Integrand size = 25, antiderivative size = 396 \[ \int \frac {(d \sec (e+f x))^{5/2}}{a+b \tan (e+f x)} \, dx=\frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f}-\frac {\sqrt [4]{a^2+b^2} d^2 \arctan \left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sqrt {d \sec (e+f x)}}{b^{3/2} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {\sqrt [4]{a^2+b^2} d^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sqrt {d \sec (e+f x)}}{b^{3/2} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {2 a d^2 \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sqrt {d \sec (e+f x)}}{b^2 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {a d^2 \cot (e+f x) \operatorname {EllipticPi}\left (-\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \sqrt {d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}}{b^2 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {a d^2 \cot (e+f x) \operatorname {EllipticPi}\left (\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \sqrt {d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}}{b^2 f \sqrt [4]{\sec ^2(e+f x)}} \]
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Time = 0.47 (sec) , antiderivative size = 396, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3593, 749, 858, 237, 761, 410, 109, 418, 1227, 551, 455, 65, 218, 214, 211} \[ \int \frac {(d \sec (e+f x))^{5/2}}{a+b \tan (e+f x)} \, dx=\frac {a d^2 \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \sqrt {d \sec (e+f x)} \operatorname {EllipticPi}\left (-\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right )}{b^2 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {a d^2 \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \sqrt {d \sec (e+f x)} \operatorname {EllipticPi}\left (\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right )}{b^2 f \sqrt [4]{\sec ^2(e+f x)}}-\frac {d^2 \sqrt [4]{a^2+b^2} \sqrt {d \sec (e+f x)} \arctan \left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{b^{3/2} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {d^2 \sqrt [4]{a^2+b^2} \sqrt {d \sec (e+f x)} \text {arctanh}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{b^{3/2} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {2 a d^2 \sqrt {d \sec (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )}{b^2 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f} \]
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Rule 65
Rule 109
Rule 211
Rule 214
Rule 218
Rule 237
Rule 410
Rule 418
Rule 455
Rule 551
Rule 749
Rule 761
Rule 858
Rule 1227
Rule 3593
Rubi steps \begin{align*} \text {integral}& = \frac {\left (d^2 \sqrt {d \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {\sqrt [4]{1+\frac {x^2}{b^2}}}{a+x} \, dx,x,b \tan (e+f x)\right )}{b f \sqrt [4]{\sec ^2(e+f x)}} \\ & = \frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f}+\frac {\left (d^2 \sqrt {d \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {1-\frac {a x}{b^2}}{(a+x) \left (1+\frac {x^2}{b^2}\right )^{3/4}} \, dx,x,b \tan (e+f x)\right )}{b f \sqrt [4]{\sec ^2(e+f x)}} \\ & = \frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f}-\frac {\left (a d^2 \sqrt {d \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {x^2}{b^2}\right )^{3/4}} \, dx,x,b \tan (e+f x)\right )}{b^3 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {\left (\left (1+\frac {a^2}{b^2}\right ) d^2 \sqrt {d \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{(a+x) \left (1+\frac {x^2}{b^2}\right )^{3/4}} \, dx,x,b \tan (e+f x)\right )}{b f \sqrt [4]{\sec ^2(e+f x)}} \\ & = \frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f}-\frac {2 a d^2 \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sqrt {d \sec (e+f x)}}{b^2 f \sqrt [4]{\sec ^2(e+f x)}}-\frac {\left (\left (1+\frac {a^2}{b^2}\right ) d^2 \sqrt {d \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {x}{\left (a^2-x^2\right ) \left (1+\frac {x^2}{b^2}\right )^{3/4}} \, dx,x,b \tan (e+f x)\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}+\frac {\left (a \left (1+\frac {a^2}{b^2}\right ) d^2 \sqrt {d \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\left (a^2-x^2\right ) \left (1+\frac {x^2}{b^2}\right )^{3/4}} \, dx,x,b \tan (e+f x)\right )}{b f \sqrt [4]{\sec ^2(e+f x)}} \\ & = \frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f}-\frac {2 a d^2 \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sqrt {d \sec (e+f x)}}{b^2 f \sqrt [4]{\sec ^2(e+f x)}}-\frac {\left (\left (1+\frac {a^2}{b^2}\right ) d^2 \sqrt {d \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\left (a^2-x\right ) \left (1+\frac {x}{b^2}\right )^{3/4}} \, dx,x,b^2 \tan ^2(e+f x)\right )}{2 b f \sqrt [4]{\sec ^2(e+f x)}}+\frac {\left (a \left (1+\frac {a^2}{b^2}\right ) d^2 \cot (e+f x) \sqrt {d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\left (a^2-x\right ) \sqrt {-\frac {x}{b^2}} \left (1+\frac {x}{b^2}\right )^{3/4}} \, dx,x,b^2 \tan ^2(e+f x)\right )}{2 b^2 f \sqrt [4]{\sec ^2(e+f x)}} \\ & = \frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f}-\frac {2 a d^2 \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sqrt {d \sec (e+f x)}}{b^2 f \sqrt [4]{\sec ^2(e+f x)}}-\frac {\left (2 \left (1+\frac {a^2}{b^2}\right ) b d^2 \sqrt {d \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{a^2+b^2-b^2 x^4} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{f \sqrt [4]{\sec ^2(e+f x)}}-\frac {\left (2 a \left (1+\frac {a^2}{b^2}\right ) d^2 \cot (e+f x) \sqrt {d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^4} \left (-1-\frac {a^2}{b^2}+x^4\right )} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{b^2 