Optimal. Leaf size=134 \[ -\frac{b (5 a-2 b) \tan (e+f x)}{3 a^2 f (a-b)^2 \sqrt{a+b \tan ^2(e+f x)}}+\frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f (a-b)^{5/2}}-\frac{b \tan (e+f x)}{3 a f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]
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Rubi [A] time = 0.112636, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {3661, 414, 527, 12, 377, 203} \[ -\frac{b (5 a-2 b) \tan (e+f x)}{3 a^2 f (a-b)^2 \sqrt{a+b \tan ^2(e+f x)}}+\frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f (a-b)^{5/2}}-\frac{b \tan (e+f x)}{3 a f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]
Antiderivative was successfully verified.
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Rule 3661
Rule 414
Rule 527
Rule 12
Rule 377
Rule 203
Rubi steps
\begin{align*} \int \frac{1}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{b \tan (e+f x)}{3 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{3 a-2 b-2 b x^2}{\left (1+x^2\right ) \left (a+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{3 a (a-b) f}\\ &=-\frac{b \tan (e+f x)}{3 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{(5 a-2 b) b \tan (e+f x)}{3 a^2 (a-b)^2 f \sqrt{a+b \tan ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{3 a^2}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{3 a^2 (a-b)^2 f}\\ &=-\frac{b \tan (e+f x)}{3 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{(5 a-2 b) b \tan (e+f x)}{3 a^2 (a-b)^2 f \sqrt{a+b \tan ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{(a-b)^2 f}\\ &=-\frac{b \tan (e+f x)}{3 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{(5 a-2 b) b \tan (e+f x)}{3 a^2 (a-b)^2 f \sqrt{a+b \tan ^2(e+f x)}}+\frac{\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 )}{(a-b)^2 f}\\ &=\frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{(a-b)^{5/2} f}-\frac{b \tan (e+f x)}{3 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{(5 a-2 b) b \tan (e+f x)}{3 a^2 (a-b)^2 f \sqrt{a+b \tan ^2(e+f x)}}\\ \end{align*}
Mathematica [C] time = 9.42009, size = 1331, normalized size = 9.93 \[ \text{result too large to display} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.021, size = 176, normalized size = 1.3 \begin{align*} -{\frac{b\tan \left ( fx+e \right ) }{f \left ( a-b \right ) ^{2}a}{\frac{1}{\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}}}+{\frac{1}{f \left ( a-b \right ) ^{3}{b}^{2}}\sqrt{{b}^{4} \left ( a-b \right ) }\arctan \left ({ \left ( a-b \right ){b}^{2}\tan \left ( fx+e \right ){\frac{1}{\sqrt{{b}^{4} \left ( a-b \right ) }}}{\frac{1}{\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}}} \right ) }-{\frac{b\tan \left ( fx+e \right ) }{3\,a \left ( a-b \right ) f} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{-{\frac{3}{2}}}}-{\frac{2\,b\tan \left ( fx+e \right ) }{3\,f \left ( a-b \right ){a}^{2}}{\frac{1}{\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}}} \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 [B] time = 2.47403, size = 1245, normalized size = 9.29 \begin{align*} \left [-\frac{3 \,{\left (a^{2} b^{2} \tan \left (f x + e\right )^{4} + 2 \, a^{3} b \tan \left (f x + e\right )^{2} + a^{4}\right )} \sqrt{-a + b} \log \left (-\frac{{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - 2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{-a + b} \tan \left (f x + e\right ) - a}{\tan \left (f x + e\right )^{2} + 1}\right ) + 2 \,{\left ({\left (5 \, a^{2} b^{2} - 7 \, a b^{3} + 2 \, b^{4}\right )} \tan \left (f x + e\right )^{3} + 3 \,{\left (2 \, a^{3} b - 3 \, a^{2} b^{2} + a b^{3}\right )} \tan \left (f x + e\right )\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a}}{6 \,{\left ({\left (a^{5} b^{2} - 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} - a^{2} b^{5}\right )} f \tan \left (f x + e\right )^{4} + 2 \,{\left (a^{6} b - 3 \, a^{5} b^{2} + 3 \, a^{4} b^{3} - a^{3} b^{4}\right )} f \tan \left (f x + e\right )^{2} +{\left (a^{7} - 3 \, a^{6} b + 3 \, a^{5} b^{2} - a^{4} b^{3}\right )} f\right )}}, \frac{3 \,{\left (a^{2} b^{2} \tan \left (f x + e\right )^{4} + 2 \, a^{3} b \tan \left (f x + e\right )^{2} + a^{4}\right )} \sqrt{a - b} \arctan \left (-\frac{\sqrt{b \tan \left (f x + e\right )^{2} + a}}{\sqrt{a - b} \tan \left (f x + e\right )}\right ) -{\left ({\left (5 \, a^{2} b^{2} - 7 \, a b^{3} + 2 \, b^{4}\right )} \tan \left (f x + e\right )^{3} + 3 \,{\left (2 \, a^{3} b - 3 \, a^{2} b^{2} + a b^{3}\right )} \tan \left (f x + e\right )\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a}}{3 \,{\left ({\left (a^{5} b^{2} - 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} - a^{2} b^{5}\right )} f \tan \left (f x + e\right )^{4} + 2 \,{\left (a^{6} b - 3 \, a^{5} b^{2} + 3 \, a^{4} b^{3} - a^{3} b^{4}\right )} f \tan \left (f x + e\right )^{2} +{\left (a^{7} - 3 \, a^{6} b + 3 \, a^{5} b^{2} - a^{4} b^{3}\right )} f\right )}}\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{1}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac{5}{2}}}\, 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{1}{{\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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