Optimal. Leaf size=367 \[ -\frac {2 e^4 (c+d x)^4 \sqrt {1+(c+d x)^2}}{b d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {e^4 e^{a/b} \sqrt {\pi } \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{8 b^{3/2} d}+\frac {3 e^4 e^{\frac {3 a}{b}} \sqrt {3 \pi } \text {Erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 b^{3/2} d}-\frac {e^4 e^{\frac {5 a}{b}} \sqrt {5 \pi } \text {Erf}\left (\frac {\sqrt {5} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 b^{3/2} d}+\frac {e^4 e^{-\frac {a}{b}} \sqrt {\pi } \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{8 b^{3/2} d}-\frac {3 e^4 e^{-\frac {3 a}{b}} \sqrt {3 \pi } \text {Erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 b^{3/2} d}+\frac {e^4 e^{-\frac {5 a}{b}} \sqrt {5 \pi } \text {Erfi}\left (\frac {\sqrt {5} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 b^{3/2} d} \]
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Rubi [A]
time = 0.44, antiderivative size = 367, normalized size of antiderivative = 1.00, number of steps
used = 19, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {5859, 12,
5778, 3389, 2211, 2236, 2235} \begin {gather*} -\frac {\sqrt {\pi } e^4 e^{a/b} \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{8 b^{3/2} d}+\frac {3 \sqrt {3 \pi } e^4 e^{\frac {3 a}{b}} \text {Erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 b^{3/2} d}-\frac {\sqrt {5 \pi } e^4 e^{\frac {5 a}{b}} \text {Erf}\left (\frac {\sqrt {5} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 b^{3/2} d}+\frac {\sqrt {\pi } e^4 e^{-\frac {a}{b}} \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{8 b^{3/2} d}-\frac {3 \sqrt {3 \pi } e^4 e^{-\frac {3 a}{b}} \text {Erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 b^{3/2} d}+\frac {\sqrt {5 \pi } e^4 e^{-\frac {5 a}{b}} \text {Erfi}\left (\frac {\sqrt {5} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 b^{3/2} d}-\frac {2 e^4 (c+d x)^4 \sqrt {(c+d x)^2+1}}{b d \sqrt {a+b \sinh ^{-1}(c+d x)}} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2211
Rule 2235
Rule 2236
Rule 3389
Rule 5778
Rule 5859
Rubi steps
\begin {align*} \int \frac {(c e+d e x)^4}{\left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}} \, dx &=\frac {\text {Subst}\left (\int \frac {e^4 x^4}{\left (a+b \sinh ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^4 \text {Subst}\left (\int \frac {x^4}{\left (a+b \sinh ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac {2 e^4 (c+d x)^4 \sqrt {1+(c+d x)^2}}{b d \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {\left (2 e^4\right ) \text {Subst}\left (\int \left (\frac {\sinh (x)}{8 \sqrt {a+b x}}-\frac {9 \sinh (3 x)}{16 \sqrt {a+b x}}+\frac {5 \sinh (5 x)}{16 \sqrt {a+b x}}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{b d}\\ &=-\frac {2 e^4 (c+d x)^4 \sqrt {1+(c+d x)^2}}{b d \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {e^4 \text {Subst}\left (\int \frac {\sinh (x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{4 b d}+\frac {\left (5 e^4\right ) \text {Subst}\left (\int \frac {\sinh (5 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 b d}-\frac {\left (9 e^4\right ) \text {Subst}\left (\int \frac {\sinh (3 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 b d}\\ &=-\frac {2 e^4 (c+d x)^4 \sqrt {1+(c+d x)^2}}{b d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {e^4 \text {Subst}\left (\int \frac {e^{-x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 b d}+\frac {e^4 \text {Subst}\left (\int \frac {e^x}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 b d}-\frac {\left (5 e^4\right ) \text {Subst}\left (\int \frac {e^{-5 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{16 b d}+\frac {\left (5 e^4\right ) \text {Subst}\left (\int \frac {e^{5 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{16 b d}+\frac {\left (9 e^4\right ) \text {Subst}\left (\int \frac {e^{-3 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{16 b d}-\frac {\left (9 e^4\right ) \text {Subst}\left (\int \frac {e^{3 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{16 b d}\\ &=-\frac {2 e^4 (c+d x)^4 \sqrt {1+(c+d x)^2}}{b d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {e^4 \text {Subst}\left (\int e^{\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{4 b^2 d}+\frac {e^4 \text {Subst}\left (\int