Optimal. Leaf size=326 \[ -\frac {2 e^3 (c+d x)^3 \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {4 e^3 (c+d x)^2}{b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {16 e^3 (c+d x)^4}{3 b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {2 e^3 e^{\frac {4 a}{b}} \sqrt {\pi } \text {Erf}\left (\frac {2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}+\frac {e^3 e^{\frac {2 a}{b}} \sqrt {2 \pi } \text {Erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}+\frac {2 e^3 e^{-\frac {4 a}{b}} \sqrt {\pi } \text {Erfi}\left (\frac {2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}-\frac {e^3 e^{-\frac {2 a}{b}} \sqrt {2 \pi } \text {Erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d} \]
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Rubi [A]
time = 0.74, antiderivative size = 326, normalized size of antiderivative = 1.00, number of steps
used = 26, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5859, 12,
5779, 5818, 5780, 5556, 3389, 2211, 2236, 2235} \begin {gather*} -\frac {2 \sqrt {\pi } e^3 e^{\frac {4 a}{b}} \text {Erf}\left (\frac {2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}+\frac {\sqrt {2 \pi } e^3 e^{\frac {2 a}{b}} \text {Erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}+\frac {2 \sqrt {\pi } e^3 e^{-\frac {4 a}{b}} \text {Erfi}\left (\frac {2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}-\frac {\sqrt {2 \pi } e^3 e^{-\frac {2 a}{b}} \text {Erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}-\frac {16 e^3 (c+d x)^4}{3 b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {4 e^3 (c+d x)^2}{b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {2 e^3 \sqrt {(c+d x)^2+1} (c+d x)^3}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2211
Rule 2235
Rule 2236
Rule 3389
Rule 5556
Rule 5779
Rule 5780
Rule 5818
Rule 5859
Rubi steps
\begin {align*} \int \frac {(c e+d e x)^3}{\left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}} \, dx &=\frac {\text {Subst}\left (\int \frac {e^3 x^3}{\left (a+b \sinh ^{-1}(x)\right )^{5/2}} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^3 \text {Subst}\left (\int \frac {x^3}{\left (a+b \sinh ^{-1}(x)\right )^{5/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac {2 e^3 (c+d x)^3 \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}+\frac {\left (2 e^3\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {1+x^2} \left (a+b \sinh ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{b d}+\frac {\left (8 e^3\right ) \text {Subst}\left (\int \frac {x^4}{\sqrt {1+x^2} \left (a+b \sinh ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{3 b d}\\ &=-\frac {2 e^3 (c+d x)^3 \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {4 e^3 (c+d x)^2}{b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {16 e^3 (c+d x)^4}{3 b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {\left (8 e^3\right ) \text {Subst}\left (\int \frac {x}{\sqrt {a+b \sinh ^{-1}(x)}} \, dx,x,c+d x\right )}{b^2 d}+\frac {\left (64 e^3\right ) \text {Subst}\left (\int \frac {x^3}{\sqrt {a+b \sinh ^{-1}(x)}} \, dx,x,c+d x\right )}{3 b^2 d}\\ &=-\frac {2 e^3 (c+d x)^3 \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {4 e^3 (c+d x)^2}{b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {16 e^3 (c+d x)^4}{3 b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {\left (8 e^3\right ) \text {Subst}\left (\int \frac {\cosh (x) \sinh (x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}+\frac {\left (64 e^3\right ) \text {Subst}\left (\int \frac {\cosh (x) \sinh ^3(x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 b^2 d}\\ &=-\frac {2 e^3 (c+d x)^3 \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {4 e^3 (c+d x)^2}{b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {16 e^3 (c+d x)^4}{3 b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {\left (8 e^3\right ) \text {Subst}\left (\int \frac {\sinh (2 x)}{2 \sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}+\frac {\left (64 e^3\right ) \text {Subst}\left (\int \left (-\frac {\sinh (2 x)}{4 \sqrt {a+b x}}+\frac {\sinh (4 x)}{8 \sqrt {a+b x}}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 b^2 d}\\ &=-\frac {2 e^3 (c+d x)^3 \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {4 e^3 (c+d x)^2}{b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {16 e^3 (c+d x)^4}{3 b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {\left (8 e^3\right ) \text {Subst}\left (\int \frac {\sinh (4 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 b^2 d}+\frac {\left (4 e^3\right ) \text {Subst}\left (\int \frac {\sinh (2 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}-\frac {\left (16 e^3\right ) \text {Subst}\left (\int \frac {\sinh (2 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 b^2 d}\\ &=-\frac {2 e^3 (c+d x)^3 \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {4 e^3 (c+d x)^2}{b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {16 e^3 (c+d x)^4}{3 b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {\left (4 e^3\right ) \text {Subst}\left (\int \frac {e^{-4 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 b^2 d}+\frac {\left (4 e^3\right ) \text {Subst}\left (\int \frac {e^{4 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 b^2 