Integrand size = 15, antiderivative size = 76 \[ \int \frac {\sec ^2(x)}{\left (a+b \sin ^2(x)\right )^2} \, dx=\frac {b (4 a+b) \arctan \left (\frac {\sqrt {a+b} \tan (x)}{\sqrt {a}}\right )}{2 a^{3/2} (a+b)^{5/2}}+\frac {\tan (x)}{(a+b)^2}+\frac {b^2 \tan (x)}{2 a (a+b)^2 \left (a+(a+b) \tan ^2(x)\right )} \]
1/2*b*(4*a+b)*arctan((a+b)^(1/2)*tan(x)/a^(1/2))/a^(3/2)/(a+b)^(5/2)+tan(x )/(a+b)^2+1/2*b^2*tan(x)/a/(a+b)^2/(a+(a+b)*tan(x)^2)
Time = 0.55 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00 \[ \int \frac {\sec ^2(x)}{\left (a+b \sin ^2(x)\right )^2} \, dx=\frac {1}{2} \left (\frac {b (4 a+b) \arctan \left (\frac {\sqrt {a+b} \tan (x)}{\sqrt {a}}\right )}{a^{3/2} (a+b)^{5/2}}+\frac {\frac {b^2 \sin (2 x)}{a (2 a+b-b \cos (2 x))}+2 \tan (x)}{(a+b)^2}\right ) \]
((b*(4*a + b)*ArcTan[(Sqrt[a + b]*Tan[x])/Sqrt[a]])/(a^(3/2)*(a + b)^(5/2) ) + ((b^2*Sin[2*x])/(a*(2*a + b - b*Cos[2*x])) + 2*Tan[x])/(a + b)^2)/2
Time = 0.30 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, 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.
\(\displaystyle \int \frac {\sec ^2(x)}{\left (a+b \sin ^2(x)\right )^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\cos (x)^2 \left (a+b \sin (x)^2\right )^2}dx\) |
\(\Big \downarrow \) 3670 |
\(\displaystyle \int \frac {\left (\tan ^2(x)+1\right )^2}{\left ((a+b) \tan ^2(x)+a\right )^2}d\tan (x)\) |
\(\Big \downarrow \) 300 |
\(\displaystyle \int \left (\frac {2 b (a+b) \tan ^2(x)+b (2 a+b)}{(a+b)^2 \left ((a+b) \tan ^2(x)+a\right )^2}+\frac {1}{(a+b)^2}\right )d\tan (x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b (4 a+b) \arctan \left (\frac {\sqrt {a+b} \tan (x)}{\sqrt {a}}\right )}{2 a^{3/2} (a+b)^{5/2}}+\frac {b^2 \tan (x)}{2 a (a+b)^2 \left ((a+b) \tan ^2(x)+a\right )}+\frac {\tan (x)}{(a+b)^2}\) |
(b*(4*a + b)*ArcTan[(Sqrt[a + b]*Tan[x])/Sqrt[a]])/(2*a^(3/2)*(a + b)^(5/2 )) + Tan[x]/(a + b)^2 + (b^2*Tan[x])/(2*a*(a + b)^2*(a + (a + b)*Tan[x]^2) )
3.4.21.3.1 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 = 1.04 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.07
method | result | size |
default | \(\frac {\tan \left (x \right )}{a^{2}+2 a b +b^{2}}+\frac {b \left (\frac {b \tan \left (x \right )}{2 a \left (a \left (\tan ^{2}\left (x \right )\right )+\left (\tan ^{2}\left (x \right )\right ) b +a \right )}+\frac {\left (4 a +b \right ) \arctan \left (\frac {\left (a +b \right ) \tan \left (x \right )}{\sqrt {a \left (a +b \right )}}\right )}{2 a \sqrt {a \left (a +b \right )}}\right )}{\left (a +b \right )^{2}}\) | \(81\) |
risch | \(\frac {i \left (-4 a b \,{\mathrm e}^{4 i x}-b^{2} {\mathrm e}^{4 i x}+8 \,{\mathrm e}^{2 i x} a^{2}+2 a b \,{\mathrm e}^{2 i x}-2 a b +b^{2}\right )}{a \left (a +b \right )^{2} \left (-b \,{\mathrm e}^{4 i x}+4 a \,{\mathrm e}^{2 i x}+2 b \,{\mathrm e}^{2 i x}-b \right ) \left ({\mathrm e}^{2 i x}+1\right )}-\frac {b \ln \left ({\mathrm e}^{2 i x}-\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{\sqrt {-a^{2}-a b}\, \left (a +b \right )^{2}}-\frac {b^{2} \ln \left ({\mathrm e}^{2 i x}-\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{4 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} a}+\frac {b \ln \left ({\mathrm e}^{2 i x}+\frac {2 i a^{2}+2 i a b -2 a \sqrt {-a^{2}-a b}-b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{\sqrt {-a^{2}-a b}\, \left (a +b \right )^{2}}+\frac {b^{2} \ln \left ({\mathrm e}^{2 i x}+\frac {2 i a^{2}+2 i a b -2 a \sqrt {-a^{2}-a b}-b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{4 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} a}\) | \(447\) |
tan(x)/(a^2+2*a*b+b^2)+b/(a+b)^2*(1/2/a*b*tan(x)/(a*tan(x)^2+tan(x)^2*b+a) +1/2*(4*a+b)/a/(a*(a+b))^(1/2)*arctan((a+b)*tan(x)/(a*(a+b))^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 210 vs. \(2 (64) = 128\).
