Integrand size = 35, antiderivative size = 247 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {(i a-b)^{5/2} (A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(i a+b)^{5/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a (8 A b+5 a B) \sqrt {a+b \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2 A-23 A b^2-35 a b B\right ) \sqrt {a+b \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)} \]
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Time = 1.38 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.00, number of steps used = 10, 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} \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 \left (15 a^2 A-35 a b B-23 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}-\frac {(-b+i a)^{5/2} (A+i B) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(b+i a)^{5/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a (5 a B+8 A b) \sqrt {a+b \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)} \]
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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}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {\sqrt {a+b \tan (c+d x)} \left (\frac {1}{2} a (8 A b+5 a B)-\frac {5}{2} \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)-\frac {1}{2} b (2 a A-5 b B) \tan ^2(c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx \\ & = -\frac {2 a (8 A b+5 a B) \sqrt {a+b \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {4}{15} \int \frac {-\frac {1}{4} a \left (15 a^2 A-23 A b^2-35 a b B\right )-\frac {15}{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 (22 a A b+10 a^2 B-15 b^2 B\right ) \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx \\ & = -\frac {2 a (8 A b+5 a B) \sqrt {a+b \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2 A-23 A b^2-35 a b B\right ) \sqrt {a+b \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {8 \int \frac {\frac {15}{8} a \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )-\frac {15}{8} a \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{15 a} \\ & = -\frac {2 a (8 A b+5 a B) \sqrt {a+b \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2 A-23 A b^2-35 a b B\right ) \sqrt {a+b \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {1}{2} \left ((a-i b)^3 (i A+B)\right ) \int \frac {1+i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx-\frac {\left (4 \left (\frac {15}{8} a \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )-\frac {15}{8} i a \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )\right )\right ) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{15 a} \\ & = -\frac {2 a (8 A b+5 a B) \sqrt {a+b \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2 A-23 A b^2-35 a b B\right ) \sqrt {a+b \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {\left ((a-i b)^3 (i A+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 (4 \left (\frac {15}{8} a \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )-\frac {15}{8} i a \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )\right )\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 d} \\ & = -\frac {2 a (8 A b+5 a B) \sqrt {a+b \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2 A-23 A b^2-35 a b B\right ) \sqrt {a+b \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {\left ((a-i b)^3 (i A+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 (8 \left (\frac {15}{8} a \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )-\frac {15}{8} i a \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )\right )\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 d} \\ & = -\frac {(i a-b)^{5/2} (A+i 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} (A-i 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 (8 A b+5 a B) \sqrt {a+b \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2 A-23 A b^2-35 a b B\right ) \sqrt {a+b \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)} \\ \end{align*}
Time = 2.29 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.30 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\frac {60 \sqrt [4]{-1} \left ((-a+i b)^{5/2} (A-i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+(a+i b)^{5/2} (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )\right ) \tan ^{\frac {5}{2}}(c+d x)+15 b (-2 A b+a B) \sqrt {a+b \tan (c+d x)}-3 \left (8 a^2 A-10 A b^2-15 a b B\right ) \sqrt {a+b \tan (c+d x)}-4 \left (22 a A b+10 a^2 B-15 b^2 B\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}+8 \left (15 a^2 A-23 A b^2-35 a b B\right ) \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}-60 b B (a+b \tan (c+d x))^{3/2}}{60 d \tan ^{\frac {5}{2}}(c+d x)} \]
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result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 1.02 (sec) , antiderivative size = 2652302, normalized size of antiderivative = 10738.06
\[\text {output too large to display}\]
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Leaf count of result is larger than twice the leaf count of optimal. 18802 vs. \(2 (201) = 402\).
Time = 5.06 (sec) , antiderivative size = 18802, normalized size of antiderivative = 76.12 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Too large to display} \]
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Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{7/2}} \,d x \]
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