Optimal. Leaf size=267 \[ \frac {2^{-n-3} e^{-\frac {2 a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \Gamma \left (n+1,-\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{d^2}-\frac {c e^{-\frac {a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \Gamma \left (n+1,-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{2 d^2}+\frac {c e^{a/b} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \Gamma \left (n+1,\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{2 d^2}+\frac {2^{-n-3} e^{\frac {2 a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \Gamma \left (n+1,\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{d^2} \]
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Rubi [A] time = 0.48, antiderivative size = 267, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {5865, 5805, 6741, 12, 6742, 3307, 2181, 5448, 3308} \[ \frac {2^{-n-3} e^{-\frac {2 a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \text {Gamma}\left (n+1,-\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{d^2}-\frac {c e^{-\frac {a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \text {Gamma}\left (n+1,-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{2 d^2}+\frac {c e^{a/b} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \text {Gamma}\left (n+1,\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{2 d^2}+\frac {2^{-n-3} e^{\frac {2 a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{-n} \text {Gamma}\left (n+1,\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{d^2} \]
Antiderivative was successfully verified.
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Rule 12
Rule 2181
Rule 3307
Rule 3308
Rule 5448
Rule 5805
Rule 5865
Rule 6741
Rule 6742
Rubi steps
\begin {align*} \int x \left (a+b \sinh ^{-1}(c+d x)\right )^n \, dx &=\frac {\operatorname {Subst}\left (\int \left (-\frac {c}{d}+\frac {x}{d}\right ) \left (a+b \sinh ^{-1}(x)\right )^n \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int (a+b x)^n \cosh (x) \left (-\frac {c}{d}+\frac {\sinh (x)}{d}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {(a+b x)^n \cosh (x) (-c+\sinh (x))}{d} \, dx,x,\sinh ^{-1}(c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int (a+b x)^n \cosh (x) (-c+\sinh (x)) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d^2}\\ &=\frac {\operatorname {Subst}\left (\int \left (-c (a+b x)^n \cosh (x)+(a+b x)^n \cosh (x) \sinh (x)\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d^2}\\ &=\frac {\operatorname {Subst}\left (\int (a+b x)^n \cosh (x) \sinh (x) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d^2}-\frac {c \operatorname {Subst}\left (\int (a+b x)^n \cosh (x) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d^2}\\ &=\frac {\operatorname {Subst}\left (\int \frac {1}{2} (a+b x)^n \sinh (2 x) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d^2}-\frac {c \operatorname {Subst}\left (\int e^{-x} (a+b x)^n \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 d^2}-\frac {c \operatorname {Subst}\left (\int e^x (a+b x)^n \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 d^2}\\ &=-\frac {c 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^2}+\frac {c 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^2}+\frac {\operatorname {Subst}\left (\int (a+b x)^n \sinh (2 x) \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 d^2}\\ &=-\frac {c 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^2}+\frac {c 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^2}-\frac {\operatorname {Subst}\left (\int e^{-2 x} (a+b x)^n \, dx,x,\sinh ^{-1}(c+d x)\right )}{4 d^2}+\frac {\operatorname {Subst}\left (\int e^{2 x} (a+b x)^n \, dx,x,\sinh ^{-1}(c+d x)\right )}{4 d^2}\\ &=\frac {2^{-3-n} 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^2}-\frac {c 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^2}+\frac {c 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^2}+\frac {2^{-3-n} 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^2}\\ \end {align*}
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Mathematica [A] time = 0.19, size = 228, normalized size = 0.85 \[ \frac {2^{-n-3} e^{-\frac {2 a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right )^n \left (-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^2}{b^2}\right )^{-n} \left (\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )^n \Gamma \left (n+1,-\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )-c 2^{n+2} e^{a/b} \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )^n \Gamma \left (n+1,-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )+c 2^{n+2} e^{\frac {3 a}{b}} \left (-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^n \Gamma \left (n+1,\frac {a}{b}+\sinh ^{-1}(c+d x)\right )+e^{\frac {4 a}{b}} \left (-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^n \Gamma \left (n+1,\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )\right )}{d^2} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{n} x, x\right ) \]
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 \[ \int {\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{n} x\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.32, size = 0, normalized size = 0.00 \[ \int x \left (a +b \arcsinh \left (d x +c \right )\right )^{n}\, dx \]
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 \[ \int {\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{n} x\,{d x} \]
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 \[ \int x\,{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^n \,d x \]
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 \[ \int x \left (a + b \operatorname {asinh}{\left (c + d x \right )}\right )^{n}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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