Optimal. Leaf size=261 \[ \frac {b^2 \text {Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^4}-\frac {b^2 \text {Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^4}-\frac {b^2 \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac {b \sqrt {(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^4}-\frac {b^3 \text {Li}_3\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac {b^3 \text {Li}_3\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac {b^3 \tanh ^{-1}\left (\sqrt {(c+d x)^2+1}\right )}{d e^4} \]
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Rubi [A] time = 0.41, antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {5865, 12, 5661, 5747, 5760, 4182, 2531, 2282, 6589, 266, 63, 207} \[ \frac {b^2 \text {PolyLog}\left (2,-e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^4}-\frac {b^2 \text {PolyLog}\left (2,e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^4}-\frac {b^3 \text {PolyLog}\left (3,-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac {b^3 \text {PolyLog}\left (3,e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac {b^2 \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac {b \sqrt {(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^4}-\frac {b^3 \tanh ^{-1}\left (\sqrt {(c+d x)^2+1}\right )}{d e^4} \]
Antiderivative was successfully verified.
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Rule 12
Rule 63
Rule 207
Rule 266
Rule 2282
Rule 2531
Rule 4182
Rule 5661
Rule 5747
Rule 5760
Rule 5865
Rule 6589
Rubi steps
\begin {align*} \int \frac {\left (a+b \sinh ^{-1}(c+d x)\right )^3}{(c e+d e x)^4} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sinh ^{-1}(x)\right )^3}{e^4 x^4} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sinh ^{-1}(x)\right )^3}{x^4} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \operatorname {Subst}\left (\int \frac {\left (a+b \sinh ^{-1}(x)\right )^2}{x^3 \sqrt {1+x^2}} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}-\frac {b \operatorname {Subst}\left (\int \frac {\left (a+b \sinh ^{-1}(x)\right )^2}{x \sqrt {1+x^2}} \, dx,x,c+d x\right )}{2 d e^4}+\frac {b^2 \operatorname {Subst}\left (\int \frac {a+b \sinh ^{-1}(x)}{x^2} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {b^2 \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac {b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}-\frac {b \operatorname {Subst}\left (\int (a+b x)^2 \text {csch}(x) \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 d e^4}+\frac {b^3 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1+x^2}} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {b^2 \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac {b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \left (a+b \sinh ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac {b^2 \operatorname {Subst}\left (\int (a+b x) \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^4}-\frac {b^2 \operatorname {Subst}\left (\int (a+b x) \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^4}+\frac {b^3 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1+x}} \, dx,x,(c+d x)^2\right )}{2 d e^4}\\ &=-\frac {b^2 \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac {b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \left (a+b \sinh ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac {b^2 \left (a+b \sinh ^{-1}(c+d x)\right ) \text {Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac {b^2 \left (a+b \sinh ^{-1}(c+d x)\right ) \text {Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac {b^3 \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sqrt {1+(c+d x)^2}\right )}{d e^4}-\frac {b^3 \operatorname {Subst}\left (\int \text {Li}_2\left (-e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^4}+\frac {b^3 \operatorname {Subst}\left (\int \text {Li}_2\left (e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^4}\\ &=-\frac {b^2 \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac {b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \left (a+b \sinh ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac {b^3 \tanh ^{-1}\left (\sqrt {1+(c+d x)^2}\right )}{d e^4}+\frac {b^2 \left (a+b \sinh ^{-1}(c+d x)\right ) \text {Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac {b^2 \left (a+b \sinh ^{-1}(c+d x)\right ) \text {Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac {b^3 \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac {b^3 \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}\\ &=-\frac {b^2 \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac {b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \left (a+b \sinh ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac {b^3 \tanh ^{-1}\left (\sqrt {1+(c+d x)^2}\right )}{d e^4}+\frac {b^2 \left (a+b \sinh ^{-1}(c+d x)\right ) \text {Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac {b^2 \left (a+b \sinh ^{-1}(c+d x)\right ) \text {Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac {b^3 \text {Li}_3\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac {b^3 \text {Li}_3\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}\\ \end {align*}
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Mathematica [B] time = 7.14, size = 694, normalized size = 2.66 \[ -\frac {a^3}{3 d e^4 (c+d x)^3}-\frac {a^2 b \sqrt {c^2+2 c d x+d^2 x^2+1}}{2 d e^4 (c+d x)^2}+\frac {a^2 b \log \left (\sqrt {c^2+2 c d x+d^2 x^2+1}+1\right )}{2 d e^4}-\frac {a^2 b \log (c+d x)}{2 d e^4}-\frac {a^2 b \sinh ^{-1}(c+d x)}{d e^4 (c+d x)^3}+\frac {a b^2 \left (-8 \text {Li}_2\left (-e^{-\sinh ^{-1}(c+d x)}\right )-\frac {2 \left (-4 (c+d x)^3 \text {Li}_2\left (e^{-\sinh ^{-1}(c+d x)}\right )+4 \sinh ^{-1}(c+d x)^2+2 \sinh ^{-1}(c+d x) \sinh \left (2 \sinh ^{-1}(c+d x)\right )-3 (c+d x) \sinh ^{-1}(c+d x) \log \left (1-e^{-\sinh ^{-1}(c+d x)}\right )+3 (c+d x) \sinh ^{-1}(c+d x) \log \left (e^{-\sinh ^{-1}(c+d x)}+1\right )+\sinh ^{-1}(c+d x) \sinh \left (3 \sinh ^{-1}(c+d x)\right ) \log \left (1-e^{-\sinh ^{-1}(c+d x)}\right )-\sinh ^{-1}(c+d x) \sinh \left (3 \sinh ^{-1}(c+d x)\right ) \log \left (e^{-\sinh ^{-1}(c+d x)}+1\right )+2 \cosh \left (2 \sinh ^{-1}(c+d x)\right )-2\right )}{(c+d x)^3}\right )}{8 d e^4}+\frac {b^3 \left (-48 \sinh ^{-1}(c+d x) \text {Li}_2\left (-e^{-\sinh ^{-1}(c+d x)}\right )+48 \sinh ^{-1}(c+d x) \text {Li}_2\left (e^{-\sinh ^{-1}(c+d x)}\right )-48 \text {Li}_3\left (-e^{-\sinh ^{-1}(c+d x)}\right )+48 \text {Li}_3\left (e^{-\sinh ^{-1}(c+d x)}\right )-\frac {16 \sinh ^{-1}(c+d x)^3 \sinh ^4\left (\frac {1}{2} \sinh ^{-1}(c+d x)\right )}{(c+d x)^3}-24 \sinh ^{-1}(c+d x)^2 \log \left (1-e^{-\sinh ^{-1}(c+d x)}\right )+24 \sinh ^{-1}(c+d x)^2 \log \left (e^{-\sinh ^{-1}(c+d x)}+1\right )-4 \sinh ^{-1}(c+d x)^3 \tanh \left (\frac {1}{2} \sinh ^{-1}(c+d x)\right )+24 \sinh ^{-1}(c+d x) \tanh \left (\frac {1}{2} \sinh ^{-1}(c+d x)\right )+4 \sinh ^{-1}(c+d x)^3 \coth \left (\frac {1}{2} \sinh ^{-1}(c+d x)\right )-24 \sinh ^{-1}(c+d x) \coth \left (\frac {1}{2} \sinh ^{-1}(c+d x)\right )-\left ((c+d x) \sinh ^{-1}(c+d x)^3 \text {csch}^4\left (\frac {1}{2} \sinh ^{-1}(c+d x)\right )\right )-6 \sinh ^{-1}(c+d x)^2 \text {csch}^2\left (\frac {1}{2} \sinh ^{-1}(c+d x)\right )-6 \sinh ^{-1}(c+d x)^2 \text {sech}^2\left (\frac {1}{2} \sinh ^{-1}(c+d x)\right )+48 \log \left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c+d x)\right )\right )\right )}{48 d e^4} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \operatorname {arsinh}\left (d x + c\right )^{3} + 3 \, a b^{2} \operatorname {arsinh}\left (d x + c\right )^{2} + 3 \, a^{2} b \operatorname {arsinh}\left (d x + c\right ) + a^{3}}{d^{4} e^{4} x^{4} + 4 \, c d^{3} e^{4} x^{3} + 6 \, c^{2} d^{2} e^{4} x^{2} + 4 \, c^{3} d e^{4} x + c^{4} e^{4}}, x\right ) \]
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 \[ \int \frac {{\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{3}}{{\left (d e x + c e\right )}^{4}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.31, size = 651, normalized size = 2.49 \[ -\frac {a^{3}}{3 d \,e^{4} \left (d x +c \right )^{3}}-\frac {b^{3} \arcsinh \left (d x +c \right )^{2} \sqrt {1+\left (d x +c \right )^{2}}}{2 d \,e^{4} \left (d x +c \right )^{2}}-\frac {b^{3} \arcsinh \left (d x +c \right )^{3}}{3 d \,e^{4} \left (d x +c \right )^{3}}-\frac {b^{3} \arcsinh \left (d x +c \right )}{d \,e^{4} \left (d x +c \right )}+\frac {b^{3} \arcsinh \left (d x +c \right )^{2} \ln \left (1+d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{2 d \,e^{4}}+\frac {b^{3} \arcsinh \left (d x +c \right ) \polylog \left (2, -d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{4}}-\frac {b^{3} \polylog \left (3, -d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{4}}-\frac {b^{3} \arcsinh \left (d x +c \right )^{2} \ln \left (1-d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{2 d \,e^{4}}-\frac {b^{3} \arcsinh \left (d x +c \right ) \polylog \left (2, d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{4}}+\frac {b^{3} \polylog \left (3, d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{4}}-\frac {2 b^{3} \arctanh \left (d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{4}}-\frac {a \,b^{2} \arcsinh \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}}{d \,e^{4} \left (d x +c \right )^{2}}-\frac {a \,b^{2} \arcsinh \left (d x +c \right )^{2}}{d \,e^{4} \left (d x +c \right )^{3}}-\frac {a \,b^{2}}{d \,e^{4} \left (d x +c \right )}+\frac {a \,b^{2} \arcsinh \left (d x +c \right ) \ln \left (1+d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{4}}+\frac {a \,b^{2} \polylog \left (2, -d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{4}}-\frac {a \,b^{2} \arcsinh \left (d x +c \right ) \ln \left (1-d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{4}}-\frac {a \,b^{2} \polylog \left (2, d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{4}}-\frac {a^{2} b \arcsinh \left (d x +c \right )}{d \,e^{4} \left (d x +c \right )^{3}}-\frac {a^{2} b \sqrt {1+\left (d x +c \right )^{2}}}{2 d \,e^{4} \left (d x +c \right )^{2}}+\frac {a^{2} b \arctanh \left (\frac {1}{\sqrt {1+\left (d x +c \right )^{2}}}\right )}{2 d \,e^{4}} \]
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 \[ -\frac {b^{3} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{3}}{3 \, {\left (d^{4} e^{4} x^{3} + 3 \, c d^{3} e^{4} x^{2} + 3 \, c^{2} d^{2} e^{4} x + c^{3} d e^{4}\right )}} - \frac {a^{3}}{3 \, {\left (d^{4} e^{4} x^{3} + 3 \, c d^{3} e^{4} x^{2} + 3 \, c^{2} d^{2} e^{4} x + c^{3} d e^{4}\right )}} + \int \frac {{\left (3 \, {\left (c^{3} + c\right )} a b^{2} + {\left (c^{3} + c\right )} b^{3} + {\left (3 \, a b^{2} d^{3} + b^{3} d^{3}\right )} x^{3} + 3 \, {\left (3 \, a b^{2} c d^{2} + b^{3} c d^{2}\right )} x^{2} + {\left (3 \, {\left (3 \, c^{2} d + d\right )} a b^{2} + {\left (3 \, c^{2} d + d\right )} b^{3}\right )} x + {\left (b^{3} c^{2} + 3 \, {\left (c^{2} + 1\right )} a b^{2} + {\left (3 \, a b^{2} d^{2} + b^{3} d^{2}\right )} x^{2} + 2 \, {\left (3 \, a b^{2} c d + b^{3} c d\right )} x\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{2} + 3 \, {\left (a^{2} b d^{3} x^{3} + 3 \, a^{2} b c d^{2} x^{2} + {\left (3 \, c^{2} d + d\right )} a^{2} b x + {\left (c^{3} + c\right )} a^{2} b + {\left (a^{2} b d^{2} x^{2} + 2 \, a^{2} b c d x + {\left (c^{2} + 1\right )} a^{2} b\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )}{d^{7} e^{4} x^{7} + 7 \, c d^{6} e^{4} x^{6} + c^{7} e^{4} + c^{5} e^{4} + {\left (21 \, c^{2} d^{5} e^{4} + d^{5} e^{4}\right )} x^{5} + 5 \, {\left (7 \, c^{3} d^{4} e^{4} + c d^{4} e^{4}\right )} x^{4} + 5 \, {\left (7 \, c^{4} d^{3} e^{4} + 2 \, c^{2} d^{3} e^{4}\right )} x^{3} + {\left (21 \, c^{5} d^{2} e^{4} + 10 \, c^{3} d^{2} e^{4}\right )} x^{2} + {\left (7 \, c^{6} d e^{4} + 5 \, c^{4} d e^{4}\right )} x + {\left (d^{6} e^{4} x^{6} + 6 \, c d^{5} e^{4} x^{5} + c^{6} e^{4} + c^{4} e^{4} + {\left (15 \, c^{2} d^{4} e^{4} + d^{4} e^{4}\right )} x^{4} + 4 \, {\left (5 \, c^{3} d^{3} e^{4} + c d^{3} e^{4}\right )} x^{3} + 3 \, {\left (5 \, c^{4} d^{2} e^{4} + 2 \, c^{2} d^{2} e^{4}\right )} x^{2} + 2 \, {\left (3 \, c^{5} d e^{4} + 2 \, c^{3} d e^{4}\right )} x\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}}\,{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 \[ \int \frac {{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^3}{{\left (c\,e+d\,e\,x\right )}^4} \,d x \]
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 \[ \frac {\int \frac {a^{3}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {b^{3} \operatorname {asinh}^{3}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {3 a b^{2} \operatorname {asinh}^{2}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {3 a^{2} b \operatorname {asinh}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx}{e^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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