Integrand size = 28, antiderivative size = 257 \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {5 (-1)^{3/4} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {19 i \tan ^{\frac {5}{2}}(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {41 \tan ^{\frac {3}{2}}(c+d x)}{12 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {21 i \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 a^3 d} \]
5*(-1)^(3/4)*arctan((-1)^(3/4)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c)) ^(1/2))/a^(5/2)/d-(1/8+1/8*I)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I* a*tan(d*x+c))^(1/2))/a^(5/2)/d+21/4*I*tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^ (1/2)/a^3/d+41/12*tan(d*x+c)^(3/2)/a^2/d/(a+I*a*tan(d*x+c))^(1/2)-1/5*tan( d*x+c)^(7/2)/d/(a+I*a*tan(d*x+c))^(5/2)+19/30*I*tan(d*x+c)^(5/2)/a/d/(a+I* a*tan(d*x+c))^(3/2)
Time = 3.10 (sec) , antiderivative size = 237, normalized size of antiderivative = 0.92 \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {600 \sqrt [4]{-1} \sqrt {a} \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) \sec ^2(c+d x) (\cos (2 (c+d x))+i \sin (2 (c+d x))) \sqrt {1+i \tan (c+d x)}-(15+15 i) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}+2 \sqrt {a} \sqrt {\tan (c+d x)} \left (-315 i+740 \tan (c+d x)+497 i \tan ^2(c+d x)-60 \tan ^3(c+d x)\right )}{120 a^{5/2} d (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \]
(600*(-1)^(1/4)*Sqrt[a]*ArcSinh[(-1)^(1/4)*Sqrt[Tan[c + d*x]]]*Sec[c + d*x ]^2*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)])*Sqrt[1 + I*Tan[c + d*x]] - (15 + 15*I)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]] + 2*Sqrt[a]*Sqrt[ Tan[c + d*x]]*(-315*I + 740*Tan[c + d*x] + (497*I)*Tan[c + d*x]^2 - 60*Tan [c + d*x]^3))/(120*a^(5/2)*d*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d* x]])
Time = 1.69 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.07, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.679, Rules used = {3042, 4041, 27, 3042, 4078, 27, 3042, 4078, 27, 3042, 4080, 3042, 4084, 3042, 4027, 218, 4082, 65, 216}
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.
| \(\displaystyle \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^{9/2}}{(a+i a \tan (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4041 |
| \(\displaystyle -\frac {\int -\frac {\tan ^{\frac {5}{2}}(c+d x) (7 a-12 i a \tan (c+d x))}{2 (i \tan (c+d x) a+a)^{3/2}}dx}{5 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\tan ^{\frac {5}{2}}(c+d x) (7 a-12 i a \tan (c+d x))}{(i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\tan (c+d x)^{5/2} (7 a-12 i a \tan (c+d x))}{(i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \frac {\frac {19 i a \tan ^{\frac {5}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\int \frac {5 \tan ^{\frac {3}{2}}(c+d x) \left (22 \tan (c+d x) a^2+19 i a^2\right )}{2 \sqrt {i \tan (c+d x) a+a}}dx}{3 a^2}}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {19 i a \tan ^{\frac {5}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {5 \int \frac {\tan ^{\frac {3}{2}}(c+d x) \left (22 \tan (c+d x) a^2+19 i a^2\right )}{\sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {19 i a \tan ^{\frac {5}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {5 \int \frac {\tan (c+d x)^{3/2} \left (22 \tan (c+d x) a^2+19 i a^2\right )}{\sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \frac {\frac {19 i a \tan ^{\frac {5}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {5 \left (-\frac {\int -\frac {3}{2} \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a} \left (41 a^3-42 i a^3 \tan (c+d x)\right )dx}{a^2}-\frac {41 a^2 \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {19 i a \tan ^{\frac {5}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {5 \left (\frac {3 \int \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a} \left (41 a^3-42 i a^3 \tan (c+d x)\right )dx}{2 a^2}-\frac {41 a^2 \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {19 i a \tan ^{\frac {5}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {5 \left (\frac {3 \int \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a} \left (41 a^3-42 i a^3 \tan (c+d x)\right )dx}{2 a^2}-\frac {41 a^2 \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4080 |
| \(\displaystyle \frac {\frac {19 i a \tan ^{\frac {5}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {5 \left (\frac {3 \left (\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (20 \tan (c+d x) a^4+21 i a^4\right )}{\sqrt {\tan (c+d x)}}dx}{a}-\frac {42 i a^3 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}-\frac {41 a^2 \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {19 i a \tan ^{\frac {5}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {5 \left (\frac {3 \left (\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (20 \tan (c+d x) a^4+21 i a^4\right )}{\sqrt {\tan (c+d x)}}dx}{a}-\frac {42 i a^3 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}-\frac {41 a^2 \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4084 |
