Optimal. Leaf size=109 \[ \frac {b^{3/2} (5 a+b) \tan ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{2 a^{3/2} (a+b)^3}-\frac {b (a-b) \sin (x)}{2 a (a+b)^2 \left (a+b \sin ^2(x)\right )}+\frac {(a+5 b) \tanh ^{-1}(\sin (x))}{2 (a+b)^3}+\frac {\tan (x) \sec (x)}{2 (a+b) \left (a+b \sin ^2(x)\right )} \]
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Rubi [A] time = 0.16, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3190, 414, 527, 522, 206, 205} \[ \frac {b^{3/2} (5 a+b) \tan ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{2 a^{3/2} (a+b)^3}-\frac {b (a-b) \sin (x)}{2 a (a+b)^2 \left (a+b \sin ^2(x)\right )}+\frac {(a+5 b) \tanh ^{-1}(\sin (x))}{2 (a+b)^3}+\frac {\tan (x) \sec (x)}{2 (a+b) \left (a+b \sin ^2(x)\right )} \]
Antiderivative was successfully verified.
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Rule 205
Rule 206
Rule 414
Rule 522
Rule 527
Rule 3190
Rubi steps
\begin {align*} \int \frac {\sec ^3(x)}{\left (a+b \sin ^2(x)\right )^2} \, dx &=\operatorname {Subst}\left (\int \frac {1}{\left (1-x^2\right )^2 \left (a+b x^2\right )^2} \, dx,x,\sin (x)\right )\\ &=\frac {\sec (x) \tan (x)}{2 (a+b) \left (a+b \sin ^2(x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {a+2 b+3 b x^2}{\left (1-x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\sin (x)\right )}{2 (a+b)}\\ &=-\frac {(a-b) b \sin (x)}{2 a (a+b)^2 \left (a+b \sin ^2(x)\right )}+\frac {\sec (x) \tan (x)}{2 (a+b) \left (a+b \sin ^2(x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {-2 \left (a^2+4 a b+b^2\right )-2 (a-b) b x^2}{\left (1-x^2\right ) \left (a+b x^2\right )} \, dx,x,\sin (x)\right )}{4 a (a+b)^2}\\ &=-\frac {(a-b) b \sin (x)}{2 a (a+b)^2 \left (a+b \sin ^2(x)\right )}+\frac {\sec (x) \tan (x)}{2 (a+b) \left (a+b \sin ^2(x)\right )}+\frac {\left (b^2 (5 a+b)\right ) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sin (x)\right )}{2 a (a+b)^3}+\frac {(a+5 b) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (x)\right )}{2 (a+b)^3}\\ &=\frac {b^{3/2} (5 a+b) \tan ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{2 a^{3/2} (a+b)^3}+\frac {(a+5 b) \tanh ^{-1}(\sin (x))}{2 (a+b)^3}-\frac {(a-b) b \sin (x)}{2 a (a+b)^2 \left (a+b \sin ^2(x)\right )}+\frac {\sec (x) \tan (x)}{2 (a+b) \left (a+b \sin ^2(x)\right )}\\ \end {align*}
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Mathematica [A] time = 1.04, size = 183, normalized size = 1.68 \[ \frac {\frac {b^{3/2} (5 a+b) \tan ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{a^{3/2}}-\frac {b^{3/2} (5 a+b) \tan ^{-1}\left (\frac {\sqrt {a} \csc (x)}{\sqrt {b}}\right )}{a^{3/2}}+\frac {4 b^2 (a+b) \sin (x)}{a (2 a-b \cos (2 x)+b)}+\frac {a+b}{\left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )^2}-\frac {a+b}{\left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )^2}-2 (a+5 b) \log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+2 (a+5 b) \log \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )}{4 (a+b)^3} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.57, size = 560, normalized size = 5.14 \[ \left [\frac {{\left ({\left (5 \, a b^{2} + b^{3}\right )} \cos \relax (x)^{4} - {\left (5 \, a^{2} b + 6 \, a b^{2} + b^{3}\right )} \cos \relax (x)^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (-\frac {b \cos \relax (x)^{2} - 2 \, a \sqrt {-\frac {b}{a}} \sin \relax (x) + a - b}{b \cos \relax (x)^{2} - a - b}\right ) + {\left ({\left (a^{2} b + 5 \, a b^{2}\right )} \cos \relax (x)^{4} - {\left (a^{3} + 6 \, a^{2} b + 5 \, a b^{2}\right )} \cos \relax (x)^{2}\right )} \log \left (\sin \relax (x) + 1\right ) - {\left ({\left (a^{2} b + 5 \, a b^{2}\right )} \cos \relax (x)^{4} - {\left (a^{3} + 6 \, a^{2} b + 5 \, a b^{2}\right )} \cos \relax (x)^{2}\right )} \log \left (-\sin \relax (x) + 1\right ) - 2 \, {\left (a^{3} + 2 \, a^{2} b + a b^{2} - {\left (a^{2} b - b^{3}\right )} \cos \relax (x)^{2}\right )} \sin \relax (x)}{4 \, {\left ({\left (a^{4} b + 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} \cos \relax (x)^{4} - {\left (a^{5} + 4 \, a^{4} b + 6 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + a b^{4}\right )} \cos \relax (x)^{2}\right )}}, \frac {2 \, {\left ({\left (5 \, a b^{2} + b^{3}\right )} \cos \relax (x)^{4} - {\left (5 \, a^{2} b + 6 \, a b^{2} + b^{3}\right )} \cos \relax (x)^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\sqrt {\frac {b}{a}} \sin \relax (x)\right ) + {\left ({\left (a^{2} b + 5 \, a b^{2}\right )} \cos \relax (x)^{4} - {\left (a^{3} + 6 \, a^{2} b + 5 \, a b^{2}\right )} \cos \relax (x)^{2}\right )} \log \left (\sin \relax (x) + 1\right ) - {\left ({\left (a^{2} b + 5 \, a b^{2}\right )} \cos \relax (x)^{4} - {\left (a^{3} + 6 \, a^{2} b + 5 \, a b^{2}\right )} \cos \relax (x)^{2}\right )} \log \left (-\sin \relax (x) + 1\right ) - 2 \, {\left (a^{3} + 2 \, a^{2} b + a b^{2} - {\left (a^{2} b - b^{3}\right )} \cos \relax (x)^{2}\right )} \sin \relax (x)}{4 \, {\left ({\left (a^{4} b + 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} \cos \relax (x)^{4} - {\left (a^{5} + 4 \, a^{4} b + 6 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + a b^{4}\right )} \cos \relax (x)^{2}\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.13, size = 194, normalized size = 1.78 \[ \frac {{\left (a + 5 \, b\right )} \log \left (\sin \relax (x) + 1\right )}{4 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} - \frac {{\left (a + 5 \, b\right )} \log \left (-\sin \relax (x) + 1\right )}{4 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} + \frac {{\left (5 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {b \sin \relax (x)}{\sqrt {a b}}\right )}{2 \, {\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} \sqrt {a b}} - \frac {a b \sin \relax (x)^{3} - b^{2} \sin \relax (x)^{3} + a^{2} \sin \relax (x) + b^{2} \sin \relax (x)}{2 \, {\left (b \sin \relax (x)^{4} + a \sin \relax (x)^{2} - b \sin \relax (x)^{2} - a\right )} {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.29, size = 180, normalized size = 1.65 \[ -\frac {1}{4 \left (a +b \right )^{2} \left (-1+\sin \relax (x )\right )}-\frac {\ln \left (-1+\sin \relax (x )\right ) a}{4 \left (a +b \right )^{3}}-\frac {5 \ln \left (-1+\sin \relax (x )\right ) b}{4 \left (a +b \right )^{3}}+\frac {b^{2} \sin \relax (x )}{2 \left (a +b \right )^{3} \left (a +b \left (\sin ^{2}\relax (x )\right )\right )}+\frac {b^{3} \sin \relax (x )}{2 \left (a +b \right )^{3} a \left (a +b \left (\sin ^{2}\relax (x )\right )\right )}+\frac {5 b^{2} \arctan \left (\frac {\sin \relax (x ) b}{\sqrt {a b}}\right )}{2 \left (a +b \right )^{3} \sqrt {a b}}+\frac {b^{3} \arctan \left (\frac {\sin \relax (x ) b}{\sqrt {a b}}\right )}{2 \left (a +b \right )^{3} a \sqrt {a b}}-\frac {1}{4 \left (a +b \right )^{2} \left (1+\sin \relax (x )\right )}+\frac {\ln \left (1+\sin \relax (x )\right ) a}{4 \left (a +b \right )^{3}}+\frac {5 \ln \left (1+\sin \relax (x )\right ) b}{4 \left (a +b \right )^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.65, size = 220, normalized size = 2.02 \[ \frac {{\left (a + 5 \, b\right )} \log \left (\sin \relax (x) + 1\right )}{4 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} - \frac {{\left (a + 5 \, b\right )} \log \left (\sin \relax (x) - 1\right )}{4 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} + \frac {{\left (5 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {b \sin \relax (x)}{\sqrt {a b}}\right )}{2 \, {\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} \sqrt {a b}} - \frac {{\left (a b - b^{2}\right )} \sin \relax (x)^{3} + {\left (a^{2} + b^{2}\right )} \sin \relax (x)}{2 \, {\left ({\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} \sin \relax (x)^{4} - a^{4} - 2 \, a^{3} b - a^{2} b^{2} + {\left (a^{4} + a^{3} b - a^{2} b^{2} - a b^{3}\right )} \sin \relax (x)^{2}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 15.06, size = 2009, normalized size = 18.43 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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