Optimal. Leaf size=389 \[ -\frac {7 b c \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \sqrt {c^2 x^2+1}}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (c^2 x^2+1\right )^{3/2}}-\frac {15 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 d^3 \left (c^2 x^2+1\right )}-\frac {5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (c^2 x^2+1\right )^2}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d^3 x \left (c^2 x^2+1\right )^2}+\frac {15 i b c \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3}-\frac {15 i b c \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3}-\frac {15 c \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3}-\frac {4 b c \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^3}+\frac {b^2 c^2 x}{12 d^3 \left (c^2 x^2+1\right )}-\frac {2 b^2 c \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{d^3}+\frac {2 b^2 c \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{d^3}-\frac {15 i b^2 c \text {Li}_3\left (-i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac {15 i b^2 c \text {Li}_3\left (i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac {11 b^2 c \tan ^{-1}(c x)}{6 d^3} \]
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Rubi [A] time = 0.77, antiderivative size = 389, normalized size of antiderivative = 1.00, number of steps used = 27, number of rules used = 15, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.577, Rules used = {5747, 5690, 5693, 4180, 2531, 2282, 6589, 5717, 203, 199, 5755, 5760, 4182, 2279, 2391} \[ \frac {15 i b c \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3}-\frac {15 i b c \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3}-\frac {2 b^2 c \text {PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{d^3}+\frac {2 b^2 c \text {PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{d^3}-\frac {15 i b^2 c \text {PolyLog}\left (3,-i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac {15 i b^2 c \text {PolyLog}\left (3,i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}-\frac {7 b c \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \sqrt {c^2 x^2+1}}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (c^2 x^2+1\right )^{3/2}}-\frac {15 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 d^3 \left (c^2 x^2+1\right )}-\frac {5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (c^2 x^2+1\right )^2}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d^3 x \left (c^2 x^2+1\right )^2}-\frac {15 c \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3}-\frac {4 b c \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^3}+\frac {b^2 c^2 x}{12 d^3 \left (c^2 x^2+1\right )}+\frac {11 b^2 c \tan ^{-1}(c x)}{6 d^3} \]
Antiderivative was successfully verified.
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Rule 199
Rule 203
Rule 2279
Rule 2282
Rule 2391
Rule 2531
Rule 4180
Rule 4182
Rule 5690
Rule 5693
Rule 5717
Rule 5747
Rule 5755
Rule 5760
Rule 6589
Rubi steps
\begin {align*} \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{x^2 \left (d+c^2 d x^2\right )^3} \, dx &=-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d^3 x \left (1+c^2 x^2\right )^2}-\left (5 c^2\right ) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{\left (d+c^2 d x^2\right )^3} \, dx+\frac {(2 b c) \int \frac {a+b \sinh ^{-1}(c x)}{x \left (1+c^2 x^2\right )^{5/2}} \, dx}{d^3}\\ &=\frac {2 b c \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d^3 x \left (1+c^2 x^2\right )^2}-\frac {5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}+\frac {(2 b c) \int \frac {a+b \sinh ^{-1}(c x)}{x \left (1+c^2 x^2\right )^{3/2}} \, dx}{d^3}-\frac {\left (2 b^2 c^2\right ) \int \frac {1}{\left (1+c^2 x^2\right )^2} \, dx}{3 d^3}+\frac {\left (5 b c^3\right ) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{2 d^3}-\frac {\left (15 c^2\right ) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{\left (d+c^2 d x^2\right )^2} \, dx}{4 d}\\ &=-\frac {b^2 c^2 x}{3 d^3 \left (1+c^2 x^2\right )}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {2 b c \left (a+b \sinh ^{-1}(c x)\right )}{d^3 \sqrt {1+c^2 x^2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d^3 x \left (1+c^2 x^2\right )^2}-\frac {5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {15 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 d^3 \left (1+c^2 x^2\right )}+\frac {(2 b c) \int \frac {a+b \sinh ^{-1}(c x)}{x \sqrt {1+c^2 x^2}} \, dx}{d^3}-\frac {\left (b^2 c^2\right ) \int \frac {1}{1+c^2 x^2} \, dx}{3 d^3}+\frac {\left (5 b^2 c^2\right ) \int \frac {1}{\left (1+c^2 x^2\right )^2} \, dx}{6 d^3}-\frac {\left (2 b^2 c^2\right ) \int \frac {1}{1+c^2 x^2} \, dx}{d^3}+\frac {\left (15 b c^3\right ) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{4 d^3}-\frac {\left (15 c^2\right ) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d+c^2 d x^2} \, dx}{8 d^2}\\ &=\frac {b^2 c^2 x}{12 d^3 \left (1+c^2 x^2\right )}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {7 b c \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \sqrt {1+c^2 x^2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d^3 x \left (1+c^2 x^2\right )^2}-\frac {5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {15 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 d^3 \left (1+c^2 x^2\right )}-\frac {7 b^2 c \tan ^{-1}(c x)}{3 d^3}-\frac {(15 c) \operatorname {Subst}\left (\int (a+b x)^2 \text {sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{8 d^3}+\frac {(2 b c) \operatorname {Subst}\left (\int (a+b x) \text {csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}+\frac {\left (5 b^2 c^2\right ) \int \frac {1}{1+c^2 x^2} \, dx}{12 d^3}+\frac {\left (15 b^2 c^2\right ) \int \frac {1}{1+c^2 x^2} \, dx}{4 d^3}\\ &=\frac {b^2 c^2 x}{12 d^3 \left (1+c^2 x^2\right )}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {7 b c \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \sqrt {1+c^2 x^2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d^3 x \left (1+c^2 x^2\right )^2}-\frac {5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {15 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 d^3 \left (1+c^2 x^2\right )}-\frac {15 c \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac {11 b^2 c \tan ^{-1}(c x)}{6 d^3}-\frac {4 b c \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d^3}+\frac {(15 i b c) \operatorname {Subst}\left (\int (a+b x) \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{4 d^3}-\frac {(15 i b c) \operatorname {Subst}\left (\int (a+b x) \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{4 d^3}-\frac {\left (2 b^2 c\right ) \operatorname {Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}+\frac {\left (2 b^2 c\right ) \operatorname {Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}\\ &=\frac {b^2 c^2 x}{12 d^3 \left (1+c^2 x^2\right )}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {7 b c \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \sqrt {1+c^2 x^2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d^3 x \left (1+c^2 x^2\right )^2}-\frac {5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {15 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 d^3 \left (1+c^2 x^2\right )}-\frac {15 c \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac {11 b^2 c \tan ^{-1}(c x)}{6 d^3}-\frac {4 b c \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d^3}+\frac {15 i b c \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}-\frac {15 i b c \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}-\frac {\left (15 i b^2 c\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{4 d^3}+\frac {\left (15 i b^2 c\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{4 d^3}-\frac {\left (2 b^2 c\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{d^3}+\frac {\left (2 b^2 c\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{d^3}\\ &=\frac {b^2 c^2 x}{12 d^3 \left (1+c^2 x^2\right )}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {7 b c \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \sqrt {1+c^2 x^2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d^3 x \left (1+c^2 x^2\right )^2}-\frac {5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {15 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 d^3 \left (1+c^2 x^2\right )}-\frac {15 c \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac {11 b^2 c \tan ^{-1}(c x)}{6 d^3}-\frac {4 b c \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d^3}-\frac {2 b^2 c \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{d^3}+\frac {15 i b c \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}-\frac {15 i b c \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac {2 b^2 c \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{d^3}-\frac {\left (15 i b^2 c\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac {\left (15 i b^2 c\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{4 d^3}\\ &=\frac {b^2 c^2 x}{12 d^3 \left (1+c^2 x^2\right )}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {7 b c \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \sqrt {1+c^2 x^2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d^3 x \left (1+c^2 x^2\right )^2}-\frac {5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {15 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 d^3 \left (1+c^2 x^2\right )}-\frac {15 c \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac {11 b^2 c \tan ^{-1}(c x)}{6 d^3}-\frac {4 b c \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d^3}-\frac {2 b^2 c \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{d^3}+\frac {15 i b c \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}-\frac {15 i b c \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac {2 b^2 c \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{d^3}-\frac {15 i b^2 c \text {Li}_3\left (-i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac {15 i b^2 c \text {Li}_3\left (i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}\\ \end {align*}
