Optimal. Leaf size=188 \[ -\frac{b \left (5 a^2+2 b^2\right ) \tan (e+f x)}{10 f (a+b)^4 \left (a+b \tan ^2(e+f x)+b\right )}-\frac{\left (5 a^2-10 a b-b^2\right ) \cot (e+f x)}{5 f (a+b)^4}-\frac{a \sqrt{b} (3 a-4 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 f (a+b)^{9/2}}-\frac{(10 a+3 b) \cot ^3(e+f x)}{15 f (a+b)^3}-\frac{\cot ^5(e+f x)}{5 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.26434, antiderivative size = 188, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {4132, 462, 456, 1261, 205} \[ -\frac{b \left (5 a^2+2 b^2\right ) \tan (e+f x)}{10 f (a+b)^4 \left (a+b \tan ^2(e+f x)+b\right )}-\frac{\left (5 a^2-10 a b-b^2\right ) \cot (e+f x)}{5 f (a+b)^4}-\frac{a \sqrt{b} (3 a-4 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 f (a+b)^{9/2}}-\frac{(10 a+3 b) \cot ^3(e+f x)}{15 f (a+b)^3}-\frac{\cot ^5(e+f x)}{5 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 4132
Rule 462
Rule 456
Rule 1261
Rule 205
Rubi steps
\begin{align*} \int \frac{\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^2}{x^6 \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cot ^5(e+f x)}{5 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{10 a+3 b+5 (a+b) x^2}{x^4 \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{5 (a+b) f}\\ &=-\frac{\cot ^5(e+f x)}{5 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{b \left (5 a^2+2 b^2\right ) \tan (e+f x)}{10 (a+b)^4 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{b \operatorname{Subst}\left (\int \frac{-\frac{2 (10 a+3 b)}{b (a+b)}-\frac{2 \left (5 a^2+2 b^2\right ) x^2}{b (a+b)^2}+\frac{\left (5 a^2+2 b^2\right ) x^4}{(a+b)^3}}{x^4 \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{10 (a+b) f}\\ &=-\frac{\cot ^5(e+f x)}{5 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{b \left (5 a^2+2 b^2\right ) \tan (e+f x)}{10 (a+b)^4 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{b \operatorname{Subst}\left (\int \left (-\frac{2 (10 a+3 b)}{b (a+b)^2 x^4}+\frac{2 \left (-5 a^2+10 a b+b^2\right )}{b (a+b)^3 x^2}+\frac{5 a (3 a-4 b)}{(a+b)^3 \left (a+b+b x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{10 (a+b) f}\\ &=-\frac{\left (5 a^2-10 a b-b^2\right ) \cot (e+f x)}{5 (a+b)^4 f}-\frac{(10 a+3 b) \cot ^3(e+f x)}{15 (a+b)^3 f}-\frac{\cot ^5(e+f x)}{5 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{b \left (5 a^2+2 b^2\right ) \tan (e+f x)}{10 (a+b)^4 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{(a (3 a-4 b) b) \operatorname{Subst}\left (\int \frac{1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{2 (a+b)^4 f}\\ &=-\frac{a (3 a-4 b) \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 (a+b)^{9/2} f}-\frac{\left (5 a^2-10 a b-b^2\right ) \cot (e+f x)}{5 (a+b)^4 f}-\frac{(10 a+3 b) \cot ^3(e+f x)}{15 (a+b)^3 f}-\frac{\cot ^5(e+f x)}{5 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{b \left (5 a^2+2 b^2\right ) \tan (e+f x)}{10 (a+b)^4 f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end{align*}
