Optimal. Leaf size=188 \[ -\frac {e^3 \cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{2 b^2 d}+\frac {e^3 \cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{2 b^2 d}+\frac {e^3 \sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{2 b^2 d}-\frac {e^3 \sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{2 b^2 d}-\frac {e^3 (c+d x)^3 \sqrt {(c+d x)^2+1}}{b d \left (a+b \sinh ^{-1}(c+d x)\right )} \]
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Rubi [A] time = 0.31, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {5865, 12, 5665, 3303, 3298, 3301} \[ -\frac {e^3 \cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 a}{b}+2 \sinh ^{-1}(c+d x)\right )}{2 b^2 d}+\frac {e^3 \cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 a}{b}+4 \sinh ^{-1}(c+d x)\right )}{2 b^2 d}+\frac {e^3 \sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 a}{b}+2 \sinh ^{-1}(c+d x)\right )}{2 b^2 d}-\frac {e^3 \sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 a}{b}+4 \sinh ^{-1}(c+d x)\right )}{2 b^2 d}-\frac {e^3 (c+d x)^3 \sqrt {(c+d x)^2+1}}{b d \left (a+b \sinh ^{-1}(c+d x)\right )} \]
Antiderivative was successfully verified.
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Rule 12
Rule 3298
Rule 3301
Rule 3303
Rule 5665
Rule 5865
Rubi steps
\begin {align*} \int \frac {(c e+d e x)^3}{\left (a+b \sinh ^{-1}(c+d x)\right )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {e^3 x^3}{\left (a+b \sinh ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^3 \operatorname {Subst}\left (\int \frac {x^3}{\left (a+b \sinh ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{d}\\ &=-\frac {e^3 (c+d x)^3 \sqrt {1+(c+d x)^2}}{b d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac {e^3 \operatorname {Subst}\left (\int \left (-\frac {\cosh (2 x)}{2 (a+b x)}+\frac {\cosh (4 x)}{2 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{b d}\\ &=-\frac {e^3 (c+d x)^3 \sqrt {1+(c+d x)^2}}{b d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {e^3 \operatorname {Subst}\left (\int \frac {\cosh (2 x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b d}+\frac {e^3 \operatorname {Subst}\left (\int \frac {\cosh (4 x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b d}\\ &=-\frac {e^3 (c+d x)^3 \sqrt {1+(c+d x)^2}}{b d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {\left (e^3 \cosh \left (\frac {2 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b d}+\frac {\left (e^3 \cosh \left (\frac {4 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b d}+\frac {\left (e^3 \sinh \left (\frac {2 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b d}-\frac {\left (e^3 \sinh \left (\frac {4 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b d}\\ &=-\frac {e^3 (c+d x)^3 \sqrt {1+(c+d x)^2}}{b d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {e^3 \cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 a}{b}+2 \sinh ^{-1}(c+d x)\right )}{2 b^2 d}+\frac {e^3 \cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 a}{b}+4 \sinh ^{-1}(c+d x)\right )}{2 b^2 d}+\frac {e^3 \sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 a}{b}+2 \sinh ^{-1}(c+d x)\right )}{2 b^2 d}-\frac {e^3 \sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 a}{b}+4 \sinh ^{-1}(c+d x)\right )}{2 b^2 d}\\ \end {align*}
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Mathematica [A] time = 0.88, size = 193, normalized size = 1.03 \[ -\frac {e^3 \left (\cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (2 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right )-\cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (4 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right )-4 \sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (2 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right )+\sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (4 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right )+3 \left (\sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (2 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right )+\log \left (a+b \sinh ^{-1}(c+d x)\right )\right )+\frac {2 b \sqrt {(c+d x)^2+1} (c+d x)^3}{a+b \sinh ^{-1}(c+d x)}-3 \log \left (a+b \sinh ^{-1}(c+d x)\right )\right )}{2 b^2 d} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {d^{3} e^{3} x^{3} + 3 \, c d^{2} e^{3} x^{2} + 3 \, c^{2} d e^{3} x + c^{3} e^{3}}{b^{2} \operatorname {arsinh}\left (d x + c\right )^{2} + 2 \, a b \operatorname {arsinh}\left (d x + c\right ) + a^{2}}, x\right ) \]
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 \[ \int \frac {{\left (d e x + c e\right )}^{3}}{{\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.30, size = 388, normalized size = 2.06 \[ \frac {\frac {\left (8 \left (d x +c \right )^{4}-8 \left (d x +c \right )^{3} \sqrt {1+\left (d x +c \right )^{2}}+8 \left (d x +c \right )^{2}-4 \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}+1\right ) e^{3}}{16 \left (a +b \arcsinh \left (d x +c \right )\right ) b}-\frac {e^{3} {\mathrm e}^{\frac {4 a}{b}} \Ei \left (1, 4 \arcsinh \left (d x +c \right )+\frac {4 a}{b}\right )}{4 b^{2}}-\frac {\left (2 \left (d x +c \right )^{2}-2 \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}+1\right ) e^{3}}{8 \left (a +b \arcsinh \left (d x +c \right )\right ) b}+\frac {e^{3} {\mathrm e}^{\frac {2 a}{b}} \Ei \left (1, 2 \arcsinh \left (d x +c \right )+\frac {2 a}{b}\right )}{4 b^{2}}+\frac {e^{3} \left (2 \left (d x +c \right )^{2}+1+2 \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}\right )}{8 b \left (a +b \arcsinh \left (d x +c \right )\right )}+\frac {e^{3} {\mathrm e}^{-\frac {2 a}{b}} \Ei \left (1, -2 \arcsinh \left (d x +c \right )-\frac {2 a}{b}\right )}{4 b^{2}}-\frac {e^{3} \left (8 \left (d x +c \right )^{4}+8 \left (d x +c \right )^{2}+8 \left (d x +c \right )^{3} \sqrt {1+\left (d x +c \right )^{2}}+4 \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}+1\right )}{16 b \left (a +b \arcsinh \left (d x +c \right )\right )}-\frac {e^{3} {\mathrm e}^{-\frac {4 a}{b}} \Ei \left (1, -4 \arcsinh \left (d x +c \right )-\frac {4 a}{b}\right )}{4 b^{2}}}{d} \]
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 \[ \text {result too large to display} \]
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 \[ \int \frac {{\left (c\,e+d\,e\,x\right )}^3}{{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^2} \,d x \]
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 \[ e^{3} \left (\int \frac {c^{3}}{a^{2} + 2 a b \operatorname {asinh}{\left (c + d x \right )} + b^{2} \operatorname {asinh}^{2}{\left (c + d x \right )}}\, dx + \int \frac {d^{3} x^{3}}{a^{2} + 2 a b \operatorname {asinh}{\left (c + d x \right )} + b^{2} \operatorname {asinh}^{2}{\left (c + d x \right )}}\, dx + \int \frac {3 c d^{2} x^{2}}{a^{2} + 2 a b \operatorname {asinh}{\left (c + d x \right )} + b^{2} \operatorname {asinh}^{2}{\left (c + d x \right )}}\, dx + \int \frac {3 c^{2} d x}{a^{2} + 2 a b \operatorname {asinh}{\left (c + d x \right )} + b^{2} \operatorname {asinh}^{2}{\left (c + d x \right )}}\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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