Optimal. Leaf size=291 \[ -\frac {b^2 (a+b \text {ArcSin}(c+d x))}{d e^4 (c+d x)}-\frac {b \sqrt {1-(c+d x)^2} (a+b \text {ArcSin}(c+d x))^2}{2 d e^4 (c+d x)^2}-\frac {(a+b \text {ArcSin}(c+d x))^3}{3 d e^4 (c+d x)^3}-\frac {b (a+b \text {ArcSin}(c+d x))^2 \tanh ^{-1}\left (e^{i \text {ArcSin}(c+d x)}\right )}{d e^4}-\frac {b^3 \tanh ^{-1}\left (\sqrt {1-(c+d x)^2}\right )}{d e^4}+\frac {i b^2 (a+b \text {ArcSin}(c+d x)) \text {PolyLog}\left (2,-e^{i \text {ArcSin}(c+d x)}\right )}{d e^4}-\frac {i b^2 (a+b \text {ArcSin}(c+d x)) \text {PolyLog}\left (2,e^{i \text {ArcSin}(c+d x)}\right )}{d e^4}-\frac {b^3 \text {PolyLog}\left (3,-e^{i \text {ArcSin}(c+d x)}\right )}{d e^4}+\frac {b^3 \text {PolyLog}\left (3,e^{i \text {ArcSin}(c+d x)}\right )}{d e^4} \]
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Rubi [A]
time = 0.29, antiderivative size = 291, 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 = {4889, 12,
4723, 4789, 4803, 4268, 2611, 2320, 6724, 272, 65, 212} \begin {gather*} \frac {i b^2 \text {Li}_2\left (-e^{i \text {ArcSin}(c+d x)}\right ) (a+b \text {ArcSin}(c+d x))}{d e^4}-\frac {i b^2 \text {Li}_2\left (e^{i \text {ArcSin}(c+d x)}\right ) (a+b \text {ArcSin}(c+d x))}{d e^4}-\frac {b^2 (a+b \text {ArcSin}(c+d x))}{d e^4 (c+d x)}-\frac {b \sqrt {1-(c+d x)^2} (a+b \text {ArcSin}(c+d x))^2}{2 d e^4 (c+d x)^2}-\frac {(a+b \text {ArcSin}(c+d x))^3}{3 d e^4 (c+d x)^3}-\frac {b \tanh ^{-1}\left (e^{i \text {ArcSin}(c+d x)}\right ) (a+b \text {ArcSin}(c+d x))^2}{d e^4}-\frac {b^3 \text {Li}_3\left (-e^{i \text {ArcSin}(c+d x)}\right )}{d e^4}+\frac {b^3 \text {Li}_3\left (e^{i \text {ArcSin}(c+d x)}\right )}{d e^4}-\frac {b^3 \tanh ^{-1}\left (\sqrt {1-(c+d x)^2}\right )}{d e^4} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 65
Rule 212
Rule 272
Rule 2320
Rule 2611
Rule 4268
Rule 4723
Rule 4789
Rule 4803
Rule 4889
Rule 6724
Rubi steps
\begin {align*} \int \frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{(c e+d e x)^4} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^3}{e^4 x^4} \, dx,x,c+d x\right )}{d}\\ &=\frac {\text {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^3}{x^4} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \text {Subst}\left (\int \frac {\left (a+b \sin ^{-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 \sin ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \text {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^2}{x \sqrt {1-x^2}} \, dx,x,c+d x\right )}{2 d e^4}+\frac {b^2 \text {Subst}\left (\int \frac {a+b \sin ^{-1}(x)}{x^2} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {b^2 \left (a+b \sin ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac {b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \text {Subst}\left (\int (a+b x)^2 \csc (x) \, dx,x,\sin ^{-1}(c+d x)\right )}{2 d e^4}+\frac {b^3 \text {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 \sin ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac {b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}-\frac {b \left (a+b \sin ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {b^2 \text {Subst}\left (\int (a+b x) \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^4}+\frac {b^2 \text {Subst}\left (\int (a+b x) \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^4}+\frac {b^3 \text {Subst}\left (\int \frac {1}{\sqrt {1-x} x} \, dx,x,(c+d x)^2\right )}{2 d e^4}\\ &=-\frac {b^2 \left (a+b \sin ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac {b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}-\frac {b \left (a+b \sin ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}+\frac {i b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {i b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {\left (i b^3\right ) \text {Subst}\left (\int \text {Li}_2\left (-e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^4}+\frac {\left (i b^3\right ) \text {Subst}\left (\int \text {Li}_2\left (e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^4}-\frac {b^3 \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {1-(c+d x)^2}\right )}{d e^4}\\ &=-\frac {b^2 \left (a+b \sin ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac {b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}-\frac {b \left (a+b \sin ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-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 {i b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {i b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {b^3 \text {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}+\frac {b^3 \text {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}\\ &=-\frac {b^2 \left (a+b \sin ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac {b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}-\frac {b \left (a+b \sin ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-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 {i b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {i b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {b^3 \text {Li}_3\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}+\frac {b^3 \text {Li}_3\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}\\ \end {align*}
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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice
the leaf count of optimal. \(732\) vs. \(2(291)=582\).
