Optimal. Leaf size=348 \[ -\frac{2 (2 a-b) \tan (e+f x)}{3 f (a+b)^2 \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{8 b (a-b) \sin (e+f x) \cos (e+f x)}{3 f (a+b)^4 \sqrt{a+b \sin ^2(e+f x)}}+\frac{b (5 a-3 b) \sin (e+f x) \cos (e+f x)}{3 f (a+b)^3 \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{\tan (e+f x) \sec ^2(e+f x)}{3 f (a+b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{(5 a-3 b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 f (a+b)^3 \sqrt{a+b \sin ^2(e+f x)}}+\frac{8 (a-b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 f (a+b)^4 \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}} \]
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Rubi [A] time = 0.426439, antiderivative size = 348, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {3196, 470, 527, 524, 426, 424, 421, 419} \[ -\frac{2 (2 a-b) \tan (e+f x)}{3 f (a+b)^2 \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{8 b (a-b) \sin (e+f x) \cos (e+f x)}{3 f (a+b)^4 \sqrt{a+b \sin ^2(e+f x)}}+\frac{b (5 a-3 b) \sin (e+f x) \cos (e+f x)}{3 f (a+b)^3 \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{\tan (e+f x) \sec ^2(e+f x)}{3 f (a+b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{(5 a-3 b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 f (a+b)^3 \sqrt{a+b \sin ^2(e+f x)}}+\frac{8 (a-b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 f (a+b)^4 \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}} \]
Antiderivative was successfully verified.
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Rule 3196
Rule 470
Rule 527
Rule 524
Rule 426
Rule 424
Rule 421
Rule 419
Rubi steps
\begin{align*} \int \frac{\tan ^4(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{x^4}{\left (1-x^2\right )^{5/2} \left (a+b x^2\right )^{5/2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\sec ^2(e+f x) \tan (e+f x)}{3 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{a+(3 a-2 b) x^2}{\left (1-x^2\right )^{3/2} \left (a+b x^2\right )^{5/2}} \, dx,x,\sin (e+f x)\right )}{3 (a+b) f}\\ &=-\frac{2 (2 a-b) \tan (e+f x)}{3 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{\sec ^2(e+f x) \tan (e+f x)}{3 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{-3 a (a-b)+6 (2 a-b) b x^2}{\sqrt{1-x^2} \left (a+b x^2\right )^{5/2}} \, dx,x,\sin (e+f x)\right )}{3 (a+b)^2 f}\\ &=\frac{(5 a-3 b) b \cos (e+f x) \sin (e+f x)}{3 (a+b)^3 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{2 (2 a-b) \tan (e+f x)}{3 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{\sec ^2(e+f x) \tan (e+f x)}{3 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{3 a^2 (3 a-5 b)-3 a (5 a-3 b) b x^2}{\sqrt{1-x^2} \left (a+b x^2\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{9 a (a+b)^3 f}\\ &=\frac{(5 a-3 b) b \cos (e+f x) \sin (e+f x)}{3 (a+b)^3 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{8 (a-b) b \cos (e+f x) \sin (e+f x)}{3 (a+b)^4 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{2 (2 a-b) \tan (e+f x)}{3 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{\sec ^2(e+f x) \tan (e+f x)}{3 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{-3 a^2 (a-3 b) (3 a-b)-24 a^2 (a-b) b x^2}{\sqrt{1-x^2} \sqrt{a+b x^2}} \, dx,x,\sin (e+f x)\right )}{9 a^2 (a+b)^4 f}\\ &=\frac{(5 a-3 b) b \cos (e+f x) \sin (e+f x)}{3 (a+b)^3 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{8 (a-b) b \cos (e+f x) \sin (e+f x)}{3 (a+b)^4 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{2 (2 a-b) \tan (e+f x)}{3 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{\sec ^2(e+f x) \tan (e+f x)}{3 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{\left (8 (a-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)^4 f}-\frac{\left ((5 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)^3 f}\\ &=\frac{(5 a-3 b) b \cos (e+f x) \sin (e+f x)}{3 (a+b)^3 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{8 (a-b) b \cos (e+f x) \sin (e+f x)}{3 (a+b)^4 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{2 (2 a-b) \tan (e+f x)}{3 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{\sec ^2(e+f x) \tan (e+f x)}{3 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{\left (8 (a-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)^4 f \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}}-\frac{\left ((5 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)^3 f \sqrt{a+b \sin ^2(e+f x)}}\\ &=\frac{(5 a-3 b) b \cos (e+f x) \sin (e+f x)}{3 (a+b)^3 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{8 (a-b) b \cos (e+f x) \sin (e+f x)}{3 (a+b)^4 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{8 (a-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)^4 f \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}}-\frac{(5 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)^3 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{2 (2 a-b) \tan (e+f x)}{3 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{\sec ^2(e+f x) \tan (e+f x)}{3 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}\\ \end{align*}
Mathematica [A] time = 3.44579, size = 235, normalized size = 0.68 \[ \frac{2 a b \left (\frac{2 a-b \cos (2 (e+f x))+b}{a}\right )^{3/2} \left (\left (-5 a^2-2 a b+3 b^2\right ) F\left (e+f x\left |-\frac{b}{a}\right .\right )+8 a (a-b) E\left (e+f x\left |-\frac{b}{a}\right .\right )\right )+\sqrt{2} b \left (2 a b (a+b) \sin (2 (e+f x))+4 b (a-b) \sin (2 (e+f x)) (2 a-b \cos (2 (e+f x))+b)-4 (a-b) \tan (e+f x) (2 a-b \cos (2 (e+f x))+b)^2+(a+b) \tan (e+f x) \sec ^2(e+f x) (2 a-b \cos (2 (e+f x))+b)^2\right )}{6 b f (a+b)^4 (2 a-b \cos (2 (e+f x))+b)^{3/2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 3.062, size = 666, normalized size = 1.9 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [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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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} \tan \left (f x + e\right )^{4}}{b^{3} \cos \left (f x + e\right )^{6} - 3 \,{\left (a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{4} - a^{3} - 3 \, a^{2} b - 3 \, a b^{2} - b^{3} + 3 \,{\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{2}}, x\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 [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (f x + e\right )^{4}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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