Optimal. Leaf size=589 \[ -\frac {2 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {d-c^2 d x^2} \sqrt {c^2 f^2-g^2}}+\frac {2 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {d-c^2 d x^2} \sqrt {c^2 f^2-g^2}}-\frac {i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {d-c^2 d x^2} \sqrt {c^2 f^2-g^2}}+\frac {i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{\sqrt {d-c^2 d x^2} \sqrt {c^2 f^2-g^2}}-\frac {2 i b^2 \sqrt {1-c^2 x^2} \text {Li}_3\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {d-c^2 d x^2} \sqrt {c^2 f^2-g^2}}+\frac {2 i b^2 \sqrt {1-c^2 x^2} \text {Li}_3\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {d-c^2 d x^2} \sqrt {c^2 f^2-g^2}} \]
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Rubi [A] time = 1.02, antiderivative size = 589, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {4777, 4773, 3323, 2264, 2190, 2531, 2282, 6589} \[ -\frac {2 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text {PolyLog}\left (2,\frac {i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {d-c^2 d x^2} \sqrt {c^2 f^2-g^2}}+\frac {2 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text {PolyLog}\left (2,\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{\sqrt {d-c^2 d x^2} \sqrt {c^2 f^2-g^2}}-\frac {2 i b^2 \sqrt {1-c^2 x^2} \text {PolyLog}\left (3,\frac {i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {d-c^2 d x^2} \sqrt {c^2 f^2-g^2}}+\frac {2 i b^2 \sqrt {1-c^2 x^2} \text {PolyLog}\left (3,\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{\sqrt {d-c^2 d x^2} \sqrt {c^2 f^2-g^2}}-\frac {i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {d-c^2 d x^2} \sqrt {c^2 f^2-g^2}}+\frac {i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{\sqrt {d-c^2 d x^2} \sqrt {c^2 f^2-g^2}} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2264
Rule 2282
Rule 2531
Rule 3323
Rule 4773
Rule 4777
Rule 6589
Rubi steps
\begin {align*} \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{(f+g x) \sqrt {d-c^2 d x^2}} \, dx &=\frac {\sqrt {1-c^2 x^2} \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{(f+g x) \sqrt {1-c^2 x^2}} \, dx}{\sqrt {d-c^2 d x^2}}\\ &=\frac {\sqrt {1-c^2 x^2} \operatorname {Subst}\left (\int \frac {(a+b x)^2}{c f+g \sin (x)} \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt {d-c^2 d x^2}}\\ &=\frac {\left (2 \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {e^{i x} (a+b x)^2}{2 c e^{i x} f+i g-i e^{2 i x} g} \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt {d-c^2 d x^2}}\\ &=-\frac {\left (2 i g \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {e^{i x} (a+b x)^2}{2 c f-2 i e^{i x} g-2 \sqrt {c^2 f^2-g^2}} \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}+\frac {\left (2 i g \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {e^{i x} (a+b x)^2}{2 c f-2 i e^{i x} g+2 \sqrt {c^2 f^2-g^2}} \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}\\ &=-\frac {i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}+\frac {i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}+\frac {\left (2 i b \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int (a+b x) \log \left (1-\frac {2 i e^{i x} g}{2 c f-2 \sqrt {c^2 f^2-g^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}-\frac {\left (2 i b \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int (a+b x) \log \left (1-\frac {2 i e^{i x} g}{2 c f+2 \sqrt {c^2 f^2-g^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}\\ &=-\frac {i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}+\frac {i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}-\frac {2 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}+\frac {2 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}+\frac {\left (2 b^2 \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (\frac {2 i e^{i x} g}{2 c f-2 \sqrt {c^2 f^2-g^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}-\frac {\left (2 b^2 \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (\frac {2 i e^{i x} g}{2 c f+2 \sqrt {c^2 f^2-g^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}\\ &=-\frac {i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}+\frac {i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}-\frac {2 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}+\frac {2 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}-\frac {\left (2 i b^2 \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {i g x}{c f-\sqrt {c^2 f^2-g^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{\sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}+\frac {\left (2 i b^2 \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {i g x}{c f+\sqrt {c^2 f^2-g^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{\sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}\\ &=-\frac {i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}+\frac {i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}-\frac {2 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}+\frac {2 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}-\frac {2 i b^2 \sqrt {1-c^2 x^2} \text {Li}_3\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}+\frac {2 i b^2 \sqrt {1-c^2 x^2} \text {Li}_3\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}\\ \end {align*}
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Mathematica [A] time = 0.35, size = 357, normalized size = 0.61 \[ -\frac {i \sqrt {1-c^2 x^2} \left (-2 i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )+2 i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )+\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1+\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}-c f}\right )-\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )+2 b^2 \text {Li}_3\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )-2 b^2 \text {Li}_3\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )\right )}{\sqrt {d-c^2 d x^2} \sqrt {c^2 f^2-g^2}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-c^{2} d x^{2} + d} {\left (b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}\right )}}{c^{2} d g x^{3} + c^{2} d f x^{2} - d g x - d f}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.09, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arcsin \left (c x \right )\right )^{2}}{\left (g x +f \right ) \sqrt {-c^{2} d \,x^{2}+d}}\, dx \]
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 \[ \int \frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{\sqrt {-c^{2} d x^{2} + d} {\left (g x + f\right )}}\,{d x} \]
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 \[ \int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{\left (f+g\,x\right )\,\sqrt {d-c^2\,d\,x^2}} \,d x \]
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 \[ \int \frac {\left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{\sqrt {- d \left (c x - 1\right ) \left (c x + 1\right )} \left (f + g x\right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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