Integrand size = 21, antiderivative size = 94 \[ \int (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx=-a^2 x-\frac {3 b^2 x}{2}+\frac {2 a b \cos (c+d x)}{d}+\frac {2 a b \sec (c+d x)}{d}+\frac {a^2 \tan (c+d x)}{d}+\frac {3 b^2 \tan (c+d x)}{2 d}-\frac {b^2 \sin ^2(c+d x) \tan (c+d x)}{2 d} \]
-a^2*x-3/2*b^2*x+2*a*b*cos(d*x+c)/d+2*a*b*sec(d*x+c)/d+a^2*tan(d*x+c)/d+3/ 2*b^2*tan(d*x+c)/d-1/2*b^2*sin(d*x+c)^2*tan(d*x+c)/d
Time = 0.55 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.82 \[ \int (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\frac {-4 \left (2 a^2+3 b^2\right ) (c+d x)+b \sec (c+d x) (24 a+8 a \cos (2 (c+d x))+b \sin (3 (c+d x)))+\left (8 a^2+9 b^2\right ) \tan (c+d x)}{8 d} \]
(-4*(2*a^2 + 3*b^2)*(c + d*x) + b*Sec[c + d*x]*(24*a + 8*a*Cos[2*(c + d*x) ] + b*Sin[3*(c + d*x)]) + (8*a^2 + 9*b^2)*Tan[c + d*x])/(8*d)
Time = 0.31 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {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.
\(\displaystyle \int \tan ^2(c+d x) (a+b \sin (c+d x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x)^2 (a+b \sin (c+d x))^2dx\) |
\(\Big \downarrow \) 3201 |
\(\displaystyle \int \left (a^2 \tan ^2(c+d x)+2 a b \sin (c+d x) \tan ^2(c+d x)+b^2 \sin ^2(c+d x) \tan ^2(c+d x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^2 \tan (c+d x)}{d}+a^2 (-x)+\frac {2 a b \cos (c+d x)}{d}+\frac {2 a b \sec (c+d x)}{d}+\frac {3 b^2 \tan (c+d x)}{2 d}-\frac {b^2 \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac {3 b^2 x}{2}\) |
-(a^2*x) - (3*b^2*x)/2 + (2*a*b*Cos[c + d*x])/d + (2*a*b*Sec[c + d*x])/d + (a^2*Tan[c + d*x])/d + (3*b^2*Tan[c + d*x])/(2*d) - (b^2*Sin[c + d*x]^2*T an[c + d*x])/(2*d)
3.2.56.3.1 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 = 2.19 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.23
method | result | size |
derivativedivides | \(\frac {a^{2} \left (\tan \left (d x +c \right )-d x -c \right )+2 a b \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+b^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )}{d}\) | \(116\) |
default | \(\frac {a^{2} \left (\tan \left (d x +c \right )-d x -c \right )+2 a b \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+b^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )}{d}\) | \(116\) |
parts | \(\frac {a^{2} \left (\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {b^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )}{d}+\frac {2 a b \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )}{d}\) | \(123\) |
risch | \(-a^{2} x -\frac {3 b^{2} x}{2}-\frac {i b^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {a b \,{\mathrm e}^{i \left (d x +c \right )}}{d}+\frac {a b \,{\mathrm e}^{-i \left (d x +c \right )}}{d}+\frac {i b^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i a^{2}+2 i b^{2}+4 a b \,{\mathrm e}^{i \left (d x +c \right )}}{d \left (1+{\mathrm e}^{2 i \left (d x +c \right )}\right )}\) | \(124\) |
1/d*(a^2*(tan(d*x+c)-d*x-c)+2*a*b*(1/cos(d*x+c)*sin(d*x+c)^4+(2+sin(d*x+c) ^2)*cos(d*x+c))+b^2*(sin(d*x+c)^5/cos(d*x+c)+(sin(d*x+c)^3+3/2*sin(d*x+c)) *cos(d*x+c)-3/2*d*x-3/2*c))
Time = 0.29 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.86 \[ \int (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx=-\frac {{\left (2 \, a^{2} + 3 \, b^{2}\right )} d x \cos \left (d x + c\right ) - 4 \, a b \cos \left (d x + c\right )^{2} - 4 \, a b - {\left (b^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} + 2 \, b^{2}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]