d}-\frac {\left (2 e^3\right ) \text {Subst}\left (\int \frac {e^{-2 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}+\frac {\left (2 e^3\right ) \text {Subst}\left (\int \frac {e^{2 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}+\frac {\left (8 e^3\right ) \text {Subst}\left (\int \frac {e^{-2 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 b^2 d}-\frac {\left (8 e^3\right ) \text {Subst}\left (\int \frac {e^{2 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 b^2 d}\\ &=-\frac {2 e^3 (c+d x)^3 \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {4 e^3 (c+d x)^2}{b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {16 e^3 (c+d x)^4}{3 b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {\left (8 e^3\right ) \text {Subst}\left (\int e^{\frac {4 a}{b}-\frac {4 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{3 b^3 d}+\frac {\left (8 e^3\right ) \text {Subst}\left (\int e^{-\frac {4 a}{b}+\frac {4 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{3 b^3 d}-\frac {\left (4 e^3\right ) \text {Subst}\left (\int e^{\frac {2 a}{b}-\frac {2 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{b^3 d}+\frac {\left (4 e^3\right ) \text {Subst}\left (\int e^{-\frac {2 a}{b}+\frac {2 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{b^3 d}+\frac {\left (16 e^3\right ) \text {Subst}\left (\int e^{\frac {2 a}{b}-\frac {2 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{3 b^3 d}-\frac {\left (16 e^3\right ) \text {Subst}\left (\int e^{-\frac {2 a}{b}+\frac {2 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{3 b^3 d}\\ &=-\frac {2 e^3 (c+d x)^3 \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {4 e^3 (c+d x)^2}{b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {16 e^3 (c+d x)^4}{3 b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {2 e^3 e^{\frac {4 a}{b}} \sqrt {\pi } \text {erf}\left (\frac {2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}+\frac {e^3 e^{\frac {2 a}{b}} \sqrt {2 \pi } \text {erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}+\frac {2 e^3 e^{-\frac {4 a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}-\frac {e^3 e^{-\frac {2 a}{b}} \sqrt {2 \pi } \text {erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}\\ \end {align*}
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Mathematica [A]
time = 1.22, size = 390, normalized size = 1.20 \begin {gather*} \frac {e^3 e^{-4 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )} \left (-8 b e^{4 \sinh ^{-1}(c+d x)} \left (-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {4 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )+4 \sqrt {2} b e^{\frac {2 a}{b}+4 \sinh ^{-1}(c+d x)} \left (-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )+\frac {1}{2} e^{\frac {4 a}{b}} \left (-\left (-1+e^{2 \sinh ^{-1}(c+d x)}\right )^2 \left (b \left (-1+e^{4 \sinh ^{-1}(c+d x)}\right )+8 a \left (1+e^{2 \sinh ^{-1}(c+d x)}+e^{4 \sinh ^{-1}(c+d x)}\right )+8 b \left (1+e^{2 \sinh ^{-1}(c+d x)}+e^{4 \sinh ^{-1}(c+d x)}\right ) \sinh ^{-1}(c+d x)\right )-8 \sqrt {2} e^{\frac {2 a}{b}+4 \sinh ^{-1}(c+d x)} \sqrt {\frac {a}{b}+\sinh ^{-1}(c+d x)} \left (a+b \sinh ^{-1}(c+d x)\right ) \Gamma \left (\frac {1}{2},\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )+16 e^{4 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )} \sqrt {\frac {a}{b}+\sinh ^{-1}(c+d x)} \left (a+b \sinh ^{-1}(c+d x)\right ) \Gamma \left (\frac {1}{2},\frac {4 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )\right )\right )}{12 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\left (d e x +c e \right )^{3}}{\left (a +b \arcsinh \left (d x +c \right )\right )^{\frac {5}{2}}}\, dx\]
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 [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \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*} e^{3} \left (\int \frac {c^{3}}{a^{2} \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} + 2 a b \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}{\left (c + d x \right )} + b^{2} \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}^{2}{\left (c + d x \right )}}\, dx + \int \frac {d^{3} x^{3}}{a^{2} \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} + 2 a b \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}{\left (c + d x \right )} + b^{2} \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}^{2}{\left (c + d x \right )}}\, dx + \int \frac {3 c d^{2} x^{2}}{a^{2} \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} + 2 a b \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}{\left (c + d x \right )} + b^{2} \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}^{2}{\left (c + d x \right )}}\, dx + \int \frac {3 c^{2} d x}{a^{2} \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} + 2 a b \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}{\left (c + d x \right )} + b^{2} \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}^{2}{\left (c + d x \right )}}\, dx\right ) \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 \frac {{\left (c\,e+d\,e\,x\right )}^3}{{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^{5/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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