Optimal. Leaf size=85 \[ \frac {a^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{b^{5/2} f (a-b)}-\frac {(a+b) \tan (e+f x)}{b^2 f}-\frac {x}{a-b}+\frac {\tan ^3(e+f x)}{3 b f} \]
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Rubi [A] time = 0.19, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3670, 479, 582, 522, 203, 205} \[ \frac {a^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{b^{5/2} f (a-b)}-\frac {(a+b) \tan (e+f x)}{b^2 f}-\frac {x}{a-b}+\frac {\tan ^3(e+f x)}{3 b f} \]
Antiderivative was successfully verified.
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Rule 203
Rule 205
Rule 479
Rule 522
Rule 582
Rule 3670
Rubi steps
\begin {align*} \int \frac {\tan ^6(e+f x)}{a+b \tan ^2(e+f x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^6}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\tan ^3(e+f x)}{3 b f}-\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 )} \, dx,x,\tan (e+f x)\right )}{3 b f}\\ &=-\frac {(a+b) \tan (e+f x)}{b^2 f}+\frac {\tan ^3(e+f x)}{3 b f}+\frac {\operatorname {Subst}\left (\int \frac {3 a (a+b)+3 \left (a^2+a b+b^2\right ) x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{3 b^2 f}\\ &=-\frac {(a+b) \tan (e+f x)}{b^2 f}+\frac {\tan ^3(e+f x)}{3 b f}-\frac {\operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{(a-b) f}+\frac {a^3 \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{(a-b) b^2 f}\\ &=-\frac {x}{a-b}+\frac {a^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{(a-b) b^{5/2} f}-\frac {(a+b) \tan (e+f x)}{b^2 f}+\frac {\tan ^3(e+f x)}{3 b f}\\ \end {align*}
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Mathematica [A] time = 0.84, size = 92, normalized size = 1.08 \[ \frac {\sqrt {b} \left ((a-b) \tan (e+f x) \left (3 a-b \sec ^2(e+f x)+4 b\right )+3 b^2 (e+f x)\right )-3 a^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{3 b^{5/2} f (b-a)} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.49, size = 278, normalized size = 3.27 \[ \left [-\frac {12 \, b^{2} f x - 4 \, {\left (a b - b^{2}\right )} \tan \left (f x + e\right )^{3} + 3 \, a^{2} \sqrt {-\frac {a}{b}} \log \left (\frac {b^{2} \tan \left (f x + e\right )^{4} - 6 \, a b \tan \left (f x + e\right )^{2} + a^{2} - 4 \, {\left (b^{2} \tan \left (f x + e\right )^{3} - a b \tan \left (f x + e\right )\right )} \sqrt {-\frac {a}{b}}}{b^{2} \tan \left (f x + e\right )^{4} + 2 \, a b \tan \left (f x + e\right )^{2} + a^{2}}\right ) + 12 \, {\left (a^{2} - b^{2}\right )} \tan \left (f x + e\right )}{12 \, {\left (a b^{2} - b^{3}\right )} f}, -\frac {6 \, b^{2} f x - 2 \, {\left (a b - b^{2}\right )} \tan \left (f x + e\right )^{3} - 3 \, a^{2} \sqrt {\frac {a}{b}} \arctan \left (\frac {{\left (b \tan \left (f x + e\right )^{2} - a\right )} \sqrt {\frac {a}{b}}}{2 \, a \tan \left (f x + e\right )}\right ) + 6 \, {\left (a^{2} - b^{2}\right )} \tan \left (f x + e\right )}{6 \, {\left (a b^{2} - b^{3}\right )} f}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 21.52, size = 118, normalized size = 1.39 \[ \frac {\frac {3 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )\right )} a^{3}}{{\left (a b^{2} - b^{3}\right )} \sqrt {a b}} - \frac {3 \, {\left (f x + e\right )}}{a - b} + \frac {b^{2} \tan \left (f x + e\right )^{3} - 3 \, a b \tan \left (f x + e\right ) - 3 \, b^{2} \tan \left (f x + e\right )}{b^{3}}}{3 