Integrand size = 35, antiderivative size = 290 \[ \int \cot ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {\sqrt {i a-b} (A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {\sqrt {i a+b} (A-i B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {2 \left (15 a^2 A+2 A b^2-5 a b B\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{15 a^2 d}-\frac {2 (A b+5 a B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 a d}-\frac {2 A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d} \]
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Time = 1.41 (sec) , antiderivative size = 290, 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 = {4326, 3689, 3730, 3697, 3696, 95, 209, 212} \[ \int \cot ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {2 \left (15 a^2 A-5 a b B+2 A b^2\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{15 a^2 d}+\frac {\sqrt {-b+i a} (A+i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {\sqrt {b+i a} (A-i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 (5 a B+A b) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 a d}-\frac {2 A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d} \]
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Rule 95
Rule 209
Rule 212
Rule 3689
Rule 3696
Rule 3697
Rule 3730
Rule 4326
Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx \\ & = -\frac {2 A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d}-\frac {1}{5} \left (2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\frac {1}{2} (-A b-5 a B)+\frac {5}{2} (a A-b B) \tan (c+d x)+2 A b \tan ^2(c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx \\ & = -\frac {2 (A b+5 a B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 a d}-\frac {2 A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d}+\frac {\left (4 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\frac {1}{4} \left (-15 a^2 A-2 A b^2+5 a b B\right )-\frac {15}{4} a (A b+a B) \tan (c+d x)-\frac {1}{2} b (A b+5 a B) \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx}{15 a} \\ & = \frac {2 \left (15 a^2 A+2 A b^2-5 a b B\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{15 a^2 d}-\frac {2 (A b+5 a B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 a d}-\frac {2 A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d}-\frac {\left (8 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\frac {15}{8} a^2 (A b+a B)-\frac {15}{8} a^2 (a A-b B) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{15 a^2} \\ & = \frac {2 \left (15 a^2 A+2 A b^2-5 a b B\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{15 a^2 d}-\frac {2 (A b+5 a B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 a d}-\frac {2 A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d}-\frac {\left (4 \left (\frac {15}{8} a^2 (A b+a B)-\frac {15}{8} i a^2 (a A-b B)\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{15 a^2}-\frac {\left (4 \left (\frac {15}{8} a^2 (A b+a B)+\frac {15}{8} i a^2 (a A-b B)\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {1+i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{15 a^2} \\ & = \frac {2 \left (15 a^2 A+2 A b^2-5 a b B\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{15 a^2 d}-\frac {2 (A b+5 a B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 a d}-\frac {2 A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d}-\frac {\left (4 \left (\frac {15}{8} a^2 (A b+a B)-\frac {15}{8} i a^2 (a A-b B)\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{(1+i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{15 a^2 d}-\frac {\left (4 \left (\frac {15}{8} a^2 (A b+a B)+\frac {15}{8} i a^2 (a A-b B)\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{(1-i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{15 a^2 d} \\ & = \frac {2 \left (15 a^2 A+2 A b^2-5 a b B\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{15 a^2 d}-\frac {2 (A b+5 a B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 a d}-\frac {2 A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d}-\frac {\left (8 \left (\frac {15}{8} a^2 (A b+a B)-\frac {15}{8} i a^2 (a A-b B)\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\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 )}{15 a^2 d}-\frac {\left (8 \left (\frac {15}{8} a^2 (A b+a B)+\frac {15}{8} i a^2 (a A-b B)\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\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 )}{15 a^2 d} \\ & = \frac {\sqrt {i a-b} (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 {i a+b} (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 {2 \left (15 a^2 A+2 A b^2-5 a b B\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{15 a^2 d}-\frac {2 (A b+5 a B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 a d}-\frac {2 A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d} \\ \end{align*}
Time = 2.48 (sec) , antiderivative size = 252, normalized size of antiderivative = 0.87 \[ \int \cot ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {\cot ^{\frac {5}{2}}(c+d x) \left (15 \sqrt [4]{-1} a^2 \sqrt {-a+i b} (A-i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \tan ^{\frac {5}{2}}(c+d x)+15 \sqrt [4]{-1} a^2 \sqrt {a+i b} (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \tan ^{\frac {5}{2}}(c+d x)+2 \sqrt {a+b \tan (c+d x)} \left (-3 a^2 A-a (A b+5 a B) \tan (c+d x)+\left (15 a^2 A+2 A b^2-5 a b B\right ) \tan ^2(c+d x)\right )\right )}{15 a^2 d} \]
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result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 1.18 (sec) , antiderivative size = 2183483, normalized size of antiderivative = 7529.25
\[\text {output too large to display}\]
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Leaf count of result is larger than twice the leaf count of optimal. 8104 vs. \(2 (236) = 472\).
Time = 1.40 (sec) , antiderivative size = 8104, normalized size of antiderivative = 27.94 \[ \int \cot ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]
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Timed out. \[ \int \cot ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]
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\[ \int \cot ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {b \tan \left (d x + c\right ) + a} \cot \left (d x + c\right )^{\frac {7}{2}} \,d x } \]
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Exception generated. \[ \int \cot ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\text {Exception raised: TypeError} \]
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Timed out. \[ \int \cot ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^{7/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )} \,d x \]
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