Integrand size = 23, antiderivative size = 78 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan (e+f x) \, dx=-\frac {a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{f}+\frac {a \sqrt {a+b \sec ^2(e+f x)}}{f}+\frac {\left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 f} \]
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Time = 0.10 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {4224, 272, 52, 65, 214} \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan (e+f x) \, dx=-\frac {a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{f}+\frac {a \sqrt {a+b \sec ^2(e+f x)}}{f}+\frac {\left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 f} \]
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Rule 52
Rule 65
Rule 214
Rule 272
Rule 4224
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (a+b x^2\right )^{3/2}}{x} \, dx,x,\sec (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {(a+b x)^{3/2}}{x} \, dx,x,\sec ^2(e+f x)\right )}{2 f} \\ & = \frac {\left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 f}+\frac {a \text {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,\sec ^2(e+f x)\right )}{2 f} \\ & = \frac {a \sqrt {a+b \sec ^2(e+f x)}}{f}+\frac {\left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 f}+\frac {a^2 \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\sec ^2(e+f x)\right )}{2 f} \\ & = \frac {a \sqrt {a+b \sec ^2(e+f x)}}{f}+\frac {\left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 f}+\frac {a^2 \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sec ^2(e+f x)}\right )}{b f} \\ & = -\frac {a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{f}+\frac {a \sqrt {a+b \sec ^2(e+f x)}}{f}+\frac {\left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 f} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.33 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.08 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan (e+f x) \, dx=\frac {2 b \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {3}{2},-\frac {1}{2},-\frac {a \cos ^2(e+f x)}{b}\right ) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 f \sqrt {1+\frac {a \cos ^2(e+f x)}{b}} (a+2 b+a \cos (2 (e+f x)))} \]
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Time = 0.14 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.04
method | result | size |
derivativedivides | \(\frac {\left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{3 f}-\frac {a^{\frac {3}{2}} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \sec \left (f x +e \right )^{2}}}{\sec \left (f x +e \right )}\right )}{f}+\frac {a \sqrt {a +b \sec \left (f x +e \right )^{2}}}{f}\) | \(81\) |
default | \(\frac {\left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{3 f}-\frac {a^{\frac {3}{2}} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \sec \left (f x +e \right )^{2}}}{\sec \left (f x +e \right )}\right )}{f}+\frac {a \sqrt {a +b \sec \left (f x +e \right )^{2}}}{f}\) | \(81\) |
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Leaf count of result is larger than twice the leaf count of optimal. 163 vs. \(2 (66) = 132\).
Time = 0.60 (sec) , antiderivative size = 373, normalized size of antiderivative = 4.78 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan (e+f x) \, dx=\left [\frac {3 \, a^{\frac {3}{2}} \cos \left (f x + e\right )^{2} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} + 256 \, a^{3} b \cos \left (f x + e\right )^{6} + 160 \, a^{2} b^{2} \cos \left (f x + e\right )^{4} + 32 \, a b^{3} \cos \left (f x + e\right )^{2} + b^{4} - 8 \, {\left (16 \, a^{3} \cos \left (f x + e\right )^{8} + 24 \, a^{2} b \cos \left (f x + e\right )^{6} + 10 \, a b^{2} \cos \left (f x + e\right )^{4} + b^{3} \cos \left (f x + e\right )^{2}\right )} \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}\right ) + 8 \, {\left (4 \, a \cos \left (f x + e\right )^{2} + b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{24 \, f \cos \left (f x + e\right )^{2}}, \frac {3 \, \sqrt {-a} a \arctan \left (\frac {{\left (8 \, a^{2} \cos \left (f x + e\right )^{4} + 8 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, a^{3} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b \cos \left (f x + e\right )^{2} + a b^{2}\right )}}\right ) \cos \left (f x + e\right )^{2} + 4 \, {\left (4 \, a \cos \left (f x + e\right )^{2} + b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{12 \, f \cos \left (f x + e\right )^{2}}\right ] \]
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\[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan (e+f x) \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}} \tan {\left (e + f x \right )}\, dx \]
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\[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan (e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \tan \left (f x + e\right ) \,d x } \]
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Leaf count of result is larger than twice the leaf count of optimal. 867 vs. \(2 (66) = 132\).
Time = 1.00 (sec) , antiderivative size = 867, normalized size of antiderivative = 11.12 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan (e+f x) \, dx=\text {Too large to display} \]
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Time = 23.60 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.85 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan (e+f x) \, dx=\frac {{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}}{3\,f}-\frac {a^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}}{\sqrt {a}}\right )}{f}+\frac {a\,\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}}{f} \]
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