Optimal. Leaf size=231 \[ -\frac{\sqrt{b} \left (15 a^2-70 a b+63 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{8 a^{11/2} f}-\frac{b \left (35 a^2-110 a b+99 b^2\right ) \tan (e+f x)}{40 a^5 f \left (a+b \tan ^2(e+f x)\right )}-\frac{b \left (5 a^2-10 a b+9 b^2\right ) \tan (e+f x)}{20 a^4 f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{\left (5 a^2-30 a b+27 b^2\right ) \cot (e+f x)}{5 a^5 f}-\frac{(10 a-9 b) \cot ^3(e+f x)}{15 a^4 f}-\frac{\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^2} \]
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Rubi [A] time = 0.302911, antiderivative size = 231, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3663, 462, 456, 1259, 1261, 205} \[ -\frac{\sqrt{b} \left (15 a^2-70 a b+63 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{8 a^{11/2} f}-\frac{b \left (35 a^2-110 a b+99 b^2\right ) \tan (e+f x)}{40 a^5 f \left (a+b \tan ^2(e+f x)\right )}-\frac{b \left (5 a^2-10 a b+9 b^2\right ) \tan (e+f x)}{20 a^4 f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{\left (5 a^2-30 a b+27 b^2\right ) \cot (e+f x)}{5 a^5 f}-\frac{(10 a-9 b) \cot ^3(e+f x)}{15 a^4 f}-\frac{\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^2} \]
Antiderivative was successfully verified.
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Rule 3663
Rule 462
Rule 456
Rule 1259
Rule 1261
Rule 205
Rubi steps
\begin{align*} \int \frac{\csc ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^2}{x^6 \left (a+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^2}+\frac{\operatorname{Subst}\left (\int \frac{10 a-9 b+5 a x^2}{x^4 \left (a+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{5 a f}\\ &=-\frac{\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{b \left (5 a^2-10 a b+9 b^2\right ) \tan (e+f x)}{20 a^4 f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{b \operatorname{Subst}\left (\int \frac{4 \left (\frac{9}{a}-\frac{10}{b}\right )+4 \left (\frac{10}{a}-\frac{5}{b}-\frac{9 b}{a^2}\right ) x^2+\frac{3 \left (5 a^2-10 a b+9 b^2\right ) x^4}{a^3}}{x^4 \left (a+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{20 a f}\\ &=-\frac{\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{b \left (5 a^2-10 a b+9 b^2\right ) \tan (e+f x)}{20 a^4 f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{b \left (35 a^2-110 a b+99 b^2\right ) \tan (e+f x)}{40 a^5 f \left (a+b \tan ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{-8 a (10 a-9 b) b-8 b \left (5 a^2-20 a b+18 b^2\right ) x^2+\frac{b^2 \left (35 a^2-110 a b+99 b^2\right ) x^4}{a}}{x^4 \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{40 a^4 b f}\\ &=-\frac{\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{b \left (5 a^2-10 a b+9 b^2\right ) \tan (e+f x)}{20 a^4 f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{b \left (35 a^2-110 a b+99 b^2\right ) \tan (e+f x)}{40 a^5 f \left (a+b \tan ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \left (-\frac{8 (10 a-9 b) b}{x^4}-\frac{8 b \left (5 a^2-30 a b+27 b^2\right )}{a x^2}+\frac{5 b^2 \left (15 a^2-70 a b+63 b^2\right )}{a \left (a+b x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{40 a^4 b f}\\ &=-\frac{\left (5 a^2-30 a b+27 b^2\right ) \cot (e+f x)}{5 a^5 f}-\frac{(10 a-9 b) \cot ^3(e+f x)}{15 a^4 f}-\frac{\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{b \left (5 a^2-10 a b+9 b^2\right ) \tan (e+f x)}{20 a^4 f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{b \left (35 a^2-110 a b+99 b^2\right ) \tan (e+f x)}{40 a^5 f \left (a+b \tan ^2(e+f x)\right )}-\frac{\left (b \left (15 a^2-70 a b+63 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{8 a^5 f}\\ &=-\frac{\sqrt{b} \left (15 a^2-70 a b+63 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{8 a^{11/2} f}-\frac{\left (5 a^2-30 a b+27 b^2\right ) \cot (e+f x)}{5 a^5 f}-\frac{(10 a-9 b) \cot ^3(e+f x)}{15 a^4 f}-\frac{\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{b \left (5 a^2-10 a b+9 b^2\right ) \tan (e+f x)}{20 a^4 f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{b \left (35 a^2-110 a b+99 b^2\right ) \tan (e+f x)}{40 a^5 f \left (a+b \tan ^2(e+f x)\right )}\\ \end{align*}
