Integrand size = 35, antiderivative size = 301 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx=\frac {(i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{5/2} d}-\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{5/2} d}-\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^2 A+4 A b^2-a b B\right ) \sqrt {\tan (c+d x)}}{3 a^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^4 A+17 a^2 A b^2+8 A b^4-8 a^3 b B-2 a b^3 B\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}} \]
(I*A-B)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/(I*a -b)^(5/2)/d-(I*A+B)*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c) )^(1/2))/(I*a+b)^(5/2)/d-2/3*b*(3*A*a^4+17*A*a^2*b^2+8*A*b^4-8*B*a^3*b-2*B *a*b^3)*tan(d*x+c)^(1/2)/a^3/(a^2+b^2)^2/d/(a+b*tan(d*x+c))^(1/2)-2*A/a/d/ tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(3/2)-2/3*b*(3*A*a^2+4*A*b^2-B*a*b)*tan( d*x+c)^(1/2)/a^2/(a^2+b^2)/d/(a+b*tan(d*x+c))^(3/2)
Time = 4.03 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.08 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx=\frac {-\frac {6 a A}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^2 A+4 A b^2-a b B\right ) \sqrt {\tan (c+d x)}}{\left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {3 \sqrt [4]{-1} a^3 \left (\frac {(a+i b)^2 (A-i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}-\frac {(a-i b)^2 (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}\right )-\frac {2 b \left (3 a^4 A+17 a^2 A b^2+8 A b^4-8 a^3 b B-2 a b^3 B\right ) \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{a \left (a^2+b^2\right )^2}}{3 a^2 d} \]
((-6*a*A)/(Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(3/2)) - (2*b*(3*a^2*A + 4*A*b^2 - a*b*B)*Sqrt[Tan[c + d*x]])/((a^2 + b^2)*(a + b*Tan[c + d*x])^( 3/2)) + (3*(-1)^(1/4)*a^3*(((a + I*b)^2*(A - I*B)*ArcTan[((-1)^(1/4)*Sqrt[ -a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[-a + I*b] - ((a - I*b)^2*(A + I*B)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]] )/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[a + I*b]) - (2*b*(3*a^4*A + 17*a^2*A*b^2 + 8*A*b^4 - 8*a^3*b*B - 2*a*b^3*B)*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]])/(a*(a^2 + b^2)^2))/(3*a^2*d)
Time = 1.99 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.20, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.457, Rules used = {3042, 4092, 27, 3042, 4132, 27, 3042, 4132, 27, 3042, 4099, 3042, 4098, 104, 216, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \tan (c+d x)}{\tan (c+d x)^{3/2} (a+b \tan (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 4092 |
\(\displaystyle -\frac {2 \int \frac {4 A b \tan ^2(c+d x)+a A \tan (c+d x)+4 A b-a B}{2 \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}dx}{a}-\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {4 A b \tan ^2(c+d x)+a A \tan (c+d x)+4 A b-a B}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}dx}{a}-\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \frac {4 A b \tan (c+d x)^2+a A \tan (c+d x)+4 A b-a B}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}dx}{a}-\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 4132 |
\(\displaystyle -\frac {\frac {2 \int \frac {-3 B a^3+9 A b a^2+3 (a A+b B) \tan (c+d x) a^2-2 b^2 B a+8 A b^3+2 b \left (3 A a^2-b B a+4 A b^2\right ) \tan ^2(c+d x)}{2 \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}dx}{3 a \left (a^2+b^2\right )}+\frac {2 b \left (3 a^2 A-a b B+4 A b^2\right ) \sqrt {\tan (c+d x)}}{3 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}}{a}-\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\frac {\int \frac {-3 B a^3+9 A b a^2+3 (a A+b B) \tan (c+d x) a^2-2 b^2 B a+8 A b^3+2 b \left (3 A a^2-b B a+4 A b^2\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}dx}{3 a \left (a^2+b^2\right )}+\frac {2 b \left (3 a^2 A-a b B+4 A b^2\right ) \sqrt {\tan (c+d x)}}{3 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}}{a}-\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {\int \frac {-3 B a^3+9 A b a^2+3 (a A+b B) \tan (c+d x) a^2-2 b^2 B a+8 A b^3+2 b \left (3 A a^2-b B a+4 A b^2\right ) \tan (c+d x)^2}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}dx}{3 a \left (a^2+b^2\right )}+\frac {2 b \left (3 a^2 A-a b B+4 A b^2\right ) \sqrt {\tan (c+d x)}}{3 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}}{a}-\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 4132 |
