Integrand size = 23, antiderivative size = 90 \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {x}{a^2}+\frac {(a-2 b) \sqrt {a+b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 a^2 b^{3/2} f}-\frac {(a+b) \tan (e+f x)}{2 a b f \left (a+b+b \tan ^2(e+f x)\right )} \]
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Time = 0.19 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {4226, 2000, 481, 536, 209, 211} \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {(a-2 b) \sqrt {a+b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 a^2 b^{3/2} f}+\frac {x}{a^2}-\frac {(a+b) \tan (e+f x)}{2 a b f \left (a+b \tan ^2(e+f x)+b\right )} \]
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Rule 209
Rule 211
Rule 481
Rule 536
Rule 2000
Rule 4226
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^4}{\left (1+x^2\right ) \left (a+b \left (1+x^2\right )\right )^2} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {x^4}{\left (1+x^2\right ) \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {(a+b) \tan (e+f x)}{2 a b f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {\text {Subst}\left (\int \frac {a+b+(a-b) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{2 a b f} \\ & = -\frac {(a+b) \tan (e+f x)}{2 a b 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^2 f}+\frac {((a-2 b) (a+b)) \text {Subst}\left (\int \frac {1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{2 a^2 b f} \\ & = \frac {x}{a^2}+\frac {(a-2 b) \sqrt {a+b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 a^2 b^{3/2} f}-\frac {(a+b) \tan (e+f x)}{2 a b f \left (a+b+b \tan ^2(e+f x)\right )} \\ \end{align*}
Result contains complex when optimal does not.
Time = 3.73 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.77 \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^4(e+f x) \left (2 x (a+2 b+a \cos (2 (e+f x)))+\frac {\left (-a^2+a b+2 b^2\right ) \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (a+2 b+a \cos (2 (e+f x))) (\cos (2 e)-i \sin (2 e))}{b \sqrt {a+b} f \sqrt {b (\cos (e)-i \sin (e))^4}}+\frac {(a+b) ((a+2 b) \sin (2 e)-a \sin (2 f x))}{b f (\cos (e)-\sin (e)) (\cos (e)+\sin (e))}\right )}{8 a^2 \left (a+b \sec ^2(e+f x)\right )^2} \]
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Time = 4.55 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.94
method | result | size |
derivativedivides | \(\frac {\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a^{2}}+\frac {\left (a +b \right ) \left (-\frac {a \tan \left (f x +e \right )}{2 b \left (a +b +b \tan \left (f x +e \right )^{2}\right )}+\frac {\left (a -2 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 b \sqrt {\left (a +b \right ) b}}\right )}{a^{2}}}{f}\) | \(85\) |
default | \(\frac {\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a^{2}}+\frac {\left (a +b \right ) \left (-\frac {a \tan \left (f x +e \right )}{2 b \left (a +b +b \tan \left (f x +e \right )^{2}\right )}+\frac {\left (a -2 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 b \sqrt {\left (a +b \right ) b}}\right )}{a^{2}}}{f}\) | \(85\) |
risch | \(\frac {x}{a^{2}}-\frac {i \left (a^{2} {\mathrm e}^{2 i \left (f x +e \right )}+3 a b \,{\mathrm e}^{2 i \left (f x +e \right )}+2 b^{2} {\mathrm e}^{2 i \left (f x +e \right )}+a^{2}+a b \right )}{a^{2} b f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )}+\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{4 b^{2} f a}-\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{2 b f \,a^{2}}-\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{4 b^{2} f a}+\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{2 b f \,a^{2}}\) | \(311\) |
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Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (78) = 156\).
Time = 0.29 (sec) , antiderivative size = 393, normalized size of antiderivative = 4.37 \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\left [\frac {8 \, a b f x \cos \left (f x + e\right )^{2} + 8 \, b^{2} f x - 4 \, {\left (a^{2} + a b\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left ({\left (a^{2} - 2 \, a b\right )} \cos \left (f x + e\right )^{2} + a b - 2 \, b^{2}\right )} \sqrt {-\frac {a + b}{b}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - b^{2} \cos \left (f x + e\right )\right )} \sqrt {-\frac {a + b}{b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right )}{8 \, {\left (a^{3} b f \cos \left (f x + e\right )^{2} + a^{2} b^{2} f\right )}}, \frac {4 \, a b f x \cos \left (f x + e\right )^{2} + 4 \, b^{2} f x - 2 \, {\left (a^{2} + a b\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left ({\left (a^{2} - 2 \, a b\right )} \cos \left (f x + e\right )^{2} + a b - 2 \, b^{2}\right )} \sqrt {\frac {a + b}{b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {a + b}{b}}}{2 \, {\left (a + b\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right )}{4 \, {\left (a^{3} b f \cos \left (f x + e\right )^{2} + a^{2} b^{2} f\right )}}\right ] \]
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\[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\int \frac {\tan ^{4}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2}}\, dx \]
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Time = 0.29 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.07 \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=-\frac {\frac {{\left (a + b\right )} \tan \left (f x + e\right )}{a b^{2} \tan \left (f x + e\right )^{2} + a^{2} b + a b^{2}} - \frac {2 \, {\left (f x + e\right )}}{a^{2}} - \frac {{\left (a^{2} - a b - 2 \, b^{2}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a^{2} b}}{2 \, f} \]
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Time = 1.02 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.33 \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\frac {2 \, {\left (f x + e\right )}}{a^{2}} + \frac {{\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} {\left (a^{2} - a b - 2 \, b^{2}\right )}}{\sqrt {a b + b^{2}} a^{2} b} - \frac {a \tan \left (f x + e\right ) + b \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )} a b}}{2 \, f} \]
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Time = 19.79 (sec) , antiderivative size = 285, normalized size of antiderivative = 3.17 \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (e+f\,x\right )}{\frac {3\,b}{2\,a}-\frac {a}{2\,b}+1}-\frac {\mathrm {tan}\left (e+f\,x\right )}{2\,\left (\frac {b}{a}+\frac {3\,b^2}{2\,a^2}-\frac {1}{2}\right )}+\frac {3\,b\,\mathrm {tan}\left (e+f\,x\right )}{2\,\left (a+\frac {3\,b}{2}-\frac {a^2}{2\,b}\right )}\right )}{a^2\,f}-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (a+b\right )}{2\,a\,b\,f\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )}-\frac {\mathrm {atanh}\left (\frac {3\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-b^4-a\,b^3}}{2\,\left (\frac {a\,b}{4}-a^2+\frac {3\,b^2}{2}+\frac {a^3}{4\,b}\right )}-\frac {5\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-b^4-a\,b^3}}{4\,\left (\frac {a^2}{4}-a\,b+\frac {b^2}{4}+\frac {3\,b^3}{2\,a}\right )}+\frac {\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-b^4-a\,b^3}}{4\,\left (\frac {a\,b}{4}-b^2+\frac {b^3}{4\,a}+\frac {3\,b^4}{2\,a^2}\right )}\right )\,\sqrt {-b^3\,\left (a+b\right )}\,\left (a-2\,b\right )}{2\,a^2\,b^3\,f} \]
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