| \(\displaystyle \frac {\frac {19 i a \tan ^{\frac {5}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {5 \left (\frac {3 \left (\frac {i a^4 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+20 i a^3 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{a}-\frac {42 i a^3 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}-\frac {41 a^2 \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {19 i a \tan ^{\frac {5}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {5 \left (\frac {3 \left (\frac {i a^4 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+20 i a^3 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{a}-\frac {42 i a^3 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}-\frac {41 a^2 \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4027 |
| \(\displaystyle \frac {\frac {19 i a \tan ^{\frac {5}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {5 \left (\frac {3 \left (\frac {20 i a^3 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+\frac {2 a^6 \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}}{a}-\frac {42 i a^3 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}-\frac {41 a^2 \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {19 i a \tan ^{\frac {5}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {5 \left (\frac {3 \left (\frac {20 i a^3 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+\frac {(1+i) a^{9/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{a}-\frac {42 i a^3 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}-\frac {41 a^2 \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4082 |
| \(\displaystyle \frac {\frac {19 i a \tan ^{\frac {5}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {5 \left (\frac {3 \left (\frac {\frac {20 i a^5 \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}+\frac {(1+i) a^{9/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{a}-\frac {42 i a^3 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}-\frac {41 a^2 \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 65 |
| \(\displaystyle \frac {\frac {19 i a \tan ^{\frac {5}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {5 \left (\frac {3 \left (\frac {\frac {40 i a^5 \int \frac {1}{1-\frac {i a \tan (c+d x)}{i \tan (c+d x) a+a}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}+\frac {(1+i) a^{9/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{a}-\frac {42 i a^3 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}-\frac {41 a^2 \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {19 i a \tan ^{\frac {5}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {5 \left (\frac {3 \left (\frac {\frac {(1+i) a^{9/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {40 (-1)^{3/4} a^{9/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{a}-\frac {42 i a^3 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}-\frac {41 a^2 \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
-1/5*Tan[c + d*x]^(7/2)/(d*(a + I*a*Tan[c + d*x])^(5/2)) + ((((19*I)/3)*a* Tan[c + d*x]^(5/2))/(d*(a + I*a*Tan[c + d*x])^(3/2)) - (5*((-41*a^2*Tan[c + d*x]^(3/2))/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (3*(((-40*(-1)^(3/4)*a^(9/2 )*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x] ]])/d + ((1 + I)*a^(9/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt [a + I*a*Tan[c + d*x]]])/d)/a - ((42*I)*a^3*Sqrt[Tan[c + d*x]]*Sqrt[a + I* a*Tan[c + d*x]])/d))/(2*a^2)))/(6*a^2))/(10*a^2)
3.3.25.3.1 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[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2 Sub st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d }, x] && !GtQ[c, 0]
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[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[-2*a*(b/f) Subst[Int[1/(a*c - b*d - 2* a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && N eQ[c^2 + d^2, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b*c - a*d))*(a + b*Tan[e + f*x])^m* ((c + d*Tan[e + f*x])^(n - 1)/(2*a*f*m)), x] + Simp[1/(2*a^2*m) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1)) - d*(b*d*(m - n + 1) - a*c*(m + n - 1))*Ta n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, 0] && GtQ[n, 1] && (In tegerQ[m] || IntegersQ[2*m, 2*n])
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-(A*b - a*B))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(2*a*f*m)), x] + Simp[1/(2*a^2*m) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f* x])^(n - 1)*Simp[A*(a*c*m + b*d*n) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a *A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && GtQ[n, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[B*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(f*(m + n))), x] + Simp[ 1/(a*(m + n)) Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Sim p[a*A*c*(m + n) - B*(b*c*m + a*d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*T an[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[n, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[b*(B/f) Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, 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] && EqQ[ a^2 + b^2, 0] && EqQ[A*b + a*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] :> Sim p[(A*b + a*B)/b Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x], x] - Simp[B/b Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[ e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1006 vs. \(2 (201 ) = 402\).
