Optimal. Leaf size=389 \[ \frac {b^2 c^2 x}{12 d^3 \left (1+c^2 x^2\right )}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {7 b c \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \sqrt {1+c^2 x^2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d^3 x \left (1+c^2 x^2\right )^2}-\frac {5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {15 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 d^3 \left (1+c^2 x^2\right )}-\frac {15 c \left (a+b \sinh ^{-1}(c x)\right )^2 \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac {11 b^2 c \text {ArcTan}(c x)}{6 d^3}-\frac {4 b c \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d^3}-\frac {2 b^2 c \text {PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{d^3}+\frac {15 i b c \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}-\frac {15 i b c \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac {2 b^2 c \text {PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{d^3}-\frac {15 i b^2 c \text {PolyLog}\left (3,-i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac {15 i b^2 c \text {PolyLog}\left (3,i e^{\sinh ^{-1}(c x)}\right )}{4 d^3} \]
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Rubi [A]
time = 0.53, 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 =
{5809, 5788, 5789, 4265, 2611, 2320, 6724, 5798, 209, 205, 5811, 5816, 4267, 2317, 2438}
\begin {gather*} -\frac {15 c \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{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 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 {4 b c \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^3}+\frac {11 b^2 c \text {ArcTan}(c x)}{6 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} \end {gather*}
Antiderivative was successfully verified.
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Rule 205
Rule 209
Rule 2317
Rule 2320
Rule 2438
Rule 2611
Rule 4265
Rule 4267
Rule 5788
Rule 5789
Rule 5798
Rule 5809
Rule 5811
Rule 5816
Rule 6724
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) \text {Subst}\left (\int (a+b x)^2 \text {sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{8 d^3}+\frac {(2 b c) \text {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) \text {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) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 = 7.09, size = 716, normalized size = 1.84 \begin {gather*} -\frac {a^2}{d^3 x}-\frac {a^2 c^2 x}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {7 a^2 c^2 x}{8 d^3 \left (1+c^2 x^2\right )}-\frac {15 a^2 c \text {ArcTan}(c x)}{8 d^3}+\frac {2 a b c \left (\frac {7 \left (\sqrt {1+c^2 x^2}+i \sinh ^{-1}(c x)\right )}{16 (-1-i c x)}-\frac {\sinh ^{-1}(c x)}{c x}-\frac {7 \left (i \sqrt {1+c^2 x^2}+\sinh ^{-1}(c x)\right )}{16 (i+c x)}+\frac {i \left ((-2 i+c x) \sqrt {1+c^2 x^2}+3 \sinh ^{-1}(c x)\right )}{48 (-i+c x)^2}-\frac {i \left ((2 i+c x) \sqrt {1+c^2 x^2}+3 \sinh ^{-1}(c x)\right )}{48 (i+c x)^2}-\tanh ^{-1}\left (\sqrt {1+c^2 x^2}\right )+\frac {15}{16} i \left (-\frac {1}{2} \sinh ^{-1}(c x)^2+2 \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )+2 \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )\right )-\frac {15}{16} i \left (-\frac {1}{2} \sinh ^{-1}(c x)^2+2 \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )+2 \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )\right )\right )}{d^3}+\frac {b^2 c \left (\frac {2 c x}{1+c^2 x^2}-\frac {4 \sinh ^{-1}(c x)}{\left (1+c^2 x^2\right )^{3/2}}-\frac {42 \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}}-\frac {6 c x \sinh ^{-1}(c x)^2}{\left (1+c^2 x^2\right )^2}-\frac {21 c x \sinh ^{-1}(c x)^2}{1+c^2 x^2}+88 \text {ArcTan}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )-12 \sinh ^{-1}(c x)^2 \coth \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+48 \sinh ^{-1}(c x) \log \left (1-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 \text {PolyLog}\left (2,-e^{-\sinh ^{-1}(c x)}\right )+90 i \sinh ^{-1}(c x) \text {PolyLog}\left (2,-i e^{-\sinh ^{-1}(c x)}\right )-90 i \sinh ^{-1}(c x) \text {PolyLog}\left (2,i e^{-\sinh ^{-1}(c x)}\right )-48 \text {PolyLog}\left (2,e^{-\sinh ^{-1}(c x)}\right )+90 i \text {PolyLog}\left (3,-i e^{-\sinh ^{-1}(c x)}\right )-90 i \text {PolyLog}\left (3,i e^{-\sinh ^{-1}(c x)}\right )+12 \sinh ^{-1}(c x)^2 \tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )}{24 d^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.10, 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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \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}} \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \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 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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