Integrand size = 21, antiderivative size = 136 \[ \int \sec ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=-\frac {4 d^2 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sec (a+b x) \sqrt {\sin (2 a+2 b x)}}{77 b \sqrt {d \tan (a+b x)}}-\frac {4 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{77 b}-\frac {2 d \sec ^3(a+b x) \sqrt {d \tan (a+b x)}}{77 b}+\frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b} \]
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Time = 0.22 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2691, 2693, 2694, 2653, 2720} \[ \int \sec ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=-\frac {4 d^2 \sqrt {\sin (2 a+2 b x)} \sec (a+b x) \operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right )}{77 b \sqrt {d \tan (a+b x)}}+\frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}-\frac {2 d \sec ^3(a+b x) \sqrt {d \tan (a+b x)}}{77 b}-\frac {4 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{77 b} \]
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Rule 2653
Rule 2691
Rule 2693
Rule 2694
Rule 2720
Rubi steps \begin{align*} \text {integral}& = \frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}-\frac {1}{11} d^2 \int \frac {\sec ^5(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx \\ & = -\frac {2 d \sec ^3(a+b x) \sqrt {d \tan (a+b x)}}{77 b}+\frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}-\frac {1}{77} \left (6 d^2\right ) \int \frac {\sec ^3(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx \\ & = -\frac {4 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{77 b}-\frac {2 d \sec ^3(a+b x) \sqrt {d \tan (a+b x)}}{77 b}+\frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}-\frac {1}{77} \left (4 d^2\right ) \int \frac {\sec (a+b x)}{\sqrt {d \tan (a+b x)}} \, dx \\ & = -\frac {4 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{77 b}-\frac {2 d \sec ^3(a+b x) \sqrt {d \tan (a+b x)}}{77 b}+\frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}-\frac {\left (4 d^2 \sqrt {\sin (a+b x)}\right ) \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}} \, dx}{77 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}} \\ & = -\frac {4 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{77 b}-\frac {2 d \sec ^3(a+b x) \sqrt {d \tan (a+b x)}}{77 b}+\frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}-\frac {\left (4 d^2 \sec (a+b x) \sqrt {\sin (2 a+2 b x)}\right ) \int \frac {1}{\sqrt {\sin (2 a+2 b x)}} \, dx}{77 \sqrt {d \tan (a+b x)}} \\ & = -\frac {4 d^2 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sec (a+b x) \sqrt {\sin (2 a+2 b x)}}{77 b \sqrt {d \tan (a+b x)}}-\frac {4 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{77 b}-\frac {2 d \sec ^3(a+b x) \sqrt {d \tan (a+b x)}}{77 b}+\frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.77 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.66 \[ \int \sec ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=-\frac {d \sec ^5(a+b x) \left (-23+6 \cos (2 (a+b x))+\cos (4 (a+b x))+16 \cos ^6(a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\tan ^2(a+b x)\right ) \sqrt {\sec ^2(a+b x)}\right ) \sqrt {d \tan (a+b x)}}{154 b} \]
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Time = 1.67 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.95
method | result | size |
default | \(\frac {d \sqrt {d \tan \left (b x +a \right )}\, \left (4 \sin \left (b x +a \right ) \cos \left (b x +a \right ) \sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \sqrt {-\csc \left (b x +a \right )+1+\cot \left (b x +a \right )}\, F\left (\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )+4 \sin \left (b x +a \right ) \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \sqrt {-\csc \left (b x +a \right )+1+\cot \left (b x +a \right )}\, \sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}\, F\left (\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )+2 \sin \left (b x +a \right ) \tan \left (b x +a \right ) \sqrt {2}+\left (\tan ^{2}\left (b x +a \right )\right ) \sec \left (b x +a \right ) \sqrt {2}-7 \left (\tan ^{2}\left (b x +a \right )\right ) \left (\sec ^{3}\left (b x +a \right )\right ) \sqrt {2}\right ) \sqrt {2}}{77 b \left (\cos ^{2}\left (b x +a \right )-1\right )}\) | \(265\) |
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Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.93 \[ \int \sec ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\frac {2 \, {\left (2 \, \sqrt {i \, d} d \cos \left (b x + a\right )^{5} F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) + 2 \, \sqrt {-i \, d} d \cos \left (b x + a\right )^{5} F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) - {\left (2 \, d \cos \left (b x + a\right )^{4} + d \cos \left (b x + a\right )^{2} - 7 \, d\right )} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}\right )}}{77 \, b \cos \left (b x + a\right )^{5}} \]
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\[ \int \sec ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\int \left (d \tan {\left (a + b x \right )}\right )^{\frac {3}{2}} \sec ^{5}{\left (a + b x \right )}\, dx \]
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\[ \int \sec ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} \sec \left (b x + a\right )^{5} \,d x } \]
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\[ \int \sec ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} \sec \left (b x + a\right )^{5} \,d x } \]
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Timed out. \[ \int \sec ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\int \frac {{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2}}{{\cos \left (a+b\,x\right )}^5} \,d x \]
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