Optimal. Leaf size=196 \[ -\frac {\left (6 a b c d-a^2 \left (2 c^2+d^2\right )-b^2 \left (c^2+2 d^2\right )\right ) \tan ^{-1}\left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\left (c^2-d^2\right )^{5/2} f}+\frac {(b c-a d)^2 \cos (e+f x)}{2 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}-\frac {(b c-a d) \left (3 a c d+b \left (c^2-4 d^2\right )\right ) \cos (e+f x)}{2 d \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))} \]
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Rubi [A]
time = 0.22, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2869, 2833, 12,
2739, 632, 210} \begin {gather*} -\frac {\left (-\left (a^2 \left (2 c^2+d^2\right )\right )+6 a b c d-b^2 \left (c^2+2 d^2\right )\right ) \text {ArcTan}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{f \left (c^2-d^2\right )^{5/2}}+\frac {(b c-a d)^2 \cos (e+f x)}{2 d f \left (c^2-d^2\right ) (c+d \sin (e+f x))^2}-\frac {\left (3 a c d+b \left (c^2-4 d^2\right )\right ) (b c-a d) \cos (e+f x)}{2 d f \left (c^2-d^2\right )^2 (c+d \sin (e+f x))} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 210
Rule 632
Rule 2739
Rule 2833
Rule 2869
Rubi steps
\begin {align*} \int \frac {(a+b \sin (e+f x))^2}{(c+d \sin (e+f x))^3} \, dx &=\frac {(b c-a d)^2 \cos (e+f x)}{2 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}+\frac {\int \frac {2 d \left (\left (a^2+b^2\right ) c-2 a b d\right )+\left (b^2 c^2+2 a b c d-\left (a^2+2 b^2\right ) d^2\right ) \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx}{2 d \left (c^2-d^2\right )}\\ &=\frac {(b c-a d)^2 \cos (e+f x)}{2 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}-\frac {(b c-a d) \left (3 a c d+b \left (c^2-4 d^2\right )\right ) \cos (e+f x)}{2 d \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}-\frac {\int \frac {d \left (6 a b c d-a^2 \left (2 c^2+d^2\right )-b^2 \left (c^2+2 d^2\right )\right )}{c+d \sin (e+f x)} \, dx}{2 d \left (c^2-d^2\right )^2}\\ &=\frac {(b c-a d)^2 \cos (e+f x)}{2 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}-\frac {(b c-a d) \left (3 a c d+b \left (c^2-4 d^2\right )\right ) \cos (e+f x)}{2 d \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}-\frac {\left (6 a b c d-a^2 \left (2 c^2+d^2\right )-b^2 \left (c^2+2 d^2\right )\right ) \int \frac {1}{c+d \sin (e+f x)} \, dx}{2 \left (c^2-d^2\right )^2}\\ &=\frac {(b c-a d)^2 \cos (e+f x)}{2 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}-\frac {(b c-a d) \left (3 a c d+b \left (c^2-4 d^2\right )\right ) \cos (e+f x)}{2 d \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}-\frac {\left (6 a b c d-a^2 \left (2 c^2+d^2\right )-b^2 \left (c^2+2 d^2\right )\right ) \text {Subst}\left (\int \frac {1}{c+2 d x+c x^2} \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{\left (c^2-d^2\right )^2 f}\\ &=\frac {(b c-a d)^2 \cos (e+f x)}{2 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}-\frac {(b c-a d) \left (3 a c d+b \left (c^2-4 d^2\right )\right ) \cos (e+f x)}{2 d \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}+\frac {\left (2 \left (6 a b c d-a^2 \left (2 c^2+d^2\right )-b^2 \left (c^2+2 d^2\right )\right )\right ) \text {Subst}\left (\int \frac {1}{-4 \left (c^2-d^2\right )-x^2} \, dx,x,2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )}{\left (c^2-d^2\right )^2 f}\\ &=-\frac {\left (6 a b c d-a^2 \left (2 c^2+d^2\right )-b^2 \left (c^2+2 d^2\right )\right ) \tan ^{-1}\left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\left (c^2-d^2\right )^{5/2} f}+\frac {(b c-a d)^2 \cos (e+f x)}{2 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}-\frac {(b c-a d) \left (3 a c d+b \left (c^2-4 d^2\right )\right ) \cos (e+f x)}{2 d \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}\\ \end {align*}
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Mathematica [A]
time = 1.00, size = 202, normalized size = 1.03 \begin {gather*} \frac {\frac {2 \left (-6 a b c d+a^2 \left (2 c^2+d^2\right )+b^2 \left (c^2+2 d^2\right )\right ) \tan ^{-1}\left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\left (c^2-d^2\right )^{5/2}}+\frac {(b c-a d)^2 \cos (e+f x)}{(c-d) d (c+d) (c+d \sin (e+f x))^2}-\frac {\left (-3 a^2 c d^2+2 a b d \left (c^2+2 d^2\right )+b^2 \left (c^3-4 c d^2\right )\right ) \cos (e+f x)}{(c-d)^2 d (c+d)^2 (c+d \sin (e+f x))}}{2 f} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(466\) vs.
