Optimal. Leaf size=256 \[ -\frac {e^4 \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{8 b^2 d}+\frac {9 e^4 \sinh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{16 b^2 d}-\frac {5 e^4 \sinh \left (\frac {5 a}{b}\right ) \text {Chi}\left (\frac {5 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{16 b^2 d}+\frac {e^4 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{8 b^2 d}-\frac {9 e^4 \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{16 b^2 d}+\frac {5 e^4 \cosh \left (\frac {5 a}{b}\right ) \text {Shi}\left (\frac {5 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{16 b^2 d}-\frac {e^4 (c+d x)^4 \sqrt {(c+d x)^2+1}}{b d \left (a+b \sinh ^{-1}(c+d x)\right )} \]
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Rubi [A] time = 0.41, antiderivative size = 252, normalized size of antiderivative = 0.98, number of steps used = 13, 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} \[ -\frac {e^4 \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )}{8 b^2 d}+\frac {9 e^4 \sinh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c+d x)\right )}{16 b^2 d}-\frac {5 e^4 \sinh \left (\frac {5 a}{b}\right ) \text {Chi}\left (\frac {5 a}{b}+5 \sinh ^{-1}(c+d x)\right )}{16 b^2 d}+\frac {e^4 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )}{8 b^2 d}-\frac {9 e^4 \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c+d x)\right )}{16 b^2 d}+\frac {5 e^4 \cosh \left (\frac {5 a}{b}\right ) \text {Shi}\left (\frac {5 a}{b}+5 \sinh ^{-1}(c+d x)\right )}{16 b^2 d}-\frac {e^4 (c+d x)^4 \sqrt {(c+d x)^2+1}}{b d \left (a+b \sinh ^{-1}(c+d x)\right )} \]
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)^4}{\left (a+b \sinh ^{-1}(c+d x)\right )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {e^4 x^4}{\left (a+b \sinh ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^4 \operatorname {Subst}\left (\int \frac {x^4}{\left (a+b \sinh ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{d}\\ &=-\frac {e^4 (c+d x)^4 \sqrt {1+(c+d x)^2}}{b d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac {e^4 \operatorname {Subst}\left (\int \left (\frac {\sinh (x)}{8 (a+b x)}-\frac {9 \sinh (3 x)}{16 (a+b x)}+\frac {5 \sinh (5 x)}{16 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{b d}\\ &=-\frac {e^4 (c+d x)^4 \sqrt {1+(c+d x)^2}}{b d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac {e^4 \operatorname {Subst}\left (\int \frac {\sinh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 b d}+\frac {\left (5 e^4\right ) \operatorname {Subst}\left (\int \frac {\sinh (5 x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{16 b d}-\frac {\left (9 e^4\right ) \operatorname {Subst}\left (\int \frac {\sinh (3 x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{16 b d}\\ &=-\frac {e^4 (c+d x)^4 \sqrt {1+(c+d x)^2}}{b d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac {\left (e^4 \cosh \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 b d}-\frac {\left (9 e^4 \cosh \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{16 b d}+\frac {\left (5 e^4 \cosh \left (\frac {5 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{16 b d}-\frac {\left (e^4 \sinh \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 b d}+\frac {\left (9 e^4 \sinh \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{16 b d}-\frac {\left (5 e^4 \sinh \left (\frac {5 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{16 b d}\\ &=-\frac {e^4 (c+d x)^4 \sqrt {1+(c+d x)^2}}{b d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {e^4 \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right ) \sinh \left (\frac {a}{b}\right )}{8 b^2 d}+\frac {9 e^4 \text {Chi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c+d x)\right ) \sinh \left (\frac {3 a}{b}\right )}{16 b^2 d}-\frac {5 e^4 \text {Chi}\left (\frac {5 a}{b}+5 \sinh ^{-1}(c+d x)\right ) \sinh \left (\frac {5 a}{b}\right )}{16 b^2 d}+\frac {e^4 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )}{8 b^2 d}-\frac {9 e^4 \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c+d x)\right )}{16 b^2 d}+\frac {5 e^4 \cosh \left (\frac {5 a}{b}\right ) \text {Shi}\left (\frac {5 a}{b}+5 \sinh ^{-1}(c+d x)\right )}{16 b^2 d}\\ \end {align*}