Time = 1.28 (sec) , antiderivative size = 1007, normalized size of antiderivative = 3.92
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(1007\) |
| default | \(\text {Expression too large to display}\) | \(1007\) |
1/240/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)/a^3*(2228*(a*tan(d*x+c )*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^3-60*I*2^(1/ 2)*(I*a)^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)) )^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*a*tan(d*x+c)+3600*I*(-I*a)^(1/ 2)*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^ (1/2)+a)/(I*a)^(1/2))*a*tan(d*x+c)^2+1260*I*(-I*a)^(1/2)*(I*a)^(1/2)*(a*ta n(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-15*2^(1/2)*(I*a)^(1/2)*ln(-(-2*2^(1/2)*(- I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan( d*x+c)+I))*a*tan(d*x+c)^4+240*I*(-I*a)^(1/2)*(I*a)^(1/2)*(a*tan(d*x+c)*(1+ I*tan(d*x+c)))^(1/2)*tan(d*x+c)^4-600*I*(-I*a)^(1/2)*ln(1/2*(2*I*a*tan(d*x +c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*a- 4948*I*(-I*a)^(1/2)*(I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan( d*x+c)^2+90*2^(1/2)*(I*a)^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c) *(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*a*tan(d*x+c)^ 2-2400*(-I*a)^(1/2)*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+ c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*a*tan(d*x+c)^3-600*I*(-I*a)^(1/2)*l n(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2 )+a)/(I*a)^(1/2))*a*tan(d*x+c)^4+60*I*2^(1/2)*(I*a)^(1/2)*ln(-(-2*2^(1/2)* (-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(ta n(d*x+c)+I))*a*tan(d*x+c)^3-15*(I*a)^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 623 vs. \(2 (189) = 378\).
Time = 0.32 (sec) , antiderivative size = 623, normalized size of antiderivative = 2.42 \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {{\left (30 \, a^{3} d \sqrt {\frac {i}{8 \, a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (i \, a^{3} d \sqrt {\frac {i}{8 \, a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \frac {1}{4} \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right ) - 30 \, a^{3} d \sqrt {\frac {i}{8 \, a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-i \, a^{3} d \sqrt {\frac {i}{8 \, a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \frac {1}{4} \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right ) - 30 \, a^{3} d \sqrt {\frac {25 i}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac {104 \, {\left (10 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (3 i \, d x + 3 i \, c\right )} + e^{\left (i \, d x + i \, c\right )}\right )} - {\left (3 i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{3} d\right )} \sqrt {\frac {25 i}{a^{5} d^{2}}}\right )}}{3025 \, {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}}\right ) + 30 \, a^{3} d \sqrt {\frac {25 i}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac {104 \, {\left (10 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (3 i \, d x + 3 i \, c\right )} + e^{\left (i \, d x + i \, c\right )}\right )} - {\left (-3 i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{3} d\right )} \sqrt {\frac {25 i}{a^{5} d^{2}}}\right )}}{3025 \, {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (403 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 252 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 28 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{120 \, a^{3} d} \]
1/120*(30*a^3*d*sqrt(1/8*I/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(I*a^3*d*sqrt (1/8*I/(a^5*d^2))*e^(I*d*x + I*c) + 1/4*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c ) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2 *I*d*x + 2*I*c) + 1)) - 30*a^3*d*sqrt(1/8*I/(a^5*d^2))*e^(5*I*d*x + 5*I*c) *log(-I*a^3*d*sqrt(1/8*I/(a^5*d^2))*e^(I*d*x + I*c) + 1/4*sqrt(2)*sqrt(a/( e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)) - 30*a^3*d*sqrt(25*I/(a^5*d^2))*e^ (5*I*d*x + 5*I*c)*log(104/3025*(10*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1 ))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*(e^(3*I*d* x + 3*I*c) + e^(I*d*x + I*c)) - (3*I*a^3*d*e^(2*I*d*x + 2*I*c) - I*a^3*d)* sqrt(25*I/(a^5*d^2)))/(e^(2*I*d*x + 2*I*c) + 1)) + 30*a^3*d*sqrt(25*I/(a^5 *d^2))*e^(5*I*d*x + 5*I*c)*log(104/3025*(10*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2 *I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*( e^(3*I*d*x + 3*I*c) + e^(I*d*x + I*c)) - (-3*I*a^3*d*e^(2*I*d*x + 2*I*c) + I*a^3*d)*sqrt(25*I/(a^5*d^2)))/(e^(2*I*d*x + 2*I*c) + 1)) + sqrt(2)*sqrt( a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*(403*I*e^(6*I*d*x + 6*I*c) + 252*I*e^(4*I*d*x + 4*I*c) - 2 8*I*e^(2*I*d*x + 2*I*c) + 3*I))*e^(-5*I*d*x - 5*I*c)/(a^3*d)
Timed out. \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]
Exception generated. \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Non regular value [0] was discarded and replaced randomly by 0=[-3]Warning, replacing -3 by -93, a substituti on variab
Timed out. \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{9/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]