Optimal. Leaf size=212 \[ \frac{2 (a+2 b) \tan (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 f (a+b)^2}+\frac{(2 a+3 b) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (e+f x\left |-\frac{b}{a}\right .\right )}{3 f (a+b) \sqrt{a+b \sin ^2(e+f x)}}-\frac{2 (a+2 b) \sqrt{a+b \sin ^2(e+f x)} E\left (e+f x\left |-\frac{b}{a}\right .\right )}{3 f (a+b)^2 \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}}+\frac{\tan (e+f x) \sec ^2(e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 f (a+b)} \]
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Rubi [A] time = 0.248134, antiderivative size = 252, normalized size of antiderivative = 1.19, 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, 527, 524, 426, 424, 421, 419} \[ \frac{2 (a+2 b) \tan (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 f (a+b)^2}+\frac{\tan (e+f x) \sec ^2(e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 f (a+b)}+\frac{(2 a+3 b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 f (a+b) \sqrt{a+b \sin ^2(e+f x)}}-\frac{2 (a+2 b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 f (a+b)^2 \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}} \]
Antiderivative was successfully verified.
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Rule 3192
Rule 414
Rule 527
Rule 524
Rule 426
Rule 424
Rule 421
Rule 419
Rubi steps
\begin{align*} \int \frac{\sec ^4(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 )^{5/2} \sqrt{a+b x^2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\sec ^2(e+f x) \sqrt{a+b \sin ^2(e+f x)} \tan (e+f x)}{3 (a+b) f}+\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{2 a+3 b+b x^2}{\left (1-x^2\right )^{3/2} \sqrt{a+b x^2}} \, dx,x,\sin (e+f x)\right )}{3 (a+b) f}\\ &=\frac{2 (a+2 b) \sqrt{a+b \sin ^2(e+f x)} \tan (e+f x)}{3 (a+b)^2 f}+\frac{\sec ^2(e+f x) \sqrt{a+b \sin ^2(e+f x)} \tan (e+f x)}{3 (a+b) f}+\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{b (a+3 b)-2 b (a+2 b) x^2}{\sqrt{1-x^2} \sqrt{a+b x^2}} \, dx,x,\sin (e+f x)\right )}{3 (a+b)^2 f}\\ &=\frac{2 (a+2 b) \sqrt{a+b \sin ^2(e+f x)} \tan (e+f x)}{3 (a+b)^2 f}+\frac{\sec ^2(e+f x) \sqrt{a+b \sin ^2(e+f x)} \tan (e+f x)}{3 (a+b) f}-\frac{\left (2 (a+2 b) \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 )}{3 (a+b)^2 f}+\frac{\left ((2 a+3 b) \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 )}{3 (a+b) f}\\ &=\frac{2 (a+2 b) \sqrt{a+b \sin ^2(e+f x)} \tan (e+f x)}{3 (a+b)^2 f}+\frac{\sec ^2(e+f x) \sqrt{a+b \sin ^2(e+f x)} \tan (e+f x)}{3 (a+b) f}-\frac{\left (2 (a+2 b) \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 )}{3 (a+b)^2 f \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}}+\frac{\left ((2 a+3 b) \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 )}{3 (a+b) f \sqrt{a+b \sin ^2(e+f x)}}\\ &=-\frac{2 (a+2 b) \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)}}{3 (a+b)^2 f \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}}+\frac{(2 a+3 b) \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}}}{3 (a+b) f \sqrt{a+b \sin ^2(e+f x)}}+\frac{2 (a+2 b) \sqrt{a+b \sin ^2(e+f x)} \tan (e+f x)}{3 (a+b)^2 f}+\frac{\sec ^2(e+f x) \sqrt{a+b \sin ^2(e+f x)} \tan (e+f x)}{3 (a+b) f}\\ \end{align*}
Mathematica [A] time = 2.17632, size = 205, normalized size = 0.97 \[ \frac{2 \left (2 a^2+5 a b+3 b^2\right ) \sqrt{\frac{2 a-b \cos (2 (e+f x))+b}{a}} F\left (e+f x\left |-\frac{b}{a}\right .\right )+\frac{\tan (e+f x) \sec ^2(e+f x) \left (\left (4 a^2+6 a b-2 b^2\right ) \cos (2 (e+f x))+8 a^2-b (a+2 b) \cos (4 (e+f x))+15 a b+4 b^2\right )}{\sqrt{2}}-4 a (a+2 b) \sqrt{\frac{2 a-b \cos (2 (e+f x))+b}{a}} E\left (e+f x\left |-\frac{b}{a}\right .\right )}{6 f (a+b)^2 \sqrt{2 a-b \cos (2 (e+f x))+b}} \]
Antiderivative was successfully verified.
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Maple [A] time = 2.41, size = 405, normalized size = 1.9 \begin{align*}{\frac{1}{ \left ( -3+3\,\sin \left ( fx+e \right ) \right ) \left ( 1+\sin \left ( fx+e \right ) \right ) \left ( a+b \right ) ^{2}\cos \left ( fx+e \right ) f} \left ( 2\,\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}b \left ( a+2\,b \right ) \sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4}-\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ( 2\,{a}^{2}+5\,ab+3\,{b}^{2} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) -\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ({a}^{2}+2\,ab+{b}^{2} \right ) \sin \left ( fx+e \right ) -\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}}}\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ( 2\,{\it EllipticF} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ){a}^{2}+5\,{\it EllipticF} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ) ab+3\,{\it EllipticF} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ){b}^{2}-2\,{\it EllipticE} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ){a}^{2}-4\,{\it EllipticE} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ) ab \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2} \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 )^{4}}{\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 )^{4}}{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 ^{4}{\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 )^{4}}{\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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