Optimal. Leaf size=178 \[ \frac {\left (5 a^3+15 a^2 b-5 a b^2+b^3\right ) x}{16 (a-b)^4}-\frac {a^{5/2} \sqrt {b} \text {ArcTan}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{(a-b)^4 f}-\frac {\left (11 a^2-4 a b+b^2\right ) \cos (e+f x) \sin (e+f x)}{16 (a-b)^3 f}+\frac {(3 a-b) \cos ^3(e+f x) \sin (e+f x)}{8 (a-b)^2 f}+\frac {\cos ^3(e+f x) \sin ^3(e+f x)}{6 (a-b) f} \]
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Rubi [A]
time = 0.20, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3744, 481, 592,
541, 536, 209, 211} \begin {gather*} -\frac {a^{5/2} \sqrt {b} \text {ArcTan}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{f (a-b)^4}-\frac {\left (11 a^2-4 a b+b^2\right ) \sin (e+f x) \cos (e+f x)}{16 f (a-b)^3}+\frac {x \left (5 a^3+15 a^2 b-5 a b^2+b^3\right )}{16 (a-b)^4}+\frac {\sin ^3(e+f x) \cos ^3(e+f x)}{6 f (a-b)}+\frac {(3 a-b) \sin (e+f x) \cos ^3(e+f x)}{8 f (a-b)^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 209
Rule 211
Rule 481
Rule 536
Rule 541
Rule 592
Rule 3744
Rubi steps
\begin {align*} \int \frac {\sin ^6(e+f x)}{a+b \tan ^2(e+f x)} \, dx &=\frac {\text {Subst}\left (\int \frac {x^6}{\left (1+x^2\right )^4 \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\cos ^3(e+f x) \sin ^3(e+f x)}{6 (a-b) f}-\frac {\text {Subst}\left (\int \frac {x^2 \left (3 a-3 (2 a-b) x^2\right )}{\left (1+x^2\right )^3 \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{6 (a-b) f}\\ &=\frac {(3 a-b) \cos ^3(e+f x) \sin (e+f x)}{8 (a-b)^2 f}+\frac {\cos ^3(e+f x) \sin ^3(e+f x)}{6 (a-b) f}-\frac {\text {Subst}\left (\int \frac {3 a (3 a-b)-3 \left (8 a^2-3 a b+b^2\right ) x^2}{\left (1+x^2\right )^2 \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{24 (a-b)^2 f}\\ &=-\frac {\left (11 a^2-4 a b+b^2\right ) \cos (e+f x) \sin (e+f x)}{16 (a-b)^3 f}+\frac {(3 a-b) \cos ^3(e+f x) \sin (e+f x)}{8 (a-b)^2 f}+\frac {\cos ^3(e+f x) \sin ^3(e+f x)}{6 (a-b) f}+\frac {\text {Subst}\left (\int \frac {3 a (5 a-b) (a+b)-3 b \left (11 a^2-4 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 )}{48 (a-b)^3 f}\\ &=-\frac {\left (11 a^2-4 a b+b^2\right ) \cos (e+f x) \sin (e+f x)}{16 (a-b)^3 f}+\frac {(3 a-b) \cos ^3(e+f x) \sin (e+f x)}{8 (a-b)^2 f}+\frac {\cos ^3(e+f x) \sin ^3(e+f x)}{6 (a-b) f}-\frac {\left (a^3 b\right ) \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{(a-b)^4 f}+\frac {\left (5 a^3+15 a^2 b-5 a b^2+b^3\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{16 (a-b)^4 f}\\ &=\frac {\left (5 a^3+15 a^2 b-5 a b^2+b^3\right ) x}{16 (a-b)^4}-\frac {a^{5/2} \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{(a-b)^4 f}-\frac {\left (11 a^2-4 a b+b^2\right ) \cos (e+f x) \sin (e+f x)}{16 (a-b)^3 f}+\frac {(3 a-b) \cos ^3(e+f x) \sin (e+f x)}{8 (a-b)^2 f}+\frac {\cos ^3(e+f x) \sin ^3(e+f x)}{6 (a-b) f}\\ \end {align*}
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Mathematica [A]
time = 0.39, size = 140, normalized size = 0.79 \begin {gather*} -\frac {-12 \left (5 a^3+15 a^2 b-5 a b^2+b^3\right ) (e+f x)+192 a^{5/2} \sqrt {b} \text {ArcTan}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )+3 (a-b) (5 a-b) (3 a+b) \sin (2 (e+f x))-3 (a-b)^2 (3 a-b) \sin (4 (e+f x))+(a-b)^3 \sin (6 (e+f x))}{192 (a-b)^4 f} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.38, size = 185, normalized size = 1.04
method | result | size |
derivativedivides | \(\frac {\frac {\frac {\left (-\frac {11}{16} a^{3}+\frac {15}{16} a^{2} b -\frac {5}{16} a \,b^{2}+\frac {1}{16} b^{3}\right ) \left (\tan ^{5}\left (f x +e \right )\right )+\left (-\frac {5}{6} a^{3}+\frac {1}{2} a^{2} b +\frac {1}{2} a \,b^{2}-\frac {1}{6} b^{3}\right ) \left (\tan ^{3}\left (f x +e \right )\right )+\left (-\frac {5}{16} a^{3}+\frac {1}{16} a^{2} b +\frac {5}{16} a \,b^{2}-\frac {1}{16} b^{3}\right ) \tan \left (f x +e \right )}{\left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}+\frac {\left (5 a^{3}+15 a^{2} b -5 a \,b^{2}+b^{3}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{16}}{\left (a -b \right )^{4}}-\frac {a^{3} b \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{\left (a -b \right )^{4} \sqrt {a b}}}{f}\) | \(185\) |
default | \(\frac {\frac {\frac {\left (-\frac {11}{16} a^{3}+\frac {15}{16} a^{2} b -\frac {5}{16} a \,b^{2}+\frac {1}{16} b^{3}\right ) \left (\tan ^{5}\left (f x +e \right )\right )+\left (-\frac {5}{6} a^{3}+\frac {1}{2} a^{2} b +\frac {1}{2} a \,b^{2}-\frac {1}{6} b^{3}\right ) \left (\tan ^{3}\left (f x +e \right )\right )+\left (-\frac {5}{16} a^{3}+\frac {1}{16} a^{2} b +\frac {5}{16} a \,b^{2}-\frac {1}{16} b^{3}\right ) \tan \left (f x +e \right )}{\left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}+\frac {\left (5 a^{3}+15 a^{2} b -5 a \,b^{2}+b^{3}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{16}}{\left (a -b \right )^{4}}-\frac {a^{3} b \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{\left (a -b \right )^{4} \sqrt {a b}}}{f}\) | \(185\) |
