Integrand size = 25, antiderivative size = 270 \[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\frac {i (i a-b)^{5/2} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {i (i a+b)^{5/2} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {6 a b \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (7 a^2-9 b^2\right ) \sqrt {a+b \tan (c+d x)}}{21 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 b \left (49 a^2-3 b^2\right ) \sqrt {a+b \tan (c+d x)}}{21 a d \sqrt {\tan (c+d x)}} \]
[Out]
Time = 1.76 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3646, 3730, 3697, 3696, 95, 209, 212} \[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 \left (7 a^2-9 b^2\right ) \sqrt {a+b \tan (c+d x)}}{21 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 b \left (49 a^2-3 b^2\right ) \sqrt {a+b \tan (c+d x)}}{21 a d \sqrt {\tan (c+d x)}}-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {i (-b+i a)^{5/2} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {i (b+i a)^{5/2} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {6 a b \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {5}{2}}(c+d x)} \]
[In]
[Out]
Rule 95
Rule 209
Rule 212
Rule 3646
Rule 3696
Rule 3697
Rule 3730
Rubi steps \begin{align*} \text {integral}& = -\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2}{7} \int \frac {\frac {15 a^2 b}{2}-\frac {7}{2} a \left (a^2-3 b^2\right ) \tan (c+d x)-\frac {1}{2} b \left (6 a^2-7 b^2\right ) \tan ^2(c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx \\ & = -\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {6 a b \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {4 \int \frac {\frac {5}{4} a^2 \left (7 a^2-9 b^2\right )+\frac {35}{4} a b \left (3 a^2-b^2\right ) \tan (c+d x)+15 a^2 b^2 \tan ^2(c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx}{35 a} \\ & = -\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {6 a b \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (7 a^2-9 b^2\right ) \sqrt {a+b \tan (c+d x)}}{21 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {8 \int \frac {-\frac {5}{8} a^2 b \left (49 a^2-3 b^2\right )+\frac {105}{8} a^3 \left (a^2-3 b^2\right ) \tan (c+d x)+\frac {5}{4} a^2 b \left (7 a^2-9 b^2\right ) \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx}{105 a^2} \\ & = -\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {6 a b \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (7 a^2-9 b^2\right ) \sqrt {a+b \tan (c+d x)}}{21 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 b \left (49 a^2-3 b^2\right ) \sqrt {a+b \tan (c+d x)}}{21 a d \sqrt {\tan (c+d x)}}-\frac {16 \int \frac {-\frac {105}{16} a^4 \left (a^2-3 b^2\right )-\frac {105}{16} a^3 b \left (3 a^2-b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{105 a^3} \\ & = -\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {6 a b \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (7 a^2-9 b^2\right ) \sqrt {a+b \tan (c+d x)}}{21 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 b \left (49 a^2-3 b^2\right ) \sqrt {a+b \tan (c+d x)}}{21 a d \sqrt {\tan (c+d x)}}+\frac {1}{2} (a-i b)^3 \int \frac {1+i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx+\frac {1}{2} (a+i b)^3 \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx \\ & = -\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {6 a b \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (7 a^2-9 b^2\right ) \sqrt {a+b \tan (c+d x)}}{21 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 b \left (49 a^2-3 b^2\right ) \sqrt {a+b \tan (c+d x)}}{21 a d \sqrt {\tan (c+d x)}}+\frac {(a-i b)^3 \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 {(a+i b)^3 \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^2 \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {6 a b \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (7 a^2-9 b^2\right ) \sqrt {a+b \tan (c+d x)}}{21 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 b \left (49 a^2-3 b^2\right ) \sqrt {a+b \tan (c+d x)}}{21 a d \sqrt {\tan (c+d x)}}+\frac {(a-i b)^3 \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)^3 \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 (i a-b)^{5/2} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {i (i a+b)^{5/2} \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^2 \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {6 a b \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (7 a^2-9 b^2\right ) \sqrt {a+b \tan (c+d x)}}{21 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 b \left (49 a^2-3 b^2\right ) \sqrt {a+b \tan (c+d x)}}{21 a d \sqrt {\tan (c+d x)}} \\ \end{align*}
Time = 3.74 (sec) , antiderivative size = 249, normalized size of antiderivative = 0.92 \[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=-\frac {-42 (-1)^{3/4} (-a+i b)^{5/2} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+42 (-1)^{3/4} (a+i b)^{5/2} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+\frac {\sec ^3(c+d x) \left (a \left (2 a^2+9 b^2\right ) \cos (c+d x)+\left (10 a^3-9 a b^2\right ) \cos (3 (c+d x))+2 b \left (-40 a^2+3 b^2+\left (58 a^2-3 b^2\right ) \cos (2 (c+d x))\right ) \sin (c+d x)\right ) \sqrt {a+b \tan (c+d x)}}{a \tan ^{\frac {7}{2}}(c+d x)}}{42 d} \]
[In]
[Out]
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 0.65 (sec) , antiderivative size = 1309621, normalized size of antiderivative = 4850.45
\[\text {output too large to display}\]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 5085 vs. \(2 (218) = 436\).
Time = 0.86 (sec) , antiderivative size = 5085, normalized size of antiderivative = 18.83 \[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\text {Too large to display} \]
[In]
[Out]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\text {Timed out} \]
[In]
[Out]
\[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\int { \frac {{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\tan \left (d x + c\right )^{\frac {9}{2}}} \,d x } \]
[In]
[Out]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\text {Timed out} \]
[In]
[Out]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{9/2}} \,d x \]
[In]
[Out]