Integrand size = 19, antiderivative size = 77 \[ \int (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\frac {a^2 x}{2}-b^2 x+\frac {2 a b \text {arctanh}(\sin (c+d x))}{d}-\frac {2 a b \sin (c+d x)}{d}-\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {b^2 \tan (c+d x)}{d} \]
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 = 0.80 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.51 \[ \int (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=-\frac {-2 \left (a^2-2 b^2\right ) (c+d x)+8 a b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-8 a b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+8 a b \sin (c+d x)+\left (a^2-4 b^2+a^2 \cos (2 (c+d x))\right ) \tan (c+d x)}{4 d} \]
-1/4*(-2*(a^2 - 2*b^2)*(c + 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 - 4*b^2 + a^2*Cos[2*(c + d*x)])*Tan[c + d*x])/d
Time = 0.34 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3042, 4897, 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.
\(\displaystyle \int (a \sin (c+d x)+b \tan (c+d x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \sin (c+d x)+b \tan (c+d x))^2dx\) |
\(\Big \downarrow \) 4897 |
\(\displaystyle \int \tan ^2(c+d x) (a \cos (c+d x)+b)^2dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (b-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )^2}{\tan \left (c+d x-\frac {\pi }{2}\right )^2}dx\) |
\(\Big \downarrow \) 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
3.3.39.3.1 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]
Time = 1.23 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00
method | result | size |
derivativedivides | \(\frac {a^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 a b \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+b^{2} \left (\tan \left (d x +c \right )-d x -c \right )}{d}\) | \(77\) |
default | \(\frac {a^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 a b \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+b^{2} \left (\tan \left (d x +c \right )-d x -c \right )}{d}\) | \(77\) |
parts | \(\frac {a^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {b^{2} \left (\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {2 a b \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(84\) |
risch | \(\frac {a^{2} x}{2}-x \,b^{2}+\frac {i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {i a b \,{\mathrm e}^{i \left (d x +c \right )}}{d}-\frac {i a b \,{\mathrm e}^{-i \left (d x +c \right )}}{d}-\frac {i a^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i b^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {2 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {2 a b \ln \left (i+{\mathrm e}^{i \left (d x +c \right )}\right )}{d}\) | \(146\) |
1/d*(a^2*(-1/2*sin(d*x+c)*cos(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 = 0.26 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.40 \[ \int (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\frac {{\left (a^{2} - 2 \, b^{2}\right )} d x \cos \left (d x + c\right ) + 2 \, a b \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 2 \, a b \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (a^{2} \cos \left (d x + c\right )^{2} + 4 \, a b \cos \left (d x + c\right ) - 2 \, b^{2}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]
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))
\[ \int (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\int \left (a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}\right )^{2}\, dx \]
Time = 0.29 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.09 \[ \int (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\frac {{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2}}{4 \, d} - \frac {{\left (d x + c - \tan \left (d x + c\right )\right )} b^{2}}{d} + \frac {a b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, \sin \left (d x + c\right )\right )}}{d} \]
1/4*(2*d*x + 2*c - sin(2*d*x + 2*c))*a^2/d - (d*x + c - tan(d*x + c))*b^2/ d + 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. 2320 vs. \(2 (73) = 146\).
Time = 0.75 (sec) , antiderivative size = 2320, normalized size of antiderivative = 30.13 \[ \int (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\text {Too large to display} \]
1/2*a^2*x - 1/4*a^2*sin(2*d*x + 2*c)/d - (b^2*d*x*tan(d*x)*tan(1/2*d*x)^2* tan(1/2*c)^2*tan(c) + a*b*log(2*(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d *x)^2*tan(1/2*c) + 2*tan(1/2*d*x)*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2* c)^2 - 2*tan(1/2*d*x) - 2*tan(1/2*c) + 1)/(tan(1/2*d*x)^2*tan(1/2*c)^2 + t an(1/2*d*x)^2 + tan(1/2*c)^2 + 1))*tan(d*x)*tan(1/2*d*x)^2*tan(1/2*c)^2*ta n(c) - a*b*log(2*(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)^2*tan(1/2*c ) - 2*tan(1/2*d*x)*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 + 2*tan(1/ 2*d*x) + 2*tan(1/2*c) + 1)/(tan(1/2*d*x)^2*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 + 1))*tan(d*x)*tan(1/2*d*x)^2*tan(1/2*c)^2*tan(c) - b^2*d*x* tan(1/2*d*x)^2*tan(1/2*c)^2 + b^2*d*x*tan(d*x)*tan(1/2*d*x)^2*tan(c) + b^2 *d*x*tan(d*x)*tan(1/2*c)^2*tan(c) - a*b*log(2*(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)^2*tan(1/2*c) + 2*tan(1/2*d*x)*tan(1/2*c)^2 + tan(1/2*d*x )^2 + tan(1/2*c)^2 - 2*tan(1/2*d*x) - 2*tan(1/2*c) + 1)/(tan(1/2*d*x)^2*ta n(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 + 1))*tan(1/2*d*x)^2*tan(1/2*c) ^2 + a*b*log(2*(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)^2*tan(1/2*c) - 2*tan(1/2*d*x)*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 + 2*tan(1/2* d*x) + 2*tan(1/2*c) + 1)/(tan(1/2*d*x)^2*tan(1/2*c)^2 + tan(1/2*d*x)^2 + t an(1/2*c)^2 + 1))*tan(1/2*d*x)^2*tan(1/2*c)^2 + b^2*tan(d*x)*tan(1/2*d*x)^ 2*tan(1/2*c)^2 + a*b*log(2*(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)^2 *tan(1/2*c) + 2*tan(1/2*d*x)*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)...
Time = 22.87 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.86 \[ \int (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\frac {a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {2\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {b^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}-\frac {2\,a\,b\,\sin \left (c+d\,x\right )}{d}+\frac {4\,a\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {a^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d} \]