Optimal. Leaf size=183 \[ -\frac{\sec (c+d x) (b-a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{d \left (a^2-b^2\right )}-\frac{a \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{d \left (a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{\sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{d \sqrt{a+b \sin (c+d x)}} \]
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Rubi [A] time = 0.203331, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {2696, 2752, 2663, 2661, 2655, 2653} \[ -\frac{\sec (c+d x) (b-a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{d \left (a^2-b^2\right )}-\frac{a \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{d \left (a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{\sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{d \sqrt{a+b \sin (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 2696
Rule 2752
Rule 2663
Rule 2661
Rule 2655
Rule 2653
Rubi steps
\begin{align*} \int \frac{\sec ^2(c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx &=-\frac{\sec (c+d x) (b-a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{\left (a^2-b^2\right ) d}+\frac{\int \frac{\frac{b^2}{2}+\frac{1}{2} a b \sin (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx}{-a^2+b^2}\\ &=-\frac{\sec (c+d x) (b-a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{\left (a^2-b^2\right ) d}+\frac{1}{2} \int \frac{1}{\sqrt{a+b \sin (c+d x)}} \, dx-\frac{a \int \sqrt{a+b \sin (c+d x)} \, dx}{2 \left (a^2-b^2\right )}\\ &=-\frac{\sec (c+d x) (b-a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{\left (a^2-b^2\right ) d}-\frac{\left (a \sqrt{a+b \sin (c+d x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}} \, dx}{2 \left (a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{\sqrt{\frac{a+b \sin (c+d x)}{a+b}} \int \frac{1}{\sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}}} \, dx}{2 \sqrt{a+b \sin (c+d x)}}\\ &=-\frac{\sec (c+d x) (b-a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{\left (a^2-b^2\right ) d}-\frac{a E\left (\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )|\frac{2 b}{a+b}\right ) \sqrt{a+b \sin (c+d x)}}{\left (a^2-b^2\right ) d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{F\left (\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )|\frac{2 b}{a+b}\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}{d \sqrt{a+b \sin (c+d x)}}\\ \end{align*}
Mathematica [A] time = 0.637521, size = 177, normalized size = 0.97 \[ \frac{-\left (a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )+a^2 \tan (c+d x)-a b \sec (c+d x)+a b \sin (c+d x) \tan (c+d x)+a (a+b) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} E\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )-b^2 \tan (c+d x)}{d (a-b) (a+b) \sqrt{a+b \sin (c+d x)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.585, size = 640, normalized size = 3.5 \begin{align*} -{\frac{1}{ \left ( a+b \right ) b \left ( a-b \right ) \cos \left ( dx+c \right ) d}\sqrt{ \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) b+a \left ( \cos \left ( dx+c \right ) \right ) ^{2}} \left ( \sqrt{{\frac{b\sin \left ( dx+c \right ) }{a-b}}+{\frac{a}{a-b}}}\sqrt{-{\frac{b\sin \left ( dx+c \right ) }{a+b}}+{\frac{b}{a+b}}}\sqrt{-{\frac{b\sin \left ( dx+c \right ) }{a-b}}-{\frac{b}{a-b}}}{\it EllipticF} \left ( \sqrt{{\frac{b\sin \left ( dx+c \right ) }{a-b}}+{\frac{a}{a-b}}},\sqrt{{\frac{a-b}{a+b}}} \right ){a}^{2}b-\sqrt{{\frac{b\sin \left ( dx+c \right ) }{a-b}}+{\frac{a}{a-b}}}\sqrt{-{\frac{b\sin \left ( dx+c \right ) }{a+b}}+{\frac{b}{a+b}}}\sqrt{-{\frac{b\sin \left ( dx+c \right ) }{a-b}}-{\frac{b}{a-b}}}{\it EllipticF} \left ( \sqrt{{\frac{b\sin \left ( dx+c \right ) }{a-b}}+{\frac{a}{a-b}}},\sqrt{{\frac{a-b}{a+b}}} \right ){b}^{3}-\sqrt{{\frac{b\sin \left ( dx+c \right ) }{a-b}}+{\frac{a}{a-b}}}\sqrt{-{\frac{b\sin \left ( dx+c \right ) }{a+b}}+{\frac{b}{a+b}}}\sqrt{-{\frac{b\sin \left ( dx+c \right ) }{a-b}}-{\frac{b}{a-b}}}{\it EllipticE} \left ( \sqrt{{\frac{b\sin \left ( dx+c \right ) }{a-b}}+{\frac{a}{a-b}}},\sqrt{{\frac{a-b}{a+b}}} \right ){a}^{3}+\sqrt{{\frac{b\sin \left ( dx+c \right ) }{a-b}}+{\frac{a}{a-b}}}\sqrt{-{\frac{b\sin \left ( dx+c \right ) }{a+b}}+{\frac{b}{a+b}}}\sqrt{-{\frac{b\sin \left ( dx+c \right ) }{a-b}}-{\frac{b}{a-b}}}{\it EllipticE} \left ( \sqrt{{\frac{b\sin \left ( dx+c \right ) }{a-b}}+{\frac{a}{a-b}}},\sqrt{{\frac{a-b}{a+b}}} \right ) a{b}^{2}+a{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}-{a}^{2}b\sin \left ( dx+c \right ) +{b}^{3}\sin \left ( dx+c \right ) \right ){\frac{1}{\sqrt{- \left ( a+b\sin \left ( dx+c \right ) \right ) \left ( \sin \left ( dx+c \right ) -1 \right ) \left ( 1+\sin \left ( dx+c \right ) \right ) }}}{\frac{1}{\sqrt{a+b\sin \left ( dx+c \right ) }}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )^{2}}{\sqrt{b \sin \left (d x + c\right ) + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sec \left (d x + c\right )^{2}}{\sqrt{b \sin \left (d x + c\right ) + a}}, x\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{\sec ^{2}{\left (c + d x \right )}}{\sqrt{a + b \sin{\left (c + d x \right )}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )^{2}}{\sqrt{b \sin \left (d x + c\right ) + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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