Optimal. Leaf size=285 \[ \frac{b^{7/2} \left (99 a^2+44 a b+8 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{8 a^3 f (a+b)^{11/2}}-\frac{\left (8 a^2-75 a b-20 b^2\right ) \cot ^5(e+f x)}{40 a^2 f (a+b)^3}+\frac{\left (32 a^2 b+8 a^3-51 a b^2-12 b^3\right ) \cot ^3(e+f x)}{24 a^2 f (a+b)^4}-\frac{\left (80 a^2 b^2+40 a^3 b+8 a^4-19 a b^3-4 b^4\right ) \cot (e+f x)}{8 a^2 f (a+b)^5}-\frac{b (13 a+4 b) \cot ^5(e+f x)}{8 a^2 f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )}-\frac{x}{a^3}-\frac{b \cot ^5(e+f x)}{4 a f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2} \]
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Rubi [A] time = 0.60672, antiderivative size = 285, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {4141, 1975, 472, 579, 583, 522, 203, 205} \[ \frac{b^{7/2} \left (99 a^2+44 a b+8 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{8 a^3 f (a+b)^{11/2}}-\frac{\left (8 a^2-75 a b-20 b^2\right ) \cot ^5(e+f x)}{40 a^2 f (a+b)^3}+\frac{\left (32 a^2 b+8 a^3-51 a b^2-12 b^3\right ) \cot ^3(e+f x)}{24 a^2 f (a+b)^4}-\frac{\left (80 a^2 b^2+40 a^3 b+8 a^4-19 a b^3-4 b^4\right ) \cot (e+f x)}{8 a^2 f (a+b)^5}-\frac{b (13 a+4 b) \cot ^5(e+f x)}{8 a^2 f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )}-\frac{x}{a^3}-\frac{b \cot ^5(e+f x)}{4 a f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2} \]
Antiderivative was successfully verified.
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Rule 4141
Rule 1975
Rule 472
Rule 579
Rule 583
Rule 522
Rule 203
Rule 205
Rubi steps
\begin{align*} \int \frac{\cot ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^6 \left (1+x^2\right ) \left (a+b \left (1+x^2\right )\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^6 \left (1+x^2\right ) \left (a+b+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{b \cot ^5(e+f x)}{4 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac{\operatorname{Subst}\left (\int \frac{4 a-5 b-9 b x^2}{x^6 \left (1+x^2\right ) \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 a (a+b) f}\\ &=-\frac{b \cot ^5(e+f x)}{4 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac{b (13 a+4 b) \cot ^5(e+f x)}{8 a^2 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{8 a^2-75 a b-20 b^2-7 b (13 a+4 b) x^2}{x^6 \left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{8 a^2 (a+b)^2 f}\\ &=-\frac{\left (8 a^2-75 a b-20 b^2\right ) \cot ^5(e+f x)}{40 a^2 (a+b)^3 f}-\frac{b \cot ^5(e+f x)}{4 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac{b (13 a+4 b) \cot ^5(e+f x)}{8 a^2 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{5 \left (8 a^3+32 a^2 b-51 a b^2-12 b^3\right )+5 b \left (8 a^2-75 a b-20 b^2\right ) x^2}{x^4 \left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{40 a^2 (a+b)^3 f}\\ &=\frac{\left (8 a^3+32 a^2 b-51 a b^2-12 b^3\right ) \cot ^3(e+f x)}{24 a^2 (a+b)^4 f}-\frac{\left (8 a^2-75 a b-20 b^2\right ) \cot ^5(e+f x)}{40 a^2 (a+b)^3 f}-\frac{b \cot ^5(e+f x)}{4 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac{b (13 a+4 b) \cot ^5(e+f x)}{8 a^2 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{15 \left (8 a^4+40 a^3 b+80 a^2 b^2-19 a b^3-4 b^4\right )+15 b \left (8 a^3+32 a^2 b-51 a b^2-12 b^3\right ) x^2}{x^2 \left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{120 a^2 (a+b)^4 f}\\ &=-\frac{\left (8 a^4+40 a^3 b+80 a^2 b^2-19 a b^3-4 b^4\right ) \cot (e+f x)}{8 a^2 (a+b)^5 f}+\frac{\left (8 a^3+32 a^2 b-51 a b^2-12 b^3\right ) \cot ^3(e+f