Optimal. Leaf size=270 \[ -\frac{b^3 \cos \left (a-\frac{b c}{d}\right ) \text{CosIntegral}\left (\frac{b c}{d}+b x\right )}{24 d^4}-\frac{9 b^3 \cos \left (3 a-\frac{3 b c}{d}\right ) \text{CosIntegral}\left (\frac{3 b c}{d}+3 b x\right )}{8 d^4}+\frac{b^3 \sin \left (a-\frac{b c}{d}\right ) \text{Si}\left (\frac{b c}{d}+b x\right )}{24 d^4}+\frac{9 b^3 \sin \left (3 a-\frac{3 b c}{d}\right ) \text{Si}\left (\frac{3 b c}{d}+3 b x\right )}{8 d^4}+\frac{b^2 \sin (a+b x)}{24 d^3 (c+d x)}+\frac{3 b^2 \sin (3 a+3 b x)}{8 d^3 (c+d x)}-\frac{b \cos (a+b x)}{24 d^2 (c+d x)^2}-\frac{b \cos (3 a+3 b x)}{8 d^2 (c+d x)^2}-\frac{\sin (a+b x)}{12 d (c+d x)^3}-\frac{\sin (3 a+3 b x)}{12 d (c+d x)^3} \]
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Rubi [A] time = 0.377452, antiderivative size = 270, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {4406, 3297, 3303, 3299, 3302} \[ -\frac{b^3 \cos \left (a-\frac{b c}{d}\right ) \text{CosIntegral}\left (\frac{b c}{d}+b x\right )}{24 d^4}-\frac{9 b^3 \cos \left (3 a-\frac{3 b c}{d}\right ) \text{CosIntegral}\left (\frac{3 b c}{d}+3 b x\right )}{8 d^4}+\frac{b^3 \sin \left (a-\frac{b c}{d}\right ) \text{Si}\left (\frac{b c}{d}+b x\right )}{24 d^4}+\frac{9 b^3 \sin \left (3 a-\frac{3 b c}{d}\right ) \text{Si}\left (\frac{3 b c}{d}+3 b x\right )}{8 d^4}+\frac{b^2 \sin (a+b x)}{24 d^3 (c+d x)}+\frac{3 b^2 \sin (3 a+3 b x)}{8 d^3 (c+d x)}-\frac{b \cos (a+b x)}{24 d^2 (c+d x)^2}-\frac{b \cos (3 a+3 b x)}{8 d^2 (c+d x)^2}-\frac{\sin (a+b x)}{12 d (c+d x)^3}-\frac{\sin (3 a+3 b x)}{12 d (c+d x)^3} \]
Antiderivative was successfully verified.
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Rule 4406
Rule 3297
Rule 3303
Rule 3299
Rule 3302
Rubi steps
\begin{align*} \int \frac{\cos ^2(a+b x) \sin (a+b x)}{(c+d x)^4} \, dx &=\int \left (\frac{\sin (a+b x)}{4 (c+d x)^4}+\frac{\sin (3 a+3 b x)}{4 (c+d x)^4}\right ) \, dx\\ &=\frac{1}{4} \int \frac{\sin (a+b x)}{(c+d x)^4} \, dx+\frac{1}{4} \int \frac{\sin (3 a+3 b x)}{(c+d x)^4} \, dx\\ &=-\frac{\sin (a+b x)}{12 d (c+d x)^3}-\frac{\sin (3 a+3 b x)}{12 d (c+d x)^3}+\frac{b \int \frac{\cos (a+b x)}{(c+d x)^3} \, dx}{12 d}+\frac{b \int \frac{\cos (3 a+3 b x)}{(c+d x)^3} \, dx}{4 d}\\ &=-\frac{b \cos (a+b x)}{24 d^2 (c+d x)^2}-\frac{b \cos (3 a+3 b x)}{8 d^2 (c+d x)^2}-\frac{\sin (a+b x)}{12 d (c+d x)^3}-\frac{\sin (3 a+3 b x)}{12 d (c+d x)^3}-\frac{b^2 \int \frac{\sin (a+b x)}{(c+d x)^2} \, dx}{24 d^2}-\frac{\left (3 b^2\right ) \int \frac{\sin (3 a+3 b x)}{(c+d x)^2} \, dx}{8 d^2}\\ &=-\frac{b \cos (a+b x)}{24 d^2 (c+d x)^2}-\frac{b \cos (3 a+3 b x)}{8 d^2 (c+d x)^2}-\frac{\sin (a+b x)}{12 d (c+d x)^3}+\frac{b^2 \sin (a+b x)}{24 d^3 (c+d x)}-\frac{\sin (3 a+3 b x)}{12 d (c+d x)^3}+\frac{3 b^2 \sin (3 a+3 b x)}{8 d^3 (c+d x)}-\frac{b^3 \int \frac{\cos (a+b x)}{c+d x} \, dx}{24 d^3}-\frac{\left (9 b^3\right ) \int \frac{\cos (3 a+3 b x)}{c+d x} \, dx}{8 d^3}\\ &=-\frac{b \cos (a+b x)}{24 d^2 (c+d x)^2}-\frac{b \cos (3 a+3 b x)}{8 d^2 (c+d x)^2}-\frac{\sin (a+b x)}{12 d (c+d x)^3}+\frac{b^2 \sin (a+b x)}{24 d^3 (c+d x)}-\frac{\sin (3 a+3 b x)}{12 d (c+d x)^3}+\frac{3 b^2 \sin (3 a+3 b x)}{8 d^3 (c+d x)}-\frac{\left (9 b^3 \cos \left (3 a-\frac{3 b c}{d}\right )\right ) \int \frac{\cos \left (\frac{3 b c}{d}+3 b x\right )}{c+d x} \, dx}{8 d^3}-\frac{\left (b^3 \cos \left (a-\frac{b