Integrand size = 23, antiderivative size = 147 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {x}{a^3}+\frac {\sqrt {a+b} \left (3 a^2-4 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 a^3 b^{5/2} f}-\frac {(a+b) \tan ^3(e+f x)}{4 a b f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {(3 a-4 b) (a+b) \tan (e+f x)}{8 a^2 b^2 f \left (a+b+b \tan ^2(e+f x)\right )} \]
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Time = 0.32 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {4226, 2000, 481, 592, 536, 209, 211} \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {x}{a^3}-\frac {(3 a-4 b) (a+b) \tan (e+f x)}{8 a^2 b^2 f \left (a+b \tan ^2(e+f x)+b\right )}+\frac {\sqrt {a+b} \left (3 a^2-4 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 a^3 b^{5/2} f}-\frac {(a+b) \tan ^3(e+f x)}{4 a b f \left (a+b \tan ^2(e+f x)+b\right )^2} \]
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Rule 209
Rule 211
Rule 481
Rule 536
Rule 592
Rule 2000
Rule 4226
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^6}{\left (1+x^2\right ) \left (a+b \left (1+x^2\right )\right )^3} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {x^6}{\left (1+x^2\right ) \left (a+b+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {(a+b) \tan ^3(e+f x)}{4 a b f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {\text {Subst}\left (\int \frac {x^2 \left (3 (a+b)+(3 a-b) x^2\right )}{\left (1+x^2\right ) \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 a b f} \\ & = -\frac {(a+b) \tan ^3(e+f x)}{4 a b f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {(3 a-4 b) (a+b) \tan (e+f x)}{8 a^2 b^2 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {\text {Subst}\left (\int \frac {-((3 a-4 b) (a+b))+\left (-3 a^2+a b-4 b^2\right ) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{8 a^2 b^2 f} \\ & = -\frac {(a+b) \tan ^3(e+f x)}{4 a b f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {(3 a-4 b) (a+b) \tan (e+f x)}{8 a^2 b^2 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{a^3 f}+\frac {\left ((a+b) \left (3 a^2-4 a b+8 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{8 a^3 b^2 f} \\ & = -\frac {x}{a^3}+\frac {\sqrt {a+b} \left (3 a^2-4 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 a^3 b^{5/2} f}-\frac {(a+b) \tan ^3(e+f x)}{4 a b f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {(3 a-4 b) (a+b) \tan (e+f x)}{8 a^2 b^2 f \left (a+b+b \tan ^2(e+f x)\right )} \\ \end{align*}
Result contains complex when optimal does not.
Time = 7.61 (sec) , antiderivative size = 760, normalized size of antiderivative = 5.17 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\left (-3 a^3+a^2 b-4 a b^2-8 b^3\right ) (a+2 b+a \cos (2 e+2 f x))^3 \sec ^6(e+f x) \left (\frac {\arctan \left (\sec (f x) \left (\frac {\cos (2 e)}{2 \sqrt {a+b} \sqrt {b \cos (4 e)-i b \sin (4 e)}}-\frac {i \sin (2 e)}{2 \sqrt {a+b} \sqrt {b \cos (4 e)-i b \sin (4 e)}}\right ) (-a \sin (f x)-2 b \sin (f x)+a \sin (2 e+f x))\right ) \cos (2 e)}{64 a^3 b^2 \sqrt {a+b} f \sqrt {b \cos (4 e)-i b \sin (4 e)}}-\frac {i \arctan \left (\sec (f x) \left (\frac {\cos (2 e)}{2 \sqrt {a+b} \sqrt {b \cos (4 e)-i b \sin (4 e)}}-\frac {i \sin (2 e)}{2 \sqrt {a+b} \sqrt {b \cos (4 e)-i b \sin (4 e)}}\right ) (-a \sin (f x)-2 b \sin (f x)+a \sin (2 e+f x))\right ) \sin (2 e)}{64 a^3 b^2 \sqrt {a+b} f \sqrt {b \cos (4 e)-i b \sin (4 e)}}\right )}{\left (a+b \sec ^2(e+f x)\right )^3}+\frac {(a+2 b+a \cos (2 e+2 f x)) \sec (2 e) \sec ^6(e+f x) \left (-24 a^2 b^2 f x \cos (2 e)-64 a b^3 f x \cos (2 e)-64 b^4 f x \cos (2 e)-16 a^2 b^2 f x \cos (2 f x)-32 a b^3 f x \cos (2 f x)-16 a^2 b^2 f x \cos (4 e+2 f x)-32 a b^3 f x \cos (4 e+2 f x)-4 a^2 b^2 f x \cos (2 e+4 f x)-4 a^2 b^2 f x \cos (6 e+4 f x)+9 a^4 \sin (2 e)+15 a^3 b \sin (2 e)-18 a^2 b^2 \sin (2 e)-72 a b^3 \sin (2 e)-48 b^4 \sin (2 e)-9 a^4 \sin (2 f x)-13 a^3 b \sin (2 f x)+28 a^2 b^2 \sin (2 f x)+32 a b^3 \sin (2 f x)+3 a^4 \sin (4 e+2 f x)-a^3 b \sin (4 e+2 f x)-20 a^2 b^2 \sin (4 e+2 f x)-16 a b^3 \sin (4 e+2 f x)-3 a^4 \sin (2 e+4 f x)+3 a^3 b \sin (2 e+4 f x)+6 a^2 b^2 \sin (2 e+4 f x)\right )}{128 a^3 b^2 f \left (a+b \sec ^2(e+f x)\right )^3} \]
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Time = 27.17 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.91
method | result | size |
derivativedivides | \(\frac {\frac {\left (a +b \right ) \left (\frac {-\frac {a \left (5 a -4 b \right ) \tan \left (f x +e \right )^{3}}{8 b}-\frac {a \left (3 a^{2}-a b -4 b^{2}\right ) \tan \left (f x +e \right )}{8 b^{2}}}{\left (a +b +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (3 a^{2}-4 a b +8 b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{8 b^{2} \sqrt {\left (a +b \right ) b}}\right )}{a^{3}}-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a^{3}}}{f}\) | \(134\) |
default | \(\frac {\frac {\left (a +b \right ) \left (\frac {-\frac {a \left (5 a -4 b \right ) \tan \left (f x +e \right )^{3}}{8 b}-\frac {a \left (3 a^{2}-a b -4 b^{2}\right ) \tan \left (f x +e \right )}{8 b^{2}}}{\left (a +b +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (3 a^{2}-4 a b +8 b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{8 b^{2} \sqrt {\left (a +b \right ) b}}\right )}{a^{3}}-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a^{3}}}{f}\) | \(134\) |
risch | \(-\frac {x}{a^{3}}+\frac {i \left (-3 a^{4} {\mathrm e}^{6 i \left (f x +e \right )}+a^{3} b \,{\mathrm e}^{6 i \left (f x +e \right )}+20 a^{2} b^{2} {\mathrm e}^{6 i \left (f x +e \right )}+16 a \,b^{3} {\mathrm e}^{6 i \left (f x +e \right )}-9 a^{4} {\mathrm e}^{4 i \left (f x +e \right )}-15 a^{3} b \,{\mathrm e}^{4 i \left (f x +e \right )}+18 a^{2} b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+72 a \,b^{3} {\mathrm e}^{4 i \left (f x +e \right )}+48 b^{4} {\mathrm e}^{4 i \left (f x +e \right )}-9 \,{\mathrm e}^{2 i \left (f x +e \right )} a^{4}-13 a^{3} b \,{\mathrm e}^{2 i \left (f x +e \right )}+28 a^{2} b^{2} {\mathrm e}^{2 i \left (f x +e \right )}+32 a \,b^{3} {\mathrm e}^{2 i \left (f x +e \right )}-3 a^{4}+3 a^{3} b +6 a^{2} b^{2}\right )}{4 a^{3} b^{2} f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )^{2}}-\frac {3 \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{16 b^{3} f a}+\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{4 b^{2} f \,a^{2}}-\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{2 b f \,a^{3}}+\frac {3 \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{16 b^{3} f a}-\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{4 b^{2} f \,a^{2}}+\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{2 b f \,a^{3}}\) | \(584\) |
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Leaf count of result is larger than twice the leaf count of optimal. 293 vs. \(2 (133) = 266\).
