Integrand size = 35, antiderivative size = 196 \[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {(i a-b)^{3/2} (A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(i a+b)^{3/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a A \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (4 A b+3 a B) \sqrt {a+b \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}} \]
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Time = 1.37 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3686, 3730, 3697, 3696, 95, 209, 212} \[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {(-b+i a)^{3/2} (A+i B) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(b+i a)^{3/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 (3 a B+4 A b) \sqrt {a+b \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}}-\frac {2 a A \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)} \]
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Rule 95
Rule 209
Rule 212
Rule 3686
Rule 3696
Rule 3697
Rule 3730
Rubi steps \begin{align*} \text {integral}& = -\frac {2 a A \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2}{3} \int \frac {\frac {1}{2} a (4 A b+3 a B)-\frac {3}{2} \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)-\frac {1}{2} b (2 a A-3 b B) \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx \\ & = -\frac {2 a A \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (4 A b+3 a B) \sqrt {a+b \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}}-\frac {4 \int \frac {\frac {3}{4} a \left (a^2 A-A b^2-2 a b B\right )+\frac {3}{4} a \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} \\ & = -\frac {2 a A \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (4 A b+3 a B) \sqrt {a+b \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}}-\frac {1}{2} \left ((a-i b)^2 (A-i B)\right ) \int \frac {1+i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx-\frac {1}{2} \left ((a+i b)^2 (A+i B)\right ) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx \\ & = -\frac {2 a A \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (4 A b+3 a B) \sqrt {a+b \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}}-\frac {\left ((a-i b)^2 (A-i B)\right ) \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 {\left ((a+i b)^2 (A+i B)\right ) \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 A \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (4 A b+3 a B) \sqrt {a+b \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}}-\frac {\left ((a-i b)^2 (A-i B)\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 )}{d}-\frac {\left ((a+i b)^2 (A+i B)\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 )}{d} \\ & = \frac {(i a-b)^{3/2} (A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(i a+b)^{3/2} (A-i B) \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 A \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (4 A b+3 a B) \sqrt {a+b \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}} \\ \end{align*}
Time = 1.00 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.21 \[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {3 \sqrt [4]{-1} \left ((-a+i b)^{3/2} (i A+B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+i (a+i b)^{3/2} (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )\right ) \tan ^{\frac {3}{2}}(c+d x)-3 b B \sqrt {a+b \tan (c+d x)}+(-2 a A+3 b B) \sqrt {a+b \tan (c+d x)}-2 (4 A b+3 a B) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)} \]
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result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 0.91 (sec) , antiderivative size = 2398858, normalized size of antiderivative = 12239.07
\[\text {output too large to display}\]
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Leaf count of result is larger than twice the leaf count of optimal. 12024 vs. \(2 (156) = 312\).
Time = 2.46 (sec) , antiderivative size = 12024, normalized size of antiderivative = 61.35 \[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Too large to display} \]
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\[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]
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\[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\tan \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
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Timed out. \[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}} \,d x \]
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