risch | \(\frac {5 x \,a^{3}}{16 \left (a -b \right )^{4}}+\frac {15 x \,a^{2} b}{16 \left (a -b \right )^{4}}-\frac {5 x a \,b^{2}}{16 \left (a -b \right )^{4}}+\frac {x \,b^{3}}{16 \left (a -b \right )^{4}}+\frac {15 i {\mathrm e}^{2 i \left (f x +e \right )} a^{2}}{128 \left (a -b \right )^{3} f}+\frac {i {\mathrm e}^{2 i \left (f x +e \right )} a b}{64 \left (a -b \right )^{3} f}-\frac {i {\mathrm e}^{2 i \left (f x +e \right )} b^{2}}{128 \left (a -b \right )^{3} f}-\frac {15 i {\mathrm e}^{-2 i \left (f x +e \right )} a^{2}}{128 \left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right ) f}-\frac {i {\mathrm e}^{-2 i \left (f x +e \right )} a b}{64 \left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right ) f}+\frac {i {\mathrm e}^{-2 i \left (f x +e \right )} b^{2}}{128 \left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right ) f}-\frac {\sqrt {-a b}\, a^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right )}{2 \left (a -b \right )^{4} f}+\frac {\sqrt {-a b}\, a^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right )}{2 \left (a -b \right )^{4} f}-\frac {\sin \left (6 f x +6 e \right )}{192 \left (a -b \right ) f}+\frac {3 \sin \left (4 f x +4 e \right ) a}{64 \left (a -b \right )^{2} f}-\frac {\sin \left (4 f x +4 e \right ) b}{64 \left (a -b \right )^{2} f}\) | \(417\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.51, size = 313, normalized size = 1.76 \begin {gather*} -\frac {\frac {48 \, a^{3} b \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )}{{\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} \sqrt {a b}} - \frac {3 \, {\left (5 \, a^{3} + 15 \, a^{2} b - 5 \, a b^{2} + b^{3}\right )} {\left (f x + e\right )}}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} + \frac {3 \, {\left (11 \, a^{2} - 4 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{5} + 8 \, {\left (5 \, a^{2} + 2 \, a b - b^{2}\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (5 \, a^{2} + 4 \, a b - b^{2}\right )} \tan \left (f x + e\right )}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \tan \left (f x + e\right )^{6} + 3 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \tan \left (f x + e\right )^{4} + a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3} + 3 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \tan \left (f x + e\right )^{2}}}{48 \, f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.81, size = 539, normalized size = 3.03 \begin {gather*} \left [\frac {12 \, \sqrt {-a b} a^{2} \log \left (\frac {{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a + b\right )} \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt {-a b} \sin \left (f x + e\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}}\right ) + 3 \, {\left (5 \, a^{3} + 15 \, a^{2} b - 5 \, a b^{2} + b^{3}\right )} f x - {\left (8 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{5} - 2 \, {\left (13 \, a^{3} - 33 \, a^{2} b + 27 \, a b^{2} - 7 \, b^{3}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (11 \, a^{3} - 15 \, a^{2} b + 5 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \, {\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} f}, \frac {24 \, \sqrt {a b} a^{2} \arctan \left (\frac {{\left ({\left (a + b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {a b}}{2 \, a b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) + 3 \, {\left (5 \, a^{3} + 15 \, a^{2} b - 5 \, a b^{2} + b^{3}\right )} f x - {\left (8 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{5} - 2 \, {\left (13 \, a^{3} - 33 \, a^{2} b + 27 \, a b^{2} - 7 \, b^{3}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (11 \, a^{3} - 15 \, a^{2} b + 5 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \, {\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} f}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 1.04, size = 292, normalized size = 1.64 \begin {gather*} -\frac {\frac {48 \, {\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 )} a^{3} b}{{\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} \sqrt {a b}} - \frac {3 \, {\left (5 \, a^{3} + 15 \, a^{2} b - 5 \, a b^{2} + b^{3}\right )} {\left (f x + e\right )}}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} + \frac {33 \, a^{2} \tan \left (f x + e\right )^{5} - 12 \, a b \tan \left (f x + e\right )^{5} + 3 \, b^{2} \tan \left (f x + e\right )^{5} + 40 \, a^{2} \tan \left (f x + e\right )^{3} + 16 \, a b \tan \left (f x + e\right )^{3} - 8 \, b^{2} \tan \left (f x + e\right )^{3} + 15 \, a^{2} \tan \left (f x + e\right ) + 12 \, a b \tan \left (f x + e\right ) - 3 \, b^{2} \tan \left (f x + e\right )}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} {\left (\tan \left (f x + e\right )^{2} + 1\right )}^{3}}}{48 \, f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 17.02, size = 2500, normalized size = 14.04 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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