Optimal. Leaf size=671 \[ -\frac {2 a d (c+d x) \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f^2 \left (a^2-b^2\right )^{3/2}}+\frac {2 a d (c+d x) \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{f^2 \left (a^2-b^2\right )^{3/2}}-\frac {2 d (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f^2 \left (a^2-b^2\right )}-\frac {2 d (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{f^2 \left (a^2-b^2\right )}-\frac {i a (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{3/2}}+\frac {i a (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{f \left (a^2-b^2\right )^{3/2}}+\frac {b (c+d x)^2 \cos (e+f x)}{f \left (a^2-b^2\right ) (a+b \sin (e+f x))}+\frac {i (c+d x)^2}{f \left (a^2-b^2\right )}+\frac {2 i d^2 \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f^3 \left (a^2-b^2\right )}+\frac {2 i d^2 \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{f^3 \left (a^2-b^2\right )}-\frac {2 i a d^2 \text {Li}_3\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f^3 \left (a^2-b^2\right )^{3/2}}+\frac {2 i a d^2 \text {Li}_3\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{f^3 \left (a^2-b^2\right )^{3/2}} \]
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Rubi [A] time = 1.21, antiderivative size = 671, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3324, 3323, 2264, 2190, 2531, 2282, 6589, 4519, 2279, 2391} \[ -\frac {2 a d (c+d x) \text {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f^2 \left (a^2-b^2\right )^{3/2}}+\frac {2 a d (c+d x) \text {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{f^2 \left (a^2-b^2\right )^{3/2}}+\frac {2 i d^2 \text {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f^3 \left (a^2-b^2\right )}+\frac {2 i d^2 \text {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{f^3 \left (a^2-b^2\right )}-\frac {2 i a d^2 \text {PolyLog}\left (3,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f^3 \left (a^2-b^2\right )^{3/2}}+\frac {2 i a d^2 \text {PolyLog}\left (3,\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{f^3 \left (a^2-b^2\right )^{3/2}}-\frac {2 d (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f^2 \left (a^2-b^2\right )}-\frac {2 d (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{f^2 \left (a^2-b^2\right )}-\frac {i a (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{3/2}}+\frac {i a (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{f \left (a^2-b^2\right )^{3/2}}+\frac {b (c+d x)^2 \cos (e+f x)}{f \left (a^2-b^2\right ) (a+b \sin (e+f x))}+\frac {i (c+d x)^2}{f \left (a^2-b^2\right )} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2264
Rule 2279
Rule 2282
Rule 2391
Rule 2531
Rule 3323
Rule 3324
Rule 4519
Rule 6589
Rubi steps
\begin {align*} \int \frac {(c+d x)^2}{(a+b \sin (e+f x))^2} \, dx &=\frac {b (c+d x)^2 \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\frac {a \int \frac {(c+d x)^2}{a+b \sin (e+f x)} \, dx}{a^2-b^2}-\frac {(2 b d) \int \frac {(c+d x) \cos (e+f x)}{a+b \sin (e+f x)} \, dx}{\left (a^2-b^2\right ) f}\\ &=\frac {i (c+d x)^2}{\left (a^2-b^2\right ) f}+\frac {b (c+d x)^2 \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\frac {(2 a) \int \frac {e^{i (e+f x)} (c+d x)^2}{i b+2 a e^{i (e+f x)}-i b e^{2 i (e+f x)}} \, dx}{a^2-b^2}-\frac {(2 b d) \int \frac {e^{i (e+f x)} (c+d x)}{a-\sqrt {a^2-b^2}-i b e^{i (e+f x)}} \, dx}{\left (a^2-b^2\right ) f}-\frac {(2 b d) \int \frac {e^{i (e+f x)} (c+d x)}{a+\sqrt {a^2-b^2}-i b e^{i (e+f x)}} \, dx}{\left (a^2-b^2\right ) f}\\ &=\frac {i (c+d x)^2}{\left (a^2-b^2\right ) f}-\frac {2 d (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^2}-\frac {2 d (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^2}+\frac {b (c+d x)^2 \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}-\frac {(2 i a b) \int \frac {e^{i (e+f x)} (c+d x)^2}{2 a-2 \sqrt {a^2-b^2}-2 i b e^{i (e+f x)}} \, dx}{\left (a^2-b^2\right )^{3/2}}+\frac {(2 i a b) \int \frac {e^{i (e+f x)} (c+d x)^2}{2 a+2 \sqrt {a^2-b^2}-2 i b e^{i (e+f x)}} \, dx}{\left (a^2-b^2\right )^{3/2}}+\frac {\left (2 d^2\right ) \int \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right ) f^2}+\frac {\left (2 d^2\right ) \int \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right ) f^2}\\ &=\frac {i (c+d x)^2}{\left (a^2-b^2\right ) f}-\frac {2 d (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^2}-\frac {i a (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}-\frac {2 d (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^2}+\frac {i a (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}+\frac {b (c+d x)^2 \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}-\frac {\left (2 i d^2\right ) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {i b x}{a-\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (e+f x)}\right )}{\left (a^2-b^2\right ) f^3}-\frac {\left (2 i d^2\right ) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {i b x}{a+\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (e+f