Optimal. Leaf size=545 \[ \frac {3^{-1-n} e^{-\frac {3 a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \Gamma \left (1+n,-\frac {3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{8 d^3}-\frac {2^{-2-n} c e^{-\frac {2 a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \Gamma \left (1+n,-\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{d^3}-\frac {e^{-\frac {a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \Gamma \left (1+n,-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{8 d^3}+\frac {c^2 e^{-\frac {a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \Gamma \left (1+n,-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{2 d^3}+\frac {e^{a/b} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \Gamma \left (1+n,\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{8 d^3}-\frac {c^2 e^{a/b} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \Gamma \left (1+n,\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{2 d^3}-\frac {2^{-2-n} c e^{\frac {2 a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \Gamma \left (1+n,\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{d^3}-\frac {3^{-1-n} e^{\frac {3 a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \Gamma \left (1+n,\frac {3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{8 d^3} \]
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Rubi [A]
time = 0.84, antiderivative size = 545, normalized size of antiderivative = 1.00, number of steps
used = 22, number of rules used = 9, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {5859, 5830,
6873, 12, 6874, 3388, 2212, 5556, 3389} \begin {gather*} \frac {c^2 e^{-\frac {a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \text {Gamma}\left (n+1,-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{2 d^3}-\frac {c^2 e^{a/b} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \text {Gamma}\left (n+1,\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{2 d^3}+\frac {3^{-n-1} e^{-\frac {3 a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \text {Gamma}\left (n+1,-\frac {3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{8 d^3}-\frac {c 2^{-n-2} e^{-\frac {2 a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \text {Gamma}\left (n+1,-\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{d^3}-\frac {e^{-\frac {a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \text {Gamma}\left (n+1,-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{8 d^3}+\frac {e^{a/b} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \text {Gamma}\left (n+1,\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{8 d^3}-\frac {c 2^{-n-2} e^{\frac {2 a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \text {Gamma}\left (n+1,\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{d^3}-\frac {3^{-n-1} e^{\frac {3 a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \text {Gamma}\left (n+1,\frac {3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{8 d^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2212
Rule 3388
Rule 3389
Rule 5556
Rule 5830
Rule 5859
Rule 6873
Rule 6874
Rubi steps
\begin {align*} \int x^2 \left (a+b \sinh ^{-1}(c+d x)\right )^n \, dx &=\frac {\text {Subst}\left (\int \left (-\frac {c}{d}+\frac {x}{d}\right )^2 \left (a+b \sinh ^{-1}(x)\right )^n \, dx,x,c+d x\right )}{d}\\ &=\frac {\text {Subst}\left (\int (a+b x)^n \cosh (x) \left (-\frac {c}{d}+\frac {\sinh (x)}{d}\right )^2 \, dx,x,\sinh ^{-1}(c+d x)\right )}{d}\\ &=\frac {\text {Subst}\left (\int \frac {(a+b x)^n \cosh (x) (c-\sinh (x))^2}{d^2} \, dx,x,\sinh ^{-1}(c+d x)\right )}{d}\\ &=\frac {\text {Subst}\left (\int (a+b x)^n \cosh (x) (c-\sinh (x))^2 \, dx,x,\sinh ^{-1}(c+d x)\right )}{d^3}\\ &=\frac {\text {Subst}\left (\int \left (c^2 (a+b x)^n \cosh (x)-2 c (a+b x)^n \cosh (x) \sinh (x)+(a+b x)^n \cosh (x) \sinh ^2(x)\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d^3}\\ &=\frac {\text {Subst}\left (\int (a+b x)^n \cosh (x) \sinh ^2(x) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d^3}-\frac {(2 c) \text {Subst}\left (\int (a+b x)^n \cosh (x) \sinh (x) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d^3}+\frac {c^2 \text {Subst}\left (\int (a+b x)^n \cosh (x) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d^3}\\ &=\frac {\text {Subst}\left (\int \left (-\frac {1}{4} (a+b x)^n \cosh (x)+\frac {1}{4} (a+b x)^n \cosh (3 x)\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d^3}-\frac {(2 c) \text {Subst}\left (\int \frac {1}{2} (a+b x)^n \sinh (2 x) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d^3}+\frac {c^2 \text {Subst}\left (\int e^{-x} (a+b x)^n \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 d^3}+\frac {c^2 \text {Subst}\left (\int e^x (a+b x)^n \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 d^3}\\ &=\frac {c^2 e^{-\frac {a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \Gamma \left (1+n,-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{2 d^3}-\frac {c^2 e^{a/b} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \Gamma \left (1+n,\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{2 d^3}-\frac {\text {Subst}\left (\int (a+b x)^n \cosh (x) \, dx,x,\sinh ^{-1}(c+d x)\right )}{4 d^3}+\frac {\text {Subst}\left (\int (a+b x)^n \cosh (3 x) \, dx,x,\sinh ^{-1}(c+d x)\right )}{4 d^3}-\frac {c \text {Subst}\left (\int (a+b x)^n \sinh (2 x) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d^3}\\ &=\frac {c^2 e^{-\frac {a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \Gamma \left (1+n,-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{2 d^3}-\frac {c^2 e^{a/b} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \Gamma \left (1+n,\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{2 d^3}+\frac {\text {Subst}\left (\int e^{-3 x} (a+b x)^n \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 d^3}-\frac {\text {Subst}\left (\int e^{-x} (a+b x)^n \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 d^3}-\frac {\text {Subst}\left (\int e^x (a+b x)^n \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 d^3}+\frac {\text {Subst}\left (\int e^{3 x} (a+b x)^n \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 d^3}+\frac {c \text {Subst}\left (\int e^{-2 x} (a+b x)^n \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 d^3}-\frac {c \text {Subst}\left (\int e^{2 x} (a+b x)^n \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 d^3}\\ &=\frac {3^{-1-n} e^{-\frac {3 a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \Gamma \left (1+n,-\frac {3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{8 d^3}-\frac {2^{-2-n} c e^{-\frac {2 a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \Gamma \left (1+n,-\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{d^3}-\frac {e^{-\frac {a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \Gamma \left (1+n,-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{8 d^3}+\frac {c^2 e^{-\frac {a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \Gamma \left (1+n,-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{2 d^3}+\frac {e^{a/b} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \Gamma \left (1+n,\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{8 d^3}-\frac {c^2 e^{a/b} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \Gamma \left (1+n,\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{2 d^3}-\frac {2^{-2-n} c e^{\frac {2 a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \Gamma \left (1+n,\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{d^3}-\frac {3^{-1-n} e^{\frac {3 a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \Gamma \left (1+n,\frac {3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{8 d^3}\\ \end {align*}
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Mathematica [F]
time = 1.01, size = 0, normalized size = 0.00 \begin {gather*} \int x^2 \left (a+b \sinh ^{-1}(c+d x)\right )^n \, dx \end {gather*}
Verification is not applicable to the result.
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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int x^{2} \left (a +b \arcsinh \left (d x +c \right )\right )^{n}\, dx\]
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 [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 x^{2} \left (a + b \operatorname {asinh}{\left (c + d x \right )}\right )^{n}\, 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 = -0.00 \begin {gather*} \int x^2\,{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^n \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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