Mathematica [A] time = 1.67383, size = 346, normalized size = 1.5 \[ \frac{-960 \sqrt{b} \left (15 a^2-70 a b+63 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )-\frac{2 \sqrt{a} \cot (e+f x) \csc ^4(e+f x) \left (-8800 a^2 b^2 \cos (6 (e+f x))+2479 a^2 b^2 \cos (8 (e+f x))+4 \left (-1400 a^2 b^2-447 a^3 b+416 a^4+13125 a b^3-13230 b^4\right ) \cos (2 (e+f x))-4 \left (-2821 a^2 b^2-257 a^3 b+32 a^4+8925 a b^3-6615 b^4\right ) \cos (4 (e+f x))+637 a^2 b^2+1788 a^3 b \cos (6 (e+f x))-863 a^3 b \cos (8 (e+f x))-165 a^3 b-128 a^4 \cos (6 (e+f x))+64 a^4 \cos (8 (e+f x))+1600 a^4+14700 a b^3 \cos (6 (e+f x))-2625 a b^3 \cos (8 (e+f x))-28875 a b^3-7560 b^4 \cos (6 (e+f x))+945 b^4 \cos (8 (e+f x))+33075 b^4\right )}{((a-b) \cos (2 (e+f x))+a+b)^2}}{7680 a^{11/2} f} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.102, size = 380, normalized size = 1.7 \begin{align*} -{\frac{1}{5\,f{a}^{3} \left ( \tan \left ( fx+e \right ) \right ) ^{5}}}-{\frac{2}{3\,f{a}^{3} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}}+{\frac{b}{f{a}^{4} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}}-{\frac{1}{f{a}^{3}\tan \left ( fx+e \right ) }}+6\,{\frac{b}{f{a}^{4}\tan \left ( fx+e \right ) }}-6\,{\frac{{b}^{2}}{f{a}^{5}\tan \left ( fx+e \right ) }}-{\frac{7\,{b}^{2} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{8\,f{a}^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{11\,{b}^{3} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{4\,f{a}^{4} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{15\,{b}^{4} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{8\,f{a}^{5} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{9\,b\tan \left ( fx+e \right ) }{8\,f{a}^{2} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{13\,{b}^{2}\tan \left ( fx+e \right ) }{4\,f{a}^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{17\,{b}^{3}\tan \left ( fx+e \right ) }{8\,f{a}^{4} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{15\,b}{8\,f{a}^{3}}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{35\,{b}^{2}}{4\,f{a}^{4}}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{63\,{b}^{3}}{8\,f{a}^{5}}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \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 = 2.7423, size = 2931, normalized size = 12.69 \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 = 1.62251, size = 355, normalized size = 1.54 \begin{align*} -\frac{\frac{15 \,{\left (15 \, a^{2} b - 70 \, a b^{2} + 63 \, 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 )}}{\sqrt{a b} a^{5}} + \frac{15 \,{\left (7 \, a^{2} b^{2} \tan \left (f x + e\right )^{3} - 22 \, a b^{3} \tan \left (f x + e\right )^{3} + 15 \, b^{4} \tan \left (f x + e\right )^{3} + 9 \, a^{3} b \tan \left (f x + e\right ) - 26 \, a^{2} b^{2} \tan \left (f x + e\right ) + 17 \, a b^{3} \tan \left (f x + e\right )\right )}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{2} a^{5}} + \frac{8 \,{\left (15 \, a^{2} \tan \left (f x + e\right )^{4} - 90 \, a b \tan \left (f x + e\right )^{4} + 90 \, b^{2} \tan \left (f x + e\right )^{4} + 10 \, a^{2} \tan \left (f x + e\right )^{2} - 15 \, a b \tan \left (f x + e\right )^{2} + 3 \, a^{2}\right )}}{a^{5} \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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