*log(-sin(d*x + c) + 1) - (a^2*cos(d*x + c)^2 + 4 *a*b*cos(d*x + c) - 2*b^2)*sin(d*x + c))/(d*cos(d*x + c))
\[ \int (a+b \sec (c+d x))^2 \sin ^2(c+d x) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{2} \sin ^{2}{\left (c + d x \right )}\, dx \]
Time = 0.29 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.04 \[ \int (a+b \sec (c+d x))^2 \sin ^2(c+d x) \, dx=\frac {{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} - 4 \, {\left (d x + c - \tan \left (d x + c\right )\right )} b^{2} + 4 \, a b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, \sin \left (d x + c\right )\right )}}{4 \, d} \]
1/4*((2*d*x + 2*c - sin(2*d*x + 2*c))*a^2 - 4*(d*x + c - tan(d*x + c))*b^2 + 4*a*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1) - 2*sin(d*x + c))) /d
Leaf count of result is larger than twice the leaf count of optimal. 159 vs. \(2 (73) = 146\).
Time = 0.33 (sec) , antiderivative size = 159, normalized size of antiderivative = 2.06 \[ \int (a+b \sec (c+d x))^2 \sin ^2(c+d x) \, dx=\frac {4 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 4 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + {\left (a^{2} - 2 \, b^{2}\right )} {\left (d x + c\right )} - \frac {4 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + \frac {2 \, {\left (a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]
1/2*(4*a*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 4*a*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + (a^2 - 2*b^2)*(d*x + c) - 4*b^2*tan(1/2*d*x + 1/2*c)/(tan (1/2*d*x + 1/2*c)^2 - 1) + 2*(a^2*tan(1/2*d*x + 1/2*c)^3 - 4*a*b*tan(1/2*d *x + 1/2*c)^3 - a^2*tan(1/2*d*x + 1/2*c) - 4*a*b*tan(1/2*d*x + 1/2*c))/(ta n(1/2*d*x + 1/2*c)^2 + 1)^2)/d
Time = 14.66 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.86 \[ \int (a+b \sec (c+d x))^2 \sin ^2(c+d x) \, dx=\frac {a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {2\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {b^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}-\frac {2\,a\,b\,\sin \left (c+d\,x\right )}{d}+\frac {4\,a\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {a^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d} \]