e^{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{4 b^2 d}-\frac {\left (5 e^4\right ) \text {Subst}\left (\int e^{\frac {5 a}{b}-\frac {5 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{8 b^2 d}+\frac {\left (5 e^4\right ) \text {Subst}\left (\int e^{-\frac {5 a}{b}+\frac {5 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{8 b^2 d}+\frac {\left (9 e^4\right ) \text {Subst}\left (\int e^{\frac {3 a}{b}-\frac {3 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{8 b^2 d}-\frac {\left (9 e^4\right ) \text {Subst}\left (\int e^{-\frac {3 a}{b}+\frac {3 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{8 b^2 d}\\ &=-\frac {2 e^4 (c+d x)^4 \sqrt {1+(c+d x)^2}}{b d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {e^4 e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{8 b^{3/2} d}+\frac {3 e^4 e^{\frac {3 a}{b}} \sqrt {3 \pi } \text {erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 b^{3/2} d}-\frac {e^4 e^{\frac {5 a}{b}} \sqrt {5 \pi } \text {erf}\left (\frac {\sqrt {5} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 b^{3/2} d}+\frac {e^4 e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{8 b^{3/2} d}-\frac {3 e^4 e^{-\frac {3 a}{b}} \sqrt {3 \pi } \text {erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 b^{3/2} d}+\frac {e^4 e^{-\frac {5 a}{b}} \sqrt {5 \pi } \text {erfi}\left (\frac {\sqrt {5} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 b^{3/2} d}\\ \end {align*}
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Mathematica [A]
time = 0.45, size = 490, normalized size = 1.34 \begin {gather*} \frac {e^4 e^{-5 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )} \left (-e^{\frac {5 a}{b}}+3 e^{\frac {5 a}{b}+2 \sinh ^{-1}(c+d x)}-2 e^{\frac {5 a}{b}+4 \sinh ^{-1}(c+d x)}-2 e^{\frac {5 a}{b}+6 \sinh ^{-1}(c+d x)}+3 e^{\frac {5 a}{b}+8 \sinh ^{-1}(c+d x)}-e^{\frac {5 a}{b}+10 \sinh ^{-1}(c+d x)}+2 e^{\frac {6 a}{b}+5 \sinh ^{-1}(c+d x)} \sqrt {\frac {a}{b}+\sinh ^{-1}(c+d x)} \Gamma \left (\frac {1}{2},\frac {a}{b}+\sinh ^{-1}(c+d x)\right )+\sqrt {5} e^{5 \sinh ^{-1}(c+d x)} \sqrt {-\frac {a+b \sinh ^{-1}(c+d x)}{b}} \Gamma \left (\frac {1}{2},-\frac {5 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )-3 \sqrt {3} e^{\frac {2 a}{b}+5 \sinh ^{-1}(c+d x)} \sqrt {-\frac {a+b \sinh ^{-1}(c+d x)}{b}} \Gamma \left (\frac {1}{2},-\frac {3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )+2 e^{\frac {4 a}{b}+5 \sinh ^{-1}(c+d x)} \sqrt {-\frac {a+b \sinh ^{-1}(c+d x)}{b}} \Gamma \left (\frac {1}{2},-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )-3 \sqrt {3} e^{\frac {8 a}{b}+5 \sinh ^{-1}(c+d x)} \sqrt {\frac {a}{b}+\sinh ^{-1}(c+d x)} \Gamma \left (\frac {1}{2},\frac {3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )+\sqrt {5} e^{5 \left (\frac {2 a}{b}+\sinh ^{-1}(c+d x)\right )} \sqrt {\frac {a}{b}+\sinh ^{-1}(c+d x)} \Gamma \left (\frac {1}{2},\frac {5 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )\right )}{16 b d \sqrt {a+b \sinh ^{-1}(c+d x)}} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\left (d e x +c e \right )^{4}}{\left (a +b \arcsinh \left (d x +c \right )\right )^{\frac {3}{2}}}\, dx\]
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^{4} \left (\int \frac {c^{4}}{a \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} + b \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}{\left (c + d x \right )}}\, dx + \int \frac {d^{4} x^{4}}{a \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} + b \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}{\left (c + d x \right )}}\, dx + \int \frac {4 c d^{3} x^{3}}{a \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} + b \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}{\left (c + d x \right )}}\, dx + \int \frac {6 c^{2} d^{2} x^{2}}{a \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} + b \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}{\left (c + d x \right )}}\, dx + \int \frac {4 c^{3} d x}{a \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} + b \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}{\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 )}^4}{{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^{3/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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