Optimal. Leaf size=245 \[ -\frac {a^2}{2 d (c+d x)^2}-\frac {a b f \cos (e+f x)}{d^2 (c+d x)}+\frac {b^2 f^2 \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Ci}\left (\frac {2 c f}{d}+2 f x\right )}{d^3}-\frac {a b f^2 \text {Ci}\left (\frac {c f}{d}+f x\right ) \sin \left (e-\frac {c f}{d}\right )}{d^3}-\frac {a b \sin (e+f x)}{d (c+d x)^2}-\frac {b^2 f \cos (e+f x) \sin (e+f x)}{d^2 (c+d x)}-\frac {b^2 \sin ^2(e+f x)}{2 d (c+d x)^2}-\frac {a b f^2 \cos \left (e-\frac {c f}{d}\right ) \text {Si}\left (\frac {c f}{d}+f x\right )}{d^3}-\frac {b^2 f^2 \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{d^3} \]
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Rubi [A]
time = 0.29, antiderivative size = 245, normalized size of antiderivative = 1.00, number of steps
used = 14, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3398, 3378,
3384, 3380, 3383, 3395, 31, 3393} \begin {gather*} -\frac {a^2}{2 d (c+d x)^2}-\frac {a b f^2 \text {CosIntegral}\left (\frac {c f}{d}+f x\right ) \sin \left (e-\frac {c f}{d}\right )}{d^3}-\frac {a b f^2 \cos \left (e-\frac {c f}{d}\right ) \text {Si}\left (x f+\frac {c f}{d}\right )}{d^3}-\frac {a b f \cos (e+f x)}{d^2 (c+d x)}-\frac {a b \sin (e+f x)}{d (c+d x)^2}+\frac {b^2 f^2 \text {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right ) \cos \left (2 e-\frac {2 c f}{d}\right )}{d^3}-\frac {b^2 f^2 \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{d^3}-\frac {b^2 f \sin (e+f x) \cos (e+f x)}{d^2 (c+d x)}-\frac {b^2 \sin ^2(e+f x)}{2 d (c+d x)^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 31
Rule 3378
Rule 3380
Rule 3383
Rule 3384
Rule 3393
Rule 3395
Rule 3398
Rubi steps
\begin {align*} \int \frac {(a+b \sin (e+f x))^2}{(c+d x)^3} \, dx &=\int \left (\frac {a^2}{(c+d x)^3}+\frac {2 a b \sin (e+f x)}{(c+d x)^3}+\frac {b^2 \sin ^2(e+f x)}{(c+d x)^3}\right ) \, dx\\ &=-\frac {a^2}{2 d (c+d x)^2}+(2 a b) \int \frac {\sin (e+f x)}{(c+d x)^3} \, dx+b^2 \int \frac {\sin ^2(e+f x)}{(c+d x)^3} \, dx\\ &=-\frac {a^2}{2 d (c+d x)^2}-\frac {a b \sin (e+f x)}{d (c+d x)^2}-\frac {b^2 f \cos (e+f x) \sin (e+f x)}{d^2 (c+d x)}-\frac {b^2 \sin ^2(e+f x)}{2 d (c+d x)^2}+\frac {(a b f) \int \frac {\cos (e+f x)}{(c+d x)^2} \, dx}{d}+\frac {\left (b^2 f^2\right ) \int \frac {1}{c+d x} \, dx}{d^2}-\frac {\left (2 b^2 f^2\right ) \int \frac {\sin ^2(e+f x)}{c+d x} \, dx}{d^2}\\ &=-\frac {a^2}{2 d (c+d x)^2}-\frac {a b f \cos (e+f x)}{d^2 (c+d x)}+\frac {b^2 f^2 \log (c+d x)}{d^3}-\frac {a b \sin (e+f x)}{d (c+d x)^2}-\frac {b^2 f \cos (e+f x) \sin (e+f x)}{d^2 (c+d x)}-\frac {b^2 \sin ^2(e+f x)}{2 d (c+d x)^2}-\frac {\left (a b f^2\right ) \int \frac {\sin (e+f x)}{c+d x} \, dx}{d^2}-\frac {\left (2 b^2 f^2\right ) \int \left (\frac {1}{2 (c+d x)}-\frac {\cos (2 e+2 f x)}{2 (c+d x)}\right ) \, dx}{d^2}\\ &=-\frac {a^2}{2 d (c+d x)^2}-\frac {a b f \cos (e+f x)}{d^2 (c+d x)}-\frac {a b \sin (e+f x)}{d (c+d x)^2}-\frac {b^2 f \cos (e+f x) \sin (e+f x)}{d^2 (c+d x)}-\frac {b^2 \sin ^2(e+f x)}{2 d (c+d x)^2}+\frac {\left (b^2 f^2\right ) \int \frac {\cos (2 e+2 f x)}{c+d x} \, dx}{d^2}-\frac {\left (a b f^2 \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^2}-\frac {\left (a b f^2 \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^2}\\ &=-\frac {a^2}{2 