\(\displaystyle -\frac {\frac {\frac {2 \int \frac {3 \left (\left (-B a^2+2 A b a+b^2 B\right ) a^3+\left (A a^2+2 b B a-A b^2\right ) \tan (c+d x) a^3\right )}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a \left (a^2+b^2\right )}+\frac {2 b \left (3 a^4 A-8 a^3 b B+17 a^2 A b^2-2 a b^3 B+8 A b^4\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{3 a \left (a^2+b^2\right )}+\frac {2 b \left (3 a^2 A-a b B+4 A b^2\right ) \sqrt {\tan (c+d x)}}{3 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}}{a}-\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\frac {\frac {3 \int \frac {\left (-B a^2+2 A b a+b^2 B\right ) a^3+\left (A a^2+2 b B a-A b^2\right ) \tan (c+d x) a^3}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a \left (a^2+b^2\right )}+\frac {2 b \left (3 a^4 A-8 a^3 b B+17 a^2 A b^2-2 a b^3 B+8 A b^4\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{3 a \left (a^2+b^2\right )}+\frac {2 b \left (3 a^2 A-a b B+4 A b^2\right ) \sqrt {\tan (c+d x)}}{3 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}}{a}-\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {\frac {3 \int \frac {\left (-B a^2+2 A b a+b^2 B\right ) a^3+\left (A a^2+2 b B a-A b^2\right ) \tan (c+d x) a^3}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a \left (a^2+b^2\right )}+\frac {2 b \left (3 a^4 A-8 a^3 b B+17 a^2 A b^2-2 a b^3 B+8 A b^4\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{3 a \left (a^2+b^2\right )}+\frac {2 b \left (3 a^2 A-a b B+4 A b^2\right ) \sqrt {\tan (c+d x)}}{3 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}}{a}-\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 4099 |
\(\displaystyle -\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {\frac {2 b \left (3 a^2 A-a b B+4 A b^2\right ) \sqrt {\tan (c+d x)}}{3 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {\frac {2 b \left (3 a^4 A-8 a^3 b B+17 a^2 A b^2-2 a b^3 B+8 A b^4\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {3 \left (\frac {1}{2} a^3 (a-i b)^2 (-B+i A) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} a^3 (a+i b)^2 (B+i A) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )}{a \left (a^2+b^2\right )}}{3 a \left (a^2+b^2\right )}}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {\frac {2 b \left (3 a^2 A-a b B+4 A b^2\right ) \sqrt {\tan (c+d x)}}{3 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {\frac {2 b \left (3 a^4 A-8 a^3 b B+17 a^2 A b^2-2 a b^3 B+8 A b^4\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {3 \left (\frac {1}{2} a^3 (a-i b)^2 (-B+i A) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} a^3 (a+i b)^2 (B+i A) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )}{a \left (a^2+b^2\right )}}{3 a \left (a^2+b^2\right )}}{a}\) |
\(\Big \downarrow \) 4098 |
\(\displaystyle -\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {\frac {2 b \left (3 a^2 A-a b B+4 A b^2\right ) \sqrt {\tan (c+d x)}}{3 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {\frac {2 b \left (3 a^4 A-8 a^3 b B+17 a^2 A b^2-2 a b^3 B+8 A b^4\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {3 \left (\frac {a^3 (a-i b)^2 (-B+i A) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}-\frac {a^3 (a+i b)^2 (B+i A) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}\right )}{a \left (a^2+b^2\right )}}{3 a \left (a^2+b^2\right )}}{a}\) |
\(\Big \downarrow \) 104 |
