Optimal. Leaf size=133 \[ \frac{\cos ^3(e+f x)}{3 f (a-b)^2}-\frac{(a+b) \cos (e+f x)}{f (a-b)^3}-\frac{a b \sec (e+f x)}{2 f (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )}-\frac{\sqrt{b} (3 a+2 b) \tan ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a-b}}\right )}{2 f (a-b)^{7/2}} \]
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Rubi [A] time = 0.180449, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3664, 456, 1261, 205} \[ \frac{\cos ^3(e+f x)}{3 f (a-b)^2}-\frac{(a+b) \cos (e+f x)}{f (a-b)^3}-\frac{a b \sec (e+f x)}{2 f (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )}-\frac{\sqrt{b} (3 a+2 b) \tan ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a-b}}\right )}{2 f (a-b)^{7/2}} \]
Antiderivative was successfully verified.
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Rule 3664
Rule 456
Rule 1261
Rule 205
Rubi steps
\begin{align*} \int \frac{\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{-1+x^2}{x^4 \left (a-b+b x^2\right )^2} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{a b \sec (e+f x)}{2 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac{b \operatorname{Subst}\left (\int \frac{\frac{2}{(a-b) b}-\frac{2 a x^2}{(a-b)^2 b}+\frac{a x^4}{(a-b)^3}}{x^4 \left (a-b+b x^2\right )} \, dx,x,\sec (e+f x)\right )}{2 f}\\ &=-\frac{a b \sec (e+f x)}{2 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac{b \operatorname{Subst}\left (\int \left (\frac{2}{(a-b)^2 b x^4}+\frac{2 (a+b)}{b (-a+b)^3 x^2}+\frac{3 a+2 b}{(a-b)^3 \left (a-b+b x^2\right )}\right ) \, dx,x,\sec (e+f x)\right )}{2 f}\\ &=-\frac{(a+b) \cos (e+f x)}{(a-b)^3 f}+\frac{\cos ^3(e+f x)}{3 (a-b)^2 f}-\frac{a b \sec (e+f x)}{2 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac{(b (3 a+2 b)) \operatorname{Subst}\left (\int \frac{1}{a-b+b x^2} \, dx,x,\sec (e+f x)\right )}{2 (a-b)^3 f}\\ &=-\frac{\sqrt{b} (3 a+2 b) \tan ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a-b}}\right )}{2 (a-b)^{7/2} f}-\frac{(a+b) \cos (e+f x)}{(a-b)^3 f}+\frac{\cos ^3(e+f x)}{3 (a-b)^2 f}-\frac{a b \sec (e+f x)}{2 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )}\\ \end{align*}
Mathematica [A] time = 2.98622, size = 182, normalized size = 1.37 \[ \frac{-\frac{\cos (e+f x) \left (\frac{12 a b}{(a-b) \cos (2 (e+f x))+a+b}+9 a+15 b\right )+(b-a) \cos (3 (e+f x))}{(a-b)^3}+\frac{6 \sqrt{b} (3 a+2 b) \tan ^{-1}\left (\frac{\sqrt{a-b}-\sqrt{a} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{b}}\right )}{(a-b)^{7/2}}+\frac{6 \sqrt{b} (3 a+2 b) \tan ^{-1}\left (\frac{\sqrt{a-b}+\sqrt{a} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{b}}\right )}{(a-b)^{7/2}}}{12 f} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.078, size = 269, normalized size = 2. \begin{align*}{\frac{a \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{3\,f \left ({a}^{2}-2\,ab+{b}^{2} \right ) \left ( a-b \right ) }}-{\frac{b \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{3\,f \left ({a}^{2}-2\,ab+{b}^{2} \right ) \left ( a-b \right ) }}-{\frac{\cos \left ( fx+e \right ) a}{f \left ({a}^{2}-2\,ab+{b}^{2} \right ) \left ( a-b \right ) }}-{\frac{\cos \left ( fx+e \right ) b}{f \left ({a}^{2}-2\,ab+{b}^{2} \right ) \left ( a-b \right ) }}-{\frac{ab\cos \left ( fx+e \right ) }{2\,f \left ( a-b \right ) ^{3} \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) }}+{\frac{3\,ab}{2\,f \left ( a-b \right ) ^{3}}\arctan \left ({ \left ( a-b \right ) \cos \left ( fx+e \right ){\frac{1}{\sqrt{b \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{b \left ( a-b \right ) }}}}+{\frac{{b}^{2}}{f \left ( a-b \right ) ^{3}}\arctan \left ({ \left ( a-b \right ) \cos \left ( fx+e \right ){\frac{1}{\sqrt{b \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{b \left ( a-b \right ) }}}} \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 = 2.20797, size = 1015, normalized size = 7.63 \begin{align*} \left [\frac{4 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - 4 \,{\left (3 \, a^{2} - a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - 3 \,{\left ({\left (3 \, a^{2} - a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a b + 2 \, b^{2}\right )} \sqrt{-\frac{b}{a - b}} \log \left (-\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \,{\left (a - b\right )} \sqrt{-\frac{b}{a - b}} \cos \left (f x + e\right ) - b}{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}\right ) - 6 \,{\left (3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )}{12 \,{\left ({\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{3} b - 3 \, a^{2} b^{2} + 3 \, a b^{3} - b^{4}\right )} f\right )}}, \frac{2 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - 2 \,{\left (3 \, a^{2} - a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - 3 \,{\left ({\left (3 \, a^{2} - a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a b + 2 \, b^{2}\right )} \sqrt{\frac{b}{a - b}} \arctan \left (-\frac{{\left (a - b\right )} \sqrt{\frac{b}{a - b}} \cos \left (f x + e\right )}{b}\right ) - 3 \,{\left (3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )}{6 \,{\left ({\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{3} b - 3 \, a^{2} b^{2} + 3 \, a b^{3} - b^{4}\right )} f\right )}}\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 [B] time = 1.38939, size = 497, normalized size = 3.74 \begin{align*} \frac{a^{4} f^{11} \cos \left (f x + e\right )^{3} - 4 \, a^{3} b f^{11} \cos \left (f x + e\right )^{3} + 6 \, a^{2} b^{2} f^{11} \cos \left (f x + e\right )^{3} - 4 \, a b^{3} f^{11} \cos \left (f x + e\right )^{3} + b^{4} f^{11} \cos \left (f x + e\right )^{3} - 3 \, a^{4} f^{11} \cos \left (f x + e\right ) + 6 \, a^{3} b f^{11} \cos \left (f x + e\right ) - 6 \, a b^{3} f^{11} \cos \left (f x + e\right ) + 3 \, b^{4} f^{11} \cos \left (f x + e\right )}{3 \,{\left (a^{6} f^{12} - 6 \, a^{5} b f^{12} + 15 \, a^{4} b^{2} f^{12} - 20 \, a^{3} b^{3} f^{12} + 15 \, a^{2} b^{4} f^{12} - 6 \, a b^{5} f^{12} + b^{6} f^{12}\right )}} - \frac{a b \cos \left (f x + e\right )}{2 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}{\left (a \cos \left (f x + e\right )^{2} - b \cos \left (f x + e\right )^{2} + b\right )} f} + \frac{{\left (3 \, a b + 2 \, b^{2}\right )} \arctan \left (\frac{a \cos \left (f x + e\right ) - b \cos \left (f x + e\right )}{\sqrt{a b - b^{2}}}\right )}{2 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt{a b - b^{2}} f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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