d (c+d x)^2}-\frac {a b f \cos (e+f x)}{d^2 (c+d x)}-\frac {a b f^2 \text {Ci}\left (\frac {c f}{d}+f x\right ) \sin \left (e-\frac {c f}{d}\right )}{d^3}-\frac {a b \sin (e+f x)}{d (c+d x)^2}-\frac {b^2 f \cos (e+f x) \sin (e+f x)}{d^2 (c+d x)}-\frac {b^2 \sin ^2(e+f x)}{2 d (c+d x)^2}-\frac {a b f^2 \cos \left (e-\frac {c f}{d}\right ) \text {Si}\left (\frac {c f}{d}+f x\right )}{d^3}+\frac {\left (b^2 f^2 \cos \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^2}-\frac {\left (b^2 f^2 \sin \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^2}\\ &=-\frac {a^2}{2 d (c+d x)^2}-\frac {a b f \cos (e+f x)}{d^2 (c+d x)}+\frac {b^2 f^2 \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Ci}\left (\frac {2 c f}{d}+2 f x\right )}{d^3}-\frac {a b f^2 \text {Ci}\left (\frac {c f}{d}+f x\right ) \sin \left (e-\frac {c f}{d}\right )}{d^3}-\frac {a b \sin (e+f x)}{d (c+d x)^2}-\frac {b^2 f \cos (e+f x) \sin (e+f x)}{d^2 (c+d x)}-\frac {b^2 \sin ^2(e+f x)}{2 d (c+d x)^2}-\frac {a b f^2 \cos \left (e-\frac {c f}{d}\right ) \text {Si}\left (\frac {c f}{d}+f x\right )}{d^3}-\frac {b^2 f^2 \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{d^3}\\ \end {align*}
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Mathematica [A]
time = 0.76, size = 395, normalized size = 1.61 \begin {gather*} -\frac {2 a^2 d^2+b^2 d^2+4 a b c d f \cos (e+f x)+4 a b d^2 f x \cos (e+f x)-b^2 d^2 \cos (2 (e+f x))-4 b^2 f^2 (c+d x)^2 \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Ci}\left (\frac {2 f (c+d x)}{d}\right )+4 a b f^2 (c+d x)^2 \text {Ci}\left (f \left (\frac {c}{d}+x\right )\right ) \sin \left (e-\frac {c f}{d}\right )+4 a b d^2 \sin (e+f x)+2 b^2 c d f \sin (2 (e+f x))+2 b^2 d^2 f x \sin (2 (e+f x))+4 a b c^2 f^2 \cos \left (e-\frac {c f}{d}\right ) \text {Si}\left (f \left (\frac {c}{d}+x\right )\right )+8 a b c d f^2 x \cos \left (e-\frac {c f}{d}\right ) \text {Si}\left (f \left (\frac {c}{d}+x\right )\right )+4 a b d^2 f^2 x^2 \cos \left (e-\frac {c f}{d}\right ) \text {Si}\left (f \left (\frac {c}{d}+x\right )\right )+4 b^2 c^2 f^2 \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )+8 b^2 c d f^2 x \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )+4 b^2 d^2 f^2 x^2 \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )}{4 d^3 (c+d x)^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.15, size = 374, normalized size = 1.53
method | result | size |
derivativedivides | \(\frac {-\frac {a^{2} f^{3}}{2 \left (c f -d e +d \left (f x +e \right )\right )^{2} d}+2 f^{3} a b \left (-\frac {\sin \left (f x +e \right )}{2 \left (c f -d e +d \left (f x +e \right )\right )^{2} d}+\frac {-\frac {\cos \left (f x +e \right )}{\left (c f -d e +d \left (f x +e \right )\right ) d}-\frac {\frac {\sinIntegral \left (f x +e +\frac {c f -d e}{d}\right ) \cos \left (\frac {c f -d e}{d}\right )}{d}-\frac {\cosineIntegral \left (f x +e +\frac {c f -d e}{d}\right ) \sin \left (\frac {c f -d e}{d}\right )}{d}}{d}}{2 d}\right )-\frac {f^{3} b^{2}}{4 \left (c f -d e +d \left (f x +e \right )\right )^{2} d}-\frac {f^{3} b^{2} \left (-\frac {\cos \left (2 f x +2 e \right )}{\left (c f -d e +d \left (f x +e \right )\right )^{2} d}-\frac {-\frac {2 \sin \left (2 f x +2 e \right )}{\left (c f -d e +d \left (f x +e \right )\right ) d}+\frac {\frac {4 \sinIntegral \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \sin \left (\frac {2 c f -2 d e}{d}\right )}{d}+\frac {4 \cosineIntegral \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \cos \left (\frac {2 c f -2 d e}{d}\right )}{d}}{d}}{d}\right )}{4}}{f}\) | \(374\) |
