Optimal. Leaf size=189 \[ \frac{b^{3/2} \left (35 a^2-42 a b+15 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{8 a^{7/2} f (a-b)^3}-\frac{\left (8 a^2-27 a b+15 b^2\right ) \cot (e+f x)}{8 a^3 f (a-b)^2}-\frac{b (9 a-5 b) \cot (e+f x)}{8 a^2 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )}-\frac{b \cot (e+f x)}{4 a f (a-b) \left (a+b \tan ^2(e+f x)\right )^2}-\frac{x}{(a-b)^3} \]
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Rubi [A] time = 0.290321, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {3670, 472, 579, 583, 522, 203, 205} \[ \frac{b^{3/2} \left (35 a^2-42 a b+15 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{8 a^{7/2} f (a-b)^3}-\frac{\left (8 a^2-27 a b+15 b^2\right ) \cot (e+f x)}{8 a^3 f (a-b)^2}-\frac{b (9 a-5 b) \cot (e+f x)}{8 a^2 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )}-\frac{b \cot (e+f x)}{4 a f (a-b) \left (a+b \tan ^2(e+f x)\right )^2}-\frac{x}{(a-b)^3} \]
Antiderivative was successfully verified.
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Rule 3670
Rule 472
Rule 579
Rule 583
Rule 522
Rule 203
Rule 205
Rubi steps
\begin{align*} \int \frac{\cot ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 \left (1+x^2\right ) \left (a+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{b \cot (e+f x)}{4 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}+\frac{\operatorname{Subst}\left (\int \frac{4 a-5 b-5 b x^2}{x^2 \left (1+x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 a (a-b) f}\\ &=-\frac{b \cot (e+f x)}{4 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{(9 a-5 b) b \cot (e+f x)}{8 a^2 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{8 a^2-27 a b+15 b^2-3 (9 a-5 b) b x^2}{x^2 \left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{8 a^2 (a-b)^2 f}\\ &=-\frac{\left (8 a^2-27 a b+15 b^2\right ) \cot (e+f x)}{8 a^3 (a-b)^2 f}-\frac{b \cot (e+f x)}{4 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{(9 a-5 b) b \cot (e+f x)}{8 a^2 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{8 a^3+8 a^2 b-27 a b^2+15 b^3+b \left (8 a^2-27 a b+15 b^2\right ) x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{8 a^3 (a-b)^2 f}\\ &=-\frac{\left (8 a^2-27 a b+15 b^2\right ) \cot (e+f x)}{8 a^3 (a-b)^2 f}-\frac{b \cot (e+f x)}{4 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{(9 a-5 b) b \cot (e+f x)}{8 a^2 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{(a-b)^3 f}+\frac{\left (b^2 \left (35 a^2-42 a b+15 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{8 a^3 (a-b)^3 f}\\ &=-\frac{x}{(a-b)^3}+\frac{b^{3/2} \left (35 a^2-42 a b+15 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{8 a^{7/2} (a-b)^3 f}-\frac{\left (8 a^2-27 a b+15 b^2\right ) \cot (e+f x)}{8 a^3 (a-b)^2 f}-\frac{b \cot (e+f x)}{4 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{(9 a-5 b) b \cot (e+f x)}{8 a^2 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}\\ \end{align*}
Mathematica [A] time = 2.01344, size = 174, normalized size = 0.92 \[ \frac{\frac{b^{3/2} \left (35 a^2-42 a b+15 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{a^{7/2} (a-b)^3}-\frac{4 b^3 \sin (2 (e+f x))}{a^2 (a-b)^2 ((a-b) \cos (2 (e+f x))+a+b)^2}+\frac{b^2 (13 a-7 b) \sin (2 (e+f x))}{a^3 (a-b)^2 ((a-b) \cos (2 (e+f x))+a+b)}-\frac{8 \cot (e+f x)}{a^3}+\frac{8 (e+f x)}{(b-a)^3}}{8 f} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.09, size = 379, normalized size = 2. \begin{align*} -{\frac{1}{f{a}^{3}\tan \left ( fx+e \right ) }}+{\frac{11\,{b}^{3} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{8\,f \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}a}}-{\frac{9\,{b}^{4} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{4\,f \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}{a}^{2}}}+{\frac{7\,{b}^{5} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{8\,f{a}^{3} \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{13\,{b}^{2}\tan \left ( fx+e \right ) }{8\,f \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{11\,{b}^{3}\tan \left ( fx+e \right ) }{4\,f \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}a}}+{\frac{9\,{b}^{4}\tan \left ( fx+e \right ) }{8\,f \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}{a}^{2}}}+{\frac{35\,{b}^{2}}{8\,f \left ( a-b \right ) ^{3}a}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{21\,{b}^{3}}{4\,f{a}^{2} \left ( a-b \right ) ^{3}}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{15\,{b}^{4}}{8\,f{a}^{3} \left ( a-b \right ) ^{3}}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) }{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 [B] time = 1.93673, size = 1989, normalized size = 10.52 \begin{align*} \text{result too large to display} \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 = 1.6405, size = 312, normalized size = 1.65 \begin{align*} \frac{\frac{{\left (35 \, a^{2} b^{2} - 42 \, a b^{3} + 15 \, b^{4}\right )}{\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 )}}{{\left (a^{6} - 3 \, a^{5} b + 3 \, a^{4} b^{2} - a^{3} b^{3}\right )} \sqrt{a b}} - \frac{8 \,{\left (f x + e\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac{11 \, a b^{3} \tan \left (f x + e\right )^{3} - 7 \, b^{4} \tan \left (f x + e\right )^{3} + 13 \, a^{2} b^{2} \tan \left (f x + e\right ) - 9 \, a b^{3} \tan \left (f x + e\right )}{{\left (a^{5} - 2 \, a^{4} b + a^{3} b^{2}\right )}{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{2}} - \frac{8}{a^{3} \tan \left (f x + e\right )}}{8 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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