Optimal. Leaf size=129 \[ \frac{x \left (3 a^2+6 a b-b^2\right )}{8 (a-b)^3}-\frac{a^{3/2} \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{f (a-b)^3}+\frac{\sin (e+f x) \cos ^3(e+f x)}{4 f (a-b)}-\frac{(5 a-b) \sin (e+f x) \cos (e+f x)}{8 f (a-b)^2} \]
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Rubi [A] time = 0.151545, antiderivative size = 129, 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 = {3663, 470, 527, 522, 203, 205} \[ \frac{x \left (3 a^2+6 a b-b^2\right )}{8 (a-b)^3}-\frac{a^{3/2} \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{f (a-b)^3}+\frac{\sin (e+f x) \cos ^3(e+f x)}{4 f (a-b)}-\frac{(5 a-b) \sin (e+f x) \cos (e+f x)}{8 f (a-b)^2} \]
Antiderivative was successfully verified.
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Rule 3663
Rule 470
Rule 527
Rule 522
Rule 203
Rule 205
Rubi steps
\begin{align*} \int \frac{\sin ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (1+x^2\right )^3 \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f}-\frac{\operatorname{Subst}\left (\int \frac{a+(-4 a+b) x^2}{\left (1+x^2\right )^2 \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{4 (a-b) f}\\ &=-\frac{(5 a-b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f}+\frac{\operatorname{Subst}\left (\int \frac{a (3 a+b)-(5 a-b) b x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{8 (a-b)^2 f}\\ &=-\frac{(5 a-b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f}-\frac{\left (a^2 b\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{(a-b)^3 f}+\frac{\left (3 a^2+6 a b-b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{8 (a-b)^3 f}\\ &=\frac{\left (3 a^2+6 a b-b^2\right ) x}{8 (a-b)^3}-\frac{a^{3/2} \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{(a-b)^3 f}-\frac{(5 a-b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f}\\ \end{align*}
Mathematica [A] time = 0.261784, size = 99, normalized size = 0.77 \[ \frac{4 \left (3 a^2+6 a b-b^2\right ) (e+f x)-32 a^{3/2} \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )+(a-b)^2 \sin (4 (e+f x))-8 a (a-b) \sin (2 (e+f x))}{32 f (a-b)^3} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.063, size = 304, normalized size = 2.4 \begin{align*} -{\frac{{a}^{2}b}{f \left ( a-b \right ) ^{3}}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{5\, \left ( \tan \left ( fx+e \right ) \right ) ^{3}{a}^{2}}{8\,f \left ( a-b \right ) ^{3} \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{3\, \left ( \tan \left ( fx+e \right ) \right ) ^{3}ab}{4\,f \left ( a-b \right ) ^{3} \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{ \left ( \tan \left ( fx+e \right ) \right ) ^{3}{b}^{2}}{8\,f \left ( a-b \right ) ^{3} \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{3\,\tan \left ( fx+e \right ){a}^{2}}{8\,f \left ( a-b \right ) ^{3} \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{\tan \left ( fx+e \right ) ab}{4\,f \left ( a-b \right ) ^{3} \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{\tan \left ( fx+e \right ){b}^{2}}{8\,f \left ( a-b \right ) ^{3} \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{3\,\arctan \left ( \tan \left ( fx+e \right ) \right ){a}^{2}}{8\,f \left ( a-b \right ) ^{3}}}+{\frac{3\,\arctan \left ( \tan \left ( fx+e \right ) \right ) ab}{4\,f \left ( a-b \right ) ^{3}}}-{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ){b}^{2}}{8\,f \left ( a-b \right ) ^{3}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.07707, size = 898, normalized size = 6.96 \begin{align*} \left [\frac{{\left (3 \, a^{2} + 6 \, a b - b^{2}\right )} f x - 2 \, \sqrt{-a b} a \log \left (\frac{{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \,{\left (3 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \,{\left ({\left (a + b\right )} \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt{-a b} \sin \left (f x + e\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \,{\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}}\right ) +{\left (2 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{3} -{\left (5 \, a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} f}, \frac{{\left (3 \, a^{2} + 6 \, a b - b^{2}\right )} f x + 4 \, \sqrt{a b} a \arctan \left (\frac{{\left ({\left (a + b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt{a b}}{2 \, a b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) +{\left (2 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{3} -{\left (5 \, a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} f}\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 [A] time = 1.44372, size = 257, normalized size = 1.99 \begin{align*} -\frac{\frac{8 \,{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (f x + e\right )}{\sqrt{a b}}\right )\right )} a^{2} b}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt{a b}} - \frac{{\left (3 \, a^{2} + 6 \, a b - b^{2}\right )}{\left (f x + e\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac{5 \, a \tan \left (f x + e\right )^{3} - b \tan \left (f x + e\right )^{3} + 3 \, a \tan \left (f x + e\right ) + b \tan \left (f x + e\right )}{{\left (a^{2} - 2 \, a b + b^{2}\right )}{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{2}}}{8 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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