Optimal. Leaf size=370 \[ -\frac {2 \sqrt {a^2-b^2} \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d^3}+\frac {2 \sqrt {a^2-b^2} \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b^2 d^3}+\frac {2 x \sqrt {a^2-b^2} \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d^2}-\frac {2 x \sqrt {a^2-b^2} \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b^2 d^2}+\frac {x^2 \sqrt {a^2-b^2} \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}+1\right )}{b^2 d}-\frac {x^2 \sqrt {a^2-b^2} \log \left (\frac {b e^{c+d x}}{\sqrt {a^2-b^2}+a}+1\right )}{b^2 d}-\frac {a x^3}{3 b^2}+\frac {2 \sinh (c+d x)}{b d^3}-\frac {2 x \cosh (c+d x)}{b d^2}+\frac {x^2 \sinh (c+d x)}{b d} \]
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Rubi [A] time = 0.70, antiderivative size = 370, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 10, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {5566, 30, 3296, 2637, 3320, 2264, 2190, 2531, 2282, 6589} \[ \frac {2 x \sqrt {a^2-b^2} \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d^2}-\frac {2 x \sqrt {a^2-b^2} \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{\sqrt {a^2-b^2}+a}\right )}{b^2 d^2}-\frac {2 \sqrt {a^2-b^2} \text {PolyLog}\left (3,-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d^3}+\frac {2 \sqrt {a^2-b^2} \text {PolyLog}\left (3,-\frac {b e^{c+d x}}{\sqrt {a^2-b^2}+a}\right )}{b^2 d^3}+\frac {x^2 \sqrt {a^2-b^2} \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}+1\right )}{b^2 d}-\frac {x^2 \sqrt {a^2-b^2} \log \left (\frac {b e^{c+d x}}{\sqrt {a^2-b^2}+a}+1\right )}{b^2 d}-\frac {a x^3}{3 b^2}+\frac {2 \sinh (c+d x)}{b d^3}-\frac {2 x \cosh (c+d x)}{b d^2}+\frac {x^2 \sinh (c+d x)}{b d} \]
Antiderivative was successfully verified.
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Rule 30
Rule 2190
Rule 2264
Rule 2282
Rule 2531
Rule 2637
Rule 3296
Rule 3320
Rule 5566
Rule 6589
Rubi steps
\begin {align*} \int \frac {x^2 \sinh ^2(c+d x)}{a+b \cosh (c+d x)} \, dx &=-\frac {a \int x^2 \, dx}{b^2}+\frac {\int x^2 \cosh (c+d x) \, dx}{b}+\frac {\left (a^2-b^2\right ) \int \frac {x^2}{a+b \cosh (c+d x)} \, dx}{b^2}\\ &=-\frac {a x^3}{3 b^2}+\frac {x^2 \sinh (c+d x)}{b d}+\frac {\left (2 \left (a^2-b^2\right )\right ) \int \frac {e^{c+d x} x^2}{b+2 a e^{c+d x}+b e^{2 (c+d x)}} \, dx}{b^2}-\frac {2 \int x \sinh (c+d x) \, dx}{b d}\\ &=-\frac {a x^3}{3 b^2}-\frac {2 x \cosh (c+d x)}{b d^2}+\frac {x^2 \sinh (c+d x)}{b d}+\frac {\left (2 \sqrt {a^2-b^2}\right ) \int \frac {e^{c+d x} x^2}{2 a-2 \sqrt {a^2-b^2}+2 b e^{c+d x}} \, dx}{b}-\frac {\left (2 \sqrt {a^2-b^2}\right ) \int \frac {e^{c+d x} x^2}{2 a+2 \sqrt {a^2-b^2}+2 b e^{c+d x}} \, dx}{b}+\frac {2 \int \cosh (c+d x) \, dx}{b d^2}\\ &=-\frac {a x^3}{3 b^2}-\frac {2 x \cosh (c+d x)}{b d^2}+\frac {\sqrt {a^2-b^2} x^2 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d}-\frac {\sqrt {a^2-b^2} x^2 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b^2 d}+\frac {2 \sinh (c+d x)}{b d^3}+\frac {x^2 \sinh (c+d x)}{b d}-\frac {\left (2 \sqrt {a^2-b^2}\right ) \int x \log \left (1+\frac {2 b e^{c+d x}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{b^2 d}+\frac {\left (2 \sqrt {a^2-b^2}\right ) \int x \log \left (1+\frac {2 b e^{c+d x}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{b^2 d}\\ &=-\frac {a x^3}{3 b^2}-\frac {2 x \cosh (c+d x)}{b d^2}+\frac {\sqrt {a^2-b^2} x^2 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d}-\frac {\sqrt {a^2-b^2} x^2 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b^2 d}+\frac {2 \sqrt {a^2-b^2} x \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d^2}-\frac {2 \sqrt {a^2-b^2} x \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b^2 d^2}+\frac {2 \sinh (c+d x)}{b d^3}+\frac {x^2 \sinh (c+d x)}{b d}-\frac {\left (2 \sqrt {a^2-b^2}\right ) \int \text {Li}_2\left (-\frac {2 b e^{c+d x}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{b^2 d^2}+\frac {\left (2 \sqrt {a^2-b^2}\right ) \int \text {Li}_2\left (-\frac {2 b e^{c+d x}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{b^2 d^2}\\ &=-\frac {a x^3}{3 b^2}-\frac {2 x \cosh (c+d x)}{b d^2}+\frac {\sqrt {a^2-b^2} x^2 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d}-\frac {\sqrt {a^2-b^2} x^2 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b^2 d}+\frac {2 \sqrt {a^2-b^2} x \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d^2}-\frac {2 \sqrt {a^2-b^2} x \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b^2 d^2}+\frac {2 \sinh (c+d x)}{b d^3}+\frac {x^2 \sinh (c+d x)}{b d}-\frac {\left (2 \sqrt {a^2-b^2}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {b x}{-a+\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{b^2 d^3}+\frac {\left (2 \sqrt {a^2-b^2}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {b x}{a+\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{b^2 d^3}\\ &=-\frac {a x^3}{3 b^2}-\frac {2 x \cosh (c+d x)}{b d^2}+\frac {\sqrt {a^2-b^2} x^2 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d}-\frac {\sqrt {a^2-b^2} x^2 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b^2 d}+\frac {2 \sqrt {a^2-b^2} x \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d^2}-\frac {2 \sqrt {a^2-b^2} x \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b^2 d^2}-\frac {2 \sqrt {a^2-b^2} \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d^3}+\frac {2 \sqrt {a^2-b^2} \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b^2 d^3}+\frac {2 \sinh (c+d x)}{b d^3}+\frac {x^2 \sinh (c+d x)}{b d}\\ \end {align*}
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Mathematica [A] time = 1.30, size = 293, normalized size = 0.79 \[ \frac {3 \sqrt {a^2-b^2} \left (d^2 x^2 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}+1\right )-d^2 x^2 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2-b^2}+a}+1\right )+2 d x \text {Li}_2\left (\frac {b e^{c+d x}}{\sqrt {a^2-b^2}-a}\right )-2 d x \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )-2 \text {Li}_3\left (\frac {b e^{c+d x}}{\sqrt {a^2-b^2}-a}\right )+2 \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )\right )-a d^3 x^3+3 b \cosh (d x) \left (\sinh (c) \left (d^2 x^2+2\right )-2 d x \cosh (c)\right )+3 b \sinh (d x) \left (\cosh (c) \left (d^2 x^2+2\right )-2 d x \sinh (c)\right )}{3 b^2 d^3} \]
Antiderivative was successfully verified.
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fricas [C] time = 1.70, size = 937, normalized size = 2.53 \[ \text {result too large to display} \]
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 \[ \int \frac {x^{2} \sinh \left (d x + c\right )^{2}}{b \cosh \left (d x + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.43, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \left (\sinh ^{2}\left (d x +c \right )\right )}{a +b \cosh \left (d x +c \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
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 \[ \int \frac {x^2\,{\mathrm {sinh}\left (c+d\,x\right )}^2}{a+b\,\mathrm {cosh}\left (c+d\,x\right )} \,d x \]
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 \[ \int \frac {x^{2} \sinh ^{2}{\left (c + d x \right )}}{a + b \cosh {\left (c + d x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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