Optimal. Leaf size=385 \[ -\frac {4 b^3 \text {Li}_3\left (-e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^4}+\frac {4 b^3 \text {Li}_3\left (e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^4}-\frac {8 b^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^4}+\frac {2 b^2 \text {Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^4}-\frac {2 b^2 \text {Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^4}-\frac {2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^4 (c+d x)}-\frac {2 b \sqrt {(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^2}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^4}{3 d e^4 (c+d x)^3}+\frac {4 b \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4}-\frac {4 b^4 \text {Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac {4 b^4 \text {Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac {4 b^4 \text {Li}_4\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac {4 b^4 \text {Li}_4\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.56, antiderivative size = 385, normalized size of antiderivative = 1.00, number of steps used = 21, 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, 6609, 2282, 6589, 2279, 2391} \[ -\frac {4 b^3 \text {PolyLog}\left (3,-e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^4}+\frac {4 b^3 \text {PolyLog}\left (3,e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^4}+\frac {2 b^2 \text {PolyLog}\left (2,-e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^4}-\frac {2 b^2 \text {PolyLog}\left (2,e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^4}-\frac {4 b^4 \text {PolyLog}\left (2,-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac {4 b^4 \text {PolyLog}\left (2,e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac {4 b^4 \text {PolyLog}\left (4,-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac {4 b^4 \text {PolyLog}\left (4,e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac {2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^4 (c+d x)}-\frac {8 b^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^4}-\frac {2 b \sqrt {(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^2}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^4}{3 d e^4 (c+d x)^3}+\frac {4 b \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 12
Rule 2279
Rule 2282
Rule 2391
Rule 2531
Rule 4182
Rule 5661
Rule 5747
Rule 5760
Rule 5865
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int \frac {\left (a+b \sinh ^{-1}(c+d x)\right )^4}{(c e+d e x)^4} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sinh ^{-1}(x)\right )^4}{e^4 x^4} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sinh ^{-1}(x)\right )^4}{x^4} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^4}{3 d e^4 (c+d x)^3}+\frac {(4 b) \operatorname {Subst}\left (\int \frac {\left (a+b \sinh ^{-1}(x)\right )^3}{x^3 \sqrt {1+x^2}} \, dx,x,c+d x\right )}{3 d e^4}\\ &=-\frac {2 b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^2}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^4}{3 d e^4 (c+d x)^3}-\frac {(2 b) \operatorname {Subst}\left (\int \frac {\left (a+b \sinh ^{-1}(x)\right )^3}{x \sqrt {1+x^2}} \, dx,x,c+d x\right )}{3 d e^4}+\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \frac {\left (a+b \sinh ^{-1}(x)\right )^2}{x^2} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^4 (c+d x)}-\frac {2 b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^2}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^4}{3 d e^4 (c+d x)^3}-\frac {(2 b) \operatorname {Subst}\left (\int (a+b x)^3 \text {csch}(x) \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 d e^4}+\frac {\left (4 b^3\right ) \operatorname {Subst}\left (\int \frac {a+b \sinh ^{-1}(x)}{x \sqrt {1+x^2}} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^4 (c+d x)}-\frac {2 b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^2}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^4}{3 d e^4 (c+d x)^3}+\frac {4 b \left (a+b \sinh ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{3 d e^4}+\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int (a+b x)^2 \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^4}-\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int (a+b x)^2 \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^4}+\frac {\left (4 b^3\right ) \operatorname {Subst}\left (\int (a+b x) \text {csch}(x) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^4}\\ &=-\frac {2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^4 (c+d x)}-\frac {2 b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^2}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^4}{3 d e^4 (c+d x)^3}-\frac {8 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac {4 b \left (a+b \sinh ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{3 d e^4}+\frac {2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text {Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac {2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text {Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac {\left (4 b^3\right ) \operatorname {Subst}\left (\int (a+b x) \text {Li}_2\left (-e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^4}+\frac {\left (4 b^3\right ) \operatorname {Subst}\left (\int (a+b x) \text {Li}_2\left (e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^4}-\frac {\left (4 b^4\right ) \operatorname {Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^4}+\frac {\left (4 b^4\right ) \operatorname {Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^4}\\ &=-\frac {2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^4 (c+d x)}-\frac {2 b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^2}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^4}{3 d e^4 (c+d x)^3}-\frac {8 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac {4 b \left (a+b \sinh ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{3 d e^4}+\frac {2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text {Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac {2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text {Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac {4 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text {Li}_3\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac {4 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text {Li}_3\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac {\left (4 b^4\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac {\left (4 b^4\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac {\left (4 b^4\right ) \operatorname {Subst}\left (\int \text {Li}_3\left (-e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^4}-\frac {\left (4 b^4\right ) \operatorname {Subst}\left (\int \text {Li}_3\left (e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^4}\\ &=-\frac {2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^4 (c+d x)}-\frac {2 b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^2}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^4}{3 d e^4 (c+d x)^3}-\frac {8 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac {4 b \left (a+b \sinh ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{3 d e^4}-\frac {4 b^4 \text {Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac {2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text {Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac {4 b^4 \text {Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac {2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text {Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac {4 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text {Li}_3\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac {4 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text {Li}_3\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac {\left (4 b^4\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac {\left (4 b^4\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}\\ &=-\frac {2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^4 (c+d x)}-\frac {2 b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^2}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^4}{3 d e^4 (c+d x)^3}-\frac {8 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac {4 b \left (a+b \sinh ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{3 d e^4}-\frac {4 b^4 \text {Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac {2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text {Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac {4 b^4 \text {Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac {2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text {Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac {4 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text {Li}_3\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac {4 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text {Li}_3\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac {4 b^4 \text {Li}_4\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac {4 b^4 \text {Li}_4\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}\\ \end {align*}
________________________________________________________________________________________
Mathematica [B] time = 8.59, size = 1182, normalized size = 3.07 \[ \text {result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
[Out]
________________________________________________________________________________________
fricas [F] time = 0.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{4} \operatorname {arsinh}\left (d x + c\right )^{4} + 4 \, a b^{3} \operatorname {arsinh}\left (d x + c\right )^{3} + 6 \, a^{2} b^{2} \operatorname {arsinh}\left (d x + c\right )^{2} + 4 \, a^{3} b \operatorname {arsinh}\left (d x + c\right ) + a^{4}}{d^{4} e^{4} x^{4} + 4 \, c d^{3} e^{4} x^{3} + 6 \, c^{2} d^{2} e^{4} x^{2} + 4 \, c^{3} d e^{4} x + c^{4} e^{4}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{4}}{{\left (d e x + c e\right )}^{4}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [B] time = 0.31, size = 1202, normalized size = 3.12 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {b^{4} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{4}}{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^{4}}{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 {2 \, {\left (2 \, {\left (3 \, {\left (c^{3} + c\right )} a b^{3} + {\left (c^{3} + c\right )} b^{4} + {\left (3 \, a b^{3} d^{3} + b^{4} d^{3}\right )} x^{3} + 3 \, {\left (3 \, a b^{3} c d^{2} + b^{4} c d^{2}\right )} x^{2} + {\left (3 \, {\left (3 \, c^{2} d + d\right )} a b^{3} + {\left (3 \, c^{2} d + d\right )} b^{4}\right )} x + {\left (b^{4} c^{2} + 3 \, {\left (c^{2} + 1\right )} a b^{3} + {\left (3 \, a b^{3} d^{2} + b^{4} d^{2}\right )} x^{2} + 2 \, {\left (3 \, a b^{3} c d + b^{4} 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 )^{3} + 9 \, {\left (a^{2} b^{2} d^{3} x^{3} + 3 \, a^{2} b^{2} c d^{2} x^{2} + {\left (3 \, c^{2} d + d\right )} a^{2} b^{2} x + {\left (c^{3} + c\right )} a^{2} b^{2} + {\left (a^{2} b^{2} d^{2} x^{2} + 2 \, a^{2} b^{2} c d x + {\left (c^{2} + 1\right )} a^{2} b^{2}\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} + 6 \, {\left (a^{3} b d^{3} x^{3} + 3 \, a^{3} b c d^{2} x^{2} + {\left (3 \, c^{2} d + d\right )} a^{3} b x + {\left (c^{3} + c\right )} a^{3} b + {\left (a^{3} b d^{2} x^{2} + 2 \, a^{3} b c d x + {\left (c^{2} + 1\right )} a^{3} 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 )\right )}}{3 \, {\left (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}\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^4}{{\left (c\,e+d\,e\,x\right )}^4} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a^{4}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {b^{4} \operatorname {asinh}^{4}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {4 a b^{3} \operatorname {asinh}^{3}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {6 a^{2} b^{2} \operatorname {asinh}^{2}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {4 a^{3} b \operatorname {asinh}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx}{e^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________