Optimal. Leaf size=331 \[ \frac {4 a x}{3 b^3}-\frac {2 a^3 x}{b^3}-\frac {3 (a+b x)^2}{32 b^4}+\frac {3 a^2 (a+b x)^2}{4 b^4}-\frac {2 a (a+b x)^3}{9 b^4}+\frac {(a+b x)^4}{32 b^4}-\frac {4 a \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{3 b^4}+\frac {2 a^3 \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{b^4}+\frac {3 (a+b x) \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{16 b^4}-\frac {3 a^2 (a+b x) \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{2 b^4}+\frac {2 a (a+b x)^2 \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{3 b^4}-\frac {(a+b x)^3 \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{8 b^4}-\frac {3 \sinh ^{-1}(a+b x)^2}{32 b^4}+\frac {3 a^2 \sinh ^{-1}(a+b x)^2}{4 b^4}-\frac {a^4 \sinh ^{-1}(a+b x)^2}{4 b^4}+\frac {1}{4} x^4 \sinh ^{-1}(a+b x)^2 \]
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Rubi [A]
time = 0.38, antiderivative size = 331, normalized size of antiderivative = 1.00, number of steps
used = 19, number of rules used = 8, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {5859, 5828,
5838, 5783, 5798, 8, 5812, 30} \begin {gather*} -\frac {a^4 \sinh ^{-1}(a+b x)^2}{4 b^4}+\frac {2 a^3 \sqrt {(a+b x)^2+1} \sinh ^{-1}(a+b x)}{b^4}-\frac {2 a^3 x}{b^3}+\frac {3 a^2 (a+b x)^2}{4 b^4}+\frac {3 a^2 \sinh ^{-1}(a+b x)^2}{4 b^4}-\frac {3 a^2 (a+b x) \sqrt {(a+b x)^2+1} \sinh ^{-1}(a+b x)}{2 b^4}-\frac {2 a (a+b x)^3}{9 b^4}+\frac {(a+b x)^4}{32 b^4}-\frac {3 (a+b x)^2}{32 b^4}+\frac {2 a (a+b x)^2 \sqrt {(a+b x)^2+1} \sinh ^{-1}(a+b x)}{3 b^4}-\frac {4 a \sqrt {(a+b x)^2+1} \sinh ^{-1}(a+b x)}{3 b^4}-\frac {3 \sinh ^{-1}(a+b x)^2}{32 b^4}-\frac {(a+b x)^3 \sqrt {(a+b x)^2+1} \sinh ^{-1}(a+b x)}{8 b^4}+\frac {3 (a+b x) \sqrt {(a+b x)^2+1} \sinh ^{-1}(a+b x)}{16 b^4}+\frac {4 a x}{3 b^3}+\frac {1}{4} x^4 \sinh ^{-1}(a+b x)^2 \end {gather*}
Antiderivative was successfully verified.
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Rule 8
Rule 30
Rule 5783
Rule 5798
Rule 5812
Rule 5828
Rule 5838
Rule 5859
Rubi steps
\begin {align*} \int x^3 \sinh ^{-1}(a+b x)^2 \, dx &=\frac {\text {Subst}\left (\int \left (-\frac {a}{b}+\frac {x}{b}\right )^3 \sinh ^{-1}(x)^2 \, dx,x,a+b x\right )}{b}\\ &=\frac {1}{4} x^4 \sinh ^{-1}(a+b x)^2-\frac {1}{2} \text {Subst}\left (\int \frac {\left (-\frac {a}{b}+\frac {x}{b}\right )^4 \sinh ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )\\ &=\frac {1}{4} x^4 \sinh ^{-1}(a+b x)^2-\frac {1}{2} \text {Subst}\left (\int \left (\frac {a^4 \sinh ^{-1}(x)}{b^4 \sqrt {1+x^2}}-\frac {4 a^3 x \sinh ^{-1}(x)}{b^4 \sqrt {1+x^2}}+\frac {6 a^2 x^2 \sinh ^{-1}(x)}{b^4 \sqrt {1+x^2}}-\frac {4 a x^3 \sinh ^{-1}(x)}{b^4 \sqrt {1+x^2}}+\frac {x^4 \sinh ^{-1}(x)}{b^4 \sqrt {1+x^2}}\right ) \, dx,x,a+b x\right )\\ &=\frac {1}{4} x^4 \sinh ^{-1}(a+b x)^2-\frac {\text {Subst}\left (\int \frac {x^4 \sinh ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{2 b^4}+\frac {(2 a) \text {Subst}\left (\int \frac {x^3 \sinh ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{b^4}-\frac {\left (3 a^2\right ) \text {Subst}\left (\int \frac {x^2 \sinh ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{b^4}+\frac {\left (2 a^3\right ) \text {Subst}\left (\int \frac {x \sinh ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{b^4}-\frac {a^4 \text {Subst}\left (\int \frac {\sinh ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{2 b^4}\\ &=\frac {2 a^3 \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{b^4}-\frac {3 a^2 (a+b x) \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{2 b^4}+\frac {2 a (a+b x)^2 \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{3 b^4}-\frac {(a+b x)^3 \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{8 b^4}-\frac {a^4 \sinh ^{-1}(a+b x)^2}{4 b^4}+\frac {1}{4} x^4 \sinh ^{-1}(a+b x)^2+\frac {\text {Subst}\left (\int x^3 \, dx,x,a+b x\right )}{8 b^4}+\frac {3 \text {Subst}\left (\int \frac {x^2 \sinh ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{8 b^4}-\frac {(2 a) \text {Subst}\left (\int x^2 \, dx,x,a+b x\right )}{3 b^4}-\frac {(4 a) \text {Subst}\left (\int \frac {x \sinh ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{3 b^4}+\frac {\left (3 a^2\right ) \text {Subst}(\int x \, dx,x,a+b x)}{2 b^4}+\frac {\left (3 a^2\right ) \text {Subst}\left (\int \frac {\sinh ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{2 b^4}-\frac {\left (2 a^3\right ) \text {Subst}(\int 1 \, dx,x,a+b x)}{b^4}\\ &=-\frac {2 a^3 x}{b^3}+\frac {3 