Optimal. Leaf size=112 \[ \frac {4 d (c+d x) \log \left (1+i e^{i (e+f x)}\right )}{a f^2}+\frac {(c+d x)^2 \tan \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{a f}-\frac {i (c+d x)^2}{a f}-\frac {4 i d^2 \text {Li}_2\left (-i e^{i (e+f x)}\right )}{a f^3} \]
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Rubi [A] time = 0.21, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3318, 4184, 3717, 2190, 2279, 2391} \[ -\frac {4 i d^2 \text {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{a f^3}+\frac {4 d (c+d x) \log \left (1+i e^{i (e+f x)}\right )}{a f^2}+\frac {(c+d x)^2 \tan \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{a f}-\frac {i (c+d x)^2}{a f} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2279
Rule 2391
Rule 3318
Rule 3717
Rule 4184
Rubi steps
\begin {align*} \int \frac {(c+d x)^2}{a-a \sin (e+f x)} \, dx &=\frac {\int (c+d x)^2 \csc ^2\left (\frac {1}{2} \left (e-\frac {\pi }{2}\right )+\frac {f x}{2}\right ) \, dx}{2 a}\\ &=\frac {(c+d x)^2 \tan \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{a f}+\frac {(2 d) \int (c+d x) \cot \left (\frac {e}{2}-\frac {\pi }{4}+\frac {f x}{2}\right ) \, dx}{a f}\\ &=-\frac {i (c+d x)^2}{a f}+\frac {(c+d x)^2 \tan \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{a f}-\frac {(4 d) \int \frac {e^{2 i \left (\frac {e}{2}+\frac {f x}{2}\right )} (c+d x)}{1+i e^{2 i \left (\frac {e}{2}+\frac {f x}{2}\right )}} \, dx}{a f}\\ &=-\frac {i (c+d x)^2}{a f}+\frac {4 d (c+d x) \log \left (1+i e^{i (e+f x)}\right )}{a f^2}+\frac {(c+d x)^2 \tan \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{a f}-\frac {\left (4 d^2\right ) \int \log \left (1+i e^{2 i \left (\frac {e}{2}+\frac {f x}{2}\right )}\right ) \, dx}{a f^2}\\ &=-\frac {i (c+d x)^2}{a f}+\frac {4 d (c+d x) \log \left (1+i e^{i (e+f x)}\right )}{a f^2}+\frac {(c+d x)^2 \tan \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{a f}+\frac {\left (4 i d^2\right ) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{2 i \left (\frac {e}{2}+\frac {f x}{2}\right )}\right )}{a f^3}\\ &=-\frac {i (c+d x)^2}{a f}+\frac {4 d (c+d x) \log \left (1+i e^{i (e+f x)}\right )}{a f^2}-\frac {4 i d^2 \text {Li}_2\left (-i e^{i (e+f x)}\right )}{a f^3}+\frac {(c+d x)^2 \tan \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{a f}\\ \end {align*}
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Mathematica [A] time = 0.80, size = 92, normalized size = 0.82 \[ \frac {f (c+d x) \left (f (c+d x) \tan \left (\frac {1}{4} (2 e+2 f x+\pi )\right )-i f (c+d x)+4 d \log \left (1+i e^{i (e+f x)}\right )\right )-4 i d^2 \text {Li}_2\left (-i e^{i (e+f x)}\right )}{a f^3} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.55, size = 496, normalized size = 4.43 \[ \frac {d^{2} f^{2} x^{2} + 2 \, c d f^{2} x + c^{2} f^{2} + {\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x + c^{2} f^{2}\right )} \cos \left (f x + e\right ) - {\left (-2 i \, d^{2} \cos \left (f x + e\right ) + 2 i \, d^{2} \sin \left (f x + e\right ) - 2 i \, d^{2}\right )} {\rm Li}_2\left (i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right )\right ) - {\left (2 i \, d^{2} \cos \left (f x + e\right ) - 2 i \, d^{2} \sin \left (f x + e\right ) + 2 i \, d^{2}\right )} {\rm Li}_2\left (-i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right )\right ) - 2 \, {\left (d^{2} e - c d f + {\left (d^{2} e - c d f\right )} \cos \left (f x + e\right ) - {\left (d^{2} e - c d f\right )} \sin \left (f x + e\right )\right )} \log \left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right ) + i\right ) + 2 \, {\left (d^{2} f x + d^{2} e + {\left (d^{2} f x + d^{2} e\right )} \cos \left (f x + e\right ) - {\left (d^{2} f x + d^{2} e\right )} \sin \left (f x + e\right )\right )} \log \left (i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (d^{2} f x + d^{2} e + {\left (d^{2} f x + d^{2} e\right )} \cos \left (f x + e\right ) - {\left (d^{2} f x + d^{2} e\right )} \sin \left (f x + e\right )\right )} \log \left (-i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (d^{2} e - c d f + {\left (d^{2} e - c d f\right )} \cos \left (f x + e\right ) - {\left (d^{2} e - c d f\right )} \sin \left (f x + e\right )\right )} \log \left (-\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right ) + i\right ) + {\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x + c^{2} f^{2}\right )} \sin \left (f x + e\right )}{a f^{3} \cos \left (f x + e\right ) - a f^{3} \sin \left (f x + e\right ) + a f^{3}} \]
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 \[ \int -\frac {{\left (d x + c\right )}^{2}}{a \sin \left (f x + e\right ) - a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.14, size = 254, normalized size = 2.27 \[ \frac {2 d^{2} x^{2}+4 c d x +2 c^{2}}{f a \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}+\frac {4 \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) c d}{a \,f^{2}}-\frac {4 \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right ) c d}{a \,f^{2}}-\frac {2 i d^{2} x^{2}}{a f}-\frac {4 i d^{2} e x}{a \,f^{2}}-\frac {2 i d^{2} e^{2}}{a \,f^{3}}+\frac {4 d^{2} \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{a \,f^{2}}+\frac {4 d^{2} \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right ) e}{a \,f^{3}}-\frac {4 i d^{2} \polylog \left (2, -i {\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{3}}-\frac {4 d^{2} e \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{a \,f^{3}}+\frac {4 d^{2} e \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.51, size = 316, normalized size = 2.82 \[ \frac {-2 i \, c^{2} f^{2} + {\left (4 \, c d f \cos \left (f x + e\right ) + 4 i \, c d f \sin \left (f x + e\right ) - 4 i \, c d f\right )} \arctan \left (\sin \left (f x + e\right ) - 1, \cos \left (f x + e\right )\right ) + {\left (4 \, d^{2} f x \cos \left (f x + e\right ) + 4 i \, d^{2} f x \sin \left (f x + e\right ) - 4 i \, d^{2} f x\right )} \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x\right )} \cos \left (f x + e\right ) - {\left (4 \, d^{2} \cos \left (f x + e\right ) + 4 i \, d^{2} \sin \left (f x + e\right ) - 4 i \, d^{2}\right )} {\rm Li}_2\left (-i \, e^{\left (i \, f x + i \, e\right )}\right ) - {\left (2 \, d^{2} f x + 2 \, c d f - {\left (-2 i \, d^{2} f x - 2 i \, c d f\right )} \cos \left (f x + e\right ) - 2 \, {\left (d^{2} f x + c d f\right )} \sin \left (f x + e\right )\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right ) + {\left (-2 i \, d^{2} f^{2} x^{2} - 4 i \, c d f^{2} x\right )} \sin \left (f x + e\right )}{-i \, a f^{3} \cos \left (f x + e\right ) + a f^{3} \sin \left (f x + e\right ) - a f^{3}} \]
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 \[ \int \frac {{\left (c+d\,x\right )}^2}{a-a\,\sin \left (e+f\,x\right )} \,d x \]
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 \[ - \frac {\int \frac {c^{2}}{\sin {\left (e + f x \right )} - 1}\, dx + \int \frac {d^{2} x^{2}}{\sin {\left (e + f x \right )} - 1}\, dx + \int \frac {2 c d x}{\sin {\left (e + f x \right )} - 1}\, dx}{a} \]
Verification of antiderivative is not currently implemented for this CAS.
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