Optimal. Leaf size=187 \[ -\frac {b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e^4 (c+d x)^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}-\frac {2 b \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e^4}+\frac {i b^2 \text {Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right )}{3 d e^4}-\frac {i b^2 \text {Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right )}{3 d e^4}-\frac {b^2}{3 d e^4 (c+d x)} \]
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Rubi [A] time = 0.25, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {4805, 12, 4627, 4701, 4709, 4183, 2279, 2391, 30} \[ \frac {i b^2 \text {PolyLog}\left (2,-e^{i \sin ^{-1}(c+d x)}\right )}{3 d e^4}-\frac {i b^2 \text {PolyLog}\left (2,e^{i \sin ^{-1}(c+d x)}\right )}{3 d e^4}-\frac {b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e^4 (c+d x)^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}-\frac {2 b \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e^4}-\frac {b^2}{3 d e^4 (c+d x)} \]
Antiderivative was successfully verified.
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Rule 12
Rule 30
Rule 2279
Rule 2391
Rule 4183
Rule 4627
Rule 4701
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)^4} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^2}{e^4 x^4} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^2}{x^4} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}+\frac {(2 b) \operatorname {Subst}\left (\int \frac {a+b \sin ^{-1}(x)}{x^3 \sqrt {1-x^2}} \, dx,x,c+d x\right )}{3 d e^4}\\ &=-\frac {b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e^4 (c+d x)^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}+\frac {b \operatorname {Subst}\left (\int \frac {a+b \sin ^{-1}(x)}{x \sqrt {1-x^2}} \, dx,x,c+d x\right )}{3 d e^4}+\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,c+d x\right )}{3 d e^4}\\ &=-\frac {b^2}{3 d e^4 (c+d x)}-\frac {b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e^4 (c+d x)^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}+\frac {b \operatorname {Subst}\left (\int (a+b x) \csc (x) \, dx,x,\sin ^{-1}(c+d x)\right )}{3 d e^4}\\ &=-\frac {b^2}{3 d e^4 (c+d x)}-\frac {b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e^4 (c+d x)^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}-\frac {2 b \left (a+b \sin ^{-1}(c+d x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{3 d e^4}-\frac {b^2 \operatorname {Subst}\left (\int \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{3 d e^4}+\frac {b^2 \operatorname {Subst}\left (\int \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{3 d e^4}\\ &=-\frac {b^2}{3 d e^4 (c+d x)}-\frac {b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e^4 (c+d x)^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}-\frac {2 b \left (a+b \sin ^{-1}(c+d x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{3 d e^4}+\frac {\left (i b^2\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{i \sin ^{-1}(c+d x)}\right )}{3 d e^4}-\frac {\left (i b^2\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{i \sin ^{-1}(c+d x)}\right )}{3 d e^4}\\ &=-\frac {b^2}{3 d e^4 (c+d x)}-\frac {b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e^4 (c+d x)^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}-\frac {2 b \left (a+b \sin ^{-1}(c+d x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{3 d e^4}+\frac {i b^2 \text {Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right )}{3 d e^4}-\frac {i b^2 \text {Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right )}{3 d e^4}\\ \end {align*}