\(2(187)=374\).
time = 0.51, size = 467, normalized size = 2.38 Too large to display
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 480 vs.
\(2 (192) = 384\).
time = 0.41, size = 1050, normalized size = 5.36 \begin {gather*} \left [\frac {2 \, {\left (b^{2} c^{5} + 2 \, a b c^{4} d + 2 \, a b c^{2} d^{3} - 4 \, a b d^{5} - {\left (3 \, a^{2} + 5 \, b^{2}\right )} c^{3} d^{2} + {\left (3 \, a^{2} + 4 \, b^{2}\right )} c d^{4}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (6 \, a b c^{3} d + 6 \, a b c d^{3} - {\left (2 \, a^{2} + b^{2}\right )} c^{4} - 3 \, {\left (a^{2} + b^{2}\right )} c^{2} d^{2} - {\left (a^{2} + 2 \, b^{2}\right )} d^{4} - {\left (6 \, a b c d^{3} - {\left (2 \, a^{2} + b^{2}\right )} c^{2} d^{2} - {\left (a^{2} + 2 \, b^{2}\right )} d^{4}\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (6 \, a b c^{2} d^{2} - {\left (2 \, a^{2} + b^{2}\right )} c^{3} d - {\left (a^{2} + 2 \, b^{2}\right )} c d^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c^{2} + d^{2}} \log \left (\frac {{\left (2 \, c^{2} - d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2} + 2 \, {\left (c \cos \left (f x + e\right ) \sin \left (f x + e\right ) + d \cos \left (f x + e\right )\right )} \sqrt {-c^{2} + d^{2}}}{d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}}\right ) + 2 \, {\left (4 \, a b c^{5} - 2 \, a b c^{3} d^{2} - 2 \, a b c d^{4} - a^{2} d^{5} - {\left (4 \, a^{2} + 3 \, b^{2}\right )} c^{4} d + {\left (5 \, a^{2} + 3 \, b^{2}\right )} c^{2} d^{3}\right )} \cos \left (f x + e\right )}{4 \, {\left ({\left (c^{6} d^{2} - 3 \, c^{4} d^{4} + 3 \, c^{2} d^{6} - d^{8}\right )} f \cos \left (f x + e\right )^{2} - 2 \, {\left (c^{7} d - 3 \, c^{5} d^{3} + 3 \, c^{3} d^{5} - c d^{7}\right )} f \sin \left (f x + e\right ) - {\left (c^{8} - 2 \, c^{6} d^{2} + 2 \, c^{2} d^{6} - d^{8}\right )} f\right )}}, \frac {{\left (b^{2} c^{5} + 2 \, a b c^{4} d + 2 \, a b c^{2} d^{3} - 4 \, a b d^{5} - {\left (3 \, a^{2} + 5 \, b^{2}\right )} c^{3} d^{2} + {\left (3 \, a^{2} + 4 \, b^{2}\right )} c d^{4}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (6 \, a b c^{3} d + 6 \, a b c d^{3} - {\left (2 \, a^{2} + b^{2}\right )} c^{4} - 3 \, {\left (a^{2} + b^{2}\right )} c^{2} d^{2} - {\left (a^{2} + 2 \, b^{2}\right )} d^{4} - {\left (6 \, a b c d^{3} - {\left (2 \, a^{2} + b^{2}\right )} c^{2} d^{2} - {\left (a^{2} + 2 \, b^{2}\right )} d^{4}\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (6 \, a b c^{2} d^{2} - {\left (2 \, a^{2} + b^{2}\right )} c^{3} d - {\left (a^{2} + 2 \, b^{2}\right )} c d^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {c^{2} - d^{2}} \arctan \left (-\frac {c \sin \left (f x + e\right ) + d}{\sqrt {c^{2} - d^{2}} \cos \left (f x + e\right )}\right ) + {\left (4 \, a b c^{5} - 2 \, a b c^{3} d^{2} - 2 \, a b c d^{4} - a^{2} d^{5} - {\left (4 \, a^{2} + 3 \, b^{2}\right )} c^{4} d + {\left (5 \, a^{2} + 3 \, b^{2}\right )} c^{2} d^{3}\right )} \cos \left (f x + e\right )}{2 \, {\left ({\left (c^{6} d^{2} - 3 \, c^{4} d^{4} + 3 \, c^{2} d^{6} - d^{8}\right )} f \cos \left (f x + e\right )^{2} - 2 \, {\left (c^{7} d - 3 \, c^{5} d^{3} + 3 \, c^{3} d^{5} - c d^{7}\right )} f \sin \left (f x + e\right ) - {\left (c^{8} - 2 \, c^{6} d^{2} + 2 \, c^{2} d^{6} - d^{8}\right )} f\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 609 vs.
