Optimal. Leaf size=212 \[ \frac {4 d (c+d x) \log \left (1+e^{i (e+f x)}\right )}{3 a^2 f^2}-\frac {d (c+d x) \sec ^2\left (\frac {e}{2}+\frac {f x}{2}\right )}{3 a^2 f^2}+\frac {(c+d x)^2 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{3 a^2 f}+\frac {(c+d x)^2 \tan \left (\frac {e}{2}+\frac {f x}{2}\right ) \sec ^2\left (\frac {e}{2}+\frac {f x}{2}\right )}{6 a^2 f}-\frac {i (c+d x)^2}{3 a^2 f}-\frac {4 i d^2 \text {Li}_2\left (-e^{i (e+f x)}\right )}{3 a^2 f^3}+\frac {2 d^2 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{3 a^2 f^3} \]
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Rubi [A] time = 0.26, antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {3318, 4186, 3767, 8, 4184, 3719, 2190, 2279, 2391} \[ \frac {4 d (c+d x) \log \left (1+e^{i (e+f x)}\right )}{3 a^2 f^2}-\frac {d (c+d x) \sec ^2\left (\frac {e}{2}+\frac {f x}{2}\right )}{3 a^2 f^2}+\frac {(c+d x)^2 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{3 a^2 f}+\frac {(c+d x)^2 \tan \left (\frac {e}{2}+\frac {f x}{2}\right ) \sec ^2\left (\frac {e}{2}+\frac {f x}{2}\right )}{6 a^2 f}-\frac {i (c+d x)^2}{3 a^2 f}-\frac {4 i d^2 \text {Li}_2\left (-e^{i (e+f x)}\right )}{3 a^2 f^3}+\frac {2 d^2 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{3 a^2 f^3} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2190
Rule 2279
Rule 2391
Rule 3318
Rule 3719
Rule 3767
Rule 4184
Rule 4186
Rubi steps
\begin {align*} \int \frac {(c+d x)^2}{(a+a \cos (e+f x))^2} \, dx &=\frac {\int (c+d x)^2 \csc ^4\left (\frac {e+\pi }{2}+\frac {f x}{2}\right ) \, dx}{4 a^2}\\ &=-\frac {d (c+d x) \sec ^2\left (\frac {e}{2}+\frac {f x}{2}\right )}{3 a^2 f^2}+\frac {(c+d x)^2 \sec ^2\left (\frac {e}{2}+\frac {f x}{2}\right ) \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{6 a^2 f}+\frac {\int (c+d x)^2 \sec ^2\left (\frac {e}{2}+\frac {f x}{2}\right ) \, dx}{6 a^2}+\frac {d^2 \int \sec ^2\left (\frac {e}{2}+\frac {f x}{2}\right ) \, dx}{3 a^2 f^2}\\ &=-\frac {d (c+d x) \sec ^2\left (\frac {e}{2}+\frac {f x}{2}\right )}{3 a^2 f^2}+\frac {(c+d x)^2 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{3 a^2 f}+\frac {(c+d x)^2 \sec ^2\left (\frac {e}{2}+\frac {f x}{2}\right ) \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{6 a^2 f}-\frac {\left (2 d^2\right ) \operatorname {Subst}\left (\int 1 \, dx,x,-\tan \left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{3 a^2 f^3}-\frac {(2 d) \int (c+d x) \tan \left (\frac {e}{2}+\frac {f x}{2}\right ) \, dx}{3 a^2 f}\\ &=-\frac {i (c+d x)^2}{3 a^2 f}-\frac {d (c+d x) \sec ^2\left (\frac {e}{2}+\frac {f x}{2}\right )}{3 a^2 f^2}+\frac {2 d^2 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{3 a^2 f^3}+\frac {(c+d x)^2 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{3 a^2 f}+\frac {(c+d x)^2 \sec ^2\left (\frac {e}{2}+\frac {f x}{2}\right ) \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{6 a^2 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}{3 a^2 f}\\ &=-\frac {i (c+d x)^2}{3 a^2 f}+\frac {4 d (c+d x) \log \left (1+e^{i (e+f x)}\right )}{3 a^2 f^2}-\frac {d (c+d x) \sec ^2\left (\frac {e}{2}+\frac {f x}{2}\right )}{3 a^2 f^2}+\frac {2 d^2 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{3 a^2 f^3}+\frac {(c+d x)^2 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{3 a^2 f}+\frac {(c+d x)^2 \sec ^2\left (\frac {e}{2}+\frac {f x}{2}\right ) \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{6 a^2 f}-\frac {\left (4 d^2\right ) \int \log \left (1+e^{2 i \left (\frac {e}{2}+\frac {f x}{2}\right )}\right ) \, dx}{3 a^2 f^2}\\ &=-\frac {i (c+d x)^2}{3 a^2 f}+\frac {4 d (c+d x) \log \left (1+e^{i (e+f x)}\right )}{3 a^2 f^2}-\frac {d (c+d x) \sec ^2\left (\frac {e}{2}+\frac {f x}{2}\right )}{3 a^2 f^2}+\frac {2 d^2 