Integrand size = 14, antiderivative size = 144 \[ \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {x}{a^3}-\frac {\sqrt {b} \left (15 a^2+20 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 a^3 (a+b)^{5/2} f}-\frac {b \tan (e+f x)}{4 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b (7 a+4 b) \tan (e+f x)}{8 a^2 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )} \]
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Time = 0.23 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {4213, 425, 541, 536, 209, 211} \[ \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {x}{a^3}-\frac {b (7 a+4 b) \tan (e+f x)}{8 a^2 f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )}-\frac {\sqrt {b} \left (15 a^2+20 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 a^3 f (a+b)^{5/2}}-\frac {b \tan (e+f x)}{4 a f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2} \]
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Rule 209
Rule 211
Rule 425
Rule 536
Rule 541
Rule 4213
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \left (a+b+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {b \tan (e+f x)}{4 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {\text {Subst}\left (\int \frac {4 a+b-3 b x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 a (a+b) f} \\ & = -\frac {b \tan (e+f x)}{4 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b (7 a+4 b) \tan (e+f x)}{8 a^2 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {\text {Subst}\left (\int \frac {8 a^2+9 a b+4 b^2-b (7 a+4 b) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{8 a^2 (a+b)^2 f} \\ & = -\frac {b \tan (e+f x)}{4 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b (7 a+4 b) \tan (e+f x)}{8 a^2 (a+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 (b \left (15 a^2+20 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 (a+b)^2 f} \\ & = \frac {x}{a^3}-\frac {\sqrt {b} \left (15 a^2+20 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 a^3 (a+b)^{5/2} f}-\frac {b \tan (e+f x)}{4 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b (7 a+4 b) \tan (e+f x)}{8 a^2 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )} \\ \end{align*}
Result contains complex when optimal does not.
Time = 6.52 (sec) , antiderivative size = 332, normalized size of antiderivative = 2.31 \[ \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^6(e+f x) \left (8 x (a+2 b+a \cos (2 (e+f x)))^2+\frac {b \left (15 a^2+20 a b+8 b^2\right ) \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (a+2 b+a \cos (2 (e+f x)))^2 (\cos (2 e)-i \sin (2 e))}{(a+b)^{5/2} f \sqrt {b (\cos (e)-i \sin (e))^4}}-\frac {4 b^2 ((a+2 b) \sin (2 e)-a \sin (2 f x))}{(a+b) f (\cos (e)-\sin (e)) (\cos (e)+\sin (e))}+\frac {b (a+2 b+a \cos (2 (e+f x))) \left (\left (9 a^2+28 a b+16 b^2\right ) \sin (2 e)-3 a (3 a+2 b) \sin (2 f x)\right )}{(a+b)^2 f (\cos (e)-\sin (e)) (\cos (e)+\sin (e))}\right )}{64 a^3 \left (a+b \sec ^2(e+f x)\right )^3} \]
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Time = 1.48 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.02
method | result | size |
derivativedivides | \(\frac {-\frac {b \left (\frac {\frac {a b \left (7 a +4 b \right ) \tan \left (f x +e \right )^{3}}{8 a^{2}+16 a b +8 b^{2}}+\frac {\left (9 a +4 b \right ) a \tan \left (f x +e \right )}{8 a +8 b}}{\left (a +b +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (15 a^{2}+20 a b +8 b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) \sqrt {\left (a +b \right ) b}}\right )}{a^{3}}+\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a^{3}}}{f}\) | \(147\) |
default | \(\frac {-\frac {b \left (\frac {\frac {a b \left (7 a +4 b \right ) \tan \left (f x +e \right )^{3}}{8 a^{2}+16 a b +8 b^{2}}+\frac {\left (9 a +4 b \right ) a \tan \left (f x +e \right )}{8 a +8 b}}{\left (a +b +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (15 a^{2}+20 a b +8 b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) \sqrt {\left (a +b \right ) b}}\right )}{a^{3}}+\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a^{3}}}{f}\) | \(147\) |
risch | \(\frac {x}{a^{3}}-\frac {i b \left (9 a^{3} {\mathrm e}^{6 i \left (f x +e \right )}+28 a^{2} b \,{\mathrm e}^{6 i \left (f x +e \right )}+16 a \,b^{2} {\mathrm e}^{6 i \left (f x +e \right )}+27 a^{3} {\mathrm e}^{4 i \left (f x +e \right )}+90 a^{2} b \,{\mathrm e}^{4 i \left (f x +e \right )}+120 a \,b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+48 b^{3} {\mathrm e}^{4 i \left (f x +e \right )}+27 a^{3} {\mathrm e}^{2 i \left (f x +e \right )}+68 a^{2} b \,{\mathrm e}^{2 i \left (f x +e \right )}+32 a \,b^{2} {\mathrm e}^{2 i \left (f x +e \right )}+9 a^{3}+6 a^{2} b \right )}{4 a^{3} \left (a +b \right )^{2} f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )^{2}}+\frac {15 \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{16 \left (a +b \right )^{3} f a}+\frac {5 \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right ) b}{4 \left (a +b \right )^{3} f \,a^{2}}+\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right ) b^{2}}{2 \left (a +b \right )^{3} f \,a^{3}}-\frac {15 \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{16 \left (a +b \right )^{3} f a}-\frac {5 \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right ) b}{4 \left (a +b \right )^{3} f \,a^{2}}-\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right ) b^{2}}{2 \left (a +b \right )^{3} f \,a^{3}}\) | \(548\) |
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Leaf count of result is larger than twice the leaf count of optimal. 366 vs. \(2 (130) = 260\).
