Optimal. Leaf size=157 \[ -\frac{\left (2 a^2+b^2\right ) \cos (d+e x)}{3 e \left (a^2-b^2\right )^2 (a \sin (d+e x)+b)}+\frac{b \cos (d+e x)}{3 e \left (a^2-b^2\right ) (a \sin (d+e x)+b)^2}+\frac{2 a b \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{a^2-b^2}}\right )}{e \left (a^2-b^2\right )^{5/2}}-\frac{\cos (d+e x)}{3 e (a \sin (d+e x)+b)^3} \]
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Rubi [A] time = 0.415349, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3288, 2754, 12, 2660, 618, 206} \[ -\frac{\left (2 a^2+b^2\right ) \cos (d+e x)}{3 e \left (a^2-b^2\right )^2 (a \sin (d+e x)+b)}+\frac{b \cos (d+e x)}{3 e \left (a^2-b^2\right ) (a \sin (d+e x)+b)^2}+\frac{2 a b \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{a^2-b^2}}\right )}{e \left (a^2-b^2\right )^{5/2}}-\frac{\cos (d+e x)}{3 e (a \sin (d+e x)+b)^3} \]
Antiderivative was successfully verified.
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Rule 3288
Rule 2754
Rule 12
Rule 2660
Rule 618
Rule 206
Rubi steps
\begin{align*} \int \frac{a+b \sin (d+e x)}{\left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^2} \, dx &=\left (16 a^4\right ) \int \frac{a+b \sin (d+e x)}{\left (2 a b+2 a^2 \sin (d+e x)\right )^4} \, dx\\ &=-\frac{\cos (d+e x)}{3 e (b+a \sin (d+e x))^3}+\frac{\left (4 a^2\right ) \int \frac{4 a \left (a^2-b^2\right ) \sin (d+e x)}{\left (2 a b+2 a^2 \sin (d+e x)\right )^3} \, dx}{3 \left (a^2-b^2\right )}\\ &=-\frac{\cos (d+e x)}{3 e (b+a \sin (d+e x))^3}+\frac{1}{3} \left (16 a^3\right ) \int \frac{\sin (d+e x)}{\left (2 a b+2 a^2 \sin (d+e x)\right )^3} \, dx\\ &=-\frac{\cos (d+e x)}{3 e (b+a \sin (d+e x))^3}+\frac{b \cos (d+e x)}{3 \left (a^2-b^2\right ) e (b+a \sin (d+e x))^2}+\frac{(2 a) \int \frac{4 a^2-2 a b \sin (d+e x)}{\left (2 a b+2 a^2 \sin (d+e x)\right )^2} \, dx}{3 \left (a^2-b^2\right )}\\ &=-\frac{\cos (d+e x)}{3 e (b+a \sin (d+e x))^3}+\frac{b \cos (d+e x)}{3 \left (a^2-b^2\right ) e (b+a \sin (d+e x))^2}-\frac{\left (2 a^2+b^2\right ) \cos (d+e x)}{3 \left (a^2-b^2\right )^2 e (b+a \sin (d+e x))}+\frac{\int -\frac{12 a^3 b}{2 a b+2 a^2 \sin (d+e x)} \, dx}{6 a \left (a^2-b^2\right )^2}\\ &=-\frac{\cos (d+e x)}{3 e (b+a \sin (d+e x))^3}+\frac{b \cos (d+e x)}{3 \left (a^2-b^2\right ) e (b+a \sin (d+e x))^2}-\frac{\left (2 a^2+b^2\right ) \cos (d+e x)}{3 \left (a^2-b^2\right )^2 e (b+a \sin (d+e x))}-\frac{\left (2 a^2 b\right ) \int \frac{1}{2 a b+2 a^2 \sin (d+e x)} \, dx}{\left (a^2-b^2\right )^2}\\ &=-\frac{\cos (d+e x)}{3 e (b+a \sin (d+e x))^3}+\frac{b \cos (d+e x)}{3 \left (a^2-b^2\right ) e (b+a \sin (d+e x))^2}-\frac{\left (2 a^2+b^2\right ) \cos (d+e x)}{3 \left (a^2-b^2\right )^2 e (b+a \sin (d+e x))}-\frac{\left (4 a^2 b\right ) \operatorname{Subst}\left (\int \frac{1}{2 a b+4 a^2 x+2 a b x^2} \, dx,x,\tan \left (\frac{1}{2} (d+e x)\right )\right )}{\left (a^2-b^2\right )^2 e}\\ &=-\frac{\cos (d+e x)}{3 e (b+a \sin (d+e x))^3}+\frac{b \cos (d+e x)}{3 \left (a^2-b^2\right ) e (b+a \sin (d+e x))^2}-\frac{\left (2 a^2+b^2\right ) \cos (d+e x)}{3 \left (a^2-b^2\right )^2 e (b+a \sin (d+e x))}+\frac{\left (8 a^2 b\right ) \operatorname{Subst}\left (\int \frac{1}{16 a^2 \left (a^2-b^2\right )-x^2} \, dx,x,4 a^2+4 a b \tan \left (\frac{1}{2} (d+e x)\right )\right )}{\left (a^2-b^2\right )^2 e}\\ &=\frac{2 a b \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2} e}-\frac{\cos (d+e x)}{3 e (b+a \sin (d+e x))^3}+\frac{b \cos (d+e x)}{3 \left (a^2-b^2\right ) e (b+a \sin (d+e x))^2}-\frac{\left (2 a^2+b^2\right ) \cos (d+e x)}{3 \left (a^2-b^2\right )^2 e (b+a \sin (d+e x))}\\ \end{align*}
Mathematica [A] time = 0.973576, size = 140, normalized size = 0.89 \[ -\frac{\frac{6 a b \tan ^{-1}\left (\frac{a+b \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}}+\frac{\cos (d+e x) \left (a^2 \left (2 a^2+b^2\right ) \sin ^2(d+e x)+3 a b \left (a^2+b^2\right ) \sin (d+e x)-a^2 b^2+a^4+3 b^4\right )}{(a-b)^2 (a+b)^2 (a \sin (d+e x)+b)^3}}{3 e} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.118, size = 1297, normalized size = 8.3 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.16481, size = 1705, normalized size = 10.86 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.27347, size = 613, normalized size = 3.9 \begin{align*} -\frac{2}{3} \,{\left (\frac{3 \,{\left (\pi \left \lfloor \frac{x e + d}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + a}{\sqrt{-a^{2} + b^{2}}}\right )\right )} a b}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{-a^{2} + b^{2}}} + \frac{3 \, a^{5} b^{2} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{5} - 6 \, a^{3} b^{4} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{5} + 6 \, a b^{6} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{5} + 6 \, a^{6} b \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{4} - 9 \, a^{4} b^{3} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{4} + 15 \, a^{2} b^{5} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{4} + 3 \, b^{7} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{4} + 4 \, a^{7} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{3} + 2 \, a^{5} b^{2} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{3} + 6 \, a^{3} b^{4} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{3} + 18 \, a b^{6} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{3} + 6 \, a^{6} b \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} + 18 \, a^{2} b^{5} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} + 6 \, b^{7} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} + 3 \, a^{5} b^{2} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + 12 \, a b^{6} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + a^{4} b^{3} - a^{2} b^{5} + 3 \, b^{7}}{{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )}{\left (b \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} + 2 \, a \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + b\right )}^{3}}\right )} e^{\left (-1\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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