Integrand size = 41, antiderivative size = 185 \[ \int (a+b \sin (d+e x)) \sqrt {b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)} \, dx=-\frac {\left (a^2+b^2\right ) \cos (d+e x) \sqrt {b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)}}{e (b+a \sin (d+e x))}+\frac {3 a^2 b x \sqrt {b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)}}{2 \left (a b+a^2 \sin (d+e x)\right )}-\frac {a^2 b \cos (d+e x) \sin (d+e x) \sqrt {b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)}}{2 e \left (a b+a^2 \sin (d+e x)\right )} \]
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Time = 0.12 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {3371, 2813} \[ \int (a+b \sin (d+e x)) \sqrt {b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)} \, dx=\frac {3 a^2 b x \sqrt {a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2}}{2 \left (a^2 \sin (d+e x)+a b\right )}-\frac {a^2 b \sin (d+e x) \cos (d+e x) \sqrt {a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2}}{2 e \left (a^2 \sin (d+e x)+a b\right )}-\frac {\left (a^2+b^2\right ) \cos (d+e x) \sqrt {a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2}}{e (a \sin (d+e x)+b)} \]
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Rule 2813
Rule 3371
Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)} \int \left (2 a b+2 a^2 \sin (d+e x)\right ) (a+b \sin (d+e x)) \, dx}{2 a b+2 a^2 \sin (d+e x)} \\ & = -\frac {\left (a^2+b^2\right ) \cos (d+e x) \sqrt {b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)}}{e (b+a \sin (d+e x))}+\frac {3 a^2 b x \sqrt {b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)}}{2 \left (a b+a^2 \sin (d+e x)\right )}-\frac {a^2 b \cos (d+e x) \sin (d+e x) \sqrt {b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)}}{2 e \left (a b+a^2 \sin (d+e x)\right )} \\ \end{align*}
Time = 0.39 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.38 \[ \int (a+b \sin (d+e x)) \sqrt {b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)} \, dx=-\frac {\sqrt {(b+a \sin (d+e x))^2} \left (4 \left (a^2+b^2\right ) \cos (d+e x)+a b (-6 (d+e x)+\sin (2 (d+e x)))\right )}{4 e (b+a \sin (d+e x))} \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.81 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.39
method | result | size |
default | \(\frac {\operatorname {csgn}\left (b +a \sin \left (e x +d \right )\right ) \left (a b \left (-\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )-a^{2} \cos \left (e x +d \right )-b^{2} \cos \left (e x +d \right )+a b \left (e x +d \right )\right )}{e}\) | \(72\) |
parts | \(-\frac {a \left (a \cos \left (e x +d \right )-\left (e x +d \right ) b +a \right ) \operatorname {csgn}\left (b +a \sin \left (e x +d \right )\right )}{e}-\frac {b \,\operatorname {csgn}\left (b +a \sin \left (e x +d \right )\right ) \left (\cos \left (e x +d \right ) \sin \left (e x +d \right ) a +2 b \cos \left (e x +d \right )-\left (e x +d \right ) a +2 b \right )}{2 e}\) | \(89\) |
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Time = 0.27 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.23 \[ \int (a+b \sin (d+e x)) \sqrt {b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)} \, dx=\frac {3 \, a b e x - a b \cos \left (e x + d\right ) \sin \left (e x + d\right ) - 2 \, {\left (a^{2} + b^{2}\right )} \cos \left (e x + d\right )}{2 \, e} \]
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\[ \int (a+b \sin (d+e x)) \sqrt {b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)} \, dx=\int \left (a + b \sin {\left (d + e x \right )}\right ) \sqrt {\left (a \sin {\left (d + e x \right )} + b\right )^{2}}\, dx \]
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Time = 0.37 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.01 \[ \int (a+b \sin (d+e x)) \sqrt {b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)} \, dx=\frac {2 \, {\left (b \arctan \left (\frac {\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right ) - \frac {a}{\frac {\sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} + 1}\right )} a + {\left (a \arctan \left (\frac {\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right ) - \frac {2 \, b + \frac {a \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + \frac {2 \, b \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} - \frac {a \sin \left (e x + d\right )^{3}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{3}}}{\frac {2 \, \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} + \frac {\sin \left (e x + d\right )^{4}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{4}} + 1}\right )} b}{e} \]
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Time = 0.27 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.51 \[ \int (a+b \sin (d+e x)) \sqrt {b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)} \, dx=\frac {3}{2} \, a b x \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right ) - \frac {a^{2} \cos \left (e x + d\right ) \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right )}{e} - \frac {b^{2} \cos \left (e x + d\right ) \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right )}{e} - \frac {a b \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right ) \sin \left (2 \, e x + 2 \, d\right )}{4 \, e} \]
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Timed out. \[ \int (a+b \sin (d+e x)) \sqrt {b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)} \, dx=\int \left (a+b\,\sin \left (d+e\,x\right )\right )\,\sqrt {a^2\,{\sin \left (d+e\,x\right )}^2+2\,a\,b\,\sin \left (d+e\,x\right )+b^2} \,d x \]
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