f \sqrt [4]{\sec ^2(e+f x)}} \\ & = \frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f}-\frac {2 a d^2 \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sqrt {d \sec (e+f x)}}{b^2 f \sqrt [4]{\sec ^2(e+f x)}}-\frac {\left (\left (1+\frac {a^2}{b^2}\right ) b d^2 \sqrt {d \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a^2+b^2}-b x^2} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{\sqrt {a^2+b^2} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {\left (\left (1+\frac {a^2}{b^2}\right ) b d^2 \sqrt {d \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a^2+b^2}+b x^2} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{\sqrt {a^2+b^2} f \sqrt [4]{\sec ^2(e+f x)}}+\frac {\left (a \left (1+\frac {a^2}{b^2}\right ) d^2 \cot (e+f x) \sqrt {d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\left (1-\frac {b x^2}{\sqrt {a^2+b^2}}\right ) \sqrt {1-x^4}} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{\left (a^2+b^2\right ) f \sqrt [4]{\sec ^2(e+f x)}}+\frac {\left (a \left (1+\frac {a^2}{b^2}\right ) d^2 \cot (e+f x) \sqrt {d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {b x^2}{\sqrt {a^2+b^2}}\right ) \sqrt {1-x^4}} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{\left (a^2+b^2\right ) f \sqrt [4]{\sec ^2(e+f x)}} \\ & = \frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f}-\frac {\sqrt [4]{a^2+b^2} d^2 \arctan \left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sqrt {d \sec (e+f x)}}{b^{3/2} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {\sqrt [4]{a^2+b^2} d^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sqrt {d \sec (e+f x)}}{b^{3/2} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {2 a d^2 \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sqrt {d \sec (e+f x)}}{b^2 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {\left (a \left (1+\frac {a^2}{b^2}\right ) d^2 \cot (e+f x) \sqrt {d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+x^2} \left (1-\frac {b x^2}{\sqrt {a^2+b^2}}\right )} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{\left (a^2+b^2\right ) f \sqrt [4]{\sec ^2(e+f x)}}+\frac {\left (a \left (1+\frac {a^2}{b^2}\right ) d^2 \cot (e+f x) \sqrt {d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+x^2} \left (1+\frac {b x^2}{\sqrt {a^2+b^2}}\right )} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{\left (a^2+b^2\right ) f \sqrt [4]{\sec ^2(e+f x)}} \\ & = \frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f}-\frac {\sqrt [4]{a^2+b^2} d^2 \arctan \left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sqrt {d \sec (e+f x)}}{b^{3/2} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {\sqrt [4]{a^2+b^2} d^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sqrt {d \sec (e+f x)}}{b^{3/2} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {2 a d^2 \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sqrt {d \sec (e+f x)}}{b^2 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {a d^2 \cot (e+f x) \operatorname {EllipticPi}\left (-\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \sqrt {d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}}{b^2 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {a d^2 \cot (e+f x) \operatorname {EllipticPi}\left (\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \sqrt {d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}}{b^2 f \sqrt [4]{\sec ^2(e+f x)}} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 27.35 (sec) , antiderivative size = 290, normalized size of antiderivative = 0.73 \[ \int \frac {(d \sec (e+f x))^{5/2}}{a+b \tan (e+f x)} \, dx=\frac {d^2 \cot (e+f x) \sqrt {d \sec (e+f x)} \left (-\sqrt {b} \sqrt [4]{a^2+b^2} \arctan \left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \tan (e+f x)-\sqrt {b} \sqrt [4]{a^2+b^2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \tan (e+f x)+2 b \sqrt [4]{\sec ^2(e+f x)} \tan (e+f x)-a \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},-\tan ^2(e+f x)\right ) \tan ^2(e+f x)+a \operatorname {EllipticPi}\left (-\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \sqrt {-\tan ^2(e+f x)}+a \operatorname {EllipticPi}\left (\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \sqrt {-\tan ^2(e+f x)}\right )}{b^2 f \sqrt [4]{\sec ^2(e+f x)}} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 7346 vs. \(2 (369 ) = 738\).
Time = 30.43 (sec) , antiderivative size = 7347, normalized size of antiderivative = 18.55
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\[ \int \frac {(d \sec (e+f x))^{5/2}}{a+b \tan (e+f x)} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}}{b \tan \left (f x + e\right ) + a} \,d x } \]
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\[ \int \frac {(d \sec (e+f x))^{5/2}}{a+b \tan (e+f x)} \, dx=\int \frac {\left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{2}}}{a + b \tan {\left (e + f x \right )}}\, dx \]
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\[ \int \frac {(d \sec (e+f x))^{5/2}}{a+b \tan (e+f x)} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}}{b \tan \left (f x + e\right ) + a} \,d x } \]
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\[ \int \frac {(d \sec (e+f x))^{5/2}}{a+b \tan (e+f x)} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}}{b \tan \left (f x + e\right ) + a} \,d x } \]
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Timed out. \[ \int \frac {(d \sec (e+f x))^{5/2}}{a+b \tan (e+f x)} \, dx=\int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{a+b\,\mathrm {tan}\left (e+f\,x\right )} \,d x \]
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