Time = 0.32 (sec) , antiderivative size = 819, normalized size of antiderivative = 5.69 \[ \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\left [\frac {32 \, {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} f x \cos \left (f x + e\right )^{4} + 64 \, {\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} f x \cos \left (f x + e\right )^{2} + 32 \, {\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} f x + {\left ({\left (15 \, a^{4} + 20 \, a^{3} b + 8 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + 15 \, a^{2} b^{2} + 20 \, a b^{3} + 8 \, b^{4} + 2 \, {\left (15 \, a^{3} b + 20 \, a^{2} b^{2} + 8 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-\frac {b}{a + 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^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a + 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 (3 \, a^{3} b + 2 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (7 \, a^{2} b^{2} + 4 \, a b^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{32 \, {\left ({\left (a^{7} + 2 \, a^{6} b + a^{5} b^{2}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{6} b + 2 \, a^{5} b^{2} + a^{4} b^{3}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} b^{2} + 2 \, a^{4} b^{3} + a^{3} b^{4}\right )} f\right )}}, \frac {16 \, {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} f x \cos \left (f x + e\right )^{4} + 32 \, {\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} f x \cos \left (f x + e\right )^{2} + 16 \, {\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} f x + {\left ({\left (15 \, a^{4} + 20 \, a^{3} b + 8 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + 15 \, a^{2} b^{2} + 20 \, a b^{3} + 8 \, b^{4} + 2 \, {\left (15 \, a^{3} b + 20 \, a^{2} b^{2} + 8 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{a + b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a + b}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) - 2 \, {\left (3 \, {\left (3 \, a^{3} b + 2 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (7 \, a^{2} b^{2} + 4 \, a b^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{16 \, {\left ({\left (a^{7} + 2 \, a^{6} b + a^{5} b^{2}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{6} b + 2 \, a^{5} b^{2} + a^{4} b^{3}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} b^{2} + 2 \, a^{4} b^{3} + a^{3} b^{4}\right )} f\right )}}\right ] \]
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\[ \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\int \frac {1}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{3}}\, dx \]
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Time = 0.28 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.60 \[ \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {{\left (15 \, a^{2} b + 20 \, a b^{2} + 8 \, b^{3}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} \sqrt {{\left (a + b\right )} b}} + \frac {{\left (7 \, a b^{2} + 4 \, b^{3}\right )} \tan \left (f x + e\right )^{3} + {\left (9 \, a^{2} b + 13 \, a b^{2} + 4 \, b^{3}\right )} \tan \left (f x + e\right )}{a^{6} + 4 \, a^{5} b + 6 \, a^{4} b^{2} + 4 \, a^{3} b^{3} + a^{2} b^{4} + {\left (a^{4} b^{2} + 2 \, a^{3} b^{3} + a^{2} b^{4}\right )} \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{5} b + 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} + a^{2} b^{4}\right )} \tan \left (f x + e\right )^{2}} - \frac {8 \, {\left (f x + e\right )}}{a^{3}}}{8 \, f} \]
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Time = 0.30 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.36 \[ \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {{\left (15 \, a^{2} b + 20 \, a b^{2} + 8 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )}}{{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} \sqrt {a b + b^{2}}} + \frac {7 \, a b^{2} \tan \left (f x + e\right )^{3} + 4 \, b^{3} \tan \left (f x + e\right )^{3} + 9 \, a^{2} b \tan \left (f x + e\right ) + 13 \, a b^{2} \tan \left (f x + e\right ) + 4 \, b^{3} \tan \left (f x + e\right )}{{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2}} - \frac {8 \, {\left (f x + e\right )}}{a^{3}}}{8 \, f} \]
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Time = 23.81 (sec) , antiderivative size = 3271, normalized size of antiderivative = 22.72 \[ \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Too large to display} \]
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