Optimal. Leaf size=140 \[ \frac{\tan (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{f (a+b)}+\frac{\sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (e+f x\left |-\frac{b}{a}\right .\right )}{f \sqrt{a+b \sin ^2(e+f x)}}-\frac{\sqrt{a+b \sin ^2(e+f x)} E\left (e+f x\left |-\frac{b}{a}\right .\right )}{f (a+b) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}} \]
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Rubi [A] time = 0.168916, antiderivative size = 180, normalized size of antiderivative = 1.29, number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {3192, 414, 21, 423, 426, 424, 421, 419} \[ \frac{\tan (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{f (a+b)}+\frac{\sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{f \sqrt{a+b \sin ^2(e+f x)}}-\frac{\sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{f (a+b) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}} \]
Antiderivative was successfully verified.
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Rule 3192
Rule 414
Rule 21
Rule 423
Rule 426
Rule 424
Rule 421
Rule 419
Rubi steps
\begin{align*} \int \frac{\sec ^2(e+f x)}{\sqrt{a+b \sin ^2(e+f x)}} \, dx &=\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right )^{3/2} \sqrt{a+b x^2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\sqrt{a+b \sin ^2(e+f x)} \tan (e+f x)}{(a+b) f}+\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{b-b x^2}{\sqrt{1-x^2} \sqrt{a+b x^2}} \, dx,x,\sin (e+f x)\right )}{(a+b) f}\\ &=\frac{\sqrt{a+b \sin ^2(e+f x)} \tan (e+f x)}{(a+b) f}+\frac{\left (b \sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1-x^2}}{\sqrt{a+b x^2}} \, dx,x,\sin (e+f x)\right )}{(a+b) f}\\ &=\frac{\sqrt{a+b \sin ^2(e+f x)} \tan (e+f x)}{(a+b) f}+\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2} \sqrt{a+b x^2}} \, dx,x,\sin (e+f x)\right )}{f}-\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x^2}}{\sqrt{1-x^2}} \, dx,x,\sin (e+f x)\right )}{(a+b) f}\\ &=\frac{\sqrt{a+b \sin ^2(e+f x)} \tan (e+f x)}{(a+b) f}-\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1+\frac{b x^2}{a}}}{\sqrt{1-x^2}} \, dx,x,\sin (e+f x)\right )}{(a+b) f \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}}+\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2} \sqrt{1+\frac{b x^2}{a}}} \, dx,x,\sin (e+f x)\right )}{f \sqrt{a+b \sin ^2(e+f x)}}\\ &=-\frac{\sqrt{\cos ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right ) \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{(a+b) f \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}}+\frac{\sqrt{\cos ^2(e+f x)} F\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right ) \sec (e+f x) \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}}{f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\sqrt{a+b \sin ^2(e+f x)} \tan (e+f x)}{(a+b) f}\\ \end{align*}
Mathematica [A] time = 0.609255, size = 141, normalized size = 1.01 \[ \frac{\sqrt{2} \tan (e+f x) (2 a-b \cos (2 (e+f x))+b)+2 (a+b) \sqrt{\frac{2 a-b \cos (2 (e+f x))+b}{a}} F\left (e+f x\left |-\frac{b}{a}\right .\right )-2 a \sqrt{\frac{2 a-b \cos (2 (e+f x))+b}{a}} E\left (e+f x\left |-\frac{b}{a}\right .\right )}{2 f (a+b) \sqrt{2 a-b \cos (2 (e+f x))+b}} \]
Antiderivative was successfully verified.
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Maple [A] time = 2.016, size = 278, normalized size = 2. \begin{align*}{\frac{1}{ \left ( a+b \right ) \cos \left ( fx+e \right ) f}\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ( a\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}\sqrt{-{\frac{b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{a}}+{\frac{a+b}{a}}}{\it EllipticF} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ) +b\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}\sqrt{-{\frac{b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{a}}+{\frac{a+b}{a}}}{\it EllipticF} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ) -a\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}\sqrt{-{\frac{b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{a}}+{\frac{a+b}{a}}}{\it EllipticE} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ) - \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) b+a\sin \left ( fx+e \right ) +b\sin \left ( fx+e \right ) \right ){\frac{1}{\sqrt{- \left ( a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( 1+\sin \left ( fx+e \right ) \right ) }}}{\frac{1}{\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}}}} \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 (f x + e\right )^{2}}{\sqrt{b \sin \left (f x + e\right )^{2} + 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{\sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sec \left (f x + e\right )^{2}}{b \cos \left (f x + e\right )^{2} - a - b}, 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 (e + f x \right )}}{\sqrt{a + b \sin ^{2}{\left (e + f 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 (f x + e\right )^{2}}{\sqrt{b \sin \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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