time = 7.23, size = 732, normalized size = 2.52 \begin {gather*} -\frac {a^3}{3 d e^4 (c+d x)^3}-\frac {a^2 b \sqrt {1-c^2-2 c d x-d^2 x^2}}{2 d e^4 (c+d x)^2}-\frac {a^2 b \text {ArcSin}(c+d x)}{d e^4 (c+d x)^3}+\frac {a^2 b \log (c+d x)}{2 d e^4}-\frac {a^2 b \log \left (1+\sqrt {1-c^2-2 c d x-d^2 x^2}\right )}{2 d e^4}+\frac {a b^2 \left (8 i \text {PolyLog}\left (2,-e^{i \text {ArcSin}(c+d x)}\right )-\frac {2 \left (2+4 \text {ArcSin}(c+d x)^2-2 \cos (2 \text {ArcSin}(c+d x))-3 (c+d x) \text {ArcSin}(c+d x) \log \left (1-e^{i \text {ArcSin}(c+d x)}\right )+3 (c+d x) \text {ArcSin}(c+d x) \log \left (1+e^{i \text {ArcSin}(c+d x)}\right )+4 i (c+d x)^3 \text {PolyLog}\left (2,e^{i \text {ArcSin}(c+d x)}\right )+2 \text {ArcSin}(c+d x) \sin (2 \text {ArcSin}(c+d x))+\text {ArcSin}(c+d x) \log \left (1-e^{i \text {ArcSin}(c+d x)}\right ) \sin (3 \text {ArcSin}(c+d x))-\text {ArcSin}(c+d x) \log \left (1+e^{i \text {ArcSin}(c+d x)}\right ) \sin (3 \text {ArcSin}(c+d x))\right )}{(c+d x)^3}\right )}{8 d e^4}+\frac {b^3 \left (-24 \text {ArcSin}(c+d x) \cot \left (\frac {1}{2} \text {ArcSin}(c+d x)\right )-4 \text {ArcSin}(c+d x)^3 \cot \left (\frac {1}{2} \text {ArcSin}(c+d x)\right )-6 \text {ArcSin}(c+d x)^2 \csc ^2\left (\frac {1}{2} \text {ArcSin}(c+d x)\right )-(c+d x) \text {ArcSin}(c+d x)^3 \csc ^4\left (\frac {1}{2} \text {ArcSin}(c+d x)\right )+24 \text {ArcSin}(c+d x)^2 \log \left (1-e^{i \text {ArcSin}(c+d x)}\right )-24 \text {ArcSin}(c+d x)^2 \log \left (1+e^{i \text {ArcSin}(c+d x)}\right )+48 \log \left (\tan \left (\frac {1}{2} \text {ArcSin}(c+d x)\right )\right )+48 i \text {ArcSin}(c+d x) \text {PolyLog}\left (2,-e^{i \text {ArcSin}(c+d x)}\right )-48 i \text {ArcSin}(c+d x) \text {PolyLog}\left (2,e^{i \text {ArcSin}(c+d x)}\right )-48 \text {PolyLog}\left (3,-e^{i \text {ArcSin}(c+d x)}\right )+48 \text {PolyLog}\left (3,e^{i \text {ArcSin}(c+d x)}\right )+6 \text {ArcSin}(c+d x)^2 \sec ^2\left (\frac {1}{2} \text {ArcSin}(c+d x)\right )-\frac {16 \text {ArcSin}(c+d x)^3 \sin ^4\left (\frac {1}{2} \text {ArcSin}(c+d x)\right )}{(c+d x)^3}-24 \text {ArcSin}(c+d x) \tan \left (\frac {1}{2} \text {ArcSin}(c+d x)\right )-4 \text {ArcSin}(c+d x)^3 \tan \left (\frac {1}{2} \text {ArcSin}(c+d x)\right )\right )}{48 d e^4} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.60, size = 646, normalized size = 2.22
method | result | size |
derivativedivides | \(\frac {-\frac {a^{3}}{3 e^{4} \left (d x +c \right )^{3}}-\frac {b^{3} \arcsin \left (d x +c \right )^{2} \sqrt {1-\left (d x +c \right )^{2}}}{2 e^{4} \left (d x +c \right )^{2}}-\frac {b^{3} \arcsin \left (d x +c \right )^{3}}{3 e^{4} \left (d x +c \right )^{3}}-\frac {b^{3} \arcsin \left (d x +c \right )}{e^{4} \left (d x +c \right )}-\frac {b^{3} \arcsin \left (d x +c \right )^{2} \ln \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{2 e^{4}}+\frac {i b^{3} \arcsin \left (d x +c \right ) \polylog \left (2, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{4}}-\frac {b^{3} \polylog \left (3, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{4}}+\frac {b^{3} \arcsin \left (d x +c \right )^{2} \ln \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{2 e^{4}}-\frac {i b^{3} \arcsin \left (d x +c \right ) \polylog \left (2, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{4}}+\frac {b^{3} \polylog \left (3, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{4}}-\frac {2 b^{3} \arctanh \left (i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{4}}-\frac {a \,b^{2} \arcsin \left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}}{e^{4} \left (d x +c \right )^{2}}-\frac {a \,b^{2} \arcsin \left (d x +c \right )^{2}}{e^{4} \left (d x +c \right )^{3}}-\frac {a \,b^{2}}{e^{4} \left (d x +c \right )}-\frac {a \,b^{2} \arcsin \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{4}}-\frac {i a \,b^{2} \polylog \left (2, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{4}}+\frac {a \,b^{2} \arcsin \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{4}}+\frac {i a \,b^{2} \polylog \left (2, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{4}}+\frac {3 a^{2} b \left (-\frac {\arcsin \left (d x +c \right )}{3 \left (d x +c \right )^{3}}-\frac {\sqrt {1-\left (d x +c \right )^{2}}}{6 \left (d x +c \right )^{2}}-\frac {\arctanh \left (\frac {1}{\sqrt {1-\left (d x +c \right )^{2}}}\right )}{6}\right )}{e^{4}}}{d}\) | \(646\) |
default | \(\frac {-\frac {a^{3}}{3 e^{4} \left (d x +c \right )^{3}}-\frac {b^{3} \arcsin \left (d x +c \right )^{2} \sqrt {1-\left (d x +c \right )^{2}}}{2 e^{4} \left (d x +c \right )^{2}}-\frac {b^{3} \arcsin \left (d x +c \right )^{3}}{3 e^{4} \left (d x +c \right )^{3}}-\frac {b^{3} \arcsin \left (d x +c \right )}{e^{4} \left (d x +c \right )}-\frac {b^{3} \arcsin \left (d x +c \right )^{2} \ln \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{2 e^{4}}+\frac {i b^{3} \arcsin \left (d x +c \right ) \polylog \left (2, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{4}}-\frac {b^{3} \polylog \left (3, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{4}}+\frac {b^{3} \arcsin \left (d x +c \right )^{2} \ln \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{2 e^{4}}-\frac {i b^{3} \arcsin \left (d x +c \right ) \polylog \left (2, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{4}}+\frac {b^{3} \polylog \left (3, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{4}}-\frac {2 b^{3} \arctanh \left (i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{4}}-\frac {a \,b^{2} \arcsin \left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}}{e^{4} \left (d x +c \right )^{2}}-\frac {a \,b^{2} \arcsin \left (d x +c \right )^{2}}{e^{4} \left (d x +c \right )^{3}}-\frac {a \,b^{2}}{e^{4} \left (d x +c \right )}-\frac {a \,b^{2} \arcsin \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{4}}-\frac {i a \,b^{2} \polylog \left (2, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{4}}+\frac {a \,b^{2} \arcsin \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{4}}+\frac {i a \,b^{2} \polylog \left (2, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{4}}+\frac {3 a^{2} b \left (-\frac {\arcsin \left (d x +c \right )}{3 \left (d x +c \right )^{3}}-\frac {\sqrt {1-\left (d x +c \right )^{2}}}{6 \left (d x +c \right )^{2}}-\frac {\arctanh \left (\frac {1}{\sqrt {1-\left (d x +c \right )^{2}}}\right )}{6}\right )}{e^{4}}}{d}\) | \(646\) |
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^{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 {asin}^{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 {asin}^{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 {asin}{\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}} \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 {asin}\left (c+d\,x\right )\right )}^3}{{\left (c\,e+d\,e\,x\right )}^4} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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