Optimal. Leaf size=204 \[ \frac {2 e \cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{3 b^4 d}-\frac {2 e \sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{3 b^4 d}-\frac {2 e \sqrt {(c+d x)^2+1} (c+d x)}{3 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {e (c+d x)^2}{3 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {e}{6 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {e \sqrt {(c+d x)^2+1} (c+d x)}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3} \]
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Rubi [A] time = 0.34, antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5865, 12, 5667, 5774, 5665, 3303, 3298, 3301, 5675} \[ \frac {2 e \cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 a}{b}+2 \sinh ^{-1}(c+d x)\right )}{3 b^4 d}-\frac {2 e \sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 a}{b}+2 \sinh ^{-1}(c+d x)\right )}{3 b^4 d}-\frac {e (c+d x)^2}{3 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {2 e \sqrt {(c+d x)^2+1} (c+d x)}{3 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {e}{6 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {e \sqrt {(c+d x)^2+1} (c+d x)}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3} \]
Antiderivative was successfully verified.
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Rule 12
Rule 3298
Rule 3301
Rule 3303
Rule 5665
Rule 5667
Rule 5675
Rule 5774
Rule 5865
Rubi steps
\begin {align*} \int \frac {c e+d e x}{\left (a+b \sinh ^{-1}(c+d x)\right )^4} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {e x}{\left (a+b \sinh ^{-1}(x)\right )^4} \, dx,x,c+d x\right )}{d}\\ &=\frac {e \operatorname {Subst}\left (\int \frac {x}{\left (a+b \sinh ^{-1}(x)\right )^4} \, dx,x,c+d x\right )}{d}\\ &=-\frac {e (c+d x) \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}+\frac {e \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+x^2} \left (a+b \sinh ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{3 b d}+\frac {(2 e) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+x^2} \left (a+b \sinh ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{3 b d}\\ &=-\frac {e (c+d x) \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}-\frac {e}{6 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {e (c+d x)^2}{3 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}+\frac {(2 e) \operatorname {Subst}\left (\int \frac {x}{\left (a+b \sinh ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{3 b^2 d}\\ &=-\frac {e (c+d x) \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}-\frac {e}{6 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {e (c+d x)^2}{3 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {2 e (c+d x) \sqrt {1+(c+d x)^2}}{3 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac {(2 e) \operatorname {Subst}\left (\int \frac {\cosh (2 x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 b^3 d}\\ &=-\frac {e (c+d x) \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}-\frac {e}{6 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {e (c+d x)^2}{3 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {2 e (c+d x) \sqrt {1+(c+d x)^2}}{3 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac {\left (2 e \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 )}{3 b^3 d}-\frac {\left (2 e \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 )}{3 b^3 d}\\ &=-\frac {e (c+d x) \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}-\frac {e}{6 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {e (c+d x)^2}{3 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {2 e (c+d x) \sqrt {1+(c+d x)^2}}{3 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac {2 e \cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 a}{b}+2 \sinh ^{-1}(c+d x)\right )}{3 b^4 d}-\frac {2 e \sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 a}{b}+2 \sinh ^{-1}(c+d x)\right )}{3 b^4 d}\\ \end {align*}
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Mathematica [A] time = 0.81, size = 181, normalized size = 0.89 \[ \frac {e \left (-\frac {2 b^3 (c+d x) \sqrt {(c+d x)^2+1}}{\left (a+b \sinh ^{-1}(c+d x)\right )^3}+\frac {b^2 \left (-2 (c+d x)^2-1\right )}{\left (a+b \sinh ^{-1}(c+d x)\right )^2}+4 \cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (2 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right )-4 \left (\sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (2 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right )+\log \left (a+b \sinh ^{-1}(c+d x)\right )\right )-\frac {4 b (c+d x) \sqrt {(c+d x)^2+1}}{a+b \sinh ^{-1}(c+d x)}+4 \log \left (a+b \sinh ^{-1}(c+d x)\right )\right )}{6 b^4 d} \]
Antiderivative was successfully verified.
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fricas [F] time = 1.29, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {d e x + c e}{b^{4} \operatorname {arsinh}\left (d x + c\right )^{4} + 4 \, a b^{3} \operatorname {arsinh}\left (d x + c\right )^{3} + 6 \, a^{2} b^{2} \operatorname {arsinh}\left (d x + c\right )^{2} + 4 \, a^{3} b \operatorname {arsinh}\left (d x + c\right ) + a^{4}}, x\right ) \]
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 \[ \int \frac {d e x + c e}{{\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{4}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 333, normalized size = 1.63 \[ \frac {\frac {\left (2 \left (d x +c \right )^{2}-2 \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}+1\right ) e \left (2 b^{2} \arcsinh \left (d x +c \right )^{2}+4 a b \arcsinh \left (d x +c \right )-\arcsinh \left (d x +c \right ) b^{2}+2 a^{2}-a b +b^{2}\right )}{12 b^{3} \left (b^{3} \arcsinh \left (d x +c \right )^{3}+3 a \,b^{2} \arcsinh \left (d x +c \right )^{2}+3 a^{2} b \arcsinh \left (d x +c \right )+a^{3}\right )}-\frac {e \,{\mathrm e}^{\frac {2 a}{b}} \Ei \left (1, 2 \arcsinh \left (d x +c \right )+\frac {2 a}{b}\right )}{3 b^{4}}-\frac {e \left (2 \left (d x +c \right )^{2}+1+2 \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}\right )}{12 b \left (a +b \arcsinh \left (d x +c \right )\right )^{3}}-\frac {e \left (2 \left (d x +c \right )^{2}+1+2 \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}\right )}{12 b^{2} \left (a +b \arcsinh \left (d x +c \right )\right )^{2}}-\frac {e \left (2 \left (d x +c \right )^{2}+1+2 \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}\right )}{6 b^{3} \left (a +b \arcsinh \left (d x +c \right )\right )}-\frac {e \,{\mathrm e}^{-\frac {2 a}{b}} \Ei \left (1, -2 \arcsinh \left (d x +c \right )-\frac {2 a}{b}\right )}{3 b^{4}}}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
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 \[ \int \frac {c\,e+d\,e\,x}{{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^4} \,d x \]
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 \[ e \left (\int \frac {c}{a^{4} + 4 a^{3} b \operatorname {asinh}{\left (c + d x \right )} + 6 a^{2} b^{2} \operatorname {asinh}^{2}{\left (c + d x \right )} + 4 a b^{3} \operatorname {asinh}^{3}{\left (c + d x \right )} + b^{4} \operatorname {asinh}^{4}{\left (c + d x \right )}}\, dx + \int \frac {d x}{a^{4} + 4 a^{3} b \operatorname {asinh}{\left (c + d x \right )} + 6 a^{2} b^{2} \operatorname {asinh}^{2}{\left (c + d x \right )} + 4 a b^{3} \operatorname {asinh}^{3}{\left (c + d x \right )} + b^{4} \operatorname {asinh}^{4}{\left (c + d x \right )}}\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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