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Mathematica [A] time = 7.70, size = 716, normalized size = 1.84 \[ -\frac {7 a^2 c^2 x}{8 d^3 \left (c^2 x^2+1\right )}-\frac {a^2 c^2 x}{4 d^3 \left (c^2 x^2+1\right )^2}-\frac {15 a^2 c \tan ^{-1}(c x)}{8 d^3}-\frac {a^2}{d^3 x}+\frac {2 a b c \left (\frac {7 \left (\sqrt {c^2 x^2+1}+i \sinh ^{-1}(c x)\right )}{16 (-1-i c x)}-\frac {7 \left (\sinh ^{-1}(c x)+i \sqrt {c^2 x^2+1}\right )}{16 (c x+i)}+\frac {i \left (3 \sinh ^{-1}(c x)+\sqrt {c^2 x^2+1} (c x-2 i)\right )}{48 (c x-i)^2}-\frac {i \left (3 \sinh ^{-1}(c x)+\sqrt {c^2 x^2+1} (c x+2 i)\right )}{48 (c x+i)^2}-\tanh ^{-1}\left (\sqrt {c^2 x^2+1}\right )+\frac {15}{16} i \left (2 \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )-\frac {1}{2} \sinh ^{-1}(c x)^2+2 \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )\right )-\frac {15}{16} i \left (2 \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )-\frac {1}{2} \sinh ^{-1}(c x)^2+2 \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )\right )-\frac {\sinh ^{-1}(c x)}{c x}\right )}{d^3}+\frac {b^2 c \left (\frac {2 c x}{c^2 x^2+1}-\frac {21 c x \sinh ^{-1}(c x)^2}{c^2 x^2+1}-\frac {6 c x \sinh ^{-1}(c x)^2}{\left (c^2 x^2+1\right )^2}-\frac {42 \sinh ^{-1}(c x)}{\sqrt {c^2 x^2+1}}-\frac {4 \sinh ^{-1}(c x)}{\left (c^2 x^2+1\right )^{3/2}}+90 i \sinh ^{-1}(c x) \text {Li}_2\left (-i e^{-\sinh ^{-1}(c x)}\right )-90 i \sinh ^{-1}(c x) \text {Li}_2\left (i e^{-\sinh ^{-1}(c x)}\right )+48 \text {Li}_2\left (-e^{-\sinh ^{-1}(c x)}\right )-48 \text {Li}_2\left (e^{-\sinh ^{-1}(c x)}\right )+90 i \text {Li}_3\left (-i e^{-\sinh ^{-1}(c x)}\right )-90 i \text {Li}_3\left (i e^{-\sinh ^{-1}(c x)}\right )+45 i \sinh ^{-1}(c x)^2 \log \left (1-i e^{-\sinh ^{-1}(c x)}\right )-45 i \sinh ^{-1}(c x)^2 \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )+48 \sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )-48 \sinh ^{-1}(c x) \log \left (e^{-\sinh ^{-1}(c x)}+1\right )+12 \sinh ^{-1}(c x)^2 \tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-12 \sinh ^{-1}(c x)^2 \coth \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+88 \tan ^{-1}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )\right )}{24 d^3} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \operatorname {arsinh}\left (c x\right )^{2} + 2 \, a b \operatorname {arsinh}\left (c x\right ) + a^{2}}{c^{6} d^{3} x^{8} + 3 \, c^{4} d^{3} x^{6} + 3 \, c^{2} d^{3} x^{4} + d^{3} x^{2}}, x\right ) \]
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 \[ \int \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}{{\left (c^{2} d x^{2} + d\right )}^{3} x^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.49, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arcsinh \left (c x \right )\right )^{2}}{x^{2} \left (c^{2} d \,x^{2}+d \right )^{3}}\, dx \]
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 \[ -\frac {1}{8} \, a^{2} {\left (\frac {15 \, c^{4} x^{4} + 25 \, c^{2} x^{2} + 8}{c^{4} d^{3} x^{5} + 2 \, c^{2} d^{3} x^{3} + d^{3} x} + \frac {15 \, c \arctan \left (c x\right )}{d^{3}}\right )} + \int \frac {b^{2} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2}}{c^{6} d^{3} x^{8} + 3 \, c^{4} d^{3} x^{6} + 3 \, c^{2} d^{3} x^{4} + d^{3} x^{2}} + \frac {2 \, a b \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{c^{6} d^{3} x^{8} + 3 \, c^{4} d^{3} x^{6} + 3 \, c^{2} d^{3} x^{4} + d^{3} x^{2}}\,{d x} \]
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 \[ \int \frac {{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2}{x^2\,{\left (d\,c^2\,x^2+d\right )}^3} \,d x \]
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 \[ \frac {\int \frac {a^{2}}{c^{6} x^{8} + 3 c^{4} x^{6} + 3 c^{2} x^{4} + x^{2}}\, dx + \int \frac {b^{2} \operatorname {asinh}^{2}{\left (c x \right )}}{c^{6} x^{8} + 3 c^{4} x^{6} + 3 c^{2} x^{4} + x^{2}}\, dx + \int \frac {2 a b \operatorname {asinh}{\left (c x \right )}}{c^{6} x^{8} + 3 c^{4} x^{6} + 3 c^{2} x^{4} + x^{2}}\, dx}{d^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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