Optimal. Leaf size=171 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{b^{5/2} f}-\frac {a (a-2 b) \tan (e+f x)}{b^2 f (a-b)^2 \sqrt {a+b \tan ^2(e+f x)}}-\frac {\tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f (a-b)^{5/2}}-\frac {a \tan ^3(e+f x)}{3 b f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]
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Rubi [A] time = 0.27, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3670, 470, 578, 523, 217, 206, 377, 203} \[ -\frac {a (a-2 b) \tan (e+f x)}{b^2 f (a-b)^2 \sqrt {a+b \tan ^2(e+f x)}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{b^{5/2} f}-\frac {a \tan ^3(e+f x)}{3 b f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f (a-b)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 203
Rule 206
Rule 217
Rule 377
Rule 470
Rule 523
Rule 578
Rule 3670
Rubi steps
\begin {align*} \int \frac {\tan ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^6}{\left (1+x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {a \tan ^3(e+f x)}{3 (a-b) b f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (3 a+3 (a-b) x^2\right )}{\left (1+x^2\right ) \left (a+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{3 (a-b) b f}\\ &=-\frac {a \tan ^3(e+f x)}{3 (a-b) b f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {a (a-2 b) \tan (e+f x)}{(a-b)^2 b^2 f \sqrt {a+b \tan ^2(e+f x)}}-\frac {\operatorname {Subst}\left (\int \frac {-3 a (a-2 b)-3 (a-b)^2 x^2}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{3 (a-b)^2 b^2 f}\\ &=-\frac {a \tan ^3(e+f x)}{3 (a-b) b f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {a (a-2 b) \tan (e+f x)}{(a-b)^2 b^2 f \sqrt {a+b \tan ^2(e+f x)}}-\frac {\operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{(a-b)^2 f}+\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{b^2 f}\\ &=-\frac {a \tan ^3(e+f x)}{3 (a-b) b f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {a (a-2 b) \tan (e+f x)}{(a-b)^2 b^2 f \sqrt {a+b \tan ^2(e+f x)}}-\frac {\operatorname {Subst}\left (\int \frac {1}{1-(-a+b) x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{(a-b)^2 f}+\frac {\operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{b^2 f}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{(a-b)^{5/2} f}+\frac {\tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{b^{5/2} f}-\frac {a \tan ^3(e+f x)}{3 (a-b) b f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {a (a-2 b) \tan (e+f x)}{(a-b)^2 b^2 f \sqrt {a+b \tan ^2(e+f x)}}\\ \end {align*}
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Mathematica [C] time = 4.58, size = 295, normalized size = 1.73 \[ -\frac {\sqrt {\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (a^2 (a-b) \sin (2 (e+f x)) ((3 a-7 b) ((a-b) \cos (2 (e+f x))+a+b)+2 a b)-\frac {3 a^2 b \sin ^2(e+f x) \sin (2 (e+f x)) \left (\frac {\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}\right )^{3/2} \left (\left (a^2-3 a b+2 b^2\right ) F\left (\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right )+b^2 \Pi \left (-\frac {b}{a-b};\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right )\right )}{\sqrt {2}}\right )}{3 \sqrt {2} a b^2 f (a-b)^3 ((a-b) \cos (2 (e+f x))+a+b)^2} \]
Antiderivative was successfully verified.
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fricas [B] time = 2.58, size = 1714, normalized size = 10.02 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [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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maple [B] time = 0.35, size = 382, normalized size = 2.23 \[ -\frac {\tan ^{3}\left (f x +e \right )}{3 f b \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}}}-\frac {\tan \left (f x +e \right )}{f \,b^{2} \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}+\frac {\ln \left (\tan \left (f x +e \right ) \sqrt {b}+\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}\right )}{f \,b^{\frac {5}{2}}}+\frac {\tan \left (f x +e \right )}{3 f b \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}}}-\frac {\tan \left (f x +e \right )}{3 f b a \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}+\frac {\tan \left (f x +e \right )}{3 f a \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}}}+\frac {2 \tan \left (f x +e \right )}{3 f \,a^{2} \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}+\frac {b \tan \left (f x +e \right )}{3 a \left (a -b \right ) f \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}}}+\frac {2 b \tan \left (f x +e \right )}{3 f \left (a -b \right ) a^{2} \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}+\frac {b \tan \left (f x +e \right )}{f \left (a -b \right )^{2} a \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}-\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {\left (a -b \right ) b^{2} \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}\right )}{f \left (a -b \right )^{3} b^{2}} \]
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] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {tan}\left (e+f\,x\right )}^6}{{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \]
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 \[ \int \frac {\tan ^{6}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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