Optimal. Leaf size=116 \[ -\frac {\left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}-\frac {4 b \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^2}+\frac {2 i b^2 \text {Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac {2 i b^2 \text {Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2} \]
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Rubi [A] time = 0.16, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {4805, 12, 4627, 4709, 4183, 2279, 2391} \[ \frac {2 i b^2 \text {PolyLog}\left (2,-e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac {2 i b^2 \text {PolyLog}\left (2,e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}-\frac {4 b \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^2} \]
Antiderivative was successfully verified.
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Rule 12
Rule 2279
Rule 2391
Rule 4183
Rule 4627
Rule 4709
Rule 4805
Rubi steps
\begin {align*} \int \frac {\left (a+b \sin ^{-1}(c+d x)\right )^2}{(c e+d e x)^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^2}{e^2 x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^2}{x^2} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}+\frac {(2 b) \operatorname {Subst}\left (\int \frac {a+b \sin ^{-1}(x)}{x \sqrt {1-x^2}} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}+\frac {(2 b) \operatorname {Subst}\left (\int (a+b x) \csc (x) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^2}\\ &=-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}-\frac {4 b \left (a+b \sin ^{-1}(c+d x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^2}+\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^2}\\ &=-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}-\frac {4 b \left (a+b \sin ^{-1}(c+d x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}+\frac {\left (2 i b^2\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac {\left (2 i b^2\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}\\ &=-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}-\frac {4 b \left (a+b \sin ^{-1}(c+d x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}+\frac {2 i b^2 \text {Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac {2 i b^2 \text {Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}\\ \end {align*}
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Mathematica [A] time = 0.64, size = 176, normalized size = 1.52 \[ \frac {-\frac {a^2}{c+d x}-2 a b \left (\frac {\sin ^{-1}(c+d x)}{c+d x}-\log \left (\sin \left (\frac {1}{2} \sin ^{-1}(c+d x)\right )\right )+\log \left (\frac {1}{2} (c+d x) \csc \left (\frac {1}{2} \sin ^{-1}(c+d x)\right )\right )\right )+b^2 \left (2 i \text {Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right )-2 i \text {Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right )+\sin ^{-1}(c+d x) \left (-\frac {\sin ^{-1}(c+d x)}{c+d x}+2 \log \left (1-e^{i \sin ^{-1}(c+d x)}\right )-2 \log \left (1+e^{i \sin ^{-1}(c+d x)}\right )\right )\right )}{d e^2} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \arcsin \left (d x + c\right )^{2} + 2 \, a b \arcsin \left (d x + c\right ) + a^{2}}{d^{2} e^{2} x^{2} + 2 \, c d e^{2} x + c^{2} e^{2}}, x\right ) \]
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 \[ \int \frac {{\left (b \arcsin \left (d x + c\right ) + a\right )}^{2}}{{\left (d e x + c e\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.09, size = 251, normalized size = 2.16 \[ -\frac {a^{2}}{d \,e^{2} \left (d x +c \right )}-\frac {b^{2} \arcsin \left (d x +c \right )^{2}}{d \,e^{2} \left (d x +c \right )}+\frac {2 b^{2} \arcsin \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{d \,e^{2}}-\frac {2 b^{2} \arcsin \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{d \,e^{2}}+\frac {2 i b^{2} \dilog \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{d \,e^{2}}-\frac {2 i b^{2} \dilog \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{d \,e^{2}}-\frac {2 a b \arcsin \left (d x +c \right )}{d \,e^{2} \left (d x +c \right )}-\frac {2 a b \arctanh \left (\frac {1}{\sqrt {1-\left (d x +c \right )^{2}}}\right )}{d \,e^{2}} \]
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 \[ -2 \, a b {\left (\frac {\arcsin \left (d x + c\right )}{d^{2} e^{2} x + c d e^{2}} + \frac {\log \left (\frac {2 \, \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1}}{{\left | d^{2} e^{2} x + c d e^{2} \right |}} + \frac {2}{{\left | d^{2} e^{2} x + c d e^{2} \right |}}\right )}{d e^{2}}\right )} - \frac {{\left (\arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )^{2} + \frac {2 \, {\left (d^{2} e^{2} x + c d e^{2}\right )} \int \frac {\sqrt {-d x - c + 1} \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )}{\sqrt {d x + c + 1} {\left (d x + c\right )} {\left (d x + c - 1\right )}}\,{d x}}{e^{2}}\right )} b^{2}}{d^{2} e^{2} x + c d e^{2}} - \frac {a^{2}}{d^{2} e^{2} x + c d e^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^2}{{\left (c\,e+d\,e\,x\right )}^2} \,d x \]
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 \[ \frac {\int \frac {a^{2}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {b^{2} \operatorname {asin}^{2}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {2 a b \operatorname {asin}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx}{e^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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