Integrand size = 25, antiderivative size = 161 \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+a \tan (e+f x))^3} \, dx=\frac {\sqrt {d} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 a^3 f}+\frac {\sqrt {d} \text {arctanh}\left (\frac {\sqrt {d}+\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{2 \sqrt {2} a^3 f}-\frac {\sqrt {d \tan (e+f x)}}{4 a f (a+a \tan (e+f x))^2}-\frac {3 \sqrt {d \tan (e+f x)}}{8 f \left (a^3+a^3 \tan (e+f x)\right )} \]
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Time = 0.60 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3649, 3730, 3735, 3613, 214, 3715, 65, 211} \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+a \tan (e+f x))^3} \, dx=\frac {\sqrt {d} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 a^3 f}+\frac {\sqrt {d} \text {arctanh}\left (\frac {\sqrt {d} \tan (e+f x)+\sqrt {d}}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{2 \sqrt {2} a^3 f}-\frac {3 \sqrt {d \tan (e+f x)}}{8 f \left (a^3 \tan (e+f x)+a^3\right )}-\frac {\sqrt {d \tan (e+f x)}}{4 a f (a \tan (e+f x)+a)^2} \]
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Rule 65
Rule 211
Rule 214
Rule 3613
Rule 3649
Rule 3715
Rule 3730
Rule 3735
Rubi steps \begin{align*} \text {integral}& = -\frac {\sqrt {d \tan (e+f x)}}{4 a f (a+a \tan (e+f x))^2}-\frac {\int \frac {-\frac {a d}{2}-2 a d \tan (e+f x)+\frac {3}{2} a d \tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^2} \, dx}{4 a^2} \\ & = -\frac {\sqrt {d \tan (e+f x)}}{4 a f (a+a \tan (e+f x))^2}-\frac {3 \sqrt {d \tan (e+f x)}}{8 f \left (a^3+a^3 \tan (e+f x)\right )}-\frac {\int \frac {-\frac {5}{2} a^3 d^2+\frac {3}{2} a^3 d^2 \tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{8 a^5 d} \\ & = -\frac {\sqrt {d \tan (e+f x)}}{4 a f (a+a \tan (e+f x))^2}-\frac {3 \sqrt {d \tan (e+f x)}}{8 f \left (a^3+a^3 \tan (e+f x)\right )}-\frac {\int \frac {-4 a^4 d^2+4 a^4 d^2 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{16 a^7 d}+\frac {d \int \frac {1+\tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{16 a^2} \\ & = -\frac {\sqrt {d \tan (e+f x)}}{4 a f (a+a \tan (e+f x))^2}-\frac {3 \sqrt {d \tan (e+f x)}}{8 f \left (a^3+a^3 \tan (e+f x)\right )}+\frac {d \text {Subst}\left (\int \frac {1}{\sqrt {d x} (a+a x)} \, dx,x,\tan (e+f x)\right )}{16 a^2 f}+\frac {\left (2 a d^3\right ) \text {Subst}\left (\int \frac {1}{-32 a^8 d^4+d x^2} \, dx,x,\frac {-4 a^4 d^2-4 a^4 d^2 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}\right )}{f} \\ & = \frac {\sqrt {d} \text {arctanh}\left (\frac {\sqrt {d}+\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{2 \sqrt {2} a^3 f}-\frac {\sqrt {d \tan (e+f x)}}{4 a f (a+a \tan (e+f x))^2}-\frac {3 \sqrt {d \tan (e+f x)}}{8 f \left (a^3+a^3 \tan (e+f x)\right )}+\frac {\text {Subst}\left (\int \frac {1}{a+\frac {a x^2}{d}} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{8 a^2 f} \\ & = \frac {\sqrt {d} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 a^3 f}+\frac {\sqrt {d} \text {arctanh}\left (\frac {\sqrt {d}+\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{2 \sqrt {2} a^3 f}-\frac {\sqrt {d \tan (e+f x)}}{4 a f (a+a \tan (e+f x))^2}-\frac {3 \sqrt {d \tan (e+f x)}}{8 f \left (a^3+a^3 \tan (e+f x)\right )} \\ \end{align*}
Time = 1.68 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.16 \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+a \tan (e+f x))^3} \, dx=\frac {d^{3/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )-\sqrt {2} d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )+\sqrt {2} d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )+\frac {2 (d \tan (e+f x))^{3/2}}{(1+\tan (e+f x))^2}-\frac {5 d \sqrt {d \tan (e+f x)}}{1+\tan (e+f x)}}{8 a^3 d f} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(348\) vs. \(2(132)=264\).
