Optimal. Leaf size=123 \[ -\frac{b \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{1}{2} \left (c+d x^2\right )\right )}{\sqrt{a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{b^2 \tanh \left (c+d x^2\right )}{2 a d \left (a^2-b^2\right ) \left (a+b \text{sech}\left (c+d x^2\right )\right )}+\frac{x^2}{2 a^2} \]
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Rubi [A] time = 0.249768, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {5436, 3785, 3919, 3831, 2659, 208} \[ -\frac{b \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{1}{2} \left (c+d x^2\right )\right )}{\sqrt{a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{b^2 \tanh \left (c+d x^2\right )}{2 a d \left (a^2-b^2\right ) \left (a+b \text{sech}\left (c+d x^2\right )\right )}+\frac{x^2}{2 a^2} \]
Antiderivative was successfully verified.
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Rule 5436
Rule 3785
Rule 3919
Rule 3831
Rule 2659
Rule 208
Rubi steps
\begin{align*} \int \frac{x}{\left (a+b \text{sech}\left (c+d x^2\right )\right )^2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{(a+b \text{sech}(c+d x))^2} \, dx,x,x^2\right )\\ &=\frac{b^2 \tanh \left (c+d x^2\right )}{2 a \left (a^2-b^2\right ) d \left (a+b \text{sech}\left (c+d x^2\right )\right )}-\frac{\operatorname{Subst}\left (\int \frac{-a^2+b^2+a b \text{sech}(c+d x)}{a+b \text{sech}(c+d x)} \, dx,x,x^2\right )}{2 a \left (a^2-b^2\right )}\\ &=\frac{x^2}{2 a^2}+\frac{b^2 \tanh \left (c+d x^2\right )}{2 a \left (a^2-b^2\right ) d \left (a+b \text{sech}\left (c+d x^2\right )\right )}-\frac{\left (b \left (2 a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{\text{sech}(c+d x)}{a+b \text{sech}(c+d x)} \, dx,x,x^2\right )}{2 a^2 \left (a^2-b^2\right )}\\ &=\frac{x^2}{2 a^2}+\frac{b^2 \tanh \left (c+d x^2\right )}{2 a \left (a^2-b^2\right ) d \left (a+b \text{sech}\left (c+d x^2\right )\right )}-\frac{\left (2 a^2-b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a \cosh (c+d x)}{b}} \, dx,x,x^2\right )}{2 a^2 \left (a^2-b^2\right )}\\ &=\frac{x^2}{2 a^2}+\frac{b^2 \tanh \left (c+d x^2\right )}{2 a \left (a^2-b^2\right ) d \left (a+b \text{sech}\left (c+d x^2\right )\right )}+\frac{\left (i \left (2 a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a}{b}+\left (1-\frac{a}{b}\right ) x^2} \, dx,x,i \tanh \left (\frac{1}{2} \left (c+d x^2\right )\right )\right )}{a^2 \left (a^2-b^2\right ) d}\\ &=\frac{x^2}{2 a^2}-\frac{b \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{1}{2} \left (c+d x^2\right )\right )}{\sqrt{a+b}}\right )}{a^2 (a-b)^{3/2} (a+b)^{3/2} d}+\frac{b^2 \tanh \left (c+d x^2\right )}{2 a \left (a^2-b^2\right ) d \left (a+b \text{sech}\left (c+d x^2\right )\right )}\\ \end{align*}
Mathematica [A] time = 0.438418, size = 220, normalized size = 1.79 \[ \frac{b \left (\left (a^2-b^2\right )^{3/2} \left (c+d x^2\right )+a b \sqrt{a^2-b^2} \sinh \left (c+d x^2\right )+\left (4 a^2 b-2 b^3\right ) \tan ^{-1}\left (\frac{(b-a) \tanh \left (\frac{1}{2} \left (c+d x^2\right )\right )}{\sqrt{a^2-b^2}}\right )\right )+a \cosh \left (c+d x^2\right ) \left (\left (a^2-b^2\right )^{3/2} \left (c+d x^2\right )+\left (4 a^2 b-2 b^3\right ) \tan ^{-1}\left (\frac{(b-a) \tanh \left (\frac{1}{2} \left (c+d x^2\right )\right )}{\sqrt{a^2-b^2}}\right )\right )}{2 a^2 d (a-b) (a+b) \sqrt{a^2-b^2} \left (a \cosh \left (c+d x^2\right )+b\right )} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.053, size = 234, normalized size = 1.9 \begin{align*} -{\frac{1}{2\,d{a}^{2}}\ln \left ( \tanh \left ({\frac{d{x}^{2}}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{{b}^{2}}{da \left ({a}^{2}-{b}^{2} \right ) }\tanh \left ({\frac{d{x}^{2}}{2}}+{\frac{c}{2}} \right ) \left ( \left ( \tanh \left ({\frac{d{x}^{2}}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a- \left ( \tanh \left ({\frac{d{x}^{2}}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}b+a+b \right ) ^{-1}}-2\,{\frac{b}{d \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tanh \left ( 1/2\,d{x}^{2}+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+{\frac{{b}^{3}}{d{a}^{2} \left ( a+b \right ) \left ( a-b \right ) }\arctan \left ({(a-b)\tanh \left ({\frac{d{x}^{2}}{2}}+{\frac{c}{2}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}}}+{\frac{1}{2\,d{a}^{2}}\ln \left ( \tanh \left ({\frac{d{x}^{2}}{2}}+{\frac{c}{2}} \right ) +1 \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.3971, size = 2867, normalized size = 23.31 \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}{\left (a + b \operatorname{sech}{\left (c + d x^{2} \right )}\right )^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.21601, size = 200, normalized size = 1.63 \begin{align*} -\frac{{\left (2 \, a^{2} b - b^{3}\right )} \arctan \left (\frac{a e^{\left (d x^{2} + c\right )} + b}{\sqrt{a^{2} - b^{2}}}\right )}{{\left (a^{4} d - a^{2} b^{2} d\right )} \sqrt{a^{2} - b^{2}}} - \frac{b^{3} e^{\left (d x^{2} + c\right )} + a b^{2}}{{\left (a^{4} d - a^{2} b^{2} d\right )}{\left (a e^{\left (2 \, d x^{2} + 2 \, c\right )} + 2 \, b e^{\left (d x^{2} + c\right )} + a\right )}} + \frac{d x^{2} + c}{2 \, a^{2} d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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