Optimal. Leaf size=203 \[ -\frac{\text{PolyLog}\left (2,\frac{b e^{2 i x}}{-2 \sqrt{a} \sqrt{a+b}+2 a+b}\right )}{4 \sqrt{a} \sqrt{a+b}}+\frac{\text{PolyLog}\left (2,\frac{b e^{2 i x}}{2 \sqrt{a} \sqrt{a+b}+2 a+b}\right )}{4 \sqrt{a} \sqrt{a+b}}-\frac{i x \log \left (1-\frac{b e^{2 i x}}{-2 \sqrt{a} \sqrt{a+b}+2 a+b}\right )}{2 \sqrt{a} \sqrt{a+b}}+\frac{i x \log \left (1-\frac{b e^{2 i x}}{2 \sqrt{a} \sqrt{a+b}+2 a+b}\right )}{2 \sqrt{a} \sqrt{a+b}} \]
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Rubi [A] time = 0.378754, antiderivative size = 203, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4585, 3321, 2264, 2190, 2279, 2391} \[ -\frac{\text{PolyLog}\left (2,\frac{b e^{2 i x}}{-2 \sqrt{a} \sqrt{a+b}+2 a+b}\right )}{4 \sqrt{a} \sqrt{a+b}}+\frac{\text{PolyLog}\left (2,\frac{b e^{2 i x}}{2 \sqrt{a} \sqrt{a+b}+2 a+b}\right )}{4 \sqrt{a} \sqrt{a+b}}-\frac{i x \log \left (1-\frac{b e^{2 i x}}{-2 \sqrt{a} \sqrt{a+b}+2 a+b}\right )}{2 \sqrt{a} \sqrt{a+b}}+\frac{i x \log \left (1-\frac{b e^{2 i x}}{2 \sqrt{a} \sqrt{a+b}+2 a+b}\right )}{2 \sqrt{a} \sqrt{a+b}} \]
Antiderivative was successfully verified.
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Rule 4585
Rule 3321
Rule 2264
Rule 2190
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int \frac{x}{a+b \sin ^2(x)} \, dx &=2 \int \frac{x}{2 a+b-b \cos (2 x)} \, dx\\ &=4 \int \frac{e^{2 i x} x}{-b+2 (2 a+b) e^{2 i x}-b e^{4 i x}} \, dx\\ &=-\frac{(2 b) \int \frac{e^{2 i x} x}{-4 \sqrt{a} \sqrt{a+b}+2 (2 a+b)-2 b e^{2 i x}} \, dx}{\sqrt{a} \sqrt{a+b}}+\frac{(2 b) \int \frac{e^{2 i x} x}{4 \sqrt{a} \sqrt{a+b}+2 (2 a+b)-2 b e^{2 i x}} \, dx}{\sqrt{a} \sqrt{a+b}}\\ &=-\frac{i x \log \left (1-\frac{b e^{2 i x}}{2 a+b-2 \sqrt{a} \sqrt{a+b}}\right )}{2 \sqrt{a} \sqrt{a+b}}+\frac{i x \log \left (1-\frac{b e^{2 i x}}{2 a+b+2 \sqrt{a} \sqrt{a+b}}\right )}{2 \sqrt{a} \sqrt{a+b}}+\frac{i \int \log \left (1-\frac{2 b e^{2 i x}}{-4 \sqrt{a} \sqrt{a+b}+2 (2 a+b)}\right ) \, dx}{2 \sqrt{a} \sqrt{a+b}}-\frac{i \int \log \left (1-\frac{2 b e^{2 i x}}{4 \sqrt{a} \sqrt{a+b}+2 (2 a+b)}\right ) \, dx}{2 \sqrt{a} \sqrt{a+b}}\\ &=-\frac{i x \log \left (1-\frac{b e^{2 i x}}{2 a+b-2 \sqrt{a} \sqrt{a+b}}\right )}{2 \sqrt{a} \sqrt{a+b}}+\frac{i x \log \left (1-\frac{b e^{2 i x}}{2 a+b+2 \sqrt{a} \sqrt{a+b}}\right )}{2 \sqrt{a} \sqrt{a+b}}+\frac{\operatorname{Subst}\left (\int \frac{\log \left (1-\frac{2 b x}{-4 \sqrt{a} \sqrt{a+b}+2 (2 a+b)}\right )}{x} \, dx,x,e^{2 i x}\right )}{4 \sqrt{a} \sqrt{a+b}}-\frac{\operatorname{Subst}\left (\int \frac{\log \left (1-\frac{2 b x}{4 \sqrt{a} \sqrt{a+b}+2 (2 a+b)}\right )}{x} \, dx,x,e^{2 i x}\right )}{4 \sqrt{a} \sqrt{a+b}}\\ &=-\frac{i x \log \left (1-\frac{b e^{2 i x}}{2 a+b-2 \sqrt{a} \sqrt{a+b}}\right )}{2 \sqrt{a} \sqrt{a+b}}+\frac{i x \log \left (1-\frac{b e^{2 i x}}{2 a+b+2 \sqrt{a} \sqrt{a+b}}\right )}{2 \sqrt{a} \sqrt{a+b}}-\frac{\text{Li}_2\left (\frac{b e^{2 i x}}{2 a+b-2 \sqrt{a} \sqrt{a+b}}\right )}{4 \sqrt{a} \sqrt{a+b}}+\frac{\text{Li}_2\left (\frac{b e^{2 i x}}{2 a+b+2 \sqrt{a} \sqrt{a+b}}\right )}{4 \sqrt{a} \sqrt{a+b}}\\ \end{align*}
Mathematica [B] time = 0.60514, size = 545, normalized size = 2.68 \[ \frac{i \left (\text{PolyLog}\left (2,\frac{\left (-2 i \sqrt{-a (a+b)}+2 a+b\right ) \left (\sqrt{-a (a+b)} \tan (x)-a\right )}{b \left (\sqrt{-a (a+b)} \tan (x)+a\right )}\right )-\text{PolyLog}\left (2,\frac{\left (2 i \sqrt{-a (a+b)}+2 a+b\right ) \left (\sqrt{-a (a+b)} \tan (x)-a\right )}{b \left (\sqrt{-a (a+b)} \tan (x)+a\right )}\right )\right )+4 x \tanh ^{-1}\left (\frac{a \cot (x)}{\sqrt{-a (a+b)}}\right )-2 \cos ^{-1}\left (\frac{2 a}{b}+1\right ) \tanh ^{-1}\left (\frac{\sqrt{-a (a+b)} \tan (x)}{a}\right )-\log \left (\frac{2 a \left (-i \sqrt{-a (a+b)}+a+b\right ) (1-i \tan (x))}{b \left (\sqrt{-a (a+b)} \tan (x)+a\right )}\right ) \left (\cos ^{-1}\left (\frac{2 a}{b}+1\right )+2 i \tanh ^{-1}\left (\frac{\sqrt{-a (a+b)} \tan (x)}{a}\right )\right )-\log \left (\frac{2 a \left (i \sqrt{-a (a+b)}+a+b\right ) (1+i \tan (x))}{b \left (\sqrt{-a (a+b)} \tan (x)+a\right )}\right ) \left (\cos ^{-1}\left (\frac{2 a}{b}+1\right )-2 i \tanh ^{-1}\left (\frac{\sqrt{-a (a+b)} \tan (x)}{a}\right )\right )+\log \left (\frac{\sqrt{2} e^{-i x} \sqrt{-a (a+b)}}{\sqrt{-b} \sqrt{2 a-b \cos (2 x)+b}}\right ) \left (2 i \tanh ^{-1}\left (\frac{\sqrt{-a (a+b)} \tan (x)}{a}\right )-2 i \tanh ^{-1}\left (\frac{a \cot (x)}{\sqrt{-a (a+b)}}\right )+\cos ^{-1}\left (\frac{2 a}{b}+1\right )\right )+\log \left (\frac{\sqrt{2} e^{i x} \sqrt{-a (a+b)}}{\sqrt{-b} \sqrt{2 a-b \cos (2 x)+b}}\right ) \left (\cos ^{-1}\left (\frac{2 a}{b}+1\right )+2 i \left (\tanh ^{-1}\left (\frac{a \cot (x)}{\sqrt{-a (a+b)}}\right )-\tanh ^{-1}\left (\frac{\sqrt{-a (a+b)} \tan (x)}{a}\right )\right )\right )}{4 \sqrt{-a (a+b)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.104, size = 465, normalized size = 2.3 \begin{align*}{ix\ln \left ( 1-{b{{\rm e}^{2\,ix}} \left ( 2\,\sqrt{ \left ( a+b \right ) a}+2\,a+b \right ) ^{-1}} \right ) \left ( 2\,\sqrt{ \left ( a+b \right ) a}+2\,a+b \right ) ^{-1}}+{iax\ln \left ( 1-{b{{\rm e}^{2\,ix}} \left ( 2\,\sqrt{ \left ( a+b \right ) a}+2\,a+b \right ) ^{-1}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) a}}} \left ( 2\,\sqrt{ \left ( a+b \right ) a}+2\,a+b \right ) ^{-1}}+{{\frac{i}{2}}bx\ln \left ( 1-{b{{\rm e}^{2\,ix}} \left ( 2\,\sqrt{ \left ( a+b \right ) a}+2\,a+b \right ) ^{-1}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) a}}} \left ( 2\,\sqrt{ \left ( a+b \right ) a}+2\,a+b \right ) ^{-1}}+{{x}^{2} \left ( 2\,\sqrt{ \left ( a+b \right ) a}+2\,a+b \right ) ^{-1}}+{a{x}^{2}{\frac{1}{\sqrt{ \left ( a+b \right ) a}}} \left ( 2\,\sqrt{ \left ( a+b \right ) a}+2\,a+b \right ) ^{-1}}+{\frac{b{x}^{2}}{2}{\frac{1}{\sqrt{ \left ( a+b \right ) a}}} \left ( 2\,\sqrt{ \left ( a+b \right ) a}+2\,a+b \right ) ^{-1}}+{\frac{1}{2}{\it polylog} \left ( 2,{b{{\rm e}^{2\,ix}} \left ( 2\,\sqrt{ \left ( a+b \right ) a}+2\,a+b \right ) ^{-1}} \right ) \left ( 2\,\sqrt{ \left ( a+b \right ) a}+2\,a+b \right ) ^{-1}}+{\frac{a}{2}{\it polylog} \left ( 2,{b{{\rm e}^{2\,ix}} \left ( 2\,\sqrt{ \left ( a+b \right ) a}+2\,a+b \right ) ^{-1}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) a}}} \left ( 2\,\sqrt{ \left ( a+b \right ) a}+2\,a+b \right ) ^{-1}}+{\frac{b}{4}{\it polylog} \left ( 2,{b{{\rm e}^{2\,ix}} \left ( 2\,\sqrt{ \left ( a+b \right ) a}+2\,a+b \right ) ^{-1}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) a}}} \left ( 2\,\sqrt{ \left ( a+b \right ) a}+2\,a+b \right ) ^{-1}}-{{\frac{i}{2}}x\ln \left ( 1-{b{{\rm e}^{2\,ix}} \left ( -2\,\sqrt{ \left ( a+b \right ) a}+2\,a+b \right ) ^{-1}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) a}}}}-{\frac{{x}^{2}}{2}{\frac{1}{\sqrt{ \left ( a+b \right ) a}}}}-{\frac{1}{4}{\it polylog} \left ( 2,{b{{\rm e}^{2\,ix}} \left ( -2\,\sqrt{ \left ( a+b \right ) a}+2\,a+b \right ) ^{-1}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) a}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{b \sin \left (x\right )^{2} + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 0.935243, size = 4247, normalized size = 20.92 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{a + b \sin ^{2}{\left (x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{b \sin \left (x\right )^{2} + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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