Integrand size = 25, antiderivative size = 182 \[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {i (i a-b)^{5/2} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {i (i a+b)^{5/2} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {14 a b \sqrt {a+b \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}} \]
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Time = 0.85 (sec) , antiderivative size = 182, 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} \[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {i (-b+i a)^{5/2} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {i (b+i a)^{5/2} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {14 a b \sqrt {a+b \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}} \]
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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 {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2}{3} \int \frac {\frac {7 a^2 b}{2}-\frac {3}{2} a \left (a^2-3 b^2\right ) \tan (c+d x)-\frac {1}{2} b \left (2 a^2-3 b^2\right ) \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx \\ & = -\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {14 a b \sqrt {a+b \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}}-\frac {4 \int \frac {\frac {3}{4} a^2 \left (a^2-3 b^2\right )+\frac {3}{4} a b \left (3 a^2-b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{3 a} \\ & = -\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {14 a b \sqrt {a+b \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}}-\frac {1}{2} (a-i b)^3 \int \frac {1+i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx-\frac {1}{2} (a+i b)^3 \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx \\ & = -\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {14 a b \sqrt {a+b \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}}-\frac {(a-i b)^3 \text {Subst}\left (\int \frac {1}{(1-i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}-\frac {(a+i b)^3 \text {Subst}\left (\int \frac {1}{(1+i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d} \\ & = -\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {14 a b \sqrt {a+b \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}}-\frac {(a-i b)^3 \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 )}{d}-\frac {(a+i b)^3 \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 )}{d} \\ & = -\frac {i (i a-b)^{5/2} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {i (i a+b)^{5/2} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {14 a b \sqrt {a+b \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}} \\ \end{align*}
Time = 0.94 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.93 \[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {-3 (-1)^{3/4} (-a+i b)^{5/2} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+3 (-1)^{3/4} (a+i b)^{5/2} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-\frac {2 a \sqrt {a+b \tan (c+d x)} (a+7 b \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)}}{3 d} \]
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result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 0.65 (sec) , antiderivative size = 1308720, normalized size of antiderivative = 7190.77
\[\text {output too large to display}\]
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Leaf count of result is larger than twice the leaf count of optimal. 5026 vs. \(2 (142) = 284\).
Time = 0.84 (sec) , antiderivative size = 5026, normalized size of antiderivative = 27.62 \[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Too large to display} \]
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\[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]
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\[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\tan \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
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Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}} \,d x \]
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