Optimal. Leaf size=218 \[ \frac{b^{7/2} (9 a-7 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{2 a^{9/2} f (a-b)^2}+\frac{\left (2 a^2+2 a b-7 b^2\right ) \cot ^3(e+f x)}{6 a^3 f (a-b)}-\frac{\left (2 a^2 b+2 a^3+2 a b^2-7 b^3\right ) \cot (e+f x)}{2 a^4 f (a-b)}-\frac{(2 a-7 b) \cot ^5(e+f x)}{10 a^2 f (a-b)}-\frac{b \cot ^5(e+f x)}{2 a f (a-b) \left (a+b \tan ^2(e+f x)\right )}-\frac{x}{(a-b)^2} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.341666, antiderivative size = 218, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3670, 472, 583, 522, 203, 205} \[ \frac{b^{7/2} (9 a-7 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{2 a^{9/2} f (a-b)^2}+\frac{\left (2 a^2+2 a b-7 b^2\right ) \cot ^3(e+f x)}{6 a^3 f (a-b)}-\frac{\left (2 a^2 b+2 a^3+2 a b^2-7 b^3\right ) \cot (e+f x)}{2 a^4 f (a-b)}-\frac{(2 a-7 b) \cot ^5(e+f x)}{10 a^2 f (a-b)}-\frac{b \cot ^5(e+f x)}{2 a f (a-b) \left (a+b \tan ^2(e+f x)\right )}-\frac{x}{(a-b)^2} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 3670
Rule 472
Rule 583
Rule 522
Rule 203
Rule 205
Rubi steps
\begin{align*} \int \frac{\cot ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^6 \left (1+x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{b \cot ^5(e+f x)}{2 a (a-b) f \left (a+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{2 a-7 b-7 b x^2}{x^6 \left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{2 a (a-b) f}\\ &=-\frac{(2 a-7 b) \cot ^5(e+f x)}{10 a^2 (a-b) f}-\frac{b \cot ^5(e+f x)}{2 a (a-b) f \left (a+b \tan ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{5 \left (2 a^2+2 a b-7 b^2\right )+5 (2 a-7 b) b x^2}{x^4 \left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{10 a^2 (a-b) f}\\ &=\frac{\left (2 a^2+2 a b-7 b^2\right ) \cot ^3(e+f x)}{6 a^3 (a-b) f}-\frac{(2 a-7 b) \cot ^5(e+f x)}{10 a^2 (a-b) f}-\frac{b \cot ^5(e+f x)}{2 a (a-b) f \left (a+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{15 \left (2 a^3+2 a^2 b+2 a b^2-7 b^3\right )+15 b \left (2 a^2+2 a b-7 b^2\right ) x^2}{x^2 \left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{30 a^3 (a-b) f}\\ &=-\frac{\left (2 a^3+2 a^2 b+2 a b^2-7 b^3\right ) \cot (e+f x)}{2 a^4 (a-b) f}+\frac{\left (2 a^2+2 a b-7 b^2\right ) \cot ^3(e+f x)}{6 a^3 (a-b) f}-\frac{(2 a-7 b) \cot ^5(e+f x)}{10 a^2 (a-b) f}-\frac{b \cot ^5(e+f x)}{2 a (a-b) f \left (a+b \tan ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{15 \left (2 a^4+2 a^3 b+2 a^2 b^2+2 a b^3-7 b^4\right )+15 b \left (2 a^3+2 a^2 b+2 a b^2-7 b^3\right ) x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{30 a^4 (a-b) f}\\ &=-\frac{\left (2 a^3+2 a^2 b+2 a b^2-7 b^3\right ) \cot (e+f x)}{2 a^4 (a-b) f}+\frac{\left (2 a^2+2 a b-7 b^2\right ) \cot ^3(e+f x)}{6 a^3 (a-b) f}-\frac{(2 a-7 b) \cot ^5(e+f x)}{10 a^2 (a-b) f}-\frac{b \cot ^5(e+f x)}{2 a (a-b) 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)^2 f}+\frac{\left ((9 a-7 b) b^4\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{2 a^4 (a-b)^2 f}\\ &=-\frac{x}{(a-b)^2}+\frac{(9 a-7 b) b^{7/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{2 a^{9/2} (a-b)^2 f}-\frac{\left (2 a^3+2 a^2 b+2 a b^2-7 b^3\right ) \cot (e+f x)}{2 a^4 (a-b) f}+\frac{\left (2 a^2+2 a b-7 b^2\right ) \cot ^3(e+f x)}{6 a^3 (a-b) f}-\frac{(2 a-7 b) \cot ^5(e+f x)}{10 a^2 (a-b) f}-\frac{b \cot ^5(e+f x)}{2 a (a-b) f \left (a+b \tan ^2(e+f x)\right )}\\ \end{align*}
