Optimal. Leaf size=358 \[ -\frac {(e f-d g) \left (a+b \sin ^{-1}(c x)\right )}{e^2 (d+e x)}+\frac {g \log (d+e x) \left (a+b \sin ^{-1}(c x)\right )}{e^2}+\frac {b c (e f-d g) \tan ^{-1}\left (\frac {c^2 d x+e}{\sqrt {1-c^2 x^2} \sqrt {c^2 d^2-e^2}}\right )}{e^2 \sqrt {c^2 d^2-e^2}}-\frac {i b g \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e^2}-\frac {i b g \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e^2}+\frac {b g \sin ^{-1}(c x) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e^2}+\frac {b g \sin ^{-1}(c x) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )}{e^2}-\frac {b g \sin ^{-1}(c x) \log (d+e x)}{e^2}-\frac {i b g \sin ^{-1}(c x)^2}{2 e^2} \]
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Rubi [A] time = 0.95, antiderivative size = 358, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 13, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {43, 4753, 12, 6742, 725, 204, 216, 2404, 4741, 4519, 2190, 2279, 2391} \[ -\frac {i b g \text {PolyLog}\left (2,\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e^2}-\frac {i b g \text {PolyLog}\left (2,\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )}{e^2}-\frac {(e f-d g) \left (a+b \sin ^{-1}(c x)\right )}{e^2 (d+e x)}+\frac {g \log (d+e x) \left (a+b \sin ^{-1}(c x)\right )}{e^2}+\frac {b c (e f-d g) \tan ^{-1}\left (\frac {c^2 d x+e}{\sqrt {1-c^2 x^2} \sqrt {c^2 d^2-e^2}}\right )}{e^2 \sqrt {c^2 d^2-e^2}}+\frac {b g \sin ^{-1}(c x) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e^2}+\frac {b g \sin ^{-1}(c x) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )}{e^2}-\frac {b g \sin ^{-1}(c x) \log (d+e x)}{e^2}-\frac {i b g \sin ^{-1}(c x)^2}{2 e^2} \]
Antiderivative was successfully verified.
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Rule 12
Rule 43
Rule 204
Rule 216
Rule 725
Rule 2190
Rule 2279
Rule 2391
Rule 2404
Rule 4519
Rule 4741
Rule 4753
Rule 6742
Rubi steps
\begin {align*} \int \frac {(f+g x) \left (a+b \sin ^{-1}(c x)\right )}{(d+e x)^2} \, dx &=-\frac {(e f-d g) \left (a+b \sin ^{-1}(c x)\right )}{e^2 (d+e x)}+\frac {g \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^2}-(b c) \int \frac {-e f \left (1-\frac {d g}{e f}\right )+g (d+e x) \log (d+e x)}{e^2 (d+e x) \sqrt {1-c^2 x^2}} \, dx\\ &=-\frac {(e f-d g) \left (a+b \sin ^{-1}(c x)\right )}{e^2 (d+e x)}+\frac {g \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^2}-\frac {(b c) \int \frac {-e f \left (1-\frac {d g}{e f}\right )+g (d+e x) \log (d+e x)}{(d+e x) \sqrt {1-c^2 x^2}} \, dx}{e^2}\\ &=-\frac {(e f-d g) \left (a+b \sin ^{-1}(c x)\right )}{e^2 (d+e x)}+\frac {g \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^2}-\frac {(b c) \int \left (\frac {-e f+d g}{(d+e x) \sqrt {1-c^2 x^2}}+\frac {g \log (d+e x)}{\sqrt {1-c^2 x^2}}\right ) \, dx}{e^2}\\ &=-\frac {(e f-d g) \left (a+b \sin ^{-1}(c x)\right )}{e^2 (d+e x)}+\frac {g \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^2}-\frac {(b c g) \int \frac {\log (d+e x)}{\sqrt {1-c^2 x^2}} \, dx}{e^2}+\frac {(b c (e f-d g)) \int \frac {1}{(d+e x) \sqrt {1-c^2 x^2}} \, dx}{e^2}\\ &=-\frac {(e f-d g) \left (a+b \sin ^{-1}(c x)\right )}{e^2 (d+e x)}-\frac {b g \sin ^{-1}(c x) \log (d+e x)}{e^2}+\frac {g \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^2}+\frac {(b c g) \int \frac {\sin ^{-1}(c x)}{c d+c e x} \, dx}{e}-\frac {(b c (e f-d g)) \operatorname {Subst}\left (\int \frac {1}{-c^2 d^2+e^2-x^2} \, dx,x,\frac {e+c^2 d x}{\sqrt {1-c^2 x^2}}\right )}{e^2}\\ &=-\frac {(e f-d g) \left (a+b \sin ^{-1}(c x)\right )}{e^2 (d+e x)}+\frac {b c (e f-d g) \tan ^{-1}\left (\frac {e+c^2 d x}{\sqrt {c^2 d^2-e^2} \sqrt {1-c^2 x^2}}\right )}{e^2 \sqrt {c^2 d^2-e^2}}-\frac {b g \sin ^{-1}(c x) \log (d+e x)}{e^2}+\frac {g \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^2}+\frac {(b c g) \operatorname {Subst}\left (\int \frac {x \cos (x)}{c^2 d+c e \sin (x)} \, dx,x,\sin ^{-1}(c x)\right )}{e}\\ &=-\frac {i b g \sin ^{-1}(c x)^2}{2 e^2}-\frac {(e f-d g) \left (a+b \sin ^{-1}(c x)\right )}{e^2 (d+e x)}+\frac {b c (e f-d g) \tan ^{-1}\left (\frac {e+c^2 d x}{\sqrt {c^2 d^2-e^2} \sqrt {1-c^2 x^2}}\right )}{e^2 \sqrt {c^2 d^2-e^2}}-\frac {b g \sin ^{-1}(c x) \log (d+e x)}{e^2}+\frac {g \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^2}+\frac {(b c g) \operatorname {Subst}\left (\int \frac {e^{i x} x}{c^2 d-c \sqrt {c^2 d^2-e^2}-i c e e^{i x}} \, dx,x,\sin ^{-1}(c x)\right )}{e}+\frac {(b c g) \operatorname {Subst}\left (\int \frac {e^{i x} x}{c^2 d+c \sqrt {c^2 d^2-e^2}-i c e e^{i x}} \, dx,x,\sin ^{-1}(c x)\right )}{e}\\ &=-\frac {i b g \sin ^{-1}(c x)^2}{2 e^2}-\frac {(e f-d g) \left (a+b \sin ^{-1}(c x)\right )}{e^2 (d+e x)}+\frac {b c (e f-d g) \tan ^{-1}\left (\frac {e+c^2 d x}{\sqrt {c^2 d^2-e^2} \sqrt {1-c^2 x^2}}\right )}{e^2 \sqrt {c^2 d^2-e^2}}+\frac {b g \sin ^{-1}(c x) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e^2}+\frac {b g \sin ^{-1}(c x) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e^2}-\frac {b g \sin ^{-1}(c x) \log (d+e x)}{e^2}+\frac {g \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^2}-\frac {(b g) \operatorname {Subst}\left (\int \log \left (1-\frac {i c e e^{i x}}{c^2 d-c \sqrt {c^2 d^2-e^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{e^2}-\frac {(b g) \operatorname {Subst}\left (\int \log \left (1-\frac {i c e e^{i x}}{c^2 d+c \sqrt {c^2 d^2-e^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{e^2}\\ &=-\frac {i b g \sin ^{-1}(c x)^2}{2 e^2}-\frac {(e f-d g) \left (a+b \sin ^{-1}(c x)\right )}{e^2 (d+e x)}+\frac {b c (e f-d g) \tan ^{-1}\left (\frac {e+c^2 d x}{\sqrt {c^2 d^2-e^2} \sqrt {1-c^2 x^2}}\right )}{e^2 \sqrt {c^2 d^2-e^2}}+\frac {b g \sin ^{-1}(c x) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e^2}+\frac {b g \sin ^{-1}(c x) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e^2}-\frac {b g \sin ^{-1}(c x) \log (d+e x)}{e^2}+\frac {g \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^2}+\frac {(i b g) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {i c e x}{c^2 d-c \sqrt {c^2 d^2-e^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{e^2}+\frac {(i b g) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {i c e x}{c^2 d+c \sqrt {c^2 d^2-e^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{e^2}\\ &=-\frac {i b g \sin ^{-1}(c x)^2}{2 e^2}-\frac {(e f-d g) \left (a+b \sin ^{-1}(c x)\right )}{e^2 (d+e x)}+\frac {b c (e f-d g) \tan ^{-1}\left (\frac {e+c^2 d x}{\sqrt {c^2 d^2-e^2} \sqrt {1-c^2 x^2}}\right )}{e^2 \sqrt {c^2 d^2-e^2}}+\frac {b g \sin ^{-1}(c x) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e^2}+\frac {b g \sin ^{-1}(c x) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e^2}-\frac {b g \sin ^{-1}(c x) \log (d+e x)}{e^2}+\frac {g \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^2}-\frac {i b g \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e^2}-\frac {i b g \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e^2}\\ \end {align*}
