Optimal. Leaf size=137 \[ \frac{b x (a \sin (d+e x)+b)}{a \sqrt{a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2}}-\frac{2 \sqrt{a^2-b^2} (a \sin (d+e x)+b) \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{a^2-b^2}}\right )}{a e \sqrt{a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2}} \]
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Rubi [A] time = 0.198201, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.122, Rules used = {3290, 2735, 2660, 618, 206} \[ \frac{b x (a \sin (d+e x)+b)}{a \sqrt{a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2}}-\frac{2 \sqrt{a^2-b^2} (a \sin (d+e x)+b) \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{a^2-b^2}}\right )}{a e \sqrt{a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2}} \]
Antiderivative was successfully verified.
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Rule 3290
Rule 2735
Rule 2660
Rule 618
Rule 206
Rubi steps
\begin{align*} \int \frac{a+b \sin (d+e x)}{\sqrt{b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)}} \, dx &=\frac{\left (2 a b+2 a^2 \sin (d+e x)\right ) \int \frac{a+b \sin (d+e x)}{2 a b+2 a^2 \sin (d+e x)} \, dx}{\sqrt{b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)}}\\ &=\frac{b x (b+a \sin (d+e x))}{a \sqrt{b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)}}-\frac{\left (\left (-2 a^3+2 a b^2\right ) \left (2 a b+2 a^2 \sin (d+e x)\right )\right ) \int \frac{1}{2 a b+2 a^2 \sin (d+e x)} \, dx}{2 a^2 \sqrt{b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)}}\\ &=\frac{b x (b+a \sin (d+e x))}{a \sqrt{b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)}}-\frac{\left (\left (-2 a^3+2 a b^2\right ) \left (2 a b+2 a^2 \sin (d+e x)\right )\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 )}{a^2 e \sqrt{b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)}}\\ &=\frac{b x (b+a \sin (d+e x))}{a \sqrt{b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)}}+\frac{\left (2 \left (-2 a^3+2 a b^2\right ) \left (2 a b+2 a^2 \sin (d+e x)\right )\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 )}{a^2 e \sqrt{b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)}}\\ &=\frac{b x (b+a \sin (d+e x))}{a \sqrt{b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)}}-\frac{2 \sqrt{a^2-b^2} \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{a^2-b^2}}\right ) (b+a \sin (d+e x))}{a e \sqrt{b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)}}\\ \end{align*}
Mathematica [A] time = 0.183296, size = 85, normalized size = 0.62 \[ \frac{(a \sin (d+e x)+b) \left (b (d+e x)-2 \sqrt{b^2-a^2} \tan ^{-1}\left (\frac{a+b \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{b^2-a^2}}\right )\right )}{a e \sqrt{(a \sin (d+e x)+b)^2}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.164, size = 176, normalized size = 1.3 \begin{align*} -{\frac{b+a\sin \left ( ex+d \right ) }{ae} \left ( 2\,\arctan \left ({\frac{b\cos \left ( ex+d \right ) -a\sin \left ( ex+d \right ) -b}{\sin \left ( ex+d \right ) \sqrt{-{a}^{2}+{b}^{2}}}} \right ){a}^{2}-2\,\arctan \left ({\frac{b\cos \left ( ex+d \right ) -a\sin \left ( ex+d \right ) -b}{\sin \left ( ex+d \right ) \sqrt{-{a}^{2}+{b}^{2}}}} \right ){b}^{2}-b \left ( ex+d \right ) \sqrt{-{a}^{2}+{b}^{2}} \right ){\frac{1}{\sqrt{-{a}^{2}+{b}^{2}}}}{\frac{1}{\sqrt{-{a}^{2} \left ( \cos \left ( ex+d \right ) \right ) ^{2}+2\,ab\sin \left ( ex+d \right ) +{a}^{2}+{b}^{2}}}}} \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 [A] time = 1.88315, size = 462, normalized size = 3.37 \begin{align*} \left [\frac{2 \, b e x + \sqrt{a^{2} - b^{2}} \log \left (-\frac{{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (e x + d\right )^{2} + 2 \, a b \sin \left (e x + d\right ) + a^{2} + b^{2} - 2 \,{\left (b \cos \left (e x + d\right ) \sin \left (e x + d\right ) + a \cos \left (e x + d\right )\right )} \sqrt{a^{2} - b^{2}}}{a^{2} \cos \left (e x + d\right )^{2} - 2 \, a b \sin \left (e x + d\right ) - a^{2} - b^{2}}\right )}{2 \, a e}, \frac{b e x - \sqrt{-a^{2} + b^{2}} \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \sin \left (e x + d\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \left (e x + d\right )}\right )}{a e}\right ] \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 [A] time = 1.34954, size = 281, normalized size = 2.05 \begin{align*}{\left (\frac{{\left (x e - 2 \, \pi \left \lfloor \frac{x e + d}{2 \, \pi } + \frac{1}{2} \right \rfloor + d\right )} b}{a \mathrm{sgn}\left (b \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{4} + 2 \, a \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{3} + 2 \, 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 )} + \frac{2 \,{\left (a^{2} - b^{2}\right )} \arctan \left (\frac{b \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + a}{\sqrt{-a^{2} + b^{2}}}\right )}{\sqrt{-a^{2} + b^{2}} a \mathrm{sgn}\left (b \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{4} + 2 \, a \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{3} + 2 \, 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 )}\right )} e^{\left (-1\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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