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Mathematica [A] time = 1.10, size = 281, normalized size = 1.10 \[ \frac {e^4 \left (16 \left (3 \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )-\sinh \left (\frac {3 a}{b}\right ) \text {Chi}\left (3 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right )-3 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )+\cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (3 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right )\right )-5 \left (10 \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )-5 \sinh \left (\frac {3 a}{b}\right ) \text {Chi}\left (3 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right )+\sinh \left (\frac {5 a}{b}\right ) \text {Chi}\left (5 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right )-10 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )+5 \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (3 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right )-\cosh \left (\frac {5 a}{b}\right ) \text {Shi}\left (5 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right )\right )-\frac {16 b \sqrt {(c+d x)^2+1} (c+d x)^4}{a+b \sinh ^{-1}(c+d x)}\right )}{16 b^2 d} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {d^{4} e^{4} x^{4} + 4 \, c d^{3} e^{4} x^{3} + 6 \, c^{2} d^{2} e^{4} x^{2} + 4 \, c^{3} d e^{4} x + c^{4} e^{4}}{b^{2} \operatorname {arsinh}\left (d x + c\right )^{2} + 2 \, a b \operatorname {arsinh}\left (d x + c\right ) + a^{2}}, x\right ) \]
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 \[ \int \frac {{\left (d e x + c e\right )}^{4}}{{\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.38, size = 602, normalized size = 2.35 \[ \frac {\frac {\left (16 \left (d x +c \right )^{5}-16 \left (d x +c \right )^{4} \sqrt {1+\left (d x +c \right )^{2}}+20 \left (d x +c \right )^{3}-12 \left (d x +c \right )^{2} \sqrt {1+\left (d x +c \right )^{2}}+5 d x +5 c -\sqrt {1+\left (d x +c \right )^{2}}\right ) e^{4}}{32 b \left (a +b \arcsinh \left (d x +c \right )\right )}+\frac {5 e^{4} {\mathrm e}^{\frac {5 a}{b}} \Ei \left (1, 5 \arcsinh \left (d x +c \right )+\frac {5 a}{b}\right )}{32 b^{2}}-\frac {3 \left (-4 \left (d x +c \right )^{2} \sqrt {1+\left (d x +c \right )^{2}}+4 \left (d x +c \right )^{3}-\sqrt {1+\left (d x +c \right )^{2}}+3 d x +3 c \right ) e^{4}}{32 b \left (a +b \arcsinh \left (d x +c \right )\right )}-\frac {9 e^{4} {\mathrm e}^{\frac {3 a}{b}} \Ei \left (1, 3 \arcsinh \left (d x +c \right )+\frac {3 a}{b}\right )}{32 b^{2}}+\frac {\left (-\sqrt {1+\left (d x +c \right )^{2}}+d x +c \right ) e^{4}}{16 b \left (a +b \arcsinh \left (d x +c \right )\right )}+\frac {e^{4} {\mathrm e}^{\frac {a}{b}} \Ei \left (1, \arcsinh \left (d x +c \right )+\frac {a}{b}\right )}{16 b^{2}}-\frac {e^{4} \left (d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{16 b \left (a +b \arcsinh \left (d x +c \right )\right )}-\frac {e^{4} {\mathrm e}^{-\frac {a}{b}} \Ei \left (1, -\arcsinh \left (d x +c \right )-\frac {a}{b}\right )}{16 b^{2}}+\frac {3 e^{4} \left (4 \left (d x +c \right )^{3}+3 d x +3 c +4 \left (d x +c \right )^{2} \sqrt {1+\left (d x +c \right )^{2}}+\sqrt {1+\left (d x +c \right )^{2}}\right )}{32 b \left (a +b \arcsinh \left (d x +c \right )\right )}+\frac {9 e^{4} {\mathrm e}^{-\frac {3 a}{b}} \Ei \left (1, -3 \arcsinh \left (d x +c \right )-\frac {3 a}{b}\right )}{32 b^{2}}-\frac {e^{4} \left (16 \left (d x +c \right )^{5}+20 \left (d x +c \right )^{3}+16 \left (d x +c \right )^{4} \sqrt {1+\left (d x +c \right )^{2}}+5 d x +5 c +12 \left (d x +c \right )^{2} \sqrt {1+\left (d x +c \right )^{2}}+\sqrt {1+\left (d x +c \right )^{2}}\right )}{32 b \left (a +b \arcsinh \left (d x +c \right )\right )}-\frac {5 e^{4} {\mathrm e}^{-\frac {5 a}{b}} \Ei \left (1, -5 \arcsinh \left (d x +c \right )-\frac {5 a}{b}\right )}{32 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 = -0.00 \[ \int \frac {{\left (c\,e+d\,e\,x\right )}^4}{{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^2} \,d x \]
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 \[ e^{4} \left (\int \frac {c^{4}}{a^{2} + 2 a b \operatorname {asinh}{\left (c + d x \right )} + b^{2} \operatorname {asinh}^{2}{\left (c + d x \right )}}\, dx + \int \frac {d^{4} x^{4}}{a^{2} + 2 a b \operatorname {asinh}{\left (c + d x \right )} + b^{2} \operatorname {asinh}^{2}{\left (c + d x \right )}}\, dx + \int \frac {4 c d^{3} x^{3}}{a^{2} + 2 a b \operatorname {asinh}{\left (c + d x \right )} + b^{2} \operatorname {asinh}^{2}{\left (c + d x \right )}}\, dx + \int \frac {6 c^{2} d^{2} x^{2}}{a^{2} + 2 a b \operatorname {asinh}{\left (c + d x \right )} + b^{2} \operatorname {asinh}^{2}{\left (c + d x \right )}}\, dx + \int \frac {4 c^{3} d x}{a^{2} + 2 a b \operatorname {asinh}{\left (c + d x \right )} + b^{2} \operatorname {asinh}^{2}{\left (c + d x \right )}}\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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