x)}{24 a^2 (a+b)^4 f}-\frac{\left (8 a^2-75 a b-20 b^2\right ) \cot ^5(e+f x)}{40 a^2 (a+b)^3 f}-\frac{b \cot ^5(e+f x)}{4 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac{b (13 a+4 b) \cot ^5(e+f x)}{8 a^2 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{15 \left (8 a^5+48 a^4 b+120 a^3 b^2+160 a^2 b^3+21 a b^4+4 b^5\right )+15 b \left (8 a^4+40 a^3 b+80 a^2 b^2-19 a b^3-4 b^4\right ) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{120 a^2 (a+b)^5 f}\\ &=-\frac{\left (8 a^4+40 a^3 b+80 a^2 b^2-19 a b^3-4 b^4\right ) \cot (e+f x)}{8 a^2 (a+b)^5 f}+\frac{\left (8 a^3+32 a^2 b-51 a b^2-12 b^3\right ) \cot ^3(e+f x)}{24 a^2 (a+b)^4 f}-\frac{\left (8 a^2-75 a b-20 b^2\right ) \cot ^5(e+f x)}{40 a^2 (a+b)^3 f}-\frac{b \cot ^5(e+f x)}{4 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac{b (13 a+4 b) \cot ^5(e+f x)}{8 a^2 (a+b)^2 f \left (a+b+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^3 f}+\frac{\left (b^4 \left (99 a^2+44 a b+8 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{8 a^3 (a+b)^5 f}\\ &=-\frac{x}{a^3}+\frac{b^{7/2} \left (99 a^2+44 a b+8 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{8 a^3 (a+b)^{11/2} f}-\frac{\left (8 a^4+40 a^3 b+80 a^2 b^2-19 a b^3-4 b^4\right ) \cot (e+f x)}{8 a^2 (a+b)^5 f}+\frac{\left (8 a^3+32 a^2 b-51 a b^2-12 b^3\right ) \cot ^3(e+f x)}{24 a^2 (a+b)^4 f}-\frac{\left (8 a^2-75 a b-20 b^2\right ) \cot ^5(e+f x)}{40 a^2 (a+b)^3 f}-\frac{b \cot ^5(e+f x)}{4 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac{b (13 a+4 b) \cot ^5(e+f x)}{8 a^2 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end{align*}
Mathematica [C] time = 8.20968, size = 976, normalized size = 3.42 \[ \frac{\left (99 a^2+44 b a+8 b^2\right ) (\cos (2 e+2 f x) a+a+2 b)^3 \left (\frac{i b^4 \tan ^{-1}\left (\sec (f x) \left (\frac{\cos (2 e)}{2 \sqrt{a+b} \sqrt{b \cos (4 e)-i b \sin (4 e)}}-\frac{i \sin (2 e)}{2 \sqrt{a+b} \sqrt{b \cos (4 e)-i b \sin (4 e)}}\right ) (-a \sin (f x)-2 b \sin (f x)+a \sin (2 e+f x))\right ) \sin (2 e)}{64 a^3 \sqrt{a+b} f \sqrt{b \cos (4 e)-i b \sin (4 e)}}-\frac{b^4 \tan ^{-1}\left (\sec (f x) \left (\frac{\cos (2 e)}{2 \sqrt{a+b} \sqrt{b \cos (4 e)-i b \sin (4 e)}}-\frac{i \sin (2 e)}{2 \sqrt{a+b} \sqrt{b \cos (4 e)-i b \sin (4 e)}}\right ) (-a \sin (f x)-2 b \sin (f x)+a \sin (2 e+f x))\right ) \cos (2 e)}{64 a^3 \sqrt{a+b} f \sqrt{b \cos (4 e)-i b \sin (4 e)}}\right ) \sec ^6(e+f x)}{(a+b)^5 \left (b \sec ^2(e+f x)+a\right )^3}+\frac{(\cos (2 e+2 f x) a+a+2 b)^3 \csc (e) \csc ^5(e+f x) \sin (f x) \sec ^6(e+f x)}{40 (a+b)^3 f \left (b \sec ^2(e+f x)+a\right )^3}+\frac{(\cos (2 e+2 f x) a+a+2 b)^3 \csc (e) \csc ^3(e+f x) (-11 a \sin (f x)-26 b \sin (f x)) \sec ^6(e+f x)}{120 (a+b)^4 f \left (b \sec ^2(e+f x)+a\right )^3}+\frac{(\cos (2 e+2 f x) a+a+2 b)^3 \csc (e) \csc (e+f x) \left (23 \sin (f x) a^2+106 b \sin (f x) a+173 b^2 \sin (f x)\right ) \sec ^6(e+f x)}{120 (a+b)^5 f \left (b \sec ^2(e+f x)+a\right )^3}+\frac{(\cos (2 e+2 f x) a+a+2 b) \sec (2 e) \left (2 \sin (2 e) b^6+a \sin (2 e) b^5-a \sin (2 f x) b^5\right ) \sec ^6(e+f x)}{16 a^3 (a+b)^4 f \left (b \sec ^2(e+f x)+a\right )^3}+\frac{(\cos (2 e+2 f x) a+a+2 b)^2 \sec (2 e) \left (-16 \sin (2 e) b^6-52 a \sin (2 e) b^5+6 a \sin (2 f x) b^5-21 a^2 \sin (2 e) b^4+21 a^2 \sin (2 f x) b^4\right ) \sec ^6(e+f x)}{64 a^3 (a+b)^5 f \left (b \sec ^2(e+f x)+a\right )^3}-\frac{(\cos (2 e+2 f x) a+a+2 b)^3 \cot (e) \csc ^4(e+f x) \sec ^6(e+f x)}{40 (a+b)^3 f \left (b \sec ^2(e+f x)+a\right )^3}-\frac{x (\cos (2 e+2 f x) a+a+2 b)^3 \sec ^6(e+f x)}{8 a^3 \left (b \sec ^2(e+f x)+a\right )^3}+\frac{(11 a \cos (e)+26 b \cos (e)) (\cos (2 e+2 f x) a+a+2 b)^3 \csc (e) \csc ^2(e+f x) \sec ^6(e+f x)}{120 (a+b)^4 f \left (b \sec ^2(e+f x)+a\right )^3} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.131, size = 437, normalized size = 1.5 \begin{align*} -{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) }{f{a}^{3}}}-{\frac{1}{5\,f \left ( a+b \right ) ^{3} \left ( \tan \left ( fx+e \right ) \right ) ^{5}}}+{\frac{a}{3\,f \left ( a+b \right ) ^{4} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}}+{\frac{4\,b}{3\,f \left ( a+b \right ) ^{4} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}}-{\frac{{a}^{2}}{f \left ( a+b \right ) ^{5}\tan \left ( fx+e \right ) }}-5\,{\frac{ab}{f \left ( a+b \right ) ^{5}\tan \left ( fx+e \right ) }}-10\,{\frac{{b}^{2}}{f \left ( a+b \right ) ^{5}\tan \left ( fx+e \right ) }}+{\frac{19\,{b}^{5} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{8\,fa \left ( a+b \right ) ^{5} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{{b}^{6} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{2\,f{a}^{2} \left ( a+b \right ) ^{5} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{21\,{b}^{4}\tan \left ( fx+e \right ) }{8\,f \left ( a+b \right ) ^{5} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{25\,{b}^{5}\tan \left ( fx+e \right ) }{8\,fa \left ( a+b \right ) ^{5} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{{b}^{6}\tan \left ( fx+e \right ) }{2\,f{a}^{2} \left ( a+b \right ) ^{5} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{99\,{b}^{4}}{8\,fa \left ( a+b \right ) ^{5}}\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}}}}+{\frac{11\,{b}^{5}}{2\,f{a}^{2} \left ( a+b \right ) ^{5}}\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}}}}+{\frac{{b}^{6}}{f{a}^{3} \left ( a+b \right ) ^{5}}\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}}}} \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 = 1.10918, size = 5045, normalized size = 17.7 \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 = 1.64043, size = 545, normalized size = 1.91 \begin{align*} \frac{\frac{15 \,{\left (99 \, a^{2} b^{4} + 44 \, a b^{5} + 8 \, b^{6}\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^{8} + 5 \, a^{7} b + 10 \, a^{6} b^{2} + 10 \, a^{5} b^{3} + 5 \, a^{4} b^{4} + a^{3} b^{5}\right )} \sqrt{a b + b^{2}}} + \frac{15 \,{\left (19 \, a b^{5} \tan \left (f x + e\right )^{3} + 4 \, b^{6} \tan \left (f x + e\right )^{3} + 21 \, a^{2} b^{4} \tan \left (f x + e\right ) + 25 \, a b^{5} \tan \left (f x + e\right ) + 4 \, b^{6} \tan \left (f x + e\right )\right )}}{{\left (a^{7} + 5 \, a^{6} b + 10 \, a^{5} b^{2} + 10 \, a^{4} b^{3} + 5 \, a^{3} b^{4} + a^{2} b^{5}\right )}{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2}} - \frac{120 \,{\left (f x + e\right )}}{a^{3}} - \frac{8 \,{\left (15 \, a^{2} \tan \left (f x + e\right )^{4} + 75 \, a b \tan \left (f x + e\right )^{4} + 150 \, b^{2} \tan \left (f x + e\right )^{4} - 5 \, a^{2} \tan \left (f x + e\right )^{2} - 25 \, a b \tan \left (f x + e\right )^{2} - 20 \, b^{2} \tan \left (f x + e\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}\right )}}{{\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} \tan \left (f x + e\right )^{5}}}{120 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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