Optimal. Leaf size=203 \[ -\frac {\text {Li}_2\left (-\frac {b e^{2 i x}}{2 a-2 \sqrt {a+b} \sqrt {a}+b}\right )}{4 \sqrt {a} \sqrt {a+b}}+\frac {\text {Li}_2\left (-\frac {b e^{2 i x}}{2 a+2 \sqrt {a+b} \sqrt {a}+b}\right )}{4 \sqrt {a} \sqrt {a+b}}-\frac {i x \log \left (1+\frac {b e^{2 i x}}{-2 \sqrt {a} \sqrt {a+b}+2 a+b}\right )}{2 \sqrt {a} \sqrt {a+b}}+\frac {i x \log \left (1+\frac {b e^{2 i x}}{2 \sqrt {a} \sqrt {a+b}+2 a+b}\right )}{2 \sqrt {a} \sqrt {a+b}} \]
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Rubi [A] time = 0.33, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4586, 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 2190
Rule 2264
Rule 2279
Rule 2391
Rule 3321
Rule 4586
Rubi steps
\begin {align*} \int \frac {x}{a+b \cos ^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*}
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Mathematica [B] time = 0.56, size = 532, normalized size = 2.62 \[ \frac {i \left (\text {Li}_2\left (\frac {\left (2 a+b-2 i \sqrt {-a (a+b)}\right ) \left (a+b-\sqrt {-a (a+b)} \tan (x)\right )}{b \left (a+b+\sqrt {-a (a+b)} \tan (x)\right )}\right )-\text {Li}_2\left (\frac {\left (2 a+b+2 i \sqrt {-a (a+b)}\right ) \left (a+b-\sqrt {-a (a+b)} \tan (x)\right )}{b \left (a+b+\sqrt {-a (a+b)} \tan (x)\right )}\right )\right )+4 x \tanh ^{-1}\left (\frac {(a+b) \cot (x)}{\sqrt {-a (a+b)}}\right )+2 \cos ^{-1}\left (-\frac {2 a}{b}-1\right ) \tanh ^{-1}\left (\frac {a \tan (x)}{\sqrt {-a (a+b)}}\right )-\log \left (\frac {2 (a+b) \left (\sqrt {-a (a+b)}-i a\right ) (\tan (x)-i)}{b \left (\sqrt {-a (a+b)} \tan (x)+a+b\right )}\right ) \left (\cos ^{-1}\left (-\frac {2 a}{b}-1\right )+2 i \tanh ^{-1}\left (\frac {a \tan (x)}{\sqrt {-a (a+b)}}\right )\right )-\log \left (\frac {2 (a+b) \left (\sqrt {-a (a+b)}+i a\right ) (\tan (x)+i)}{b \left (\sqrt {-a (a+b)} \tan (x)+a+b\right )}\right ) \left (\cos ^{-1}\left (-\frac {2 a}{b}-1\right )-2 i \tanh ^{-1}\left (\frac {a \tan (x)}{\sqrt {-a (a+b)}}\right )\right )+\log \left (\frac {\sqrt {2} e^{-i x} \sqrt {-a (a+b)}}{\sqrt {b} \sqrt {2 a+b \cos (2 x)+b}}\right ) \left (\cos ^{-1}\left (-\frac {2 a}{b}-1\right )-2 i \left (\tanh ^{-1}\left (\frac {a \tan (x)}{\sqrt {-a (a+b)}}\right )+\tanh ^{-1}\left (\frac {(a+b) \cot (x)}{\sqrt {-a (a+b)}}\right )\right )\right )+\log \left (\frac {\sqrt {2} e^{i x} \sqrt {-a (a+b)}}{\sqrt {b} \sqrt {2 a+b \cos (2 x)+b}}\right ) \left (\cos ^{-1}\left (-\frac {2 a}{b}-1\right )+2 i \left (\tanh ^{-1}\left (\frac {a \tan (x)}{\sqrt {-a (a+b)}}\right )+\tanh ^{-1}\left (\frac {(a+b) \cot (x)}{\sqrt {-a (a+b)}}\right )\right )\right )}{4 \sqrt {-a (a+b)}} \]
Antiderivative was successfully verified.
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fricas [B] time = 1.09, size = 1652, normalized size = 8.14 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{b \cos \relax (x)^{2} + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.09, size = 501, normalized size = 2.47 \[ -\frac {i x \ln \left (1-\frac {b \,{\mathrm e}^{2 i x}}{2 \sqrt {a \left (a +b \right )}-2 a -b}\right )}{2 \sqrt {a \left (a +b \right )}}-\frac {x^{2}}{2 \sqrt {a \left (a +b \right )}}-\frac {\polylog \left (2, \frac {b \,{\mathrm e}^{2 i x}}{2 \sqrt {a \left (a +b \right )}-2 a -b}\right )}{4 \sqrt {a \left (a +b \right )}}-\frac {i \ln \left (1-\frac {b \,{\mathrm e}^{2 i x}}{-2 \sqrt {a \left (a +b \right )}-2 a -b}\right ) x}{-2 \sqrt {a \left (a +b \right )}-2 a -b}-\frac {i \ln \left (1-\frac {b \,{\mathrm e}^{2 i x}}{-2 \sqrt {a \left (a +b \right )}-2 a -b}\right ) a x}{\sqrt {a \left (a +b \right )}\, \left (-2 \sqrt {a \left (a +b \right )}-2 a -b \right )}-\frac {i \ln \left (1-\frac {b \,{\mathrm e}^{2 i x}}{-2 \sqrt {a \left (a +b \right )}-2 a -b}\right ) b x}{2 \sqrt {a \left (a +b \right )}\, \left (-2 \sqrt {a \left (a +b \right )}-2 a -b \right )}-\frac {x^{2}}{-2 \sqrt {a \left (a +b \right )}-2 a -b}-\frac {a \,x^{2}}{\sqrt {a \left (a +b \right )}\, \left (-2 \sqrt {a \left (a +b \right )}-2 a -b \right )}-\frac {b \,x^{2}}{2 \sqrt {a \left (a +b \right )}\, \left (-2 \sqrt {a \left (a +b \right )}-2 a -b \right )}-\frac {\polylog \left (2, \frac {b \,{\mathrm e}^{2 i x}}{-2 \sqrt {a \left (a +b \right )}-2 a -b}\right )}{2 \left (-2 \sqrt {a \left (a +b \right )}-2 a -b \right )}-\frac {\polylog \left (2, \frac {b \,{\mathrm e}^{2 i x}}{-2 \sqrt {a \left (a +b \right )}-2 a -b}\right ) a}{2 \sqrt {a \left (a +b \right )}\, \left (-2 \sqrt {a \left (a +b \right )}-2 a -b \right )}-\frac {\polylog \left (2, \frac {b \,{\mathrm e}^{2 i x}}{-2 \sqrt {a \left (a +b \right )}-2 a -b}\right ) b}{4 \sqrt {a \left (a +b \right )}\, \left (-2 \sqrt {a \left (a +b \right )}-2 a -b \right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{b \cos \relax (x)^{2} + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x}{b\,{\cos \relax (x)}^2+a} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{a + b \cos ^{2}{\relax (x )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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