Integrand size = 35, antiderivative size = 264 \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=-\frac {(i a-b)^{3/2} (A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {\sqrt {b} (2 A b+3 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {(i a+b)^{3/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {b B \sqrt {a+b \tan (c+d x)}}{d \sqrt {\cot (c+d x)}} \]
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Time = 2.13 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {4326, 3688, 3736, 6857, 65, 223, 212, 95, 211, 214} \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=-\frac {(-b+i a)^{3/2} (A+i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\sqrt {b} (3 a B+2 A b) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {(b+i a)^{3/2} (A-i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {b B \sqrt {a+b \tan (c+d x)}}{d \sqrt {\cot (c+d x)}} \]
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Rule 65
Rule 95
Rule 211
Rule 212
Rule 214
Rule 223
Rule 3688
Rule 3736
Rule 4326
Rule 6857
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))^{3/2} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx \\ & = \frac {b B \sqrt {a+b \tan (c+d x)}}{d \sqrt {\cot (c+d x)}}+\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\frac {1}{2} a (2 a A-b B)+\left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)+\frac {1}{2} b (2 A b+3 a B) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx \\ & = \frac {b B \sqrt {a+b \tan (c+d x)}}{d \sqrt {\cot (c+d x)}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {\frac {1}{2} a (2 a A-b B)+\left (2 a A b+a^2 B-b^2 B\right ) x+\frac {1}{2} b (2 A b+3 a B) x^2}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {b B \sqrt {a+b \tan (c+d x)}}{d \sqrt {\cot (c+d x)}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \left (\frac {b (2 A b+3 a B)}{2 \sqrt {x} \sqrt {a+b x}}+\frac {a^2 A-A b^2-2 a b B+\left (2 a A b+a^2 B-b^2 B\right ) x}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {b B \sqrt {a+b \tan (c+d x)}}{d \sqrt {\cot (c+d x)}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {a^2 A-A b^2-2 a b B+\left (2 a A b+a^2 B-b^2 B\right ) x}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}+\frac {\left (b (2 A b+3 a B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d} \\ & = \frac {b B \sqrt {a+b \tan (c+d x)}}{d \sqrt {\cot (c+d x)}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \left (\frac {-2 a A b-a^2 B+b^2 B+i \left (a^2 A-A b^2-2 a b B\right )}{2 (i-x) \sqrt {x} \sqrt {a+b x}}+\frac {2 a A b+a^2 B-b^2 B+i \left (a^2 A-A b^2-2 a b B\right )}{2 \sqrt {x} (i+x) \sqrt {a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{d}+\frac {\left (b (2 A b+3 a B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d} \\ & = \frac {b B \sqrt {a+b \tan (c+d x)}}{d \sqrt {\cot (c+d x)}}+\frac {\left ((a+i b)^2 (i A-B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{(i-x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {\left ((a-i b)^2 (i A+B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} (i+x) \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {\left (b (2 A b+3 a B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \\ & = \frac {\sqrt {b} (2 A b+3 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {b B \sqrt {a+b \tan (c+d x)}}{d \sqrt {\cot (c+d x)}}+\frac {\left ((a+i b)^2 (i A-B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{i-(a+i 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 (i A+B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{i-(-a+i 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 ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {\sqrt {b} (2 A b+3 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{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 ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {b B \sqrt {a+b \tan (c+d x)}}{d \sqrt {\cot (c+d x)}} \\ \end{align*}
Time = 1.41 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.00 \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (-\sqrt [4]{-1} (-a+i b)^{3/2} (i A+B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-(-1)^{3/4} (a+i b)^{3/2} (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+b B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}+\frac {\sqrt {a} \sqrt {b} (2 A b+3 a B) \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \]
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result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 1.50 (sec) , antiderivative size = 2398750, normalized size of antiderivative = 9086.17
\[\text {output too large to display}\]
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Leaf count of result is larger than twice the leaf count of optimal. 12219 vs. \(2 (212) = 424\).
Time = 4.89 (sec) , antiderivative size = 24471, normalized size of antiderivative = 92.69 \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]
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\[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\int \left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}} \sqrt {\cot {\left (c + d x \right )}}\, dx \]
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\[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sqrt {\cot \left (d x + c\right )} \,d x } \]
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Timed out. \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]
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Timed out. \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\int \sqrt {\mathrm {cot}\left (c+d\,x\right )}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2} \,d x \]
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