Integrand size = 23, antiderivative size = 33 \[ \int \frac {\tan (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{\sqrt {a} f} \]
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Time = 0.06 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {4224, 272, 65, 214} \[ \int \frac {\tan (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{\sqrt {a} f} \]
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Rule 65
Rule 214
Rule 272
Rule 4224
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{x \sqrt {a+b x^2}} \, dx,x,\sec (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\sec ^2(e+f x)\right )}{2 f} \\ & = \frac {\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 {\text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{\sqrt {a} f} \\ \end{align*}
Time = 0.14 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int \frac {\tan (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{\sqrt {a} f} \]
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Time = 0.22 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.27
method | result | size |
derivativedivides | \(-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \sec \left (f x +e \right )^{2}}}{\sec \left (f x +e \right )}\right )}{f \sqrt {a}}\) | \(42\) |
default | \(-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \sec \left (f x +e \right )^{2}}}{\sec \left (f x +e \right )}\right )}{f \sqrt {a}}\) | \(42\) |
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Leaf count of result is larger than twice the leaf count of optimal. 108 vs. \(2 (27) = 54\).
Time = 0.32 (sec) , antiderivative size = 261, normalized size of antiderivative = 7.91 \[ \int \frac {\tan (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\left [\frac {\log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} + 256 \, a^{3} b \cos \left (f x + e\right )^{6} + 160 \, a^{2} b^{2} \cos \left (f x + e\right )^{4} + 32 \, a b^{3} \cos \left (f x + e\right )^{2} + b^{4} - 8 \, {\left (16 \, a^{3} \cos \left (f x + e\right )^{8} + 24 \, a^{2} b \cos \left (f x + e\right )^{6} + 10 \, a b^{2} \cos \left (f x + e\right )^{4} + b^{3} \cos \left (f x + e\right )^{2}\right )} \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}\right )}{8 \, \sqrt {a} f}, \frac {\sqrt {-a} \arctan \left (\frac {{\left (8 \, a^{2} \cos \left (f x + e\right )^{4} + 8 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, a^{3} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b \cos \left (f x + e\right )^{2} + a b^{2}\right )}}\right )}{4 \, a f}\right ] \]
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\[ \int \frac {\tan (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {\tan {\left (e + f x \right )}}{\sqrt {a + b \sec ^{2}{\left (e + f x \right )}}}\, dx \]
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\[ \int \frac {\tan (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int { \frac {\tan \left (f x + e\right )}{\sqrt {b \sec \left (f x + e\right )^{2} + a}} \,d x } \]
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\[ \int \frac {\tan (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int { \frac {\tan \left (f x + e\right )}{\sqrt {b \sec \left (f x + e\right )^{2} + a}} \,d x } \]
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Time = 20.53 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int \frac {\tan (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}}{\sqrt {a}}\right )}{\sqrt {a}\,f} \]
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