Optimal. Leaf size=246 \[ \frac{\left (3 a^2+24 a b+8 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{8 f (a-b)^{9/2}}-\frac{5 b (11 a+10 b) \tan (e+f x)}{24 f (a-b)^4 \sqrt{a+b \tan ^2(e+f x)}}-\frac{b (23 a+12 b) \tan (e+f x)}{24 f (a-b)^3 \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\sin (e+f x) \cos ^3(e+f x)}{4 f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{(5 a+2 b) \sin (e+f x) \cos (e+f x)}{8 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]
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Rubi [A] time = 0.333422, antiderivative size = 246, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3663, 470, 527, 12, 377, 203} \[ \frac{\left (3 a^2+24 a b+8 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{8 f (a-b)^{9/2}}-\frac{5 b (11 a+10 b) \tan (e+f x)}{24 f (a-b)^4 \sqrt{a+b \tan ^2(e+f x)}}-\frac{b (23 a+12 b) \tan (e+f x)}{24 f (a-b)^3 \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\sin (e+f x) \cos ^3(e+f x)}{4 f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{(5 a+2 b) \sin (e+f x) \cos (e+f x)}{8 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]
Antiderivative was successfully verified.
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Rule 3663
Rule 470
Rule 527
Rule 12
Rule 377
Rule 203
Rubi steps
\begin{align*} \int \frac{\sin ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (1+x^2\right )^3 \left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{\operatorname{Subst}\left (\int \frac{a-2 (2 a+b) x^2}{\left (1+x^2\right )^2 \left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{4 (a-b) f}\\ &=-\frac{(5 a+2 b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{a (3 a+4 b)-4 b (5 a+2 b) x^2}{\left (1+x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{8 (a-b)^2 f}\\ &=-\frac{(5 a+2 b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{b (23 a+12 b) \tan (e+f x)}{24 (a-b)^3 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{a^2 (9 a+26 b)-2 a b (23 a+12 b) x^2}{\left (1+x^2\right ) \left (a+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{24 a (a-b)^3 f}\\ &=-\frac{(5 a+2 b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{b (23 a+12 b) \tan (e+f x)}{24 (a-b)^3 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{5 b (11 a+10 b) \tan (e+f x)}{24 (a-b)^4 f \sqrt{a+b \tan ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{3 a^2 \left (3 a^2+24 a b+8 b^2\right )}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{24 a^2 (a-b)^4 f}\\ &=-\frac{(5 a+2 b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{b (23 a+12 b) \tan (e+f x)}{24 (a-b)^3 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{5 b (11 a+10 b) \tan (e+f x)}{24 (a-b)^4 f \sqrt{a+b \tan ^2(e+f x)}}+\frac{\left (3 a^2+24 a b+8 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{8 (a-b)^4 f}\\ &=-\frac{(5 a+2 b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{b (23 a+12 b) \tan (e+f x)}{24 (a-b)^3 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{5 b (11 a+10 b) \tan (e+f x)}{24 (a-b)^4 f \sqrt{a+b \tan ^2(e+f x)}}+\frac{\left (3 a^2+24 a b+8 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-(-a+b) x^2} \, dx,x,\frac{\tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{8 (a-b)^4 f}\\ &=\frac{\left (3 a^2+24 a b+8 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{8 (a-b)^{9/2} f}-\frac{(5 a+2 b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{b (23 a+12 b) \tan (e+f x)}{24 (a-b)^3 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{5 b (11 a+10 b) \tan (e+f x)}{24 (a-b)^4 f \sqrt{a+b \tan ^2(e+f x)}}\\ \end{align*}
Mathematica [C] time = 5.65234, size = 378, normalized size = 1.54 \[ -\frac{\sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (-3 \sqrt{2} a b \left (3 a^2+24 a b+8 b^2\right ) \sin (2 (e+f x)) \sin ^2(e+f x) \left (\frac{\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}\right )^{3/2} \left (2 (a-b) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}}}{\sqrt{2}}\right ),1\right )-2 a \Pi \left (-\frac{b}{a-b};\left .\sin ^{-1}\left (\frac{\sqrt{\frac{(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt{2}}\right )\right |1\right )\right )-a (a-b) \left (64 a b^2 \sin (2 (e+f x))-64 b (3 a+2 b) \sin (2 (e+f x)) ((a-b) \cos (2 (e+f x))+a+b)-6 (4 a+7 b) \sin (2 (e+f x)) ((a-b) \cos (2 (e+f x))+a+b)^2+3 (a-b) \sin (4 (e+f x)) ((a-b) \cos (2 (e+f x))+a+b)^2\right )\right )}{96 \sqrt{2} a f (a-b)^5 ((a-b) \cos (2 (e+f x))+a+b)^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 1.49, size = 7943, normalized size = 32.3 \begin{align*} \text{output too large to display} \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{\sin \left (f x + e\right )^{4}}{{\left (b \tan \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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Fricas [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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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{\sin \left (f x + e\right )^{4}}{{\left (b \tan \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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