default | \(\frac {-\frac {a^{2} f^{3}}{2 \left (c f -d e +d \left (f x +e \right )\right )^{2} d}+2 f^{3} a b \left (-\frac {\sin \left (f x +e \right )}{2 \left (c f -d e +d \left (f x +e \right )\right )^{2} d}+\frac {-\frac {\cos \left (f x +e \right )}{\left (c f -d e +d \left (f x +e \right )\right ) d}-\frac {\frac {\sinIntegral \left (f x +e +\frac {c f -d e}{d}\right ) \cos \left (\frac {c f -d e}{d}\right )}{d}-\frac {\cosineIntegral \left (f x +e +\frac {c f -d e}{d}\right ) \sin \left (\frac {c f -d e}{d}\right )}{d}}{d}}{2 d}\right )-\frac {f^{3} b^{2}}{4 \left (c f -d e +d \left (f x +e \right )\right )^{2} d}-\frac {f^{3} b^{2} \left (-\frac {\cos \left (2 f x +2 e \right )}{\left (c f -d e +d \left (f x +e \right )\right )^{2} d}-\frac {-\frac {2 \sin \left (2 f x +2 e \right )}{\left (c f -d e +d \left (f x +e \right )\right ) d}+\frac {\frac {4 \sinIntegral \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \sin \left (\frac {2 c f -2 d e}{d}\right )}{d}+\frac {4 \cosineIntegral \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \cos \left (\frac {2 c f -2 d e}{d}\right )}{d}}{d}}{d}\right )}{4}}{f}\) | \(374\) |
risch | \(\frac {i f^{2} a b \,{\mathrm e}^{\frac {i \left (c f -d e \right )}{d}} \expIntegral \left (1, i f x +i e +\frac {i \left (c f -d e \right )}{d}\right )}{2 d^{3}}-\frac {a^{2}}{2 d \left (d x +c \right )^{2}}-\frac {b^{2}}{4 d \left (d x +c \right )^{2}}-\frac {b^{2} f^{2} {\mathrm e}^{\frac {2 i \left (c f -d e \right )}{d}} \expIntegral \left (1, 2 i f x +2 i e +\frac {2 i \left (c f -d e \right )}{d}\right )}{2 d^{3}}-\frac {b^{2} f^{2} {\mathrm e}^{-\frac {2 i \left (c f -d e \right )}{d}} \expIntegral \left (1, -2 i f x -2 i e -\frac {2 \left (i c f -i d e \right )}{d}\right )}{2 d^{3}}-\frac {i a b \,f^{2} {\mathrm e}^{-\frac {i \left (c f -d e \right )}{d}} \expIntegral \left (1, -i f x -i e -\frac {i c f -i d e}{d}\right )}{2 d^{3}}+\frac {i a b \left (-2 i d^{3} f^{3} x^{3}-6 i c \,d^{2} f^{3} x^{2}-6 i c^{2} d \,f^{3} x -2 i c^{3} f^{3}\right ) \cos \left (f x +e \right )}{2 d^{2} \left (d x +c \right )^{2} \left (-d^{2} x^{2} f^{2}-2 c d \,f^{2} x -c^{2} f^{2}\right )}-\frac {a b \left (-2 d^{2} x^{2} f^{2}-4 c d \,f^{2} x -2 c^{2} f^{2}\right ) \sin \left (f x +e \right )}{2 d \left (d x +c \right )^{2} \left (-d^{2} x^{2} f^{2}-2 c d \,f^{2} x -c^{2} f^{2}\right )}+\frac {b^{2} \left (-2 d^{3} f^{2} x^{2}-4 c \,d^{2} f^{2} x -2 c^{2} d \,f^{2}\right ) \cos \left (2 f x +2 e \right )}{8 d^{2} \left (d x +c \right )^{2} \left (-d^{2} x^{2} f^{2}-2 c d \,f^{2} x -c^{2} f^{2}\right )}-\frac {i b^{2} \left (4 i d^{3} f^{3} x^{3}+12 i c \,d^{2} f^{3} x^{2}+12 i c^{2} d \,f^{3} x +4 i c^{3} f^{3}\right ) \sin \left (2 f x +2 e \right )}{8 d^{2} \left (d x +c \right )^{2} \left (-d^{2} x^{2} f^{2}-2 c d \,f^{2} x -c^{2} f^{2}\right )}\) | \(605\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [C] Result contains complex when optimal does not.
