Integrand size = 23, antiderivative size = 51 \[ \int \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=-\frac {1}{2} b (4 a+3 b) x+\frac {b^2 \cos (e+f x) \sin (e+f x)}{2 f}+\frac {(a+b)^2 \tan (e+f x)}{f} \]
Time = 0.91 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.94 \[ \int \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {-2 b (4 a+3 b) (e+f x)+b^2 \sin (2 (e+f x))+4 (a+b)^2 \tan (e+f x)}{4 f} \]
Time = 0.27 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.20, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3670, 300, 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.
\(\displaystyle \int \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \sin (e+f x)^2\right )^2}{\cos (e+f x)^2}dx\) |
\(\Big \downarrow \) 3670 |
\(\displaystyle \frac {\int \frac {\left ((a+b) \tan ^2(e+f x)+a\right )^2}{\left (\tan ^2(e+f x)+1\right )^2}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 300 |
\(\displaystyle \frac {\int \left ((a+b)^2-\frac {2 b (a+b) \tan ^2(e+f x)+b (2 a+b)}{\left (\tan ^2(e+f x)+1\right )^2}\right )d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{2} b (4 a+3 b) \arctan (\tan (e+f x))+(a+b)^2 \tan (e+f x)+\frac {b^2 \tan (e+f x)}{2 \left (\tan ^2(e+f x)+1\right )}}{f}\) |
(-1/2*(b*(4*a + 3*b)*ArcTan[Tan[e + f*x]]) + (a + b)^2*Tan[e + f*x] + (b^2 *Tan[e + f*x])/(2*(1 + Tan[e + f*x]^2)))/f
3.3.96.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int [PolynomialDivide[(a + b*x^2)^p, (c + d*x^2)^(-q), x], x] /; FreeQ[{a, b, c , d}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^( p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Su bst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]
Time = 1.33 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.27
method | result | size |
parallelrisch | \(\frac {b^{2} \sin \left (3 f x +3 e \right )-16 b f \left (a +\frac {3 b}{4}\right ) x \cos \left (f x +e \right )+8 \sin \left (f x +e \right ) \left (a^{2}+2 a b +\frac {9}{8} b^{2}\right )}{8 f \cos \left (f x +e \right )}\) | \(65\) |
derivativedivides | \(\frac {a^{2} \tan \left (f x +e \right )+2 a b \left (\tan \left (f x +e \right )-f x -e \right )+b^{2} \left (\frac {\sin ^{5}\left (f x +e \right )}{\cos \left (f x +e \right )}+\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )-\frac {3 f x}{2}-\frac {3 e}{2}\right )}{f}\) | \(87\) |
default | \(\frac {a^{2} \tan \left (f x +e \right )+2 a b \left (\tan \left (f x +e \right )-f x -e \right )+b^{2} \left (\frac {\sin ^{5}\left (f x +e \right )}{\cos \left (f x +e \right )}+\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )-\frac {3 f x}{2}-\frac {3 e}{2}\right )}{f}\) | \(87\) |
risch | \(-2 a b x -\frac {3 b^{2} x}{2}-\frac {i b^{2} {\mathrm e}^{2 i \left (f x +e \right )}}{8 f}+\frac {i b^{2} {\mathrm e}^{-2 i \left (f x +e \right )}}{8 f}+\frac {2 i a^{2}}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}+\frac {4 i a b}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}+\frac {2 i b^{2}}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}\) | \(114\) |
norman | \(\frac {\left (2 a b +\frac {3}{2} b^{2}\right ) x +\left (-6 a b -\frac {9}{2} b^{2}\right ) x \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (-2 a b -\frac {3}{2} b^{2}\right ) x \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (6 a b +\frac {9}{2} b^{2}\right ) x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (-4 a b -3 b^{2}\right ) x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (4 a b +3 b^{2}\right ) x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {8 \left (a^{2}+2 a b +b^{2}\right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}-\frac {8 \left (a^{2}+2 a b +b^{2}\right ) \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}-\frac {\left (2 a^{2}+4 a b +3 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}-\frac {\left (2 a^{2}+4 a b +3 b^{2}\right ) \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}-\frac {2 \left (6 a^{2}+12 a b +5 b^{2}\right ) \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{4}}\) | \(305\) |
1/8*(b^2*sin(3*f*x+3*e)-16*b*f*(a+3/4*b)*x*cos(f*x+e)+8*sin(f*x+e)*(a^2+2* a*b+9/8*b^2))/f/cos(f*x+e)
Time = 0.29 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.33 \[ \int \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=-\frac {{\left (4 \, a b + 3 \, b^{2}\right )} f x \cos \left (f x + e\right ) - {\left (b^{2} \cos \left (f x + e\right )^{2} + 2 \, a^{2} + 4 \, a b + 2 \, b^{2}\right )} \sin \left (f x + e\right )}{2 \, f \cos \left (f x + e\right )} \]
-1/2*((4*a*b + 3*b^2)*f*x*cos(f*x + e) - (b^2*cos(f*x + e)^2 + 2*a^2 + 4*a *b + 2*b^2)*sin(f*x + e))/(f*cos(f*x + e))
\[ \int \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\int \left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{2} \sec ^{2}{\left (e + f x \right )}\, dx \]
Time = 0.33 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.45 \[ \int \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=-\frac {4 \, {\left (f x + e - \tan \left (f x + e\right )\right )} a b + {\left (3 \, f x + 3 \, e - \frac {\tan \left (f x + e\right )}{\tan \left (f x + e\right )^{2} + 1} - 2 \, \tan \left (f x + e\right )\right )} b^{2} - 2 \, a^{2} \tan \left (f x + e\right )}{2 \, f} \]
-1/2*(4*(f*x + e - tan(f*x + e))*a*b + (3*f*x + 3*e - tan(f*x + e)/(tan(f* x + e)^2 + 1) - 2*tan(f*x + e))*b^2 - 2*a^2*tan(f*x + e))/f
Time = 0.31 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.51 \[ \int \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {2 \, a^{2} \tan \left (f x + e\right ) + 4 \, a b \tan \left (f x + e\right ) + 2 \, b^{2} \tan \left (f x + e\right ) - {\left (4 \, a b + 3 \, b^{2}\right )} {\left (f x + e\right )} + \frac {b^{2} \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{2} + 1}}{2 \, f} \]
1/2*(2*a^2*tan(f*x + e) + 4*a*b*tan(f*x + e) + 2*b^2*tan(f*x + e) - (4*a*b + 3*b^2)*(f*x + e) + b^2*tan(f*x + e)/(tan(f*x + e)^2 + 1))/f
Time = 13.84 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.45 \[ \int \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {\mathrm {tan}\left (e+f\,x\right )\,{\left (a+b\right )}^2}{f}+\frac {b^2\,\sin \left (2\,e+2\,f\,x\right )}{4\,f}-\frac {b\,\mathrm {atan}\left (\frac {b\,\mathrm {tan}\left (e+f\,x\right )\,\left (4\,a+3\,b\right )}{2\,\left (\frac {3\,b^2}{2}+2\,a\,b\right )}\right )\,\left (4\,a+3\,b\right )}{2\,f} \]