c}{d}\right )\right ) \int \frac{\cos \left (\frac{b c}{d}+b x\right )}{c+d x} \, dx}{24 d^3}+\frac{\left (9 b^3 \sin \left (3 a-\frac{3 b c}{d}\right )\right ) \int \frac{\sin \left (\frac{3 b c}{d}+3 b x\right )}{c+d x} \, dx}{8 d^3}+\frac{\left (b^3 \sin \left (a-\frac{b c}{d}\right )\right ) \int \frac{\sin \left (\frac{b c}{d}+b x\right )}{c+d x} \, dx}{24 d^3}\\ &=-\frac{b \cos (a+b x)}{24 d^2 (c+d x)^2}-\frac{b \cos (3 a+3 b x)}{8 d^2 (c+d x)^2}-\frac{b^3 \cos \left (a-\frac{b c}{d}\right ) \text{Ci}\left (\frac{b c}{d}+b x\right )}{24 d^4}-\frac{9 b^3 \cos \left (3 a-\frac{3 b c}{d}\right ) \text{Ci}\left (\frac{3 b c}{d}+3 b x\right )}{8 d^4}-\frac{\sin (a+b x)}{12 d (c+d x)^3}+\frac{b^2 \sin (a+b x)}{24 d^3 (c+d x)}-\frac{\sin (3 a+3 b x)}{12 d (c+d x)^3}+\frac{3 b^2 \sin (3 a+3 b x)}{8 d^3 (c+d x)}+\frac{b^3 \sin \left (a-\frac{b c}{d}\right ) \text{Si}\left (\frac{b c}{d}+b x\right )}{24 d^4}+\frac{9 b^3 \sin \left (3 a-\frac{3 b c}{d}\right ) \text{Si}\left (\frac{3 b c}{d}+3 b x\right )}{8 d^4}\\ \end{align*}
Mathematica [A] time = 1.93298, size = 300, normalized size = 1.11 \[ -\frac{b^3 (c+d x)^3 \left (\cos \left (a-\frac{b c}{d}\right ) \text{CosIntegral}\left (b \left (\frac{c}{d}+x\right )\right )-\sin \left (a-\frac{b c}{d}\right ) \text{Si}\left (b \left (\frac{c}{d}+x\right )\right )\right )+27 b^3 (c+d x)^3 \left (\cos \left (3 a-\frac{3 b c}{d}\right ) \text{CosIntegral}\left (\frac{3 b (c+d x)}{d}\right )-\sin \left (3 a-\frac{3 b c}{d}\right ) \text{Si}\left (\frac{3 b (c+d x)}{d}\right )\right )+d \cos (b x) \left (b d \cos (a) (c+d x)-\sin (a) \left (b^2 (c+d x)^2-2 d^2\right )\right )+d \cos (3 b x) \left (3 b d \cos (3 a) (c+d x)-\sin (3 a) \left (9 b^2 (c+d x)^2-2 d^2\right )\right )-d \sin (b x) \left (\cos (a) \left (b^2 (c+d x)^2-2 d^2\right )+b d \sin (a) (c+d x)\right )-d \sin (3 b x) \left (\cos (3 a) \left (9 b^2 (c+d x)^2-2 d^2\right )+3 b d \sin (3 a) (c+d x)\right )}{24 d^4 (c+d x)^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.023, size = 381, normalized size = 1.4 \begin{align*}{\frac{1}{b} \left ({\frac{{b}^{4}}{12} \left ( -{\frac{\sin \left ( 3\,bx+3\,a \right ) }{ \left ( \left ( bx+a \right ) d-ad+bc \right ) ^{3}d}}+{\frac{1}{d} \left ( -{\frac{3\,\cos \left ( 3\,bx+3\,a \right ) }{2\, \left ( \left ( bx+a \right ) d-ad+bc \right ) ^{2}d}}-{\frac{3}{2\,d} \left ( -3\,{\frac{\sin \left ( 3\,bx+3\,a \right ) }{ \left ( \left ( bx+a \right ) d-ad+bc \right ) d}}+3\,{\frac{1}{d} \left ( 3\,{\frac{1}{d}{\it Si} \left ( 3\,bx+3\,a+3\,{\frac{-ad+bc}{d}} \right ) \sin \left ( 3\,{\frac{-ad+bc}{d}} \right ) }+3\,{\frac{1}{d}{\it Ci} \left ( 3\,bx+3\,a+3\,{\frac{-ad+bc}{d}} \right ) \cos \left ( 3\,{\frac{-ad+bc}{d}} \right ) } \right ) } \right ) } \right ) } \right ) }+{\frac{{b}^{4}}{4} \left ( -{\frac{\sin \left ( bx+a \right ) }{3\, \left ( \left ( bx+a \right ) d-ad+bc \right ) ^{3}d}}+{\frac{1}{3\,d} \left ( -{\frac{\cos \left ( bx+a \right ) }{2\, \left ( \left ( bx+a \right ) d-ad+bc \right ) ^{2}d}}-{\frac{1}{2\,d} \left ( -{\frac{\sin \left ( bx+a \right ) }{ \left ( \left ( bx+a \right ) d-ad+bc \right ) d}}+{\frac{1}{d} \left ({\frac{1}{d}{\it Si} \left ( bx+a+{\frac{-ad+bc}{d}} \right ) \sin \left ({\frac{-ad+bc}{d}} \right ) }+{\frac{1}{d}{\it Ci} \left ( bx+a+{\frac{-ad+bc}{d}} \right ) \cos \left ({\frac{-ad+bc}{d}} \right ) } \right ) } \right ) } \right ) } \right ) } \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [C] time = 2.78691, size = 520, normalized size = 1.93 \begin{align*} -\frac{b^{4}{\left (i \, E_{4}\left (\frac{i \, b c + i \,{\left (b x + a\right )} d - i \, a d}{d}\right ) - i \, E_{4}\left (-\frac{i \, b c + i \,{\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \cos \left (-\frac{b c - a d}{d}\right ) + b^{4}{\left (i \, E_{4}\left (\frac{3 i \, b c + 3 i \,{\left (b x + a\right )} d - 3 i \, a d}{d}\right ) - i \, E_{4}\left (-\frac{3 i \, b c + 3 i \,{\left (b x + a\right )} d - 3 i \, a d}{d}\right )\right )} \cos \left (-\frac{3 \,{\left (b c - a d\right )}}{d}\right ) + b^{4}{\left (E_{4}\left (\frac{i \, b c + i \,{\left (b x + a\right )} d - i \, a d}{d}\right ) + E_{4}\left (-\frac{i \, b c + i \,{\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \sin \left (-\frac{b c - a d}{d}\right ) + b^{4}{\left (E_{4}\left (\frac{3 i \, b c + 3 i \,{\left (b x + a\right )} d - 3 i \, a d}{d}\right ) + E_{4}\left (-\frac{3 i \, b c + 3 i \,{\left (b x + a\right )} d - 3 i \, a d}{d}\right )\right )} \sin \left (-\frac{3 \,{\left (b c - a d\right )}}{d}\right )}{8 \,{\left (b^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} +{\left (b x + a\right )}^{3} d^{4} - a^{3} d^{4} + 3 \,{\left (b c d^{3} - a d^{4}\right )}{\left (b x + a\right )}^{2} + 3 \,{\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )}{\left (b x + a\right )}\right )} b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 0.625508, size = 1231, normalized size = 4.56 \begin{align*} -\frac{24 \,{\left (b d^{3} x + b c d^{2}\right )} \cos \left (b x + a\right )^{3} - 54 \,{\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \sin \left (-\frac{3 \,{\left (b c - a d\right )}}{d}\right ) \operatorname{Si}\left (\frac{3 \,{\left (b d x + b c\right )}}{d}\right ) - 2 \,{\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \sin \left (-\frac{b c - a d}{d}\right ) \operatorname{Si}\left (\frac{b d x + b c}{d}\right ) - 16 \,{\left (b d^{3} x + b c d^{2}\right )} \cos \left (b x + a\right ) +{\left ({\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \operatorname{Ci}\left (\frac{b d x + b c}{d}\right ) +{\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \operatorname{Ci}\left (-\frac{b d x + b c}{d}\right )\right )} \cos \left (-\frac{b c - a d}{d}\right ) + 27 \,{\left ({\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \operatorname{Ci}\left (\frac{3 \,{\left (b d x + b c\right )}}{d}\right ) +{\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \operatorname{Ci}\left (-\frac{3 \,{\left (b d x + b c\right )}}{d}\right )\right )} \cos \left (-\frac{3 \,{\left (b c - a d\right )}}{d}\right ) + 8 \,{\left (2 \, b^{2} d^{3} x^{2} + 4 \, b^{2} c d^{2} x + 2 \, b^{2} c^{2} d -{\left (9 \, b^{2} d^{3} x^{2} + 18 \, b^{2} c d^{2} x + 9 \, b^{2} c^{2} d - 2 \, d^{3}\right )} \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )}{48 \,{\left (d^{7} x^{3} + 3 \, c d^{6} x^{2} + 3 \, c^{2} d^{5} x + c^{3} d^{4}\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*} \int \frac{\sin{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{\left (c + d x\right )^{4}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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