Integrand size = 19, antiderivative size = 78 \[ \int \cos (a+b x) (d \tan (a+b x))^{3/2} \, dx=\frac {d^2 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sec (a+b x) \sqrt {\sin (2 a+2 b x)}}{2 b \sqrt {d \tan (a+b x)}}-\frac {d \cos (a+b x) \sqrt {d \tan (a+b x)}}{b} \]
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Time = 0.12 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {2690, 2694, 2653, 2720} \[ \int \cos (a+b x) (d \tan (a+b x))^{3/2} \, dx=\frac {d^2 \sqrt {\sin (2 a+2 b x)} \sec (a+b x) \operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right )}{2 b \sqrt {d \tan (a+b x)}}-\frac {d \cos (a+b x) \sqrt {d \tan (a+b x)}}{b} \]
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Rule 2653
Rule 2690
Rule 2694
Rule 2720
Rubi steps \begin{align*} \text {integral}& = -\frac {d \cos (a+b x) \sqrt {d \tan (a+b x)}}{b}+\frac {1}{2} d^2 \int \frac {\sec (a+b x)}{\sqrt {d \tan (a+b x)}} \, dx \\ & = -\frac {d \cos (a+b x) \sqrt {d \tan (a+b x)}}{b}+\frac {\left (d^2 \sqrt {\sin (a+b x)}\right ) \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}} \, dx}{2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}} \\ & = -\frac {d \cos (a+b x) \sqrt {d \tan (a+b x)}}{b}+\frac {\left (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}{2 \sqrt {d \tan (a+b x)}} \\ & = \frac {d^2 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sec (a+b x) \sqrt {\sin (2 a+2 b x)}}{2 b \sqrt {d \tan (a+b x)}}-\frac {d \cos (a+b x) \sqrt {d \tan (a+b x)}}{b} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.21 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.74 \[ \int \cos (a+b x) (d \tan (a+b x))^{3/2} \, dx=\frac {d \cos (a+b x) \left (-1+\operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\tan ^2(a+b x)\right ) \sqrt {\sec ^2(a+b x)}\right ) \sqrt {d \tan (a+b x)}}{b} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(218\) vs. \(2(95)=190\).
Time = 3.02 (sec) , antiderivative size = 219, normalized size of antiderivative = 2.81
method | result | size |
default | \(\frac {\sin \left (b x +a \right ) \left (-\sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \sqrt {-\csc \left (b x +a \right )+1+\cot \left (b x +a \right )}\, \sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}\, F\left (\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right ) \cos \left (b x +a \right )-\sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \sqrt {-\csc \left (b x +a \right )+1+\cot \left (b x +a \right )}\, \sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}\, F\left (\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )+\sin \left (b x +a \right ) \sqrt {2}\, \cos \left (b x +a \right )\right ) \sqrt {d \tan \left (b x +a \right )}\, d \sqrt {2}}{2 b \left (\cos ^{2}\left (b x +a \right )-1\right )}\) | \(219\) |
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\[ \int \cos (a+b x) (d \tan (a+b x))^{3/2} \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} \cos \left (b x + a\right ) \,d x } \]
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\[ \int \cos (a+b x) (d \tan (a+b x))^{3/2} \, dx=\int \left (d \tan {\left (a + b x \right )}\right )^{\frac {3}{2}} \cos {\left (a + b x \right )}\, dx \]
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\[ \int \cos (a+b x) (d \tan (a+b x))^{3/2} \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} \cos \left (b x + a\right ) \,d x } \]
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Exception generated. \[ \int \cos (a+b x) (d \tan (a+b x))^{3/2} \, dx=\text {Exception raised: TypeError} \]
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Timed out. \[ \int \cos (a+b x) (d \tan (a+b x))^{3/2} \, dx=\int \cos \left (a+b\,x\right )\,{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2} \,d x \]
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