Integrand size = 35, antiderivative size = 397 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=-\frac {(i a-b)^{5/2} (i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\left (40 a^3 A b-320 a A b^3-5 a^4 B-240 a^2 b^2 B+128 b^4 B\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{64 b^{3/2} d}-\frac {(i a+b)^{5/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\left (40 a^2 A b-64 A b^3-5 a^3 B-112 a b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{64 b d}+\frac {\left (40 a A b-5 a^2 B-48 b^2 B\right ) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{96 b d}+\frac {(8 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{24 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d} \]
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Time = 4.00 (sec) , antiderivative size = 397, normalized size of antiderivative = 1.00, number of steps used = 16, 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} \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {\left (-5 a^2 B+40 a A b-48 b^2 B\right ) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{96 b d}+\frac {\left (-5 a^3 B+40 a^2 A b-112 a b^2 B-64 A b^3\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{64 b d}+\frac {\left (-5 a^4 B+40 a^3 A b-240 a^2 b^2 B-320 a A b^3+128 b^4 B\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{64 b^{3/2} d}-\frac {(-b+i a)^{5/2} (-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {(b+i a)^{5/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(8 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{24 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d} \]
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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))^{7/2}}{4 b d}+\frac {\int \frac {(a+b \tan (c+d x))^{5/2} \left (-\frac {a B}{2}-4 b B \tan (c+d x)+\frac {1}{2} (8 A b-a B) \tan ^2(c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{4 b} \\ & = \frac {(8 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{24 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}+\frac {\int \frac {(a+b \tan (c+d x))^{3/2} \left (-\frac {1}{4} a (8 A b+5 a B)-12 b (A b+a B) \tan (c+d x)+\frac {1}{4} \left (40 a A b-5 a^2 B-48 b^2 B\right ) \tan ^2(c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{12 b} \\ & = \frac {\left (40 a A b-5 a^2 B-48 b^2 B\right ) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{96 b d}+\frac {(8 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{24 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}+\frac {\int \frac {\sqrt {a+b \tan (c+d x)} \left (-\frac {3}{8} a \left (24 a A b+5 a^2 B-16 b^2 B\right )-24 b \left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)+\frac {3}{8} \left (40 a^2 A b-64 A b^3-5 a^3 B-112 a b^2 B\right ) \tan ^2(c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{24 b} \\ & = \frac {\left (40 a^2 A b-64 A b^3-5 a^3 B-112 a b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{64 b d}+\frac {\left (40 a A b-5 a^2 B-48 b^2 B\right ) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{96 b d}+\frac {(8 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{24 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}+\frac {\int \frac {-\frac {3}{16} a \left (88 a^2 A b-64 A b^3+5 a^3 B-144 a b^2 B\right )-24 b \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)+\frac {3}{16} \left (40 a^3 A b-320 a A b^3-5 a^4 B-240 a^2 b^2 B+128 b^4 B\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{24 b} \\ & = \frac {\left (40 a^2 A b-64 A b^3-5 a^3 B-112 a b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{64 b d}+\frac {\left (40 a A b-5 a^2 B-48 b^2 B\right ) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{96 b d}+\frac {(8 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{24 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}+\frac {\text {Subst}\left (\int \frac {-\frac {3}{16} a \left (88 a^2 A b-64 A b^3+5 a^3 B-144 a b^2 B\right )-24 b \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) x+\frac {3}{16} \left (40 a^3 A b-320 a A b^3-5 a^4 B-240 a^2 b^2 B+128 b^4 B\right ) x^2}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{24 b d} \\ & = \frac {\left (40 a^2 A b-64 A b^3-5 a^3 B-112 a b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{64 b d}+\frac {\left (40 a A b-5 a^2 B-48 b^2 B\right ) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{96 b d}+\frac {(8 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{24 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}+\frac {\text {Subst}\left (\int \left (\frac {3 \left (40 a^3 A b-320 a A b^3-5 a^4 B-240 a^2 b^2 B+128 b^4 B\right )}{16 \sqrt {x} \sqrt {a+b x}}-\frac {24 \left (b \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )+b \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) x\right )}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{24 b d} \\ & = \frac {\left (40 a^2 A b-64 A b^3-5 a^3 B-112 a b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{64 b d}+\frac {\left (40 a A b-5 a^2 B-48 b^2 B\right ) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{96 b d}+\frac {(8 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{24 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}-\frac {\text {Subst}\left (\int \frac {b \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )+b \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) x}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{b d}+\frac {\left (40 a^3 A b-320 a A b^3-5 a^4 B-240 a^2 b^2 B+128 b^4 B\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{128 b d} \\ & = \frac {\left (40 a^2 A b-64 A b^3-5 a^3 B-112 a b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{64 b d}+\frac {\left (40 a A b-5 a^2 B-48 b^2 B\right ) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{96 b d}+\frac {(8 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{24 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}-\frac {\text {Subst}\left (\int \left (\frac {-b \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )+i b \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )}{2 (i-x) \sqrt {x} \sqrt {a+b x}}+\frac {b \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )+i b \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )}{2 \sqrt {x} (i+x) \sqrt {a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{b d}+\frac {\left (40 a^3 A b-320 a A b^3-5 a^4 B-240 a^2 b^2 B+128 b^4 B\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{64 b d} \\ & = \frac {\left (40 a^2 A b-64 A b^3-5 a^3 B-112 a b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{64 b d}+\frac {\left (40 a A b-5 a^2 B-48 b^2 B\right ) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{96 b d}+\frac {(8 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{24 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}+\frac {\left ((i a+b)^3 (A-i B)\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 (40 a^3 A b-320 a A b^3-5 a^4 B-240 a^2 b^2 B+128 b^4 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 )}{64 b d}-\frac {\left (-b \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )+i b \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )\right ) \text {Subst}\left (\int \frac {1}{(i-x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 b d} \\ & = \frac {\left (40 a^3 A b-320 a A b^3-5 a^4 B-240 a^2 b^2 B+128 b^4 B\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{64 b^{3/2} d}+\frac {\left (40 a^2 A b-64 A b^3-5 a^3 B-112 a b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{64 b d}+\frac {\left (40 a A b-5 a^2 B-48 b^2 B\right ) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{96 b d}+\frac {(8 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{24 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}+\frac {\left ((i a+b)^3 (A-i B)\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 (-b \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )+i b \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )\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 )}{b d} \\ & = -\frac {(i a-b)^{5/2} (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 (40 a^3 A b-320 a A b^3-5 a^4 B-240 a^2 b^2 B+128 b^4 B\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{64 b^{3/2} d}-\frac {(i a+b)^{5/2} (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 {\left (40 a^2 A b-64 A b^3-5 a^3 B-112 a b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{64 b d}+\frac {\left (40 a A b-5 a^2 B-48 b^2 B\right ) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{96 b d}+\frac {(8 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{24 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d} \\ \end{align*}
Time = 4.98 (sec) , antiderivative size = 411, normalized size of antiderivative = 1.04 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {-192 \sqrt [4]{-1} (-a+i b)^{5/2} b (i A+B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+192 (-1)^{3/4} (a+i b)^{5/2} b (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-3 \left (-40 a^2 A b+64 A b^3+5 a^3 B+112 a b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}-2 \left (-40 a A b+5 a^2 B+48 b^2 B\right ) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}+8 (8 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}+48 B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{7/2}-\frac {3 \sqrt {a} \left (-40 a^3 A b+320 a A b^3+5 a^4 B+240 a^2 b^2 B-128 b^4 B\right ) \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {b} \sqrt {a+b \tan (c+d x)}}}{192 b d} \]
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result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 1.06 (sec) , antiderivative size = 2659561, normalized size of antiderivative = 6699.15
\[\text {output too large to display}\]
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Leaf count of result is larger than twice the leaf count of optimal. 17728 vs. \(2 (334) = 668\).
Time = 9.41 (sec) , antiderivative size = 35458, normalized size of antiderivative = 89.31 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]
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Timed out. \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]
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\[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \tan \left (d x + c\right )^{\frac {3}{2}} \,d x } \]
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Timed out. \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]
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Timed out. \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2} \,d x \]
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