Mathematica [C] time = 3.53815, size = 777, normalized size = 4.13 \[ \frac{\sec ^4(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (-\csc (e) \sec (2 e) \csc ^5(e+f x) \left (10 a \left (16 a^2+34 a b+123 b^2\right ) \sin (f x)-a \left (16 a^2-223 a b+1336 b^2\right ) \sin (3 f x)+640 a^2 b \sin (2 e-f x)-715 a^2 b \sin (2 e+f x)+415 a^2 b \sin (4 e+f x)+165 a^2 b \sin (2 e+3 f x)+208 a^2 b \sin (4 e+3 f x)+180 a^2 b \sin (6 e+3 f x)-268 a^2 b \sin (2 e+5 f x)-223 a^2 b \sin (6 e+5 f x)-45 a^2 b \sin (8 e+5 f x)+83 a^2 b \sin (4 e+7 f x)-15 a^2 b \sin (6 e+7 f x)+68 a^2 b \sin (8 e+7 f x)+240 a^3 \sin (2 e-f x)-240 a^3 \sin (2 e+f x)+160 a^3 \sin (4 e+f x)-16 a^3 \sin (4 e+3 f x)+48 a^3 \sin (2 e+5 f x)+48 a^3 \sin (6 e+5 f x)-16 a^3 \sin (4 e+7 f x)-16 a^3 \sin (8 e+7 f x)-1460 a b^2 \sin (2 e-f x)+860 a b^2 \sin (2 e+f x)+1830 a b^2 \sin (4 e+f x)-30 a b^2 \sin (2 e+3 f x)-1036 a b^2 \sin (4 e+3 f x)-330 a b^2 \sin (6 e+3 f x)+290 a b^2 \sin (2 e+5 f x)+230 a b^2 \sin (6 e+5 f x)+60 a b^2 \sin (8 e+5 f x)-6 a b^2 \sin (4 e+7 f x)-6 a b^2 \sin (8 e+7 f x)+240 b^3 \sin (2 e-f x)-240 b^3 \sin (2 e+f x)+120 b^3 \sin (2 e+3 f x)+120 b^3 \sin (6 e+3 f x)-24 b^3 \sin (2 e+5 f x)-24 b^3 \sin (6 e+5 f x)\right )+\frac{960 a b (3 a-4 b) (\cos (2 e)-i \sin (2 e)) (a \cos (2 (e+f x))+a+2 b) \tan ^{-1}\left (\frac{(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4}}\right )}{\sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4}}\right )}{7680 f (a+b)^4 \left (a+b \sec ^2(e+f x)\right )^2} \]
Warning: Unable to verify antiderivative.
[In]
[Out]
________________________________________________________________________________________
Maple [A] time = 0.118, size = 189, normalized size = 1. \begin{align*} -{\frac{1}{5\,f \left ( a+b \right ) ^{2} \left ( \tan \left ( fx+e \right ) \right ) ^{5}}}-{\frac{{a}^{2}}{f \left ( a+b \right ) ^{4}\tan \left ( fx+e \right ) }}+2\,{\frac{ab}{f \left ( a+b \right ) ^{4}\tan \left ( fx+e \right ) }}-{\frac{2\,a}{3\,f \left ( a+b \right ) ^{3} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}}-{\frac{{a}^{2}b\tan \left ( fx+e \right ) }{2\,f \left ( a+b \right ) ^{4} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{3\,{a}^{2}b}{2\,f \left ( a+b \right ) ^{4}}\arctan \left ({\tan \left ( fx+e \right ) b{\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}}+2\,{\frac{a{b}^{2}}{f \left ( a+b \right ) ^{4}\sqrt{ \left ( a+b \right ) b}}\arctan \left ({\frac{\tan \left ( fx+e \right ) b}{\sqrt{ \left ( a+b \right ) b}}} \right ) } \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 [B] time = 0.746461, size = 2260, normalized size = 12.02 \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 = 1.3106, size = 355, normalized size = 1.89 \begin{align*} -\frac{\frac{15 \, a^{2} b \tan \left (f x + e\right )}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )}{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}} + \frac{15 \,{\left (3 \, a^{2} b - 4 \, a b^{2}\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 + b^{2}}}\right )\right )}}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \sqrt{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} + 10 \, a^{2} \tan \left (f x + e\right )^{2} + 10 \, a b \tan \left (f x + e\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}\right )}}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \tan \left (f x + e\right )^{5}}}{30 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]