Time = 0.32 (sec) , antiderivative size = 505, normalized size of antiderivative = 6.64 \[ \int \frac {\sec ^2(x)}{\left (a+b \sin ^2(x)\right )^2} \, dx=\left [-\frac {{\left ({\left (4 \, a b^{2} + b^{3}\right )} \cos \left (x\right )^{3} - {\left (4 \, a^{2} b + 5 \, a b^{2} + b^{3}\right )} \cos \left (x\right )\right )} \sqrt {-a^{2} - a b} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (x\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (x\right )^{2} + 4 \, {\left ({\left (2 \, a + b\right )} \cos \left (x\right )^{3} - {\left (a + b\right )} \cos \left (x\right )\right )} \sqrt {-a^{2} - a b} \sin \left (x\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (x\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (x\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) + 4 \, {\left (2 \, a^{4} + 4 \, a^{3} b + 2 \, a^{2} b^{2} - {\left (2 \, a^{3} b + a^{2} b^{2} - a b^{3}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{8 \, {\left ({\left (a^{5} b + 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} + a^{2} b^{4}\right )} \cos \left (x\right )^{3} - {\left (a^{6} + 4 \, a^{5} b + 6 \, a^{4} b^{2} + 4 \, a^{3} b^{3} + a^{2} b^{4}\right )} \cos \left (x\right )\right )}}, -\frac {{\left ({\left (4 \, a b^{2} + b^{3}\right )} \cos \left (x\right )^{3} - {\left (4 \, a^{2} b + 5 \, a b^{2} + b^{3}\right )} \cos \left (x\right )\right )} \sqrt {a^{2} + a b} \arctan \left (\frac {{\left (2 \, a + b\right )} \cos \left (x\right )^{2} - a - b}{2 \, \sqrt {a^{2} + a b} \cos \left (x\right ) \sin \left (x\right )}\right ) + 2 \, {\left (2 \, a^{4} + 4 \, a^{3} b + 2 \, a^{2} b^{2} - {\left (2 \, a^{3} b + a^{2} b^{2} - a b^{3}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{4 \, {\left ({\left (a^{5} b + 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} + a^{2} b^{4}\right )} \cos \left (x\right )^{3} - {\left (a^{6} + 4 \, a^{5} b + 6 \, a^{4} b^{2} + 4 \, a^{3} b^{3} + a^{2} b^{4}\right )} \cos \left (x\right )\right )}}\right ] \]
[-1/8*(((4*a*b^2 + b^3)*cos(x)^3 - (4*a^2*b + 5*a*b^2 + b^3)*cos(x))*sqrt( -a^2 - a*b)*log(((8*a^2 + 8*a*b + b^2)*cos(x)^4 - 2*(4*a^2 + 5*a*b + b^2)* cos(x)^2 + 4*((2*a + b)*cos(x)^3 - (a + b)*cos(x))*sqrt(-a^2 - a*b)*sin(x) + a^2 + 2*a*b + b^2)/(b^2*cos(x)^4 - 2*(a*b + b^2)*cos(x)^2 + a^2 + 2*a*b + b^2)) + 4*(2*a^4 + 4*a^3*b + 2*a^2*b^2 - (2*a^3*b + a^2*b^2 - a*b^3)*co s(x)^2)*sin(x))/((a^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*cos(x)^3 - (a^6 + 4*a^5*b + 6*a^4*b^2 + 4*a^3*b^3 + a^2*b^4)*cos(x)), -1/4*(((4*a*b^2 + b ^3)*cos(x)^3 - (4*a^2*b + 5*a*b^2 + b^3)*cos(x))*sqrt(a^2 + a*b)*arctan(1/ 2*((2*a + b)*cos(x)^2 - a - b)/(sqrt(a^2 + a*b)*cos(x)*sin(x))) + 2*(2*a^4 + 4*a^3*b + 2*a^2*b^2 - (2*a^3*b + a^2*b^2 - a*b^3)*cos(x)^2)*sin(x))/((a ^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*cos(x)^3 - (a^6 + 4*a^5*b + 6*a^4* b^2 + 4*a^3*b^3 + a^2*b^4)*cos(x))]