x)}\right )}{\left (a^2-b^2\right ) f^3}+\frac {(2 i a d) \int (c+d x) \log \left (1-\frac {2 i b e^{i (e+f x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right )^{3/2} f}-\frac {(2 i a d) \int (c+d x) \log \left (1-\frac {2 i b e^{i (e+f x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right )^{3/2} f}\\ &=\frac {i (c+d x)^2}{\left (a^2-b^2\right ) f}-\frac {2 d (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^2}-\frac {i a (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}-\frac {2 d (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^2}+\frac {i a (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}+\frac {2 i d^2 \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^3}-\frac {2 a d (c+d x) \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^2}+\frac {2 i d^2 \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^3}+\frac {2 a d (c+d x) \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^2}+\frac {b (c+d x)^2 \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\frac {\left (2 a d^2\right ) \int \text {Li}_2\left (\frac {2 i b e^{i (e+f x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right )^{3/2} f^2}-\frac {\left (2 a d^2\right ) \int \text {Li}_2\left (\frac {2 i b e^{i (e+f x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right )^{3/2} f^2}\\ &=\frac {i (c+d x)^2}{\left (a^2-b^2\right ) f}-\frac {2 d (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^2}-\frac {i a (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}-\frac {2 d (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^2}+\frac {i a (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}+\frac {2 i d^2 \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^3}-\frac {2 a d (c+d x) \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^2}+\frac {2 i d^2 \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^3}+\frac {2 a d (c+d x) \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^2}+\frac {b (c+d x)^2 \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}-\frac {\left (2 i a d^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {i b x}{a-\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (e+f x)}\right )}{\left (a^2-b^2\right )^{3/2} f^3}+\frac {\left (2 i a d^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {i b x}{a+\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (e+f x)}\right )}{\left (a^2-b^2\right )^{3/2} f^3}\\ &=\frac {i (c+d x)^2}{\left (a^2-b^2\right ) f}-\frac {2 d (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^2}-\frac {i a (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}-\frac {2 d (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^2}+\frac {i a (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}+\frac {2 i d^2 \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^3}-\frac {2 a d (c+d x) \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^2}+\frac {2 i d^2 \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^3}+\frac {2 a d (c+d x) \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^2}-\frac {2 i a d^2 \text {Li}_3\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^3}+\frac {2 i a d^2 \text {Li}_3\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^3}+\frac {b (c+d x)^2 \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}\\ \end {align*}
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Mathematica [A] time = 1.77, size = 530, normalized size = 0.79 \[ \frac {-\frac {i a \left (f^2 (c+d x)^2 \log \left (1+\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}-a}\right )-f^2 (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )-2 i d f (c+d x) \text {Li}_2\left (-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}-a}\right )+2 i d f (c+d x) \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )+2 d^2 \text {Li}_3\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )-2 d^2 \text {Li}_3\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )\right )}{\sqrt {a^2-b^2}}-2 d f (c+d x) \log \left (1+\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}-a}\right )-2 d f (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )+2 i d^2 \text {Li}_2\left (-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}-a}\right )+2 i d^2 \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )+\frac {b f^2 (c+d x)^2 \cos (e+f x)}{a+b \sin (e+f x)}+i f^2 (c+d x)^2}{f^3 \left (a^2-b^2\right )} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.76, size = 3113, normalized size = 4.64 \[ \text {result too large to display} \]
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 \[ \int \frac {{\left (d x + c\right )}^{2}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 2.33, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x +c \right )^{2}}{\left (a +b \sin \left (f x +e \right )\right )^{2}}\, dx \]
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(-1)] time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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