Integrand size = 35, antiderivative size = 301 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx=\frac {(i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{5/2} d}-\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{5/2} d}-\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^2 A+4 A b^2-a b B\right ) \sqrt {\tan (c+d x)}}{3 a^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^4 A+17 a^2 A b^2+8 A b^4-8 a^3 b B-2 a b^3 B\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}} \]
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Time = 1.57 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3690, 3730, 3697, 3696, 95, 209, 212} \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx=-\frac {2 b \left (3 a^2 A-a b B+4 A b^2\right ) \sqrt {\tan (c+d x)}}{3 a^2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^4 A-8 a^3 b B+17 a^2 A b^2-2 a b^3 B+8 A b^4\right ) \sqrt {\tan (c+d x)}}{3 a^3 d \left (a^2+b^2\right )^2 \sqrt {a+b \tan (c+d x)}}+\frac {(-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (-b+i a)^{5/2}}-\frac {(B+i A) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (b+i a)^{5/2}}-\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}} \]
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Rule 95
Rule 209
Rule 212
Rule 3690
Rule 3696
Rule 3697
Rule 3730
Rubi steps \begin{align*} \text {integral}& = -\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {2 \int \frac {\frac {1}{2} (4 A b-a B)+\frac {1}{2} a A \tan (c+d x)+2 A b \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}} \, dx}{a} \\ & = -\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^2 A+4 A b^2-a b B\right ) \sqrt {\tan (c+d x)}}{3 a^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {4 \int \frac {\frac {1}{4} \left (9 a^2 A b+8 A b^3-3 a^3 B-2 a b^2 B\right )+\frac {3}{4} a^2 (a A+b B) \tan (c+d x)+\frac {1}{2} b \left (3 a^2 A+4 A b^2-a b B\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx}{3 a^2 \left (a^2+b^2\right )} \\ & = -\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^2 A+4 A b^2-a b B\right ) \sqrt {\tan (c+d x)}}{3 a^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^4 A+17 a^2 A b^2+8 A b^4-8 a^3 b B-2 a b^3 B\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}-\frac {8 \int \frac {\frac {3}{8} a^3 \left (2 a A b-a^2 B+b^2 B\right )+\frac {3}{8} a^3 \left (a^2 A-A b^2+2 a b B\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{3 a^3 \left (a^2+b^2\right )^2} \\ & = -\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^2 A+4 A b^2-a b B\right ) \sqrt {\tan (c+d x)}}{3 a^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^4 A+17 a^2 A b^2+8 A b^4-8 a^3 b B-2 a b^3 B\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}-\frac {(i A-B) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{2 (a+i b)^2}+\frac {(i A+B) \int \frac {1+i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{2 (a-i b)^2} \\ & = -\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^2 A+4 A b^2-a b B\right ) \sqrt {\tan (c+d x)}}{3 a^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^4 A+17 a^2 A b^2+8 A b^4-8 a^3 b B-2 a b^3 B\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}-\frac {(i A-B) \text {Subst}\left (\int \frac {1}{(1+i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a+i b)^2 d}+\frac {(i A+B) \text {Subst}\left (\int \frac {1}{(1-i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a-i b)^2 d} \\ & = -\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^2 A+4 A b^2-a b B\right ) \sqrt {\tan (c+d x)}}{3 a^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^4 A+17 a^2 A b^2+8 A b^4-8 a^3 b B-2 a b^3 B\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}-\frac {(i A-B) \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 )}{(a+i b)^2 d}+\frac {(i A+B) \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 )}{(a-i b)^2 d} \\ & = -\frac {(i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a-b} (a+i b)^2 d}-\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{5/2} d}-\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^2 A+4 A b^2-a b B\right ) \sqrt {\tan (c+d x)}}{3 a^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^4 A+17 a^2 A b^2+8 A b^4-8 a^3 b B-2 a b^3 B\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}} \\ \end{align*}
Time = 4.03 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.08 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx=\frac {-\frac {6 a A}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^2 A+4 A b^2-a b B\right ) \sqrt {\tan (c+d x)}}{\left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {3 \sqrt [4]{-1} a^3 \left (\frac {(a+i b)^2 (A-i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}-\frac {(a-i b)^2 (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}\right )-\frac {2 b \left (3 a^4 A+17 a^2 A b^2+8 A b^4-8 a^3 b B-2 a b^3 B\right ) \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{a \left (a^2+b^2\right )^2}}{3 a^2 d} \]
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result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 2.10 (sec) , antiderivative size = 2976756, normalized size of antiderivative = 9889.55
\[\text {output too large to display}\]
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Leaf count of result is larger than twice the leaf count of optimal. 27749 vs. \(2 (257) = 514\).
Time = 14.45 (sec) , antiderivative size = 27749, normalized size of antiderivative = 92.19 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
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Timed out. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx=\int \frac {A+B\,\mathrm {tan}\left (c+d\,x\right )}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}} \,d x \]
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