Integrand size = 35, antiderivative size = 259 \[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\frac {(a+i b)^2 (i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a-b} d}+\frac {(i a+b)^{3/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a A \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 (6 A b+5 a B) \sqrt {a+b \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2 A-3 A b^2-20 a b B\right ) \sqrt {a+b \tan (c+d x)}}{15 a d \sqrt {\tan (c+d x)}} \]
[Out]
Time = 1.43 (sec) , antiderivative size = 259, 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 = {3686, 3730, 3697, 3696, 95, 209, 212} \[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 \left (15 a^2 A-20 a b B-3 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 a d \sqrt {\tan (c+d x)}}+\frac {(a+i b)^2 (-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}+\frac {(b+i a)^{3/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 (5 a B+6 A b) \sqrt {a+b \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 a A \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)} \]
[In]
[Out]
Rule 95
Rule 209
Rule 212
Rule 3686
Rule 3696
Rule 3697
Rule 3730
Rubi steps \begin{align*} \text {integral}& = -\frac {2 a A \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {\frac {1}{2} a (6 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 (4 a A-5 b B) \tan ^2(c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx \\ & = -\frac {2 a A \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 (6 A b+5 a B) \sqrt {a+b \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {4 \int \frac {\frac {1}{4} a \left (15 a^2 A-3 A b^2-20 a b B\right )+\frac {15}{4} a \left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)+\frac {1}{2} a b (6 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 a A \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 (6 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-3 A b^2-20 a b B\right ) \sqrt {a+b \tan (c+d x)}}{15 a d \sqrt {\tan (c+d x)}}+\frac {8 \int \frac {-\frac {15}{8} a^2 \left (2 a A b+a^2 B-b^2 B\right )+\frac {15}{8} a^2 \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}{15 a^2} \\ & = -\frac {2 a A \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 (6 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-3 A b^2-20 a b B\right ) \sqrt {a+b \tan (c+d x)}}{15 a d \sqrt {\tan (c+d x)}}+\frac {1}{2} \left ((a+i b)^2 (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 {1}{2} \left ((a-i b)^2 (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 {2 a A \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 (6 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-3 A b^2-20 a b B\right ) \sqrt {a+b \tan (c+d x)}}{15 a d \sqrt {\tan (c+d x)}}+\frac {\left ((a+i b)^2 (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 ((a-i b)^2 (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 {2 a A \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 (6 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-3 A b^2-20 a b B\right ) \sqrt {a+b \tan (c+d x)}}{15 a d \sqrt {\tan (c+d x)}}+\frac {\left ((a+i b)^2 (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 ((a-i b)^2 (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 {(a+i b)^2 (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} d}+\frac {(i a+b)^{3/2} (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 A \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 (6 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-3 A b^2-20 a b B\right ) \sqrt {a+b \tan (c+d x)}}{15 a d \sqrt {\tan (c+d x)}} \\ \end{align*}
Time = 3.34 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.10 \[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\frac {-30 \sqrt [4]{-1} a \left ((-a+i b)^{3/2} (A-i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-(a+i b)^{3/2} (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )\right ) \tan ^{\frac {5}{2}}(c+d x)-15 a b B \sqrt {a+b \tan (c+d x)}-3 a (4 a A-5 b B) \sqrt {a+b \tan (c+d x)}-4 a (6 A b+5 a B) \tan (c+d x) \sqrt {a+b \tan (c+d x)}+4 \left (15 a^2 A-3 A b^2-20 a b B\right ) \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{30 a d \tan ^{\frac {5}{2}}(c+d x)} \]
[In]
[Out]
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 0.88 (sec) , antiderivative size = 2400946, normalized size of antiderivative = 9270.06
\[\text {output too large to display}\]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 12093 vs. \(2 (211) = 422\).
Time = 2.32 (sec) , antiderivative size = 12093, normalized size of antiderivative = 46.69 \[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Too large to display} \]
[In]
[Out]
\[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\tan ^{\frac {7}{2}}{\left (c + d x \right )}}\, dx \]
[In]
[Out]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]
[In]
[Out]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]
[In]
[Out]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{7/2}} \,d x \]
[In]
[Out]