Time = 0.30 (sec) , antiderivative size = 664, normalized size of antiderivative = 4.52 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\left [-\frac {32 \, a^{2} b^{2} f x \cos \left (f x + e\right )^{4} + 64 \, a b^{3} f x \cos \left (f x + e\right )^{2} + 32 \, b^{4} f x - {\left ({\left (3 \, a^{4} - 4 \, a^{3} b + 8 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b^{2} - 4 \, a b^{3} + 8 \, b^{4} + 2 \, {\left (3 \, a^{3} b - 4 \, a^{2} b^{2} + 8 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-\frac {a + b}{b}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - b^{2} \cos \left (f x + e\right )\right )} \sqrt {-\frac {a + b}{b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) + 4 \, {\left (3 \, {\left (a^{4} - a^{3} b - 2 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (5 \, a^{3} b + a^{2} b^{2} - 4 \, a b^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{32 \, {\left (a^{5} b^{2} f \cos \left (f x + e\right )^{4} + 2 \, a^{4} b^{3} f \cos \left (f x + e\right )^{2} + a^{3} b^{4} f\right )}}, -\frac {16 \, a^{2} b^{2} f x \cos \left (f x + e\right )^{4} + 32 \, a b^{3} f x \cos \left (f x + e\right )^{2} + 16 \, b^{4} f x + {\left ({\left (3 \, a^{4} - 4 \, a^{3} b + 8 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b^{2} - 4 \, a b^{3} + 8 \, b^{4} + 2 \, {\left (3 \, a^{3} b - 4 \, a^{2} b^{2} + 8 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {a + b}{b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {a + b}{b}}}{2 \, {\left (a + b\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) + 2 \, {\left (3 \, {\left (a^{4} - a^{3} b - 2 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (5 \, a^{3} b + a^{2} b^{2} - 4 \, a b^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{16 \, {\left (a^{5} b^{2} f \cos \left (f x + e\right )^{4} + 2 \, a^{4} b^{3} f \cos \left (f x + e\right )^{2} + a^{3} b^{4} f\right )}}\right ] \]
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\[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\int \frac {\tan ^{6}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{3}}\, dx \]
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Time = 0.29 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.31 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {{\left (5 \, a^{2} b + a b^{2} - 4 \, b^{3}\right )} \tan \left (f x + e\right )^{3} + {\left (3 \, a^{3} + 2 \, a^{2} b - 5 \, a b^{2} - 4 \, b^{3}\right )} \tan \left (f x + e\right )}{a^{2} b^{4} \tan \left (f x + e\right )^{4} + a^{4} b^{2} + 2 \, a^{3} b^{3} + a^{2} b^{4} + 2 \, {\left (a^{3} b^{3} + a^{2} b^{4}\right )} \tan \left (f x + e\right )^{2}} + \frac {8 \, {\left (f x + e\right )}}{a^{3}} - \frac {{\left (3 \, a^{3} - a^{2} b + 4 \, a b^{2} + 8 \, b^{3}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a^{3} b^{2}}}{8 \, f} \]
