Optimal. Leaf size=131 \[ -\frac {\left (4 a \left (16-19 a^2\right )-\left (9-26 a^2\right ) (a+b x)\right ) \sqrt {(a+b x)^2+1}}{96 b^4}-\frac {\left (8 a^4-24 a^2+3\right ) \sinh ^{-1}(a+b x)}{32 b^4}+\frac {7 a x^2 \sqrt {(a+b x)^2+1}}{48 b^2}+\frac {1}{4} x^4 \sinh ^{-1}(a+b x)-\frac {x^3 \sqrt {(a+b x)^2+1}}{16 b} \]
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Rubi [A] time = 0.17, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {5865, 5801, 743, 833, 780, 215} \[ -\frac {\left (4 a \left (16-19 a^2\right )-\left (9-26 a^2\right ) (a+b x)\right ) \sqrt {(a+b x)^2+1}}{96 b^4}-\frac {\left (8 a^4-24 a^2+3\right ) \sinh ^{-1}(a+b x)}{32 b^4}+\frac {7 a x^2 \sqrt {(a+b x)^2+1}}{48 b^2}-\frac {x^3 \sqrt {(a+b x)^2+1}}{16 b}+\frac {1}{4} x^4 \sinh ^{-1}(a+b x) \]
Antiderivative was successfully verified.
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Rule 215
Rule 743
Rule 780
Rule 833
Rule 5801
Rule 5865
Rubi steps
\begin {align*} \int x^3 \sinh ^{-1}(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int \left (-\frac {a}{b}+\frac {x}{b}\right )^3 \sinh ^{-1}(x) \, dx,x,a+b x\right )}{b}\\ &=\frac {1}{4} x^4 \sinh ^{-1}(a+b x)-\frac {1}{4} \operatorname {Subst}\left (\int \frac {\left (-\frac {a}{b}+\frac {x}{b}\right )^4}{\sqrt {1+x^2}} \, dx,x,a+b x\right )\\ &=-\frac {x^3 \sqrt {1+(a+b x)^2}}{16 b}+\frac {1}{4} x^4 \sinh ^{-1}(a+b x)-\frac {1}{16} \operatorname {Subst}\left (\int \frac {\left (-\frac {3-4 a^2}{b^2}-\frac {7 a x}{b^2}\right ) \left (-\frac {a}{b}+\frac {x}{b}\right )^2}{\sqrt {1+x^2}} \, dx,x,a+b x\right )\\ &=\frac {7 a x^2 \sqrt {1+(a+b x)^2}}{48 b^2}-\frac {x^3 \sqrt {1+(a+b x)^2}}{16 b}+\frac {1}{4} x^4 \sinh ^{-1}(a+b x)-\frac {1}{48} \operatorname {Subst}\left (\int \frac {\left (\frac {a \left (23-12 a^2\right )}{b^3}-\frac {\left (9-26 a^2\right ) x}{b^3}\right ) \left (-\frac {a}{b}+\frac {x}{b}\right )}{\sqrt {1+x^2}} \, dx,x,a+b x\right )\\ &=\frac {7 a x^2 \sqrt {1+(a+b x)^2}}{48 b^2}-\frac {x^3 \sqrt {1+(a+b x)^2}}{16 b}-\frac {\left (4 a \left (16-19 a^2\right )-\left (9-26 a^2\right ) (a+b x)\right ) \sqrt {1+(a+b x)^2}}{96 b^4}+\frac {1}{4} x^4 \sinh ^{-1}(a+b x)-\frac {\left (3-24 a^2+8 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{32 b^4}\\ &=\frac {7 a x^2 \sqrt {1+(a+b x)^2}}{48 b^2}-\frac {x^3 \sqrt {1+(a+b x)^2}}{16 b}-\frac {\left (4 a \left (16-19 a^2\right )-\left (9-26 a^2\right ) (a+b x)\right ) \sqrt {1+(a+b x)^2}}{96 b^4}-\frac {\left (3-24 a^2+8 a^4\right ) \sinh ^{-1}(a+b x)}{32 b^4}+\frac {1}{4} x^4 \sinh ^{-1}(a+b x)\\ \end {align*}
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Mathematica [A] time = 0.09, size = 95, normalized size = 0.73 \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2+1} \left (50 a^3-26 a^2 b x+a \left (14 b^2 x^2-55\right )-6 b^3 x^3+9 b x\right )-3 \left (8 a^4-24 a^2-8 b^4 x^4+3\right ) \sinh ^{-1}(a+b x)}{96 b^4} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.46, size = 110, normalized size = 0.84 \[ \frac {3 \, {\left (8 \, b^{4} x^{4} - 8 \, a^{4} + 24 \, a^{2} - 3\right )} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - {\left (6 \, b^{3} x^{3} - 14 \, a b^{2} x^{2} - 50 \, a^{3} + {\left (26 \, a^{2} - 9\right )} b x + 55 \, a\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{96 \, b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.00, size = 162, normalized size = 1.24 \[ \frac {1}{4} \, x^{4} \log \left (b x + a + \sqrt {{\left (b x + a\right )}^{2} + 1}\right ) - \frac {1}{96} \, {\left (\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left ({\left (2 \, x {\left (\frac {3 \, x}{b^{2}} - \frac {7 \, a}{b^{3}}\right )} + \frac {26 \, a^{2} b^{3} - 9 \, b^{3}}{b^{7}}\right )} x - \frac {5 \, {\left (10 \, a^{3} b^{2} - 11 \, a b^{2}\right )}}{b^{7}}\right )} - \frac {3 \, {\left (8 \, a^{4} - 24 \, a^{2} + 3\right )} \log \left (-a b - {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} {\left | b \right |}\right )}{b^{4} {\left | b \right |}}\right )} b \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.02, size = 200, normalized size = 1.53 \[ \frac {\frac {\arcsinh \left (b x +a \right ) \left (b x +a \right )^{4}}{4}-\arcsinh \left (b x +a \right ) \left (b x +a \right )^{3} a +\frac {3 \arcsinh \left (b x +a \right ) \left (b x +a \right )^{2} a^{2}}{2}-\arcsinh \left (b x +a \right ) \left (b x +a \right ) a^{3}-\frac {\left (b x +a \right )^{3} \sqrt {1+\left (b x +a \right )^{2}}}{16}+\frac {3 \left (b x +a \right ) \sqrt {1+\left (b x +a \right )^{2}}}{32}-\frac {3 \arcsinh \left (b x +a \right )}{32}+a \left (\frac {\left (b x +a \right )^{2} \sqrt {1+\left (b x +a \right )^{2}}}{3}-\frac {2 \sqrt {1+\left (b x +a \right )^{2}}}{3}\right )-\frac {3 a^{2} \left (\frac {\left (b x +a \right ) \sqrt {1+\left (b x +a \right )^{2}}}{2}-\frac {\arcsinh \left (b x +a \right )}{2}\right )}{2}+a^{3} \sqrt {1+\left (b x +a \right )^{2}}}{b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.34, size = 318, normalized size = 2.43 \[ \frac {1}{4} \, x^{4} \operatorname {arsinh}\left (b x + a\right ) - \frac {1}{96} \, {\left (\frac {6 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} x^{3}}{b^{2}} - \frac {14 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a x^{2}}{b^{3}} + \frac {105 \, a^{4} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{b^{5}} + \frac {35 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a^{2} x}{b^{4}} - \frac {90 \, {\left (a^{2} + 1\right )} a^{2} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{b^{5}} - \frac {105 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a^{3}}{b^{5}} - \frac {9 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (a^{2} + 1\right )} x}{b^{4}} + \frac {9 \, {\left (a^{2} + 1\right )}^{2} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{b^{5}} + \frac {55 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (a^{2} + 1\right )} a}{b^{5}}\right )} b \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,\mathrm {asinh}\left (a+b\,x\right ) \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.44, size = 255, normalized size = 1.95 \[ \begin {cases} - \frac {a^{4} \operatorname {asinh}{\left (a + b x \right )}}{4 b^{4}} + \frac {25 a^{3} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{48 b^{4}} - \frac {13 a^{2} x \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{48 b^{3}} + \frac {3 a^{2} \operatorname {asinh}{\left (a + b x \right )}}{4 b^{4}} + \frac {7 a x^{2} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{48 b^{2}} - \frac {55 a \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{96 b^{4}} + \frac {x^{4} \operatorname {asinh}{\left (a + b x \right )}}{4} - \frac {x^{3} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{16 b} + \frac {3 x \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{32 b^{3}} - \frac {3 \operatorname {asinh}{\left (a + b x \right )}}{32 b^{4}} & \text {for}\: b \neq 0 \\\frac {x^{4} \operatorname {asinh}{\relax (a )}}{4} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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