Integrand size = 35, antiderivative size = 287 \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=-\frac {(A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{(i a-b)^{5/2} d}-\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{(i a+b)^{5/2} d}+\frac {2 b (A b-a B)}{3 a \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (8 a^2 A b+2 A b^3-5 a^3 B+a b^2 B\right )}{3 a^2 \left (a^2+b^2\right )^2 d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}} \]
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Time = 1.19 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {4326, 3690, 3730, 3697, 3696, 95, 209, 212} \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {2 b (A b-a B)}{3 a d \left (a^2+b^2\right ) \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (-5 a^3 B+8 a^2 A b+a b^2 B+2 A b^3\right )}{3 a^2 d \left (a^2+b^2\right )^2 \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {(A+i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (-b+i a)^{5/2}}-\frac {(A-i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (b+i a)^{5/2}} \]
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Rule 95
Rule 209
Rule 212
Rule 3690
Rule 3696
Rule 3697
Rule 3730
Rule 4326
Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}} \, dx \\ & = \frac {2 b (A b-a B)}{3 a \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {\left (2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\frac {1}{2} \left (3 a^2 A+2 A b^2+a b B\right )-\frac {3}{2} a (A b-a B) \tan (c+d x)+b (A b-a B) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx}{3 a \left (a^2+b^2\right )} \\ & = \frac {2 b (A b-a B)}{3 a \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (8 a^2 A b+2 A b^3-5 a^3 B+a b^2 B\right )}{3 a^2 \left (a^2+b^2\right )^2 d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}+\frac {\left (4 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\frac {3}{4} a^2 \left (a^2 A-A b^2+2 a b B\right )-\frac {3}{4} a^2 \left (2 a A b-a^2 B+b^2 B\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{3 a^2 \left (a^2+b^2\right )^2} \\ & = \frac {2 b (A b-a B)}{3 a \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (8 a^2 A b+2 A b^3-5 a^3 B+a b^2 B\right )}{3 a^2 \left (a^2+b^2\right )^2 d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}+\frac {\left ((a+i b)^2 (A-i B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {1+i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{2 \left (a^2+b^2\right )^2}+\frac {\left ((a-i b)^2 (A+i B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{2 \left (a^2+b^2\right )^2} \\ & = \frac {2 b (A b-a B)}{3 a \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (8 a^2 A b+2 A b^3-5 a^3 B+a b^2 B\right )}{3 a^2 \left (a^2+b^2\right )^2 d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}+\frac {\left ((a+i b)^2 (A-i B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{(1-i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 \left (a^2+b^2\right )^2 d}+\frac {\left ((a-i b)^2 (A+i B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{(1+i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 \left (a^2+b^2\right )^2 d} \\ & = \frac {2 b (A b-a B)}{3 a \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (8 a^2 A b+2 A b^3-5 a^3 B+a b^2 B\right )}{3 a^2 \left (a^2+b^2\right )^2 d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}+\frac {\left ((a+i b)^2 (A-i B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \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 )}{\left (a^2+b^2\right )^2 d}+\frac {\left ((a-i b)^2 (A+i B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \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 )}{\left (a^2+b^2\right )^2 d} \\ & = -\frac {(A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{(i a-b)^{5/2} d}-\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{(i a+b)^{5/2} d}+\frac {2 b (A b-a B)}{3 a \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (8 a^2 A b+2 A b^3-5 a^3 B+a b^2 B\right )}{3 a^2 \left (a^2+b^2\right )^2 d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}} \\ \end{align*}
Time = 3.90 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.02 \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (-3 \sqrt [4]{-1} \left (\frac {(a+i b)^2 (i A+B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}+\frac {i (a-i b)^2 (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}\right )+\frac {2 b \left (a^2+b^2\right ) (A b-a B) \sqrt {\tan (c+d x)}}{a (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (8 a^2 A b+2 A b^3-5 a^3 B+a b^2 B\right ) \sqrt {\tan (c+d x)}}{a^2 \sqrt {a+b \tan (c+d x)}}\right )}{3 \left (a^2+b^2\right )^2 d} \]
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result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 2.74 (sec) , antiderivative size = 2978354, normalized size of antiderivative = 10377.54
\[\text {output too large to display}\]
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Leaf count of result is larger than twice the leaf count of optimal. 27790 vs. \(2 (240) = 480\).
Time = 14.62 (sec) , antiderivative size = 27790, normalized size of antiderivative = 96.83 \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
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\[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \sqrt {\cot {\left (c + d x \right )}}}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]
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\[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {\cot \left (d x + c\right )}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
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Exception generated. \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \]
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Timed out. \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\int \frac {\sqrt {\mathrm {cot}\left (c+d\,x\right )}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}} \,d x \]
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