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Mathematica [A] time = 2.30, size = 246, normalized size = 1.32 \[ -\frac {4 a^2+8 a b \sin ^{-1}(c+d x)+2 a b \sin \left (2 \sin ^{-1}(c+d x)\right )+a b \left (3 (c+d x)-\sin \left (3 \sin ^{-1}(c+d x)\right )\right ) \left (\log \left (\cos \left (\frac {1}{2} \sin ^{-1}(c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} \sin ^{-1}(c+d x)\right )\right )\right )-4 i b^2 (c+d x)^3 \text {Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right )+b^2 \left (4 i (c+d x)^3 \text {Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right )+4 (c+d x)^2+4 \sin ^{-1}(c+d x)^2+\sin ^{-1}(c+d x) \left (2 \sin \left (2 \sin ^{-1}(c+d x)\right )+\left (\sin \left (3 \sin ^{-1}(c+d x)\right )-3 (c+d x)\right ) \left (\log \left (1-e^{i \sin ^{-1}(c+d x)}\right )-\log \left (1+e^{i \sin ^{-1}(c+d x)}\right )\right )\right )\right )}{12 d e^4 (c+d x)^3} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \arcsin \left (d x + c\right )^{2} + 2 \, a b \arcsin \left (d x + c\right ) + a^{2}}{d^{4} e^{4} x^{4} + 4 \, c d^{3} e^{4} x^{3} + 6 \, c^{2} d^{2} e^{4} x^{2} + 4 \, c^{3} d e^{4} x + c^{4} e^{4}}, x\right ) \]
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 \[ \int \frac {{\left (b \arcsin \left (d x + c\right ) + a\right )}^{2}}{{\left (d e x + c e\right )}^{4}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.43, size = 336, normalized size = 1.80 \[ -\frac {a^{2}}{3 d \,e^{4} \left (d x +c \right )^{3}}-\frac {b^{2} \arcsin \left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}}{3 d \,e^{4} \left (d x +c \right )^{2}}-\frac {b^{2} \arcsin \left (d x +c \right )^{2}}{3 d \,e^{4} \left (d x +c \right )^{3}}-\frac {b^{2}}{3 d \,e^{4} \left (d x +c \right )}-\frac {b^{2} \arcsin \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{3 d \,e^{4}}+\frac {i b^{2} \polylog \left (2, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{3 d \,e^{4}}+\frac {b^{2} \arcsin \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{3 d \,e^{4}}-\frac {i b^{2} \polylog \left (2, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{3 d \,e^{4}}-\frac {2 a b \arcsin \left (d x +c \right )}{3 d \,e^{4} \left (d x +c \right )^{3}}-\frac {a b \sqrt {1-\left (d x +c \right )^{2}}}{3 d \,e^{4} \left (d x +c \right )^{2}}-\frac {a b \arctanh \left (\frac {1}{\sqrt {1-\left (d x +c \right )^{2}}}\right )}{3 d \,e^{4}} \]
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 \[ -\frac {a^{2}}{3 \, {\left (d^{4} e^{4} x^{3} + 3 \, c d^{3} e^{4} x^{2} + 3 \, c^{2} d^{2} e^{4} x + c^{3} d e^{4}\right )}} - \frac {b^{2} \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )^{2} + 2 \, {\left (d^{4} e^{4} x^{3} + 3 \, c d^{3} e^{4} x^{2} + 3 \, c^{2} d^{2} e^{4} x + c^{3} d e^{4}\right )} \int \frac {{\left (b^{2} d x + b^{2} c\right )} \sqrt {d x + c + 1} \sqrt {-d x - c + 1} \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right ) - 3 \, {\left (a b d^{2} x^{2} + 2 \, a b c d x + a b c^{2} - a b\right )} \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )}{d^{6} e^{4} x^{6} + 6 \, c d^{5} e^{4} x^{5} + {\left (15 \, c^{2} - 1\right )} d^{4} e^{4} x^{4} + 4 \, {\left (5 \, c^{3} - c\right )} d^{3} e^{4} x^{3} + 3 \, {\left (5 \, c^{4} - 2 \, c^{2}\right )} d^{2} e^{4} x^{2} + 2 \, {\left (3 \, c^{5} - 2 \, c^{3}\right )} d e^{4} x + {\left (c^{6} - c^{4}\right )} e^{4}}\,{d x}}{3 \, {\left (d^{4} e^{4} x^{3} + 3 \, c d^{3} e^{4} x^{2} + 3 \, c^{2} d^{2} e^{4} x + c^{3} d e^{4}\right )}} \]
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 \[ \int \frac {{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^2}{{\left (c\,e+d\,e\,x\right )}^4} \,d x \]
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 \[ \frac {\int \frac {a^{2}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {b^{2} \operatorname {asin}^{2}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {2 a b \operatorname {asin}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx}{e^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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