a^2 (a+b x)^2}{4 b^4}-\frac {2 a (a+b x)^3}{9 b^4}+\frac {(a+b x)^4}{32 b^4}-\frac {4 a \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{3 b^4}+\frac {2 a^3 \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{b^4}+\frac {3 (a+b x) \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{16 b^4}-\frac {3 a^2 (a+b x) \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{2 b^4}+\frac {2 a (a+b x)^2 \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{3 b^4}-\frac {(a+b x)^3 \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{8 b^4}+\frac {3 a^2 \sinh ^{-1}(a+b x)^2}{4 b^4}-\frac {a^4 \sinh ^{-1}(a+b x)^2}{4 b^4}+\frac {1}{4} x^4 \sinh ^{-1}(a+b x)^2-\frac {3 \text {Subst}(\int x \, dx,x,a+b x)}{16 b^4}-\frac {3 \text {Subst}\left (\int \frac {\sinh ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{16 b^4}+\frac {(4 a) \text {Subst}(\int 1 \, dx,x,a+b x)}{3 b^4}\\ &=\frac {4 a x}{3 b^3}-\frac {2 a^3 x}{b^3}-\frac {3 (a+b x)^2}{32 b^4}+\frac {3 a^2 (a+b x)^2}{4 b^4}-\frac {2 a (a+b x)^3}{9 b^4}+\frac {(a+b x)^4}{32 b^4}-\frac {4 a \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{3 b^4}+\frac {2 a^3 \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{b^4}+\frac {3 (a+b x) \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{16 b^4}-\frac {3 a^2 (a+b x) \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{2 b^4}+\frac {2 a (a+b x)^2 \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{3 b^4}-\frac {(a+b x)^3 \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{8 b^4}-\frac {3 \sinh ^{-1}(a+b x)^2}{32 b^4}+\frac {3 a^2 \sinh ^{-1}(a+b x)^2}{4 b^4}-\frac {a^4 \sinh ^{-1}(a+b x)^2}{4 b^4}+\frac {1}{4} x^4 \sinh ^{-1}(a+b x)^2\\ \end {align*}
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Mathematica [A]
time = 0.12, size = 145, normalized size = 0.44 \begin {gather*} \frac {b x \left (-300 a^3+78 a^2 b x+a \left (330-28 b^2 x^2\right )+9 b x \left (-3+b^2 x^2\right )\right )+6 \sqrt {1+a^2+2 a b x+b^2 x^2} \left (50 a^3+9 b x-26 a^2 b x-6 b^3 x^3+a \left (-55+14 b^2 x^2\right )\right ) \sinh ^{-1}(a+b x)-9 \left (3-24 a^2+8 a^4-8 b^4 x^4\right ) \sinh ^{-1}(a+b x)^2}{288 b^4} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int x^{3} \arcsinh \left (b x +a \right )^{2}\, dx\]
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 [A]
time = 0.37, size = 182, normalized size = 0.55 \begin {gather*} \frac {9 \, b^{4} x^{4} - 28 \, a b^{3} x^{3} + 3 \, {\left (26 \, a^{2} - 9\right )} b^{2} x^{2} - 30 \, {\left (10 \, a^{3} - 11 \, a\right )} b x + 9 \, {\left (8 \, b^{4} x^{4} - 8 \, a^{4} + 24 \, a^{2} - 3\right )} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )^{2} - 6 \, {\left (6 \, b^{3} x^{3} - 14 \, a b^{2} x^{2} - 50 \, a^{3} + {\left (26 \, a^{2} - 9\right )} b x + 55 \, a\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}{288 \, b^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.46, size = 366, normalized size = 1.11 \begin {gather*} \begin {cases} - \frac {a^{4} \operatorname {asinh}^{2}{\left (a + b x \right )}}{4 b^{4}} - \frac {25 a^{3} x}{24 b^{3}} + \frac {25 a^{3} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname {asinh}{\left (a + b x \right )}}{24 b^{4}} + \frac {13 a^{2} x^{2}}{48 b^{2}} - \frac {13 a^{2} x \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname {asinh}{\left (a + b x \right )}}{24 b^{3}} + \frac {3 a^{2} \operatorname {asinh}^{2}{\left (a + b x \right )}}{4 b^{4}} - \frac {7 a x^{3}}{72 b} + \frac {7 a x^{2} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname {asinh}{\left (a + b x \right )}}{24 b^{2}} + \frac {55 a x}{48 b^{3}} - \frac {55 a \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname {asinh}{\left (a + b x \right )}}{48 b^{4}} + \frac {x^{4} \operatorname {asinh}^{2}{\left (a + b x \right )}}{4} + \frac {x^{4}}{32} - \frac {x^{3} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname {asinh}{\left (a + b x \right )}}{8 b} - \frac {3 x^{2}}{32 b^{2}} + \frac {3 x \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname {asinh}{\left (a + b x \right )}}{16 b^{3}} - \frac {3 \operatorname {asinh}^{2}{\left (a + b x \right )}}{32 b^{4}} & \text {for}\: b \neq 0 \\\frac {x^{4} \operatorname {asinh}^{2}{\left (a \right )}}{4} & \text {otherwise} \end {cases} \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 = -0.00 \begin {gather*} \int x^3\,{\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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