Integrand size = 25, antiderivative size = 214 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {i \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{5/2} d}-\frac {i \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{5/2} d}-\frac {2 a^2 \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 a \left (a^2+7 b^2\right ) \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}} \]
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Time = 0.87 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3646, 3730, 3697, 3696, 95, 209, 212} \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=-\frac {2 a^2 \sqrt {\tan (c+d x)}}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {2 a \left (a^2+7 b^2\right ) \sqrt {\tan (c+d x)}}{3 b d \left (a^2+b^2\right )^2 \sqrt {a+b \tan (c+d x)}}+\frac {i \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (-b+i a)^{5/2}}-\frac {i \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (b+i a)^{5/2}} \]
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Rule 95
Rule 209
Rule 212
Rule 3646
Rule 3696
Rule 3697
Rule 3730
Rubi steps \begin{align*} \text {integral}& = -\frac {2 a^2 \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 \int \frac {\frac {a^2}{2}-\frac {3}{2} a b \tan (c+d x)+\frac {1}{2} \left (a^2+3 b^2\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx}{3 b \left (a^2+b^2\right )} \\ & = -\frac {2 a^2 \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 a \left (a^2+7 b^2\right ) \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {4 \int \frac {-\frac {3}{2} a^2 b^2-\frac {3}{4} a b \left (a^2-b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{3 a b \left (a^2+b^2\right )^2} \\ & = -\frac {2 a^2 \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 a \left (a^2+7 b^2\right ) \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {i \int \frac {1+i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{2 (a-i b)^2}-\frac {i \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{2 (a+i b)^2} \\ & = -\frac {2 a^2 \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 a \left (a^2+7 b^2\right ) \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {i \text {Subst}\left (\int \frac {1}{(1-i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a-i b)^2 d}-\frac {i \text {Subst}\left (\int \frac {1}{(1+i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a+i b)^2 d} \\ & = -\frac {2 a^2 \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 a \left (a^2+7 b^2\right ) \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {i \text {Subst}\left (\int \frac {1}{1-(i a+b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(a-i b)^2 d}-\frac {i \text {Subst}\left (\int \frac {1}{1-(-i a+b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(a+i b)^2 d} \\ & = \frac {i \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{5/2} d}-\frac {i \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{5/2} d}-\frac {2 a^2 \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 a \left (a^2+7 b^2\right ) \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}} \\ \end{align*}
Time = 2.90 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.88 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {\frac {3 \sqrt [4]{-1} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(-a+i b)^{5/2}}-\frac {3 \sqrt [4]{-1} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(a+i b)^{5/2}}+\frac {2 a \sqrt {\tan (c+d x)} \left (6 a b+\left (a^2+7 b^2\right ) \tan (c+d x)\right )}{\left (a^2+b^2\right )^2 (a+b \tan (c+d x))^{3/2}}}{3 d} \]
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result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 1.27 (sec) , antiderivative size = 1488448, normalized size of antiderivative = 6955.36
\[\text {output too large to display}\]
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Leaf count of result is larger than twice the leaf count of optimal. 11334 vs. \(2 (174) = 348\).
Time = 2.37 (sec) , antiderivative size = 11334, normalized size of antiderivative = 52.96 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
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\[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\int \frac {\tan ^{\frac {5}{2}}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]
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\[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\int { \frac {\tan \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
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Timed out. \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}} \,d x \]
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