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{3 a^2 f^3}+\frac {(c+d x)^2 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{3 a^2 f}+\frac {(c+d x)^2 \sec ^2\left (\frac {e}{2}+\frac {f x}{2}\right ) \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{6 a^2 f}+\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 )}{3 a^2 f^3}\\ &=-\frac {i (c+d x)^2}{3 a^2 f}+\frac {4 d (c+d x) \log \left (1+e^{i (e+f x)}\right )}{3 a^2 f^2}-\frac {4 i d^2 \text {Li}_2\left (-e^{i (e+f x)}\right )}{3 a^2 f^3}-\frac {d (c+d x) \sec ^2\left (\frac {e}{2}+\frac {f x}{2}\right )}{3 a^2 f^2}+\frac {2 d^2 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{3 a^2 f^3}+\frac {(c+d x)^2 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{3 a^2 f}+\frac {(c+d x)^2 \sec ^2\left (\frac {e}{2}+\frac {f x}{2}\right ) \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{6 a^2 f}\\ \end {align*}
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Mathematica [A] time = 1.08, size = 212, normalized size = 1.00 \[ \frac {2 \cos \left (\frac {1}{2} (e+f x)\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right ) \left (\left (c^2 f^2+2 c d f^2 x+d^2 \left (f^2 x^2+2\right )\right ) \cos (e+f x)+2 \left (c^2 f^2+2 c d f^2 x+d^2 \left (f^2 x^2+1\right )\right )\right )-2 d f (c+d x) \cos \left (\frac {1}{2} (e+f x)\right )-2 i f (c+d x) \cos ^3\left (\frac {1}{2} (e+f x)\right ) \left (f (c+d x)+4 i d \log \left (1+e^{i (e+f x)}\right )\right )-8 i d^2 \text {Li}_2\left (-e^{i (e+f x)}\right ) \cos ^3\left (\frac {1}{2} (e+f x)\right )\right )}{3 a^2 f^3 (\cos (e+f x)+1)^2} \]
Antiderivative was successfully verified.
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fricas [B] time = 1.22, size = 390, normalized size = 1.84 \[ -\frac {2 \, d^{2} f x + 2 \, c d f + 2 \, {\left (d^{2} f x + c d f\right )} \cos \left (f x + e\right ) - {\left (2 i \, d^{2} \cos \left (f x + e\right )^{2} + 4 i \, d^{2} \cos \left (f x + e\right ) + 2 i \, d^{2}\right )} {\rm Li}_2\left (-\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) - {\left (-2 i \, d^{2} \cos \left (f x + e\right )^{2} - 4 i \, d^{2} \cos \left (f x + e\right ) - 2 i \, d^{2}\right )} {\rm Li}_2\left (-\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) - 2 \, {\left (d^{2} f x + c d f + {\left (d^{2} f x + c d f\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (d^{2} f x + c d f\right )} \cos \left (f x + e\right )\right )} \log \left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (d^{2} f x + c d f + {\left (d^{2} f x + c d f\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (d^{2} f x + c d f\right )} \cos \left (f x + e\right )\right )} \log \left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right ) + 1\right ) - {\left (2 \, d^{2} f^{2} x^{2} + 4 \, c d f^{2} x + 2 \, c^{2} f^{2} + 2 \, d^{2} + {\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x + c^{2} f^{2} + 2 \, d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{3 \, {\left (a^{2} f^{3} \cos \left (f x + e\right )^{2} + 2 \, a^{2} f^{3} \cos \left (f x + e\right ) + a^{2} f^{3}\right )}} \]
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 \[ \int \frac {{\left (d x + c\right )}^{2}}{{\left (a \cos \left (f x + e\right ) + a\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.35, size = 358, normalized size = 1.69 \[ \frac {2 i \left (2 i d^{2} f x \,{\mathrm e}^{2 i \left (f x +e \right )}+3 d^{2} f^{2} x^{2} {\mathrm e}^{i \left (f x +e \right )}+2 i c d f \,{\mathrm e}^{2 i \left (f x +e \right )}+2 i d^{2} f x \,{\mathrm e}^{i \left (f x +e \right )}+6 c d \,f^{2} x \,{\mathrm e}^{i \left (f x +e \right )}+d^{2} f^{2} x^{2}+2 i c d f \,{\mathrm e}^{i \left (f x +e \right )}+3 c^{2} f^{2} {\mathrm e}^{i \left (f x +e \right )}+2 c d \,f^{2} x +c^{2} f^{2}+2 d^{2} {\mathrm e}^{2 i \left (f x +e \right )}+4 d^{2} {\mathrm e}^{i \left (f x +e \right )}+2 d^{2}\right )}{3 f^{3} a^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{3}}+\frac {4 d c \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{3 a^{2} f^{2}}-\frac {4 d c \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{3 a^{2} f^{2}}-\frac {2 i d^{2} x^{2}}{3 a^{2} f}-\frac {4 i d^{2} e x}{3 a^{2} f^{2}}-\frac {2 i d^{2} e^{2}}{3 a^{2} f^{3}}+\frac {4 d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) x}{3 a^{2} f^{2}}-\frac {4 i d^{2} \polylog \left (2, -{\mathrm e}^{i \left (f x +e \right )}\right )}{3 a^{2} f^{3}}+\frac {4 d^{2} e \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{3 a^{2} f^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 2.47, size = 772, normalized size = 3.64 \[ \frac {2 \, c^{2} f^{2} + 4 \, d^{2} + {\left (4 \, d^{2} f x + 4 \, c d f + 4 \, {\left (d^{2} f x + c d f\right )} \cos \left (3 \, f x + 3 \, e\right ) + 12 \, {\left (d^{2} f x + c d f\right )} \cos \left (2 \, f x + 2 \, e\right ) + 12 \, {\left (d^{2} f x + c d f\right )} \cos \left (f x + e\right ) + {\left (4 i \, d^{2} f x + 4 i \, c d f\right )} \sin \left (3 \, f x + 3 \, e\right ) + {\left (12 i \, d^{2} f x + 12 i \, c d f\right )} \sin \left (2 \, f x + 2 \, e\right ) + {\left (12 i \, d^{2} f x + 12 i \, c d f\right )} \sin \left (f x + e\right )\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 (3 \, f x + 3 \, e\right ) - {\left (6 \, d^{2} f^{2} x^{2} - 4 i \, c d f - 4 \, d^{2} + 4 \, {\left (3 \, c d f^{2} - i \, d^{2} f\right )} x\right )} \cos \left (2 \, f x + 2 \, e\right ) + {\left (6 \, c^{2} f^{2} + 4 i \, d^{2} f x + 4 i \, c d f + 8 \, d^{2}\right )} \cos \left (f x + e\right ) - {\left (4 \, d^{2} \cos \left (3 \, f x + 3 \, e\right ) + 12 \, d^{2} \cos \left (2 \, f x + 2 \, e\right ) + 12 \, d^{2} \cos \left (f x + e\right ) + 4 i \, d^{2} \sin \left (3 \, f x + 3 \, e\right ) + 12 i \, d^{2} \sin \left (2 \, f x + 2 \, e\right ) + 12 i \, d^{2} \sin \left (f x + e\right ) + 4 \, 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 (3 \, f x + 3 \, e\right ) + {\left (-6 i \, d^{2} f x - 6 i \, c d f\right )} \cos \left (2 \, f x + 2 \, e\right ) + {\left (-6 i \, d^{2} f x - 6 i \, c d f\right )} \cos \left (f x + e\right ) + 2 \, {\left (d^{2} f x + c d f\right )} \sin \left (3 \, f x + 3 \, e\right ) + 6 \, {\left (d^{2} f x + c d f\right )} \sin \left (2 \, f x + 2 \, e\right ) + 6 \, {\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 (3 \, f x + 3 \, e\right ) + {\left (-6 i \, d^{2} f^{2} x^{2} - 4 \, c d f + 4 i \, d^{2} + {\left (-12 i \, c d f^{2} - 4 \, d^{2} f\right )} x\right )} \sin \left (2 \, f x + 2 \, e\right ) + {\left (6 i \, c^{2} f^{2} - 4 \, d^{2} f x - 4 \, c d f + 8 i \, d^{2}\right )} \sin \left (f x + e\right )}{-3 i \, a^{2} f^{3} \cos \left (3 \, f x + 3 \, e\right ) - 9 i \, a^{2} f^{3} \cos \left (2 \, f x + 2 \, e\right ) - 9 i \, a^{2} f^{3} \cos \left (f x + e\right ) + 3 \, a^{2} f^{3} \sin \left (3 \, f x + 3 \, e\right ) + 9 \, a^{2} f^{3} \sin \left (2 \, f x + 2 \, e\right ) + 9 \, a^{2} f^{3} \sin \left (f x + e\right ) - 3 i \, a^{2} f^{3}} \]
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] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {c^{2}}{\cos ^{2}{\left (e + f x \right )} + 2 \cos {\left (e + f x \right )} + 1}\, dx + \int \frac {d^{2} x^{2}}{\cos ^{2}{\left (e + f x \right )} + 2 \cos {\left (e + f x \right )} + 1}\, dx + \int \frac {2 c d x}{\cos ^{2}{\left (e + f x \right )} + 2 \cos {\left (e + f x \right )} + 1}\, dx}{a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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