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Time = 2.54 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.36 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {8 \, {\left (f x + e\right )}}{a^{3}} - \frac {{\left (3 \, a^{3} - a^{2} b + 4 \, a b^{2} + 8 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )}}{\sqrt {a b + b^{2}} a^{3} b^{2}} + \frac {5 \, a^{2} b \tan \left (f x + e\right )^{3} + a b^{2} \tan \left (f x + e\right )^{3} - 4 \, b^{3} \tan \left (f x + e\right )^{3} + 3 \, a^{3} \tan \left (f x + e\right ) + 2 \, a^{2} b \tan \left (f x + e\right ) - 5 \, a b^{2} \tan \left (f x + e\right ) - 4 \, b^{3} \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2} a^{2} b^{2}}}{8 \, f} \]
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Time = 20.95 (sec) , antiderivative size = 615, normalized size of antiderivative = 4.18 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\mathrm {atan}\left (\frac {25\,\mathrm {tan}\left (e+f\,x\right )}{32\,\left (\frac {5\,b}{4\,a}-\frac {3\,a}{16\,b}+\frac {9\,a^2}{32\,b^2}+\frac {25}{32}\right )}-\frac {3\,\mathrm {tan}\left (e+f\,x\right )}{16\,\left (\frac {9\,a}{32\,b}+\frac {25\,b}{32\,a}+\frac {5\,b^2}{4\,a^2}-\frac {3}{16}\right )}+\frac {9\,\mathrm {tan}\left (e+f\,x\right )}{32\,\left (\frac {25\,b^2}{32\,a^2}-\frac {3\,b}{16\,a}+\frac {5\,b^3}{4\,a^3}+\frac {9}{32}\right )}+\frac {5\,\mathrm {tan}\left (e+f\,x\right )}{4\,\left (\frac {25\,a}{32\,b}-\frac {3\,a^2}{16\,b^2}+\frac {9\,a^3}{32\,b^3}+\frac {5}{4}\right )}\right )}{a^3\,f}-\frac {\frac {{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (5\,a^2+a\,b-4\,b^2\right )}{8\,a^2\,b}-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (a+b\right )\,\left (-3\,a^2+a\,b+4\,b^2\right )}{8\,a^2\,b^2}}{f\,\left (2\,a\,b+a^2+b^2+{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (2\,b^2+2\,a\,b\right )+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^4\right )}-\frac {\mathrm {atanh}\left (\frac {27\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-b^6-a\,b^5}}{256\,\left (\frac {27\,a\,b^2}{256}-\frac {27\,b^3}{128}+\frac {171\,b^4}{256\,a}-\frac {7\,b^5}{64\,a^2}+\frac {5\,b^6}{32\,a^3}+\frac {5\,b^7}{4\,a^4}\right )}-\frac {81\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-b^6-a\,b^5}}{256\,\left (\frac {27\,a^2\,b}{256}-\frac {27\,a\,b^2}{128}+\frac {171\,b^3}{256}-\frac {7\,b^4}{64\,a}+\frac {5\,b^5}{32\,a^2}+\frac {5\,b^6}{4\,a^3}\right )}-\frac {35\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-b^6-a\,b^5}}{32\,\left (\frac {171\,a^2\,b}{256}-\frac {7\,a\,b^2}{64}-\frac {27\,a^3}{128}+\frac {5\,b^3}{32}+\frac {5\,b^4}{4\,a}+\frac {27\,a^4}{256\,b}\right )}+\frac {5\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-b^6-a\,b^5}}{4\,\left (\frac {5\,a\,b^2}{32}-\frac {7\,a^2\,b}{64}+\frac {171\,a^3}{256}+\frac {5\,b^3}{4}-\frac {27\,a^4}{128\,b}+\frac {27\,a^5}{256\,b^2}\right )}+\frac {63\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-b^6-a\,b^5}}{64\,\left (\frac {171\,a\,b^2}{256}-\frac {27\,a^2\,b}{128}+\frac {27\,a^3}{256}-\frac {7\,b^3}{64}+\frac {5\,b^4}{32\,a}+\frac {5\,b^5}{4\,a^2}\right )}\right )\,\sqrt {-b^5\,\left (a+b\right )}\,\left (3\,a^2-4\,a\,b+8\,b^2\right )}{8\,a^3\,b^5\,f} \]
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