time = 0.51, size = 503, normalized size = 2.05 \begin {gather*} -\frac {\frac {2 \, a^{2} f^{3}}{{\left (f x + e\right )}^{2} d^{3} + c^{2} d f^{2} - 2 \, c d^{2} f e + d^{3} e^{2} + 2 \, {\left (c d^{2} f - d^{3} e\right )} {\left (f x + e\right )}} - \frac {4 \, {\left (f^{3} {\left (-i \, E_{3}\left (\frac {i \, {\left (f x + e\right )} d + i \, c f - i \, d e}{d}\right ) + i \, E_{3}\left (-\frac {i \, {\left (f x + e\right )} d + i \, c f - i \, d e}{d}\right )\right )} \cos \left (\frac {c f - d e}{d}\right ) + f^{3} {\left (E_{3}\left (\frac {i \, {\left (f x + e\right )} d + i \, c f - i \, d e}{d}\right ) + E_{3}\left (-\frac {i \, {\left (f x + e\right )} d + i \, c f - i \, d e}{d}\right )\right )} \sin \left (\frac {c f - d e}{d}\right )\right )} a b}{{\left (f x + e\right )}^{2} d^{3} + c^{2} d f^{2} - 2 \, c d^{2} f e + d^{3} e^{2} + 2 \, {\left (c d^{2} f - d^{3} e\right )} {\left (f x + e\right )}} - \frac {{\left (f^{3} {\left (E_{3}\left (\frac {2 \, {\left (-i \, {\left (f x + e\right )} d - i \, c f + i \, d e\right )}}{d}\right ) + E_{3}\left (-\frac {2 \, {\left (-i \, {\left (f x + e\right )} d - i \, c f + i \, d e\right )}}{d}\right )\right )} \cos \left (\frac {2 \, {\left (c f - d e\right )}}{d}\right ) - f^{3} {\left (i \, E_{3}\left (\frac {2 \, {\left (-i \, {\left (f x + e\right )} d - i \, c f + i \, d e\right )}}{d}\right ) - i \, E_{3}\left (-\frac {2 \, {\left (-i \, {\left (f x + e\right )} d - i \, c f + i \, d e\right )}}{d}\right )\right )} \sin \left (\frac {2 \, {\left (c f - d e\right )}}{d}\right ) - f^{3}\right )} b^{2}}{{\left (f x + e\right )}^{2} d^{3} + c^{2} d f^{2} - 2 \, c d^{2} f e + d^{3} e^{2} + 2 \, {\left (c d^{2} f - d^{3} e\right )} {\left (f x + e\right )}}}{4 \, f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.39, size = 476, normalized size = 1.94 \begin {gather*} \frac {b^{2} d^{2} \cos \left (f x + e\right )^{2} - {\left (a^{2} + b^{2}\right )} d^{2} - 2 \, {\left (b^{2} d^{2} f^{2} x^{2} + 2 \, b^{2} c d f^{2} x + b^{2} c^{2} f^{2}\right )} \sin \left (-\frac {2 \, {\left (c f - d e\right )}}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) - 2 \, {\left (a b d^{2} f^{2} x^{2} + 2 \, a b c d f^{2} x + a b c^{2} f^{2}\right )} \cos \left (-\frac {c f - d e}{d}\right ) \operatorname {Si}\left (\frac {d f x + c f}{d}\right ) - 2 \, {\left (a b d^{2} f x + a b c d f\right )} \cos \left (f x + e\right ) + {\left ({\left (b^{2} d^{2} f^{2} x^{2} + 2 \, b^{2} c d f^{2} x + b^{2} c^{2} f^{2}\right )} \operatorname {Ci}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) + {\left (b^{2} d^{2} f^{2} x^{2} + 2 \, b^{2} c d f^{2} x + b^{2} c^{2} f^{2}\right )} \operatorname {Ci}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (c f - d e\right )}}{d}\right ) - 2 \, {\left (a b d^{2} + {\left (b^{2} d^{2} f x + b^{2} c d f\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right ) - {\left ({\left (a b d^{2} f^{2} x^{2} + 2 \, a b c d f^{2} x + a b c^{2} f^{2}\right )} \operatorname {Ci}\left (\frac {d f x + c f}{d}\right ) + {\left (a b d^{2} f^{2} x^{2} + 2 \, a b c d f^{2} x + a b c^{2} f^{2}\right )} \operatorname {Ci}\left (-\frac {d f x + c f}{d}\right )\right )} \sin \left (-\frac {c f - d e}{d}\right )}{2 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \end {gather*}
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 \begin {gather*} \int \frac {\left (a + b \sin {\left (e + f x \right )}\right )^{2}}{\left (c + d x\right )^{3}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [C] Result contains higher order function than in optimal. Order 9 vs. order
4.
time = 9.92, size = 123654, normalized size = 504.71 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\sin \left (e+f\,x\right )\right )}^2}{{\left (c+d\,x\right )}^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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