Time = 0.92 (sec) , antiderivative size = 349, normalized size of antiderivative = 2.17
method | result | size |
derivativedivides | \(\frac {2 d^{4} \left (\frac {\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}}{4 d^{3}}-\frac {\frac {\frac {3 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{4}+\frac {5 d \sqrt {d \tan \left (f x +e \right )}}{4}}{\left (d \tan \left (f x +e \right )+d \right )^{2}}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{4 \sqrt {d}}}{4 d^{3}}\right )}{f \,a^{3}}\) | \(349\) |
default | \(\frac {2 d^{4} \left (\frac {\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}}{4 d^{3}}-\frac {\frac {\frac {3 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{4}+\frac {5 d \sqrt {d \tan \left (f x +e \right )}}{4}}{\left (d \tan \left (f x +e \right )+d \right )^{2}}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{4 \sqrt {d}}}{4 d^{3}}\right )}{f \,a^{3}}\) | \(349\) |
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Time = 0.27 (sec) , antiderivative size = 392, normalized size of antiderivative = 2.43 \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+a \tan (e+f x))^3} \, dx=\left [-\frac {4 \, {\left (\sqrt {2} \tan \left (f x + e\right )^{2} + 2 \, \sqrt {2} \tan \left (f x + e\right ) + \sqrt {2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )} {\left (\sqrt {2} \tan \left (f x + e\right ) + \sqrt {2}\right )} \sqrt {-d}}{2 \, d \tan \left (f x + e\right )}\right ) - {\left (\tan \left (f x + e\right )^{2} + 2 \, \tan \left (f x + e\right ) + 1\right )} \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right ) + 2 \, \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} - d}{\tan \left (f x + e\right ) + 1}\right ) + 2 \, \sqrt {d \tan \left (f x + e\right )} {\left (3 \, \tan \left (f x + e\right ) + 5\right )}}{16 \, {\left (a^{3} f \tan \left (f x + e\right )^{2} + 2 \, a^{3} f \tan \left (f x + e\right ) + a^{3} f\right )}}, \frac {{\left (\tan \left (f x + e\right )^{2} + 2 \, \tan \left (f x + e\right ) + 1\right )} \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right ) + {\left (\sqrt {2} \tan \left (f x + e\right )^{2} + 2 \, \sqrt {2} \tan \left (f x + e\right ) + \sqrt {2}\right )} \sqrt {d} \log \left (\frac {d \tan \left (f x + e\right )^{2} + 2 \, \sqrt {d \tan \left (f x + e\right )} {\left (\sqrt {2} \tan \left (f x + e\right ) + \sqrt {2}\right )} \sqrt {d} + 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) - \sqrt {d \tan \left (f x + e\right )} {\left (3 \, \tan \left (f x + e\right ) + 5\right )}}{8 \, {\left (a^{3} f \tan \left (f x + e\right )^{2} + 2 \, a^{3} f \tan \left (f x + e\right ) + a^{3} f\right )}}\right ] \]
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\[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+a \tan (e+f x))^3} \, dx=\frac {\int \frac {\sqrt {d \tan {\left (e + f x \right )}}}{\tan ^{3}{\left (e + f x \right )} + 3 \tan ^{2}{\left (e + f x \right )} + 3 \tan {\left (e + f x \right )} + 1}\, dx}{a^{3}} \]
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Time = 0.33 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.14 \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+a \tan (e+f x))^3} \, dx=-\frac {\frac {3 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} d^{2} + 5 \, \sqrt {d \tan \left (f x + e\right )} d^{3}}{a^{3} d^{2} \tan \left (f x + e\right )^{2} + 2 \, a^{3} d^{2} \tan \left (f x + e\right ) + a^{3} d^{2}} - \frac {d^{2} {\left (\frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )}}{a^{3}} - \frac {d^{\frac {3}{2}} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a^{3}}}{8 \, d f} \]
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\[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+a \tan (e+f x))^3} \, dx=\int { \frac {\sqrt {d \tan \left (f x + e\right )}}{{\left (a \tan \left (f x + e\right ) + a\right )}^{3}} \,d x } \]
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Time = 5.64 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+a \tan (e+f x))^3} \, dx=\frac {\sqrt {d}\,\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{8\,a^3\,f}-\frac {\frac {3\,d\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{8}+\frac {5\,d^2\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{8}}{f\,a^3\,d^2\,{\mathrm {tan}\left (e+f\,x\right )}^2+2\,f\,a^3\,d^2\,\mathrm {tan}\left (e+f\,x\right )+f\,a^3\,d^2}+\frac {\sqrt {2}\,\sqrt {d}\,\mathrm {atanh}\left (\frac {9\,\sqrt {2}\,d^{17/2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{32\,\left (\frac {9\,d^9\,\mathrm {tan}\left (e+f\,x\right )}{32}+\frac {9\,d^9}{32}\right )}\right )}{4\,a^3\,f} \]
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