Optimal. Leaf size=54 \[ -\frac {\left (1+(a+b x)^2\right )^2}{b \sinh ^{-1}(a+b x)}+\frac {\text {Shi}\left (2 \sinh ^{-1}(a+b x)\right )}{b}+\frac {\text {Shi}\left (4 \sinh ^{-1}(a+b x)\right )}{2 b} \]
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Rubi [A]
time = 0.11, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5860, 5790,
5819, 5556, 3379} \begin {gather*} \frac {\text {Shi}\left (2 \sinh ^{-1}(a+b x)\right )}{b}+\frac {\text {Shi}\left (4 \sinh ^{-1}(a+b x)\right )}{2 b}-\frac {\left ((a+b x)^2+1\right )^2}{b \sinh ^{-1}(a+b x)} \end {gather*}
Antiderivative was successfully verified.
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Rule 3379
Rule 5556
Rule 5790
Rule 5819
Rule 5860
Rubi steps
\begin {align*} \int \frac {\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}}{\sinh ^{-1}(a+b x)^2} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (1+x^2\right )^{3/2}}{\sinh ^{-1}(x)^2} \, dx,x,a+b x\right )}{b}\\ &=-\frac {\left (1+(a+b x)^2\right )^2}{b \sinh ^{-1}(a+b x)}+\frac {4 \text {Subst}\left (\int \frac {x \left (1+x^2\right )}{\sinh ^{-1}(x)} \, dx,x,a+b x\right )}{b}\\ &=-\frac {\left (1+(a+b x)^2\right )^2}{b \sinh ^{-1}(a+b x)}+\frac {4 \text {Subst}\left (\int \frac {\cosh ^3(x) \sinh (x)}{x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=-\frac {\left (1+(a+b x)^2\right )^2}{b \sinh ^{-1}(a+b x)}+\frac {4 \text {Subst}\left (\int \left (\frac {\sinh (2 x)}{4 x}+\frac {\sinh (4 x)}{8 x}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=-\frac {\left (1+(a+b x)^2\right )^2}{b \sinh ^{-1}(a+b x)}+\frac {\text {Subst}\left (\int \frac {\sinh (4 x)}{x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b}+\frac {\text {Subst}\left (\int \frac {\sinh (2 x)}{x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=-\frac {\left (1+(a+b x)^2\right )^2}{b \sinh ^{-1}(a+b x)}+\frac {\text {Shi}\left (2 \sinh ^{-1}(a+b x)\right )}{b}+\frac {\text {Shi}\left (4 \sinh ^{-1}(a+b x)\right )}{2 b}\\ \end {align*}
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Mathematica [A]
time = 0.17, size = 70, normalized size = 1.30 \begin {gather*} \frac {-2 \left (1+a^2+2 a b x+b^2 x^2\right )^2+2 \sinh ^{-1}(a+b x) \text {Shi}\left (2 \sinh ^{-1}(a+b x)\right )+\sinh ^{-1}(a+b x) \text {Shi}\left (4 \sinh ^{-1}(a+b x)\right )}{2 b \sinh ^{-1}(a+b x)} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 3.86, size = 72, normalized size = 1.33
method | result | size |
default | \(\frac {8 \hyperbolicSineIntegral \left (2 \arcsinh \left (b x +a \right )\right ) \arcsinh \left (b x +a \right )+4 \hyperbolicSineIntegral \left (4 \arcsinh \left (b x +a \right )\right ) \arcsinh \left (b x +a \right )-4 \cosh \left (2 \arcsinh \left (b x +a \right )\right )-\cosh \left (4 \arcsinh \left (b x +a \right )\right )-3}{8 b \arcsinh \left (b x +a \right )}\) | \(72\) |
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \frac {\left (a^{2} + 2 a b x + b^{2} x^{2} + 1\right )^{\frac {3}{2}}}{\operatorname {asinh}^{2}{\left (a + b x \right )}}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}^{3/2}}{{\mathrm {asinh}\left (a+b\,x\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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