Integrand size = 35, antiderivative size = 309 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\frac {(i a-b)^{5/2} (i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(i a+b)^{5/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a (10 A b+7 a B) \sqrt {a+b \tan (c+d x)}}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (35 a^2 A-45 A b^2-77 a b B\right ) \sqrt {a+b \tan (c+d x)}}{105 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (245 a^2 A b-15 A b^3+105 a^3 B-161 a b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{105 a d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)} \]
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Time = 2.20 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {3686, 3726, 3730, 3697, 3696, 95, 209, 212} \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 \left (35 a^2 A-77 a b B-45 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{105 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (105 a^3 B+245 a^2 A b-161 a b^2 B-15 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{105 a d \sqrt {\tan (c+d x)}}+\frac {(-b+i a)^{5/2} (-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(b+i a)^{5/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a (7 a B+10 A b) \sqrt {a+b \tan (c+d x)}}{35 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)} \]
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Rule 95
Rule 209
Rule 212
Rule 3686
Rule 3696
Rule 3697
Rule 3726
Rule 3730
Rubi steps \begin{align*} \text {integral}& = -\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2}{7} \int \frac {\sqrt {a+b \tan (c+d x)} \left (\frac {1}{2} a (10 A b+7 a B)-\frac {7}{2} \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)-\frac {1}{2} b (4 a A-7 b B) \tan ^2(c+d x)\right )}{\tan ^{\frac {7}{2}}(c+d x)} \, dx \\ & = -\frac {2 a (10 A b+7 a B) \sqrt {a+b \tan (c+d x)}}{35 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {4}{35} \int \frac {-\frac {1}{4} a \left (35 a^2 A-45 A b^2-77 a b B\right )-\frac {35}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)-\frac {1}{4} b \left (60 a A b+28 a^2 B-35 b^2 B\right ) \tan ^2(c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx \\ & = -\frac {2 a (10 A b+7 a B) \sqrt {a+b \tan (c+d x)}}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (35 a^2 A-45 A b^2-77 a b B\right ) \sqrt {a+b \tan (c+d x)}}{105 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {8 \int \frac {\frac {1}{8} a \left (245 a^2 A b-15 A b^3+105 a^3 B-161 a b^2 B\right )-\frac {105}{8} a \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \tan (c+d x)-\frac {1}{4} a b \left (35 a^2 A-45 A b^2-77 a b B\right ) \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx}{105 a} \\ & = -\frac {2 a (10 A b+7 a B) \sqrt {a+b \tan (c+d x)}}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (35 a^2 A-45 A b^2-77 a b B\right ) \sqrt {a+b \tan (c+d x)}}{105 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (245 a^2 A b-15 A b^3+105 a^3 B-161 a b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{105 a d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {16 \int \frac {\frac {105}{16} a^2 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )+\frac {105}{16} a^2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{105 a^2} \\ & = -\frac {2 a (10 A b+7 a B) \sqrt {a+b \tan (c+d x)}}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (35 a^2 A-45 A b^2-77 a b B\right ) \sqrt {a+b \tan (c+d x)}}{105 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (245 a^2 A b-15 A b^3+105 a^3 B-161 a b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{105 a d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {1}{2} \left ((a-i b)^3 (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)^3 (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 (10 A b+7 a B) \sqrt {a+b \tan (c+d x)}}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (35 a^2 A-45 A b^2-77 a b B\right ) \sqrt {a+b \tan (c+d x)}}{105 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (245 a^2 A b-15 A b^3+105 a^3 B-161 a b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{105 a d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {\left ((a-i b)^3 (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)^3 (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 (10 A b+7 a B) \sqrt {a+b \tan (c+d x)}}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (35 a^2 A-45 A b^2-77 a b B\right ) \sqrt {a+b \tan (c+d x)}}{105 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (245 a^2 A b-15 A b^3+105 a^3 B-161 a b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{105 a d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {\left ((a-i b)^3 (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)^3 (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)^{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 {(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 {2 a (10 A b+7 a B) \sqrt {a+b \tan (c+d x)}}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (35 a^2 A-45 A b^2-77 a b B\right ) \sqrt {a+b \tan (c+d x)}}{105 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (245 a^2 A b-15 A b^3+105 a^3 B-161 a b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{105 a d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)} \\ \end{align*}
Time = 6.80 (sec) , antiderivative size = 381, normalized size of antiderivative = 1.23 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\frac {420 \sqrt [4]{-1} a \left ((-a+i b)^{5/2} (i A+B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+(a+i b)^{5/2} (-i A+B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )\right ) \tan ^{\frac {7}{2}}(c+d x)-35 a b (4 A b+a B) \sqrt {a+b \tan (c+d x)}-5 a \left (24 a^2 A-28 A b^2-49 a b B\right ) \sqrt {a+b \tan (c+d x)}-6 a \left (60 a A b+28 a^2 B-35 b^2 B\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}+8 a \left (35 a^2 A-45 A b^2-77 a b B\right ) \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}+8 \left (245 a^2 A b-15 A b^3+105 a^3 B-161 a b^2 B\right ) \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}-210 a b B (a+b \tan (c+d x))^{3/2}}{420 a d \tan ^{\frac {7}{2}}(c+d x)} \]
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result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 1.06 (sec) , antiderivative size = 2654465, normalized size of antiderivative = 8590.50
\[\text {output too large to display}\]
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Leaf count of result is larger than twice the leaf count of optimal. 18856 vs. \(2 (257) = 514\).
Time = 4.85 (sec) , antiderivative size = 18856, normalized size of antiderivative = 61.02 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\text {Too large to display} \]
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Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\text {Timed out} \]
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\[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\tan \left (d x + c\right )^{\frac {9}{2}}} \,d x } \]
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Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{9/2}} \,d x \]
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