Optimal. Leaf size=159 \[ -\frac {a^2 f \text {Ci}\left (2 x f+\frac {2 c f}{d}\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{d^2}-\frac {2 a^2 f \text {Ci}\left (x f+\frac {c f}{d}\right ) \sin \left (e-\frac {c f}{d}\right )}{d^2}-\frac {2 a^2 f \cos \left (e-\frac {c f}{d}\right ) \text {Si}\left (x f+\frac {c f}{d}\right )}{d^2}-\frac {a^2 f \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{d^2}-\frac {4 a^2 \cos ^4\left (\frac {e}{2}+\frac {f x}{2}\right )}{d (c+d x)} \]
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Rubi [A] time = 0.32, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3318, 3313, 3303, 3299, 3302} \[ -\frac {a^2 f \text {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{d^2}-\frac {2 a^2 f \text {CosIntegral}\left (\frac {c f}{d}+f x\right ) \sin \left (e-\frac {c f}{d}\right )}{d^2}-\frac {2 a^2 f \cos \left (e-\frac {c f}{d}\right ) \text {Si}\left (x f+\frac {c f}{d}\right )}{d^2}-\frac {a^2 f \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{d^2}-\frac {4 a^2 \cos ^4\left (\frac {e}{2}+\frac {f x}{2}\right )}{d (c+d x)} \]
Antiderivative was successfully verified.
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Rule 3299
Rule 3302
Rule 3303
Rule 3313
Rule 3318
Rubi steps
\begin {align*} \int \frac {(a+a \cos (e+f x))^2}{(c+d x)^2} \, dx &=\left (4 a^2\right ) \int \frac {\sin ^4\left (\frac {e+\pi }{2}+\frac {f x}{2}\right )}{(c+d x)^2} \, dx\\ &=-\frac {4 a^2 \cos ^4\left (\frac {e}{2}+\frac {f x}{2}\right )}{d (c+d x)}+\frac {\left (8 a^2 f\right ) \int \left (-\frac {\sin (e+f x)}{4 (c+d x)}-\frac {\sin (2 e+2 f x)}{8 (c+d x)}\right ) \, dx}{d}\\ &=-\frac {4 a^2 \cos ^4\left (\frac {e}{2}+\frac {f x}{2}\right )}{d (c+d x)}-\frac {\left (a^2 f\right ) \int \frac {\sin (2 e+2 f x)}{c+d x} \, dx}{d}-\frac {\left (2 a^2 f\right ) \int \frac {\sin (e+f x)}{c+d x} \, dx}{d}\\ &=-\frac {4 a^2 \cos ^4\left (\frac {e}{2}+\frac {f x}{2}\right )}{d (c+d x)}-\frac {\left (a^2 f \cos \left (2 e-\frac {2 c f}{d}\right )\right ) \int \frac {\sin \left (\frac {2 c f}{d}+2 f x\right )}{c+d x} \, dx}{d}-\frac {\left (2 a^2 f \cos \left (e-\frac {c f}{d}\right )\right ) \int \frac {\sin \left (\frac {c f}{d}+f x\right )}{c+d x} \, dx}{d}-\frac {\left (a^2 f \sin \left (2 e-\frac {2 c f}{d}\right )\right ) \int \frac {\cos \left (\frac {2 c f}{d}+2 f x\right )}{c+d x} \, dx}{d}-\frac {\left (2 a^2 f \sin \left (e-\frac {c f}{d}\right )\right ) \int \frac {\cos \left (\frac {c f}{d}+f x\right )}{c+d x} \, dx}{d}\\ &=-\frac {4 a^2 \cos ^4\left (\frac {e}{2}+\frac {f x}{2}\right )}{d (c+d x)}-\frac {a^2 f \text {Ci}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{d^2}-\frac {2 a^2 f \text {Ci}\left (\frac {c f}{d}+f x\right ) \sin \left (e-\frac {c f}{d}\right )}{d^2}-\frac {2 a^2 f \cos \left (e-\frac {c f}{d}\right ) \text {Si}\left (\frac {c f}{d}+f x\right )}{d^2}-\frac {a^2 f \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{d^2}\\ \end {align*}
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Mathematica [A] time = 0.50, size = 206, normalized size = 1.30 \[ -\frac {a^2 \left (2 f (c+d x) \text {Ci}\left (\frac {2 f (c+d x)}{d}\right ) \sin \left (2 e-\frac {2 c f}{d}\right )+4 f (c+d x) \text {Ci}\left (f \left (\frac {c}{d}+x\right )\right ) \sin \left (e-\frac {c f}{d}\right )+4 d f x \cos \left (e-\frac {c f}{d}\right ) \text {Si}\left (f \left (\frac {c}{d}+x\right )\right )+2 d f x \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )+4 c f \cos \left (e-\frac {c f}{d}\right ) \text {Si}\left (f \left (\frac {c}{d}+x\right )\right )+2 c f \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )+4 d \cos (e+f x)+d \cos (2 (e+f x))+3 d\right )}{2 d^2 (c+d x)} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.69, size = 284, normalized size = 1.79 \[ -\frac {2 \, a^{2} d \cos \left (f x + e\right )^{2} + 4 \, a^{2} d \cos \left (f x + e\right ) + 2 \, a^{2} d + 2 \, {\left (a^{2} d f x + a^{2} c f\right )} \cos \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) + 4 \, {\left (a^{2} d f x + a^{2} c f\right )} \cos \left (-\frac {d e - c f}{d}\right ) \operatorname {Si}\left (\frac {d f x + c f}{d}\right ) - 2 \, {\left ({\left (a^{2} d f x + a^{2} c f\right )} \operatorname {Ci}\left (\frac {d f x + c f}{d}\right ) + {\left (a^{2} d f x + a^{2} c f\right )} \operatorname {Ci}\left (-\frac {d f x + c f}{d}\right )\right )} \sin \left (-\frac {d e - c f}{d}\right ) - {\left ({\left (a^{2} d f x + a^{2} c f\right )} \operatorname {Ci}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) + {\left (a^{2} d f x + a^{2} c f\right )} \operatorname {Ci}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right )}{2 \, {\left (d^{3} x + c d^{2}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 1.00, size = 1133, normalized size = 7.13 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 276, normalized size = 1.74 \[ \frac {\frac {f^{2} a^{2} \left (-\frac {2 \cos \left (2 f x +2 e \right )}{\left (\left (f x +e \right ) d +c f -d e \right ) d}-\frac {2 \left (\frac {2 \Si \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \cos \left (\frac {2 c f -2 d e}{d}\right )}{d}-\frac {2 \Ci \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \sin \left (\frac {2 c f -2 d e}{d}\right )}{d}\right )}{d}\right )}{4}-\frac {3 f^{2} a^{2}}{2 \left (\left (f x +e \right ) d +c f -d e \right ) d}+2 f^{2} a^{2} \left (-\frac {\cos \left (f x +e \right )}{\left (\left (f x +e \right ) d +c f -d e \right ) d}-\frac {\frac {\Si \left (f x +e +\frac {c f -d e}{d}\right ) \cos \left (\frac {c f -d e}{d}\right )}{d}-\frac {\Ci \left (f x +e +\frac {c f -d e}{d}\right ) \sin \left (\frac {c f -d e}{d}\right )}{d}}{d}\right )}{f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 1.35, size = 370, normalized size = 2.33 \[ -\frac {\frac {64 \, a^{2} f^{2}}{{\left (f x + e\right )} d^{2} - d^{2} e + c d f} + \frac {8 \, {\left (8 \, f^{2} {\left (E_{2}\left (\frac {i \, {\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right ) + E_{2}\left (-\frac {i \, {\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right )\right )} \cos \left (-\frac {d e - c f}{d}\right ) + f^{2} {\left (8 i \, E_{2}\left (\frac {i \, {\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right ) - 8 i \, E_{2}\left (-\frac {i \, {\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right )\right )} \sin \left (-\frac {d e - c f}{d}\right )\right )} a^{2}}{{\left (f x + e\right )} d^{2} - d^{2} e + c d f} + \frac {{\left (16 \, f^{2} {\left (E_{2}\left (\frac {2 i \, {\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right ) + E_{2}\left (-\frac {2 i \, {\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) + f^{2} {\left (16 i \, E_{2}\left (\frac {2 i \, {\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right ) - 16 i \, E_{2}\left (-\frac {2 i \, {\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) + 32 \, f^{2}\right )} a^{2}}{{\left (f x + e\right )} d^{2} - d^{2} e + c d f}}{64 \, f} \]
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 \[ \int \frac {{\left (a+a\,\cos \left (e+f\,x\right )\right )}^2}{{\left (c+d\,x\right )}^2} \,d x \]
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 \[ a^{2} \left (\int \frac {2 \cos {\left (e + f x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {\cos ^{2}{\left (e + f x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {1}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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