Integrand size = 35, antiderivative size = 264 \[ \int \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {\sqrt {i a-b} (i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\left (4 a A b-a^2 B-8 b^2 B\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 b^{3/2} d}+\frac {\sqrt {i a+b} (i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(4 A b-a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 b d} \]
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Time = 2.32 (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 = {3688, 3728, 3736, 6857, 65, 223, 212, 95, 211, 214} \[ \int \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {\left (a^2 (-B)+4 a A b-8 b^2 B\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 b^{3/2} d}+\frac {\sqrt {-b+i a} (-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\sqrt {b+i a} (B+i A) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(4 A b-a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 b d} \]
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Rule 65
Rule 95
Rule 211
Rule 212
Rule 214
Rule 223
Rule 3688
Rule 3728
Rule 3736
Rule 6857
Rubi steps \begin{align*} \text {integral}& = \frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 b d}+\frac {\int \frac {\sqrt {a+b \tan (c+d x)} \left (-\frac {a B}{2}-2 b B \tan (c+d x)+\frac {1}{2} (4 A b-a B) \tan ^2(c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{2 b} \\ & = \frac {(4 A b-a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 b d}+\frac {\int \frac {-\frac {1}{4} a (4 A b+a B)-2 b (A b+a B) \tan (c+d x)+\frac {1}{4} \left (4 a A b-a^2 B-8 b^2 B\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{2 b} \\ & = \frac {(4 A b-a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 b d}+\frac {\text {Subst}\left (\int \frac {-\frac {1}{4} a (4 A b+a B)-2 b (A b+a B) x+\frac {1}{4} \left (4 a A b-a^2 B-8 b^2 B\right ) x^2}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{2 b d} \\ & = \frac {(4 A b-a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 b d}+\frac {\text {Subst}\left (\int \left (\frac {4 a A b-a^2 B-8 b^2 B}{4 \sqrt {x} \sqrt {a+b x}}-\frac {2 (b (a A-b B)+b (A b+a B) x)}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{2 b d} \\ & = \frac {(4 A b-a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 b d}-\frac {\text {Subst}\left (\int \frac {b (a A-b B)+b (A b+a B) x}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{b d}+\frac {\left (4 a A b-a^2 B-8 b^2 B\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{8 b d} \\ & = \frac {(4 A b-a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 b d}-\frac {\text {Subst}\left (\int \left (\frac {-b (A b+a B)+i b (a A-b B)}{2 (i-x) \sqrt {x} \sqrt {a+b x}}+\frac {b (A b+a B)+i b (a A-b B)}{2 \sqrt {x} (i+x) \sqrt {a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{b d}+\frac {\left (4 a A b-a^2 B-8 b^2 B\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{4 b d} \\ & = \frac {(4 A b-a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 b d}-\frac {((i a+b) (A-i B)) \text {Subst}\left (\int \frac {1}{\sqrt {x} (i+x) \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}-\frac {((i a-b) (A+i B)) \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 (4 a A b-a^2 B-8 b^2 B\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 )}{4 b d} \\ & = \frac {\left (4 a A b-a^2 B-8 b^2 B\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 b^{3/2} d}+\frac {(4 A b-a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 b d}-\frac {((i a+b) (A-i B)) \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) (A+i B)) \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 {\sqrt {i a-b} (i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\left (4 a A b-a^2 B-8 b^2 B\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 b^{3/2} d}+\frac {\sqrt {i a+b} (i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(4 A b-a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 b d} \\ \end{align*}
Time = 3.73 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.15 \[ \int \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {-\sqrt {a} \left (-4 a A b+a^2 B+8 b^2 B\right ) \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}+\sqrt {b} \left (-4 \sqrt [4]{-1} \sqrt {-a+i b} b (i A+B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {a+b \tan (c+d x)}+4 (-1)^{3/4} \sqrt {a+i b} b (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {a+b \tan (c+d x)}+\sqrt {\tan (c+d x)} (a+b \tan (c+d x)) (4 A b+a B+2 b B \tan (c+d x))\right )}{4 b^{3/2} d \sqrt {a+b \tan (c+d x)}} \]
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result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 1.76 (sec) , antiderivative size = 2183234, normalized size of antiderivative = 8269.83
\[\text {output too large to display}\]
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Leaf count of result is larger than twice the leaf count of optimal. 8056 vs. \(2 (212) = 424\).
Time = 3.48 (sec) , antiderivative size = 16118, normalized size of antiderivative = 61.05 \[ \int \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]
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\[ \int \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int \left (A + B \tan {\left (c + d x \right )}\right ) \sqrt {a + b \tan {\left (c + d x \right )}} \tan ^{\frac {3}{2}}{\left (c + d x \right )}\, dx \]
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\[ \int \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {b \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{\frac {3}{2}} \,d x } \]
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Timed out. \[ \int \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]
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Timed out. \[ \int \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )} \,d x \]
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