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.18, size = 102, normalized size = 1.20 \[ \frac {\tan ^{3}\left (f x +e \right )}{3 b f}-\frac {a \tan \left (f x +e \right )}{f \,b^{2}}-\frac {\tan \left (f x +e \right )}{b f}+\frac {a^{3} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {a b}}\right )}{f \,b^{2} \left (a -b \right ) \sqrt {a b}}-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{f \left (a -b \right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.90, size = 83, normalized size = 0.98 \[ \frac {\frac {3 \, a^{3} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )}{{\left (a b^{2} - b^{3}\right )} \sqrt {a b}} - \frac {3 \, {\left (f x + e\right )}}{a - b} + \frac {b \tan \left (f x + e\right )^{3} - 3 \, {\left (a + b\right )} \tan \left (f x + e\right )}{b^{2}}}{3 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 12.00, size = 1310, normalized size = 15.41 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 35.00, size = 685, normalized size = 8.06 \[ \begin {cases} \tilde {\infty } x \tan ^{4}{\relax (e )} & \text {for}\: a = 0 \wedge b = 0 \wedge f = 0 \\\frac {- x + \frac {\tan ^{5}{\left (e + f x \right )}}{5 f} - \frac {\tan ^{3}{\left (e + f x \right )}}{3 f} + \frac {\tan {\left (e + f x \right )}}{f}}{a} & \text {for}\: b = 0 \\\frac {x + \frac {\tan ^{3}{\left (e + f x \right )}}{3 f} - \frac {\tan {\left (e + f x \right )}}{f}}{b} & \text {for}\: a = 0 \\\frac {15 f x \tan ^{2}{\left (e + f x \right )}}{6 b f \tan ^{2}{\left (e + f x \right )} + 6 b f} + \frac {15 f x}{6 b f \tan ^{2}{\left (e + f x \right )} + 6 b f} + \frac {2 \tan ^{5}{\left (e + f x \right )}}{6 b f \tan ^{2}{\left (e + f x \right )} + 6 b f} - \frac {10 \tan ^{3}{\left (e + f x \right )}}{6 b f \tan ^{2}{\left (e + f x \right )} + 6 b f} - \frac {15 \tan {\left (e + f x \right )}}{6 b f \tan ^{2}{\left (e + f x \right )} + 6 b f} & \text {for}\: a = b \\\frac {x \tan ^{6}{\relax (e )}}{a + b \tan ^{2}{\relax (e )}} & \text {for}\: f = 0 \\- \frac {6 i a^{\frac {5}{2}} b \sqrt {\frac {1}{b}} \tan {\left (e + f x \right )}}{6 i a^{\frac {3}{2}} b^{3} f \sqrt {\frac {1}{b}} - 6 i \sqrt {a} b^{4} f \sqrt {\frac {1}{b}}} + \frac {2 i a^{\frac {3}{2}} b^{2} \sqrt {\frac {1}{b}} \tan ^{3}{\left (e + f x \right )}}{6 i a^{\frac {3}{2}} b^{3} f \sqrt {\frac {1}{b}} - 6 i \sqrt {a} b^{4} f \sqrt {\frac {1}{b}}} - \frac {6 i \sqrt {a} b^{3} f x \sqrt {\frac {1}{b}}}{6 i a^{\frac {3}{2}} b^{3} f \sqrt {\frac {1}{b}} - 6 i \sqrt {a} b^{4} f \sqrt {\frac {1}{b}}} - \frac {2 i \sqrt {a} b^{3} \sqrt {\frac {1}{b}} \tan ^{3}{\left (e + f x \right )}}{6 i a^{\frac {3}{2}} b^{3} f \sqrt {\frac {1}{b}} - 6 i \sqrt {a} b^{4} f \sqrt {\frac {1}{b}}} + \frac {6 i \sqrt {a} b^{3} \sqrt {\frac {1}{b}} \tan {\left (e + f x \right )}}{6 i a^{\frac {3}{2}} b^{3} f \sqrt {\frac {1}{b}} - 6 i \sqrt {a} b^{4} f \sqrt {\frac {1}{b}}} + \frac {3 a^{3} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \tan {\left (e + f x \right )} \right )}}{6 i a^{\frac {3}{2}} b^{3} f \sqrt {\frac {1}{b}} - 6 i \sqrt {a} b^{4} f \sqrt {\frac {1}{b}}} - \frac {3 a^{3} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \tan {\left (e + f x \right )} \right )}}{6 i a^{\frac {3}{2}} b^{3} f \sqrt {\frac {1}{b}} - 6 i \sqrt {a} b^{4} f \sqrt {\frac {1}{b}}} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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