Optimal. Leaf size=150 \[ -\frac{\sqrt{b} \left (15 a^2-10 a b+3 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{8 a^{5/2} f (a-b)^3}-\frac{b (7 a-3 b) \tan (e+f x)}{8 a^2 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )}-\frac{b \tan (e+f x)}{4 a f (a-b) \left (a+b \tan ^2(e+f x)\right )^2}+\frac{x}{(a-b)^3} \]
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Rubi [A] time = 0.151704, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {3661, 414, 527, 522, 203, 205} \[ -\frac{\sqrt{b} \left (15 a^2-10 a b+3 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{8 a^{5/2} f (a-b)^3}-\frac{b (7 a-3 b) \tan (e+f x)}{8 a^2 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )}-\frac{b \tan (e+f x)}{4 a f (a-b) \left (a+b \tan ^2(e+f x)\right )^2}+\frac{x}{(a-b)^3} \]
Antiderivative was successfully verified.
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Rule 3661
Rule 414
Rule 527
Rule 522
Rule 203
Rule 205
Rubi steps
\begin{align*} \int \frac{1}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \left (a+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{b \tan (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-3 b-3 b 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 \tan (e+f x)}{4 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{(7 a-3 b) b \tan (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^2-7 a b+3 b^2-(7 a-3 b) b 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{b \tan (e+f x)}{4 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{(7 a-3 b) b \tan (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 \left (15 a^2-10 a b+3 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{8 a^2 (a-b)^3 f}\\ &=\frac{x}{(a-b)^3}-\frac{\sqrt{b} \left (15 a^2-10 a b+3 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{8 a^{5/2} (a-b)^3 f}-\frac{b \tan (e+f x)}{4 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{(7 a-3 b) b \tan (e+f x)}{8 a^2 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}\\ \end{align*}
Mathematica [A] time = 1.78068, size = 138, normalized size = 0.92 \[ -\frac{\frac{\sqrt{b} \left (15 a^2-10 a b+3 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{a^{5/2}}+\frac{b (7 a-3 b) (a-b) \tan (e+f x)}{a^2 \left (a+b \tan ^2(e+f x)\right )}+\frac{2 b (a-b)^2 \tan (e+f x)}{a \left (a+b \tan ^2(e+f x)\right )^2}-8 \tan ^{-1}(\tan (e+f x))}{8 f (a-b)^3} \]
Antiderivative was successfully verified.
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Maple [B] time = 0., size = 350, normalized size = 2.3 \begin{align*} -{\frac{7\,{b}^{2} \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}}}+{\frac{5\,{b}^{3} \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}}-{\frac{3\,{b}^{4} \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}^{2}}}-{\frac{9\,\tan \left ( fx+e \right ) ab}{8\,f \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{7\,{b}^{2}\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}}}-{\frac{5\,{b}^{3}\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}}-{\frac{15\,b}{8\,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\,{b}^{2}}{4\,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{3\,{b}^{3}}{8\,f \left ( a-b \right ) ^{3}{a}^{2}}\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 = 1.71621, size = 1643, normalized size = 10.95 \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 = 1.2679, size = 288, normalized size = 1.92 \begin{align*} -\frac{\frac{{\left (15 \, a^{2} b - 10 \, a b^{2} + 3 \, b^{3}\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^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} 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{7 \, a b^{2} \tan \left (f x + e\right )^{3} - 3 \, b^{3} \tan \left (f x + e\right )^{3} + 9 \, a^{2} b \tan \left (f x + e\right ) - 5 \, a b^{2} \tan \left (f x + e\right )}{{\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )}{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{2}}}{8 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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