\[ \int \frac {\sec ^2(x)}{\left (a+b \sin ^2(x)\right )^2} \, dx=\int \frac {\sec ^{2}{\left (x \right )}}{\left (a + b \sin ^{2}{\left (x \right )}\right )^{2}}\, dx \]
Time = 0.47 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.57 \[ \int \frac {\sec ^2(x)}{\left (a+b \sin ^2(x)\right )^2} \, dx=\frac {b^{2} \tan \left (x\right )}{2 \, {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2} + {\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} \tan \left (x\right )^{2}\right )}} + \frac {{\left (4 \, a b + b^{2}\right )} \arctan \left (\frac {{\left (a + b\right )} \tan \left (x\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{2 \, {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \sqrt {{\left (a + b\right )} a}} + \frac {\tan \left (x\right )}{a^{2} + 2 \, a b + b^{2}} \]
1/2*b^2*tan(x)/(a^4 + 2*a^3*b + a^2*b^2 + (a^4 + 3*a^3*b + 3*a^2*b^2 + a*b ^3)*tan(x)^2) + 1/2*(4*a*b + b^2)*arctan((a + b)*tan(x)/sqrt((a + b)*a))/( (a^3 + 2*a^2*b + a*b^2)*sqrt((a + b)*a)) + tan(x)/(a^2 + 2*a*b + b^2)
Time = 0.32 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.49 \[ \int \frac {\sec ^2(x)}{\left (a+b \sin ^2(x)\right )^2} \, dx=\frac {b^{2} \tan \left (x\right )}{2 \, {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} {\left (a \tan \left (x\right )^{2} + b \tan \left (x\right )^{2} + a\right )}} + \frac {{\left (4 \, a b + b^{2}\right )} \arctan \left (\frac {a \tan \left (x\right ) + b \tan \left (x\right )}{\sqrt {a^{2} + a b}}\right )}{2 \, {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \sqrt {a^{2} + a b}} + \frac {\tan \left (x\right )}{a^{2} + 2 \, a b + b^{2}} \]
1/2*b^2*tan(x)/((a^3 + 2*a^2*b + a*b^2)*(a*tan(x)^2 + b*tan(x)^2 + a)) + 1 /2*(4*a*b + b^2)*arctan((a*tan(x) + b*tan(x))/sqrt(a^2 + a*b))/((a^3 + 2*a ^2*b + a*b^2)*sqrt(a^2 + a*b)) + tan(x)/(a^2 + 2*a*b + b^2)
Time = 13.91 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.62 \[ \int \frac {\sec ^2(x)}{\left (a+b \sin ^2(x)\right )^2} \, dx=\frac {\mathrm {tan}\left (x\right )}{{\left (a+b\right )}^2}+\frac {b^2\,\mathrm {tan}\left (x\right )}{2\,a\,\left (a\,b^2+2\,a^2\,b+{\mathrm {tan}\left (x\right )}^2\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )+a^3\right )}+\frac {b\,\mathrm {atan}\left (\frac {b\,\mathrm {tan}\left (x\right )\,\left (4\,a+b\right )\,\left (2\,a+2\,b\right )\,\left (a^2+2\,a\,b+b^2\right )}{2\,\sqrt {a}\,{\left (a+b\right )}^{5/2}\,\left (b^2+4\,a\,b\right )}\right )\,\left (4\,a+b\right )}{2\,a^{3/2}\,{\left (a+b\right )}^{5/2}} \]