Optimal. Leaf size=213 \[ -\frac {e^4 \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \cosh ^{-1}(c+d x)}{b}\right )}{8 b d}-\frac {3 e^4 \sinh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 \left (a+b \cosh ^{-1}(c+d x)\right )}{b}\right )}{16 b d}-\frac {e^4 \sinh \left (\frac {5 a}{b}\right ) \text {Chi}\left (\frac {5 \left (a+b \cosh ^{-1}(c+d x)\right )}{b}\right )}{16 b d}+\frac {e^4 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \cosh ^{-1}(c+d x)}{b}\right )}{8 b d}+\frac {3 e^4 \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 \left (a+b \cosh ^{-1}(c+d x)\right )}{b}\right )}{16 b d}+\frac {e^4 \cosh \left (\frac {5 a}{b}\right ) \text {Shi}\left (\frac {5 \left (a+b \cosh ^{-1}(c+d x)\right )}{b}\right )}{16 b d} \]
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Rubi [A] time = 0.43, antiderivative size = 209, normalized size of antiderivative = 0.98, number of steps used = 14, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {5866, 12, 5670, 5448, 3303, 3298, 3301} \[ -\frac {e^4 \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\cosh ^{-1}(c+d x)\right )}{8 b d}-\frac {3 e^4 \sinh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 a}{b}+3 \cosh ^{-1}(c+d x)\right )}{16 b d}-\frac {e^4 \sinh \left (\frac {5 a}{b}\right ) \text {Chi}\left (\frac {5 a}{b}+5 \cosh ^{-1}(c+d x)\right )}{16 b d}+\frac {e^4 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\cosh ^{-1}(c+d x)\right )}{8 b d}+\frac {3 e^4 \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 a}{b}+3 \cosh ^{-1}(c+d x)\right )}{16 b d}+\frac {e^4 \cosh \left (\frac {5 a}{b}\right ) \text {Shi}\left (\frac {5 a}{b}+5 \cosh ^{-1}(c+d x)\right )}{16 b d} \]
Antiderivative was successfully verified.
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Rule 12
Rule 3298
Rule 3301
Rule 3303
Rule 5448
Rule 5670
Rule 5866
Rubi steps
\begin {align*} \int \frac {(c e+d e x)^4}{a+b \cosh ^{-1}(c+d x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {e^4 x^4}{a+b \cosh ^{-1}(x)} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^4 \operatorname {Subst}\left (\int \frac {x^4}{a+b \cosh ^{-1}(x)} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^4 \operatorname {Subst}\left (\int \frac {\cosh ^4(x) \sinh (x)}{a+b x} \, dx,x,\cosh ^{-1}(c+d x)\right )}{d}\\ &=\frac {e^4 \operatorname {Subst}\left (\int \left (\frac {\sinh (x)}{8 (a+b x)}+\frac {3 \sinh (3 x)}{16 (a+b x)}+\frac {\sinh (5 x)}{16 (a+b x)}\right ) \, dx,x,\cosh ^{-1}(c+d x)\right )}{d}\\ &=\frac {e^4 \operatorname {Subst}\left (\int \frac {\sinh (5 x)}{a+b x} \, dx,x,\cosh ^{-1}(c+d x)\right )}{16 d}+\frac {e^4 \operatorname {Subst}\left (\int \frac {\sinh (x)}{a+b x} \, dx,x,\cosh ^{-1}(c+d x)\right )}{8 d}+\frac {\left (3 e^4\right ) \operatorname {Subst}\left (\int \frac {\sinh (3 x)}{a+b x} \, dx,x,\cosh ^{-1}(c+d x)\right )}{16 d}\\ &=\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,\cosh ^{-1}(c+d x)\right )}{8 d}+\frac {\left (3 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,\cosh ^{-1}(c+d x)\right )}{16 d}+\frac {\left (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,\cosh ^{-1}(c+d x)\right )}{16 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,\cosh ^{-1}(c+d x)\right )}{8 d}-\frac {\left (3 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,\cosh ^{-1}(c+d x)\right )}{16 d}-\frac {\left (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,\cosh ^{-1}(c+d x)\right )}{16 d}\\ &=-\frac {e^4 \text {Chi}\left (\frac {a}{b}+\cosh ^{-1}(c+d x)\right ) \sinh \left (\frac {a}{b}\right )}{8 b d}-\frac {3 e^4 \text {Chi}\left (\frac {3 a}{b}+3 \cosh ^{-1}(c+d x)\right ) \sinh \left (\frac {3 a}{b}\right )}{16 b d}-\frac {e^4 \text {Chi}\left (\frac {5 a}{b}+5 \cosh ^{-1}(c+d x)\right ) \sinh \left (\frac {5 a}{b}\right )}{16 b d}+\frac {e^4 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\cosh ^{-1}(c+d x)\right )}{8 b d}+\frac {3 e^4 \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 a}{b}+3 \cosh ^{-1}(c+d x)\right )}{16 b d}+\frac {e^4 \cosh \left (\frac {5 a}{b}\right ) \text {Shi}\left (\frac {5 a}{b}+5 \cosh ^{-1}(c+d x)\right )}{16 b d}\\ \end {align*}
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Mathematica [A] time = 0.25, size = 151, normalized size = 0.71 \[ \frac {e^4 \left (-2 \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\cosh ^{-1}(c+d x)\right )-3 \sinh \left (\frac {3 a}{b}\right ) \text {Chi}\left (3 \left (\frac {a}{b}+\cosh ^{-1}(c+d x)\right )\right )-\sinh \left (\frac {5 a}{b}\right ) \text {Chi}\left (5 \left (\frac {a}{b}+\cosh ^{-1}(c+d x)\right )\right )+2 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\cosh ^{-1}(c+d x)\right )+3 \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (3 \left (\frac {a}{b}+\cosh ^{-1}(c+d x)\right )\right )+\cosh \left (\frac {5 a}{b}\right ) \text {Shi}\left (5 \left (\frac {a}{b}+\cosh ^{-1}(c+d x)\right )\right )\right )}{16 b d} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {d^{4} e^{4} x^{4} + 4 \, c d^{3} e^{4} x^{3} + 6 \, c^{2} d^{2} e^{4} x^{2} + 4 \, c^{3} d e^{4} x + c^{4} e^{4}}{b \operatorname {arcosh}\left (d x + c\right ) + a}, x\right ) \]
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 \[ \int \frac {{\left (d e x + c e\right )}^{4}}{b \operatorname {arcosh}\left (d x + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.40, size = 194, normalized size = 0.91 \[ \frac {\frac {e^{4} {\mathrm e}^{\frac {5 a}{b}} \Ei \left (1, 5 \,\mathrm {arccosh}\left (d x +c \right )+\frac {5 a}{b}\right )}{32 b}+\frac {3 e^{4} {\mathrm e}^{\frac {3 a}{b}} \Ei \left (1, 3 \,\mathrm {arccosh}\left (d x +c \right )+\frac {3 a}{b}\right )}{32 b}+\frac {e^{4} {\mathrm e}^{\frac {a}{b}} \Ei \left (1, \mathrm {arccosh}\left (d x +c \right )+\frac {a}{b}\right )}{16 b}-\frac {e^{4} {\mathrm e}^{-\frac {a}{b}} \Ei \left (1, -\mathrm {arccosh}\left (d x +c \right )-\frac {a}{b}\right )}{16 b}-\frac {3 e^{4} {\mathrm e}^{-\frac {3 a}{b}} \Ei \left (1, -3 \,\mathrm {arccosh}\left (d x +c \right )-\frac {3 a}{b}\right )}{32 b}-\frac {e^{4} {\mathrm e}^{-\frac {5 a}{b}} \Ei \left (1, -5 \,\mathrm {arccosh}\left (d x +c \right )-\frac {5 a}{b}\right )}{32 b}}{d} \]
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 \[ \int \frac {{\left (d e x + c e\right )}^{4}}{b \operatorname {arcosh}\left (d x + c\right ) + a}\,{d x} \]
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 \[ \int \frac {{\left (c\,e+d\,e\,x\right )}^4}{a+b\,\mathrm {acosh}\left (c+d\,x\right )} \,d x \]
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 \[ e^{4} \left (\int \frac {c^{4}}{a + b \operatorname {acosh}{\left (c + d x \right )}}\, dx + \int \frac {d^{4} x^{4}}{a + b \operatorname {acosh}{\left (c + d x \right )}}\, dx + \int \frac {4 c d^{3} x^{3}}{a + b \operatorname {acosh}{\left (c + d x \right )}}\, dx + \int \frac {6 c^{2} d^{2} x^{2}}{a + b \operatorname {acosh}{\left (c + d x \right )}}\, dx + \int \frac {4 c^{3} d x}{a + b \operatorname {acosh}{\left (c + d x \right )}}\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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