\(2 (192) = 384\).
time = 0.49, size = 609, normalized size = 3.11 \begin {gather*} \frac {\frac {{\left (2 \, a^{2} c^{2} + b^{2} c^{2} - 6 \, a b c d + a^{2} d^{2} + 2 \, b^{2} d^{2}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )}}{{\left (c^{4} - 2 \, c^{2} d^{2} + d^{4}\right )} \sqrt {c^{2} - d^{2}}} + \frac {b^{2} c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 6 \, a b c^{4} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 5 \, a^{2} c^{3} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2 \, b^{2} c^{3} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, a^{2} c d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 4 \, a b c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, a^{2} c^{4} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, b^{2} c^{4} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 10 \, a b c^{3} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 7 \, a^{2} c^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 6 \, b^{2} c^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 4 \, a b c d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, a^{2} d^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - b^{2} c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 10 \, a b c^{4} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 11 \, a^{2} c^{3} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 10 \, b^{2} c^{3} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 8 \, a b c^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, a^{2} c d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 4 \, a b c^{5} + 4 \, a^{2} c^{4} d + 3 \, b^{2} c^{4} d - 2 \, a b c^{3} d^{2} - a^{2} c^{2} d^{3}}{{\left (c^{6} - 2 \, c^{4} d^{2} + c^{2} d^{4}\right )} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + c\right )}^{2}}}{f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 10.26, size = 641, normalized size = 3.27 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\left (\frac {\left (2\,c^4\,d-4\,c^2\,d^3+2\,d^5\right )\,\left (2\,a^2\,c^2+a^2\,d^2-6\,a\,b\,c\,d+b^2\,c^2+2\,b^2\,d^2\right )}{2\,{\left (c+d\right )}^{5/2}\,{\left (c-d\right )}^{5/2}\,\left (c^4-2\,c^2\,d^2+d^4\right )}+\frac {c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,a^2\,c^2+a^2\,d^2-6\,a\,b\,c\,d+b^2\,c^2+2\,b^2\,d^2\right )}{{\left (c+d\right )}^{5/2}\,{\left (c-d\right )}^{5/2}}\right )\,\left (c^4-2\,c^2\,d^2+d^4\right )}{2\,a^2\,c^2+a^2\,d^2-6\,a\,b\,c\,d+b^2\,c^2+2\,b^2\,d^2}\right )\,\left (2\,a^2\,c^2+a^2\,d^2-6\,a\,b\,c\,d+b^2\,c^2+2\,b^2\,d^2\right )}{f\,{\left (c+d\right )}^{5/2}\,{\left (c-d\right )}^{5/2}}-\frac {\frac {-4\,a^2\,c^2\,d+a^2\,d^3+4\,a\,b\,c^3+2\,a\,b\,c\,d^2-3\,b^2\,c^2\,d}{c^4-2\,c^2\,d^2+d^4}+\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (-11\,a^2\,c^2\,d^2+2\,a^2\,d^4+10\,a\,b\,c^3\,d+8\,a\,b\,c\,d^3+b^2\,c^4-10\,b^2\,c^2\,d^2\right )}{c\,\left (c^4-2\,c^2\,d^2+d^4\right )}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (5\,a^2\,c^2\,d^2-2\,a^2\,d^4-6\,a\,b\,c^3\,d+b^2\,c^4+2\,b^2\,c^2\,d^2\right )}{c\,\left (c^4-2\,c^2\,d^2+d^4\right )}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (c^2+2\,d^2\right )\,\left (-4\,a^2\,c^2\,d+a^2\,d^3+4\,a\,b\,c^3+2\,a\,b\,c\,d^2-3\,b^2\,c^2\,d\right )}{c^2\,\left (c^4-2\,c^2\,d^2+d^4\right )}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (2\,c^2+4\,d^2\right )+c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+c^2+4\,c\,d\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+4\,c\,d\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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