-1/2*((2*a^2 + 3*b^2)*d*x*cos(d*x + c) - 4*a*b*cos(d*x + c)^2 - 4*a*b - (b ^2*cos(d*x + c)^2 + 2*a^2 + 2*b^2)*sin(d*x + c))/(d*cos(d*x + c))
\[ \int (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right )^{2} \tan ^{2}{\left (c + d x \right )}\, dx \]
Time = 0.28 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.88 \[ \int (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx=-\frac {2 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{2} + {\left (3 \, d x + 3 \, c - \frac {\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} b^{2} - 4 \, a b {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )}}{2 \, d} \]
-1/2*(2*(d*x + c - tan(d*x + c))*a^2 + (3*d*x + 3*c - tan(d*x + c)/(tan(d* x + c)^2 + 1) - 2*tan(d*x + c))*b^2 - 4*a*b*(1/cos(d*x + c) + cos(d*x + c) ))/d
Leaf count of result is larger than twice the leaf count of optimal. 7670 vs. \(2 (88) = 176\).
Time = 5.68 (sec) , antiderivative size = 7670, normalized size of antiderivative = 81.60 \[ \int (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\text {Too large to display} \]
-1/2*(2*a^2*d*x*tan(d*x)^3*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c)^3 + 3*b^2*d* x*tan(d*x)^3*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c)^3 + 2*a^2*d*x*tan(d*x)^3*t an(1/2*d*x)^4*tan(1/2*c)^4*tan(c) + 3*b^2*d*x*tan(d*x)^3*tan(1/2*d*x)^4*ta n(1/2*c)^4*tan(c) - 2*a^2*d*x*tan(d*x)^2*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c )^2 - 3*b^2*d*x*tan(d*x)^2*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c)^2 - 8*a^2*d* x*tan(d*x)^3*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c)^3 - 12*b^2*d*x*tan(d*x)^3* tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c)^3 + 2*a^2*d*x*tan(d*x)*tan(1/2*d*x)^4*t an(1/2*c)^4*tan(c)^3 + 3*b^2*d*x*tan(d*x)*tan(1/2*d*x)^4*tan(1/2*c)^4*tan( c)^3 - 8*a*b*tan(d*x)^3*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c)^3 + 2*a^2*tan(d *x)^3*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c)^2 + 3*b^2*tan(d*x)^3*tan(1/2*d*x) ^4*tan(1/2*c)^4*tan(c)^2 + 2*a^2*tan(d*x)^2*tan(1/2*d*x)^4*tan(1/2*c)^4*ta n(c)^3 + 3*b^2*tan(d*x)^2*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c)^3 - 2*a^2*d*x *tan(d*x)^2*tan(1/2*d*x)^4*tan(1/2*c)^4 - 3*b^2*d*x*tan(d*x)^2*tan(1/2*d*x )^4*tan(1/2*c)^4 - 8*a^2*d*x*tan(d*x)^3*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c) - 12*b^2*d*x*tan(d*x)^3*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c) + 2*a^2*d*x*ta n(d*x)*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c) + 3*b^2*d*x*tan(d*x)*tan(1/2*d*x )^4*tan(1/2*c)^4*tan(c) - 8*a*b*tan(d*x)^3*tan(1/2*d*x)^4*tan(1/2*c)^4*tan (c) + 8*a^2*d*x*tan(d*x)^2*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c)^2 + 12*b^2*d *x*tan(d*x)^2*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c)^2 - 2*a^2*d*x*tan(1/2*d*x )^4*tan(1/2*c)^4*tan(c)^2 - 3*b^2*d*x*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c...
Time = 8.61 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.56 \[ \int (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\frac {\left (2\,a^2+3\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (4\,a^2+2\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+8\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (2\,a^2+3\,b^2\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+8\,a\,b}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {x\,\left (2\,a^2+3\,b^2\right )}{2} \]