Integrand size = 25, antiderivative size = 134 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {\left (8 a^2+8 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{8 (a+b)^{5/2} f}-\frac {(8 a+5 b) \sec ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{8 (a+b)^2 f}+\frac {\sec ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{4 (a+b) f} \]
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Time = 0.25 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3273, 91, 79, 65, 214} \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {\left (8 a^2+8 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{8 f (a+b)^{5/2}}+\frac {\sec ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{4 f (a+b)}-\frac {(8 a+5 b) \sec ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{8 f (a+b)^2} \]
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Rule 65
Rule 79
Rule 91
Rule 214
Rule 3273
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^2}{(1-x)^3 \sqrt {a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{2 f} \\ & = \frac {\sec ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{4 (a+b) f}-\frac {\text {Subst}\left (\int \frac {\frac {1}{2} (4 a+b)+2 (a+b) x}{(1-x)^2 \sqrt {a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{4 (a+b) f} \\ & = -\frac {(8 a+5 b) \sec ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{8 (a+b)^2 f}+\frac {\sec ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{4 (a+b) f}+\frac {\left (8 a^2+8 a b+3 b^2\right ) \text {Subst}\left (\int \frac {1}{(1-x) \sqrt {a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{16 (a+b)^2 f} \\ & = -\frac {(8 a+5 b) \sec ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{8 (a+b)^2 f}+\frac {\sec ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{4 (a+b) f}+\frac {\left (8 a^2+8 a b+3 b^2\right ) \text {Subst}\left (\int \frac {1}{1+\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sin ^2(e+f x)}\right )}{8 b (a+b)^2 f} \\ & = \frac {\left (8 a^2+8 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{8 (a+b)^{5/2} f}-\frac {(8 a+5 b) \sec ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{8 (a+b)^2 f}+\frac {\sec ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{4 (a+b) f} \\ \end{align*}
Time = 0.50 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.81 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {\left (8 a^2+8 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )+\sqrt {a+b} \sec ^2(e+f x) \left (-8 a-5 b+2 (a+b) \sec ^2(e+f x)\right ) \sqrt {a+b \sin ^2(e+f x)}}{8 (a+b)^{5/2} f} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(643\) vs. \(2(118)=236\).
Time = 1.58 (sec) , antiderivative size = 644, normalized size of antiderivative = 4.81
method | result | size |
default | \(\frac {\left (8 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a^{4}+24 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a^{3} b +27 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a^{2} b^{2}+14 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a \,b^{3}+3 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) b^{4}+8 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a^{4}+24 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a^{3} b +27 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a^{2} b^{2}+14 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a \,b^{3}+3 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) b^{4}\right ) \left (\cos ^{4}\left (f x +e \right )\right )-2 \left (a +b \right )^{\frac {5}{2}} \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (8 a +5 b \right ) \left (\cos ^{2}\left (f x +e \right )\right )+4 a \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (a +b \right )^{\frac {5}{2}}+4 b \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (a +b \right )^{\frac {5}{2}}}{16 \left (a +b \right )^{\frac {5}{2}} \cos \left (f x +e \right )^{4} \left (a^{2}+2 a b +b^{2}\right ) f}\) | \(644\) |
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Time = 0.39 (sec) , antiderivative size = 328, normalized size of antiderivative = 2.45 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\left [\frac {{\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \sqrt {a + b} \cos \left (f x + e\right )^{4} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} - 2 \, a - 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) - 2 \, {\left ({\left (8 \, a^{2} + 13 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, a^{2} - 4 \, a b - 2 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{16 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{4}}, -\frac {{\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{a + b}\right ) \cos \left (f x + e\right )^{4} + {\left ({\left (8 \, a^{2} + 13 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, a^{2} - 4 \, a b - 2 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{8 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{4}}\right ] \]
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\[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int \frac {\tan ^{5}{\left (e + f x \right )}}{\sqrt {a + b \sin ^{2}{\left (e + f x \right )}}}\, dx \]
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Leaf count of result is larger than twice the leaf count of optimal. 248 vs. \(2 (118) = 236\).
Time = 0.30 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.85 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=-\frac {\frac {{\left (8 \, a^{2} b^{3} + 8 \, a b^{4} + 3 \, b^{5}\right )} \log \left (\frac {\sqrt {b \sin \left (f x + e\right )^{2} + a} - \sqrt {a + b}}{\sqrt {b \sin \left (f x + e\right )^{2} + a} + \sqrt {a + b}}\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a + b}} - \frac {2 \, {\left ({\left (8 \, a b^{4} + 5 \, b^{5}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} - {\left (8 \, a^{2} b^{4} + 11 \, a b^{5} + 3 \, b^{6}\right )} \sqrt {b \sin \left (f x + e\right )^{2} + a}\right )}}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4} + {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{2} {\left (a^{2} + 2 \, a b + b^{2}\right )} - 2 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}}}{16 \, b^{3} f} \]
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\[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int { \frac {\tan \left (f x + e\right )^{5}}{\sqrt {b \sin \left (f x + e\right )^{2} + a}} \,d x } \]
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Timed out. \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int \frac {{\mathrm {tan}\left (e+f\,x\right )}^5}{\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}} \,d x \]
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