Optimal. Leaf size=315 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{a^{5/2} f}-\frac {\left (a^2-20 a b-5 b^2\right ) \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{5 a^2 f (a+b)^3}-\frac {b (11 a+3 b) \cot ^5(e+f x)}{3 a^2 f (a+b)^2 \sqrt {a+b \tan ^2(e+f x)+b}}+\frac {\left (5 a^3+19 a^2 b-65 a b^2-15 b^3\right ) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{15 a^2 f (a+b)^4}-\frac {\left (15 a^4+70 a^3 b+128 a^2 b^2-70 a b^3-15 b^4\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{15 a^2 f (a+b)^5}-\frac {b \cot ^5(e+f x)}{3 a f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}} \]
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Rubi [A] time = 0.60, antiderivative size = 315, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {4141, 1975, 472, 579, 583, 12, 377, 203} \[ -\frac {\left (a^2-20 a b-5 b^2\right ) \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{5 a^2 f (a+b)^3}+\frac {\left (19 a^2 b+5 a^3-65 a b^2-15 b^3\right ) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{15 a^2 f (a+b)^4}-\frac {\left (128 a^2 b^2+70 a^3 b+15 a^4-70 a b^3-15 b^4\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{15 a^2 f (a+b)^5}-\frac {\tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{a^{5/2} f}-\frac {b (11 a+3 b) \cot ^5(e+f x)}{3 a^2 f (a+b)^2 \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {b \cot ^5(e+f x)}{3 a f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}} \]
Antiderivative was successfully verified.
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Rule 12
Rule 203
Rule 377
Rule 472
Rule 579
Rule 583
Rule 1975
Rule 4141
Rubi steps
\begin {align*} \int \frac {\cot ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{x^6 \left (1+x^2\right ) \left (a+b \left (1+x^2\right )\right )^{5/2}} \, 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 )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {b \cot ^5(e+f x)}{3 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac {\operatorname {Subst}\left (\int \frac {3 a-5 b-8 b x^2}{x^6 \left (1+x^2\right ) \left (a+b+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{3 a (a+b) f}\\ &=-\frac {b \cot ^5(e+f x)}{3 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {b (11 a+3 b) \cot ^5(e+f x)}{3 a^2 (a+b)^2 f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\operatorname {Subst}\left (\int \frac {3 \left (a^2-20 a b-5 b^2\right )-6 b (11 a+3 b) x^2}{x^6 \left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{3 a^2 (a+b)^2 f}\\ &=-\frac {b \cot ^5(e+f x)}{3 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {b (11 a+3 b) \cot ^5(e+f x)}{3 a^2 (a+b)^2 f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {\left (a^2-20 a b-5 b^2\right ) \cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 a^2 (a+b)^3 f}-\frac {\operatorname {Subst}\left (\int \frac {3 \left (5 a^3+19 a^2 b-65 a b^2-15 b^3\right )+12 b \left (a^2-20 a b-5 b^2\right ) x^2}{x^4 \left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{15 a^2 (a+b)^3 f}\\ &=-\frac {b \cot ^5(e+f x)}{3 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {b (11 a+3 b) \cot ^5(e+f x)}{3 a^2 (a+b)^2 f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\left (5 a^3+19 a^2 b-65 a b^2-15 b^3\right ) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 a^2 (a+b)^4 f}-\frac {\left (a^2-20 a b-5 b^2\right ) \cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 a^2 (a+b)^3 f}+\frac {\operatorname {Subst}\left (\int \frac {3 \left (15 a^4+70 a^3 b+128 a^2 b^2-70 a b^3-15 b^4\right )+6 b \left (5 a^3+19 a^2 b-65 a b^2-15 b^3\right ) x^2}{x^2 \left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{45 a^2 (a+b)^4 f}\\ &=-\frac {b \cot ^5(e+f x)}{3 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {b (11 a+3 b) \cot ^5(e+f x)}{3 a^2 (a+b)^2 f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {\left (15 a^4+70 a^3 b+128 a^2 b^2-70 a b^3-15 b^4\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 a^2 (a+b)^5 f}+\frac {\left (5 a^3+19 a^2 b-65 a b^2-15 b^3\right ) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 a^2 (a+b)^4 f}-\frac {\left (a^2-20 a b-5 b^2\right ) \cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 a^2 (a+b)^3 f}-\frac {\operatorname {Subst}\left (\int \frac {45 (a+b)^5}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{45 a^2 (a+b)^5 f}\\ &=-\frac {b \cot ^5(e+f x)}{3 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {b (11 a+3 b) \cot ^5(e+f x)}{3 a^2 (a+b)^2 f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {\left (15 a^4+70 a^3 b+128 a^2 b^2-70 a b^3-15 b^4\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 a^2 (a+b)^5 f}+\frac {\left (5 a^3+19 a^2 b-65 a b^2-15 b^3\right ) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 a^2 (a+b)^4 f}-\frac {\left (a^2-20 a b-5 b^2\right ) \cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 a^2 (a+b)^3 f}-\frac {\operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{a^2 f}\\ &=-\frac {b \cot ^5(e+f x)}{3 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {b (11 a+3 b) \cot ^5(e+f x)}{3 a^2 (a+b)^2 f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {\left (15 a^4+70 a^3 b+128 a^2 b^2-70 a b^3-15 b^4\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 a^2 (a+b)^5 f}+\frac {\left (5 a^3+19 a^2 b-65 a b^2-15 b^3\right ) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 a^2 (a+b)^4 f}-\frac {\left (a^2-20 a b-5 b^2\right ) \cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 a^2 (a+b)^3 f}-\frac {\operatorname {Subst}\left (\int \frac {1}{1+a x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{a^2 f}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{a^{5/2} f}-\frac {b \cot ^5(e+f x)}{3 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {b (11 a+3 b) \cot ^5(e+f x)}{3 a^2 (a+b)^2 f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {\left (15 a^4+70 a^3 b+128 a^2 b^2-70 a b^3-15 b^4\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 a^2 (a+b)^5 f}+\frac {\left (5 a^3+19 a^2 b-65 a b^2-15 b^3\right ) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 a^2 (a+b)^4 f}-\frac {\left (a^2-20 a b-5 b^2\right ) \cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 a^2 (a+b)^3 f}\\ \end {align*}
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Mathematica [A] time = 24.43, size = 272, normalized size = 0.86 \[ \frac {\tan (e+f x) \sec ^4(e+f x) (a \cos (2 (e+f x))+a+2 b)^3 \left (-\frac {20 b^5 (a+b)}{a^2 (a \cos (2 (e+f x))+a+2 b)^2}+\frac {10 b^4 (15 a+4 b)}{a^2 (a \cos (2 (e+f x))+a+2 b)}-\left (23 a^2+100 a b+150 b^2\right ) \csc ^2(e+f x)-3 (a+b)^2 \csc ^6(e+f x)+(a+b) (11 a+25 b) \csc ^4(e+f x)\right )}{120 f (a+b)^5 \left (a+b \sec ^2(e+f x)\right )^{5/2}}-\frac {\sec ^5(e+f x) (a \cos (2 e+2 f x)+a+2 b)^{5/2} \tan ^{-1}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {-a \sin ^2(e+f x)+a+b}}\right )}{4 \sqrt {2} a^{5/2} f \left (a+b \sec ^2(e+f x)\right )^{5/2}} \]
Antiderivative was successfully verified.
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fricas [B] time = 43.49, size = 2059, normalized size = 6.54 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot \left (f x + e\right )^{6}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 2.52, size = 22712, normalized size = 72.10 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F(-1)] time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot ^{6}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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