\(\displaystyle -\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {\frac {2 b \left (3 a^2 A-a b B+4 A b^2\right ) \sqrt {\tan (c+d x)}}{3 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {\frac {2 b \left (3 a^4 A-8 a^3 b B+17 a^2 A b^2-2 a b^3 B+8 A b^4\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {3 \left (\frac {a^3 (a-i b)^2 (-B+i A) \int \frac {1}{\frac {(i a-b) \tan (c+d x)}{a+b \tan (c+d x)}+1}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}-\frac {a^3 (a+i b)^2 (B+i A) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}\right )}{a \left (a^2+b^2\right )}}{3 a \left (a^2+b^2\right )}}{a}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {\frac {2 b \left (3 a^2 A-a b B+4 A b^2\right ) \sqrt {\tan (c+d x)}}{3 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {\frac {2 b \left (3 a^4 A-8 a^3 b B+17 a^2 A b^2-2 a b^3 B+8 A b^4\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {3 \left (\frac {a^3 (a-i b)^2 (-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}-\frac {a^3 (a+i b)^2 (B+i A) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}\right )}{a \left (a^2+b^2\right )}}{3 a \left (a^2+b^2\right )}}{a}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {\frac {2 b \left (3 a^2 A-a b B+4 A b^2\right ) \sqrt {\tan (c+d x)}}{3 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {\frac {2 b \left (3 a^4 A-8 a^3 b B+17 a^2 A b^2-2 a b^3 B+8 A b^4\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {3 \left (\frac {a^3 (a-i b)^2 (-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}-\frac {a^3 (a+i b)^2 (B+i A) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}}\right )}{a \left (a^2+b^2\right )}}{3 a \left (a^2+b^2\right )}}{a}\) |
(-2*A)/(a*d*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(3/2)) - ((2*b*(3*a^2* A + 4*A*b^2 - a*b*B)*Sqrt[Tan[c + d*x]])/(3*a*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^(3/2)) + ((3*((a^3*(a - I*b)^2*(I*A - B)*ArcTan[(Sqrt[I*a - b]*Sqrt [Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a - b]*d) - (a^3*(a + I *b)^2*(I*A + B)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[ c + d*x]]])/(Sqrt[I*a + b]*d)))/(a*(a^2 + b^2)) + (2*b*(3*a^4*A + 17*a^2*A *b^2 + 8*A*b^4 - 8*a^3*b*B - 2*a*b^3*B)*Sqrt[Tan[c + d*x]])/(a*(a^2 + b^2) *d*Sqrt[a + b*Tan[c + d*x]]))/(3*a*(a^2 + b^2)))/a
3.5.67.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x _)), x_] :> With[{q = Denominator[m]}, Simp[q Subst[Int[x^(q*(m + 1) - 1) /(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] ] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[b*(A*b - a*B)*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1) /(f*(m + 1)*(b*c - a*d)*(a^2 + b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^ 2 + b^2)) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b* B*(b*c*(m + 1) + a*d*(n + 1)) + A*(a*(b*c - a*d)*(m + 1) - b^2*d*(m + n + 2 )) - (A*b - a*B)*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b*d*(A*b - a*B)*(m + n + 2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) && !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[A^2/f Subst[Int[(a + b*x)^m*((c + d*x)^n/(A - B*x)), x], x, Tan[e + f* x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[A^2 + B^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(A + I*B)/2 Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 - I*T an[e + f*x]), x], x] + Simp[(A - I*B)/2 Int[(a + b*Tan[e + f*x])^m*(c + d *Tan[e + f*x])^n*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A , B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[A^2 + B^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 - a*(b*B - a*C))*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 + b^2)) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1) - b^2*d* (m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d )*(A*b - a*B - b*C)*Tan[e + f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Ta n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ [b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] && !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 2.10 (sec) , antiderivative size = 2976756, normalized size of antiderivative = 9889.55
\[\text {output too large to display}\]
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 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx=\int \frac {A+B\,\mathrm {tan}\left (c+d\,x\right )}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}} \,d x \]