Mathematica [A] time = 5.02335, size = 165, normalized size = 0.76 \[ \frac{\frac{15 b^{7/2} (9 a-7 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{a^{9/2} (a-b)^2}+\frac{15 \left (\frac{b^4 (a-b) \sin (2 (e+f x))}{a^4 ((a-b) \cos (2 (e+f x))+a+b)}-2 (e+f x)\right )}{(a-b)^2}-\frac{2 \cot (e+f x) \left (3 a^2 \csc ^4(e+f x)+23 a^2-a (11 a+10 b) \csc ^2(e+f x)+40 a b+45 b^2\right )}{a^4}}{30 f} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [A] time = 0.102, size = 272, normalized size = 1.3 \begin{align*} -{\frac{1}{5\,f{a}^{2} \left ( \tan \left ( fx+e \right ) \right ) ^{5}}}+{\frac{1}{3\,f{a}^{2} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}}+{\frac{2\,b}{3\,f{a}^{3} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}}-{\frac{1}{f{a}^{2}\tan \left ( fx+e \right ) }}-2\,{\frac{b}{f{a}^{3}\tan \left ( fx+e \right ) }}-3\,{\frac{{b}^{2}}{f{a}^{4}\tan \left ( fx+e \right ) }}+{\frac{{b}^{4}\tan \left ( fx+e \right ) }{2\,f{a}^{3} \left ( a-b \right ) ^{2} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{{b}^{5}\tan \left ( fx+e \right ) }{2\,f{a}^{4} \left ( a-b \right ) ^{2} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}+{\frac{9\,{b}^{4}}{2\,f{a}^{3} \left ( a-b \right ) ^{2}}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{7\,{b}^{5}}{2\,f{a}^{4} \left ( a-b \right ) ^{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 ) ^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
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.
[In]
[Out]
________________________________________________________________________________________
Fricas [A] time = 1.92831, size = 1520, normalized size = 6.97 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
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.
[In]
[Out]
________________________________________________________________________________________
Giac [A] time = 1.44007, size = 304, normalized size = 1.39 \begin{align*} \frac{\frac{15 \, b^{4} \tan \left (f x + e\right )}{{\left (a^{5} - a^{4} b\right )}{\left (b \tan \left (f x + e\right )^{2} + a\right )}} + \frac{15 \,{\left (9 \, a b^{4} - 7 \, b^{5}\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} - 2 \, a^{5} b + a^{4} b^{2}\right )} \sqrt{a b}} - \frac{30 \,{\left (f x + e\right )}}{a^{2} - 2 \, a b + b^{2}} - \frac{2 \,{\left (15 \, a^{2} \tan \left (f x + e\right )^{4} + 30 \, a b \tan \left (f x + e\right )^{4} + 45 \, b^{2} \tan \left (f x + e\right )^{4} - 5 \, a^{2} \tan \left (f x + e\right )^{2} - 10 \, a b \tan \left (f x + e\right )^{2} + 3 \, a^{2}\right )}}{a^{4} \tan \left (f x + e\right )^{5}}}{30 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]