Optimal. Leaf size=102 \[ \frac{4 d (c+d x) \log \left (1-e^{i (e+f x)}\right )}{a f^2}-\frac{(c+d x)^2 \cot \left (\frac{e}{2}+\frac{f x}{2}\right )}{a f}-\frac{i (c+d x)^2}{a f}-\frac{4 i d^2 \text{Li}_2\left (e^{i (e+f x)}\right )}{a f^3} \]
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Rubi [A] time = 0.202976, antiderivative size = 102, normalized size of antiderivative = 1., 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} \[ \frac{4 d (c+d x) \log \left (1-e^{i (e+f x)}\right )}{a f^2}-\frac{(c+d x)^2 \cot \left (\frac{e}{2}+\frac{f x}{2}\right )}{a f}-\frac{i (c+d x)^2}{a f}-\frac{4 i d^2 \text{Li}_2\left (e^{i (e+f x)}\right )}{a f^3} \]
Antiderivative was successfully verified.
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Rule 3318
Rule 4184
Rule 3717
Rule 2190
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int \frac{(c+d x)^2}{a-a \cos (e+f x)} \, dx &=\frac{\int (c+d x)^2 \csc ^2\left (\frac{e}{2}+\frac{f x}{2}\right ) \, dx}{2 a}\\ &=-\frac{(c+d x)^2 \cot \left (\frac{e}{2}+\frac{f x}{2}\right )}{a f}+\frac{(2 d) \int (c+d x) \cot \left (\frac{e}{2}+\frac{f x}{2}\right ) \, dx}{a f}\\ &=-\frac{i (c+d x)^2}{a f}-\frac{(c+d x)^2 \cot \left (\frac{e}{2}+\frac{f x}{2}\right )}{a f}-\frac{(4 i d) \int \frac{e^{2 i \left (\frac{e}{2}+\frac{f x}{2}\right )} (c+d x)}{1-e^{2 i \left (\frac{e}{2}+\frac{f x}{2}\right )}} \, dx}{a f}\\ &=-\frac{i (c+d x)^2}{a f}-\frac{(c+d x)^2 \cot \left (\frac{e}{2}+\frac{f x}{2}\right )}{a f}+\frac{4 d (c+d x) \log \left (1-e^{i (e+f x)}\right )}{a f^2}-\frac{\left (4 d^2\right ) \int \log \left (1-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{(c+d x)^2 \cot \left (\frac{e}{2}+\frac{f x}{2}\right )}{a f}+\frac{4 d (c+d x) \log \left (1-e^{i (e+f x)}\right )}{a f^2}+\frac{\left (4 i d^2\right ) \operatorname{Subst}\left (\int \frac{\log (1-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{(c+d x)^2 \cot \left (\frac{e}{2}+\frac{f x}{2}\right )}{a f}+\frac{4 d (c+d x) \log \left (1-e^{i (e+f x)}\right )}{a f^2}-\frac{4 i d^2 \text{Li}_2\left (e^{i (e+f x)}\right )}{a f^3}\\ \end{align*}
Mathematica [B] time = 5.48236, size = 292, normalized size = 2.86 \[ \frac{2 \csc \left (\frac{e}{2}\right ) \sin \left (\frac{1}{2} (e+f x)\right ) \left (-2 c d f \sin \left (\frac{1}{2} (e+f x)\right ) \left (f x \cos \left (\frac{e}{2}\right )-2 \sin \left (\frac{e}{2}\right ) \log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right )\right )+f^2 (c+d x)^2 \sin \left (\frac{f x}{2}\right )+d^2 \sin \left (\frac{1}{2} (e+f x)\right ) \left (f^2 x^2 \cos \left (\frac{e}{2}\right ) \left (-e^{i \tan ^{-1}\left (\tan \left (\frac{e}{2}\right )\right )}\right ) \sqrt{\sec ^2\left (\frac{e}{2}\right )}-4 \sin \left (\frac{e}{2}\right ) \left (i \text{Li}_2\left (e^{i \left (f x+2 \tan ^{-1}\left (\tan \left (\frac{e}{2}\right )\right )\right )}\right )-\frac{1}{2} i f x \left (\pi -2 \tan ^{-1}\left (\tan \left (\frac{e}{2}\right )\right )\right )-\left (2 \tan ^{-1}\left (\tan \left (\frac{e}{2}\right )\right )+f x\right ) \log \left (1-e^{i \left (2 \tan ^{-1}\left (\tan \left (\frac{e}{2}\right )\right )+f x\right )}\right )+2 \tan ^{-1}\left (\tan \left (\frac{e}{2}\right )\right ) \log \left (\sin \left (\tan ^{-1}\left (\tan \left (\frac{e}{2}\right )\right )+\frac{f x}{2}\right )\right )-\pi \log \left (1+e^{-i f x}\right )+\pi \log \left (\cos \left (\frac{f x}{2}\right )\right )\right )\right )\right )}{f^3 (a-a \cos (e+f x))} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.398, size = 247, normalized size = 2.4 \begin{align*}{\frac{-2\,i \left ({d}^{2}{x}^{2}+2\,cdx+{c}^{2} \right ) }{af \left ({{\rm e}^{i \left ( fx+e \right ) }}-1 \right ) }}-4\,{\frac{cd\ln \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{a{f}^{2}}}+4\,{\frac{cd\ln \left ({{\rm e}^{i \left ( fx+e \right ) }}-1 \right ) }{a{f}^{2}}}-{\frac{2\,i{d}^{2}{x}^{2}}{af}}-{\frac{4\,i{d}^{2}ex}{a{f}^{2}}}-{\frac{2\,i{d}^{2}{e}^{2}}{{f}^{3}a}}+4\,{\frac{{d}^{2}\ln \left ( 1-{{\rm e}^{i \left ( fx+e \right ) }} \right ) x}{a{f}^{2}}}+4\,{\frac{{d}^{2}\ln \left ( 1-{{\rm e}^{i \left ( fx+e \right ) }} \right ) e}{{f}^{3}a}}-{\frac{4\,i{d}^{2}{\it polylog} \left ( 2,{{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{3}a}}+4\,{\frac{{d}^{2}e\ln \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{3}a}}-4\,{\frac{{d}^{2}e\ln \left ({{\rm e}^{i \left ( fx+e \right ) }}-1 \right ) }{{f}^{3}a}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.95884, size = 425, normalized size = 4.17 \begin{align*} -\frac{2 \, 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 \, c d f\right )} \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right ) - 1\right ) + 4 \,{\left (d^{2} f x \cos \left (f x + e\right ) + i \, d^{2} f x \sin \left (f x + e\right ) - d^{2} f x\right )} \arctan \left (\sin \left (f x + e\right ), -\cos \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 ) + 4 \,{\left (d^{2} \cos \left (f x + e\right ) + i \, d^{2} \sin \left (f x + e\right ) - d^{2}\right )}{\rm Li}_2\left (e^{\left (i \, f x + i \, e\right )}\right ) -{\left (2 i \, d^{2} f x + 2 i \, 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 \, \cos \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 ) + i \, a f^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.71963, size = 743, normalized size = 7.28 \begin{align*} -\frac{d^{2} f^{2} x^{2} + 2 \, c d f^{2} x + c^{2} f^{2} + 2 i \, d^{2}{\rm Li}_2\left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) \sin \left (f x + e\right ) - 2 i \, d^{2}{\rm Li}_2\left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) \sin \left (f x + e\right ) + 2 \,{\left (d^{2} e - c d f\right )} \log \left (-\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2} i \, \sin \left (f x + e\right ) + \frac{1}{2}\right ) \sin \left (f x + e\right ) + 2 \,{\left (d^{2} e - c d f\right )} \log \left (-\frac{1}{2} \, \cos \left (f x + e\right ) - \frac{1}{2} i \, \sin \left (f x + e\right ) + \frac{1}{2}\right ) \sin \left (f x + e\right ) - 2 \,{\left (d^{2} f x + d^{2} e\right )} \log \left (-\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) - 2 \,{\left (d^{2} f x + d^{2} e\right )} \log \left (-\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) +{\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x + c^{2} f^{2}\right )} \cos \left (f x + e\right )}{a f^{3} \sin \left (f x + e\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{\int \frac{c^{2}}{\cos{\left (e + f x \right )} - 1}\, dx + \int \frac{d^{2} x^{2}}{\cos{\left (e + f x \right )} - 1}\, dx + \int \frac{2 c d x}{\cos{\left (e + f x \right )} - 1}\, dx}{a} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{{\left (d x + c\right )}^{2}}{a \cos \left (f x + e\right ) - a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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