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Mathematica [A] time = 0.49, size = 322, normalized size = 0.90 \[ \frac {-\frac {(e f-d g) \left (a+b \sin ^{-1}(c x)\right )}{d+e x}+g \log (d+e x) \left (a+b \sin ^{-1}(c x)\right )+\frac {b c (e f-d g) \tan ^{-1}\left (\frac {c^2 d x+e}{\sqrt {1-c^2 x^2} \sqrt {c^2 d^2-e^2}}\right )}{\sqrt {c^2 d^2-e^2}}-\frac {1}{2} i b g \left (2 \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )+2 \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )+\sin ^{-1}(c x) \left (\sin ^{-1}(c x)+2 i \left (\log \left (1+\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}-c d}\right )+\log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )\right )\right )\right )-b g \sin ^{-1}(c x) \log (d+e x)}{e^2} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a g x + a f + {\left (b g x + b f\right )} \arcsin \left (c x\right )}{e^{2} x^{2} + 2 \, d e x + d^{2}}, x\right ) \]
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 \[ \int \frac {{\left (g x + f\right )} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (e x + d\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 1.42, size = 976, normalized size = 2.73 \[ \frac {c a d g}{e^{2} \left (c e x +d c \right )}-\frac {c a f}{e \left (c e x +d c \right )}+\frac {a g \ln \left (c e x +d c \right )}{e^{2}}+\frac {i b g \dilog \left (\frac {i d c +\left (i c x +\sqrt {-c^{2} x^{2}+1}\right ) e +\sqrt {-c^{2} d^{2}+e^{2}}}{i d c +\sqrt {-c^{2} d^{2}+e^{2}}}\right )}{c^{2} d^{2}-e^{2}}+\frac {c b \arcsin \left (c x \right ) d g}{e^{2} \left (c e x +d c \right )}-\frac {c b \arcsin \left (c x \right ) f}{e \left (c e x +d c \right )}-\frac {2 c b d g \arctan \left (\frac {2 \left (i c x +\sqrt {-c^{2} x^{2}+1}\right ) e +2 i d c}{2 \sqrt {c^{2} d^{2}-e^{2}}}\right )}{e^{2} \sqrt {c^{2} d^{2}-e^{2}}}+\frac {i b g \dilog \left (\frac {i d c +\left (i c x +\sqrt {-c^{2} x^{2}+1}\right ) e -\sqrt {-c^{2} d^{2}+e^{2}}}{i d c -\sqrt {-c^{2} d^{2}+e^{2}}}\right )}{c^{2} d^{2}-e^{2}}+\frac {c^{2} b g \arcsin \left (c x \right ) \ln \left (\frac {-i d c -\left (i c x +\sqrt {-c^{2} x^{2}+1}\right ) e +\sqrt {-c^{2} d^{2}+e^{2}}}{-i d c +\sqrt {-c^{2} d^{2}+e^{2}}}\right ) d^{2}}{e^{2} \left (c^{2} d^{2}-e^{2}\right )}-\frac {i b g \arcsin \left (c x \right )^{2}}{2 e^{2}}-\frac {i c^{2} b g \dilog \left (\frac {i d c +\left (i c x +\sqrt {-c^{2} x^{2}+1}\right ) e -\sqrt {-c^{2} d^{2}+e^{2}}}{i d c -\sqrt {-c^{2} d^{2}+e^{2}}}\right ) d^{2}}{e^{2} \left (c^{2} d^{2}-e^{2}\right )}+\frac {c^{2} b g \arcsin \left (c x \right ) \ln \left (\frac {i d c +\left (i c x +\sqrt {-c^{2} x^{2}+1}\right ) e +\sqrt {-c^{2} d^{2}+e^{2}}}{i d c +\sqrt {-c^{2} d^{2}+e^{2}}}\right ) d^{2}}{e^{2} \left (c^{2} d^{2}-e^{2}\right )}-\frac {i c^{2} b g \dilog \left (\frac {i d c +\left (i c x +\sqrt {-c^{2} x^{2}+1}\right ) e +\sqrt {-c^{2} d^{2}+e^{2}}}{i d c +\sqrt {-c^{2} d^{2}+e^{2}}}\right ) d^{2}}{e^{2} \left (c^{2} d^{2}-e^{2}\right )}+\frac {2 c b f \arctan \left (\frac {2 \left (i c x +\sqrt {-c^{2} x^{2}+1}\right ) e +2 i d c}{2 \sqrt {c^{2} d^{2}-e^{2}}}\right )}{e \sqrt {c^{2} d^{2}-e^{2}}}-\frac {b \ln \left (\frac {-i d c -\left (i c x +\sqrt {-c^{2} x^{2}+1}\right ) e +\sqrt {-c^{2} d^{2}+e^{2}}}{-i d c +\sqrt {-c^{2} d^{2}+e^{2}}}\right ) \arcsin \left (c x \right ) g}{c^{2} d^{2}-e^{2}}-\frac {b \ln \left (\frac {i d c +\left (i c x +\sqrt {-c^{2} x^{2}+1}\right ) e +\sqrt {-c^{2} d^{2}+e^{2}}}{i d c +\sqrt {-c^{2} d^{2}+e^{2}}}\right ) \arcsin \left (c x \right ) g}{c^{2} d^{2}-e^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
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 \[ \int \frac {\left (f+g\,x\right )\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{{\left (d+e\,x\right )}^2} \,d x \]
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 \[ \int \frac {\left (a + b \operatorname {asin}{\left (c x \right )}\right ) \left (f + g x\right )}{\left (d + e x\right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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