Integrand size = 26, antiderivative size = 205 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}-\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {7 i \tan ^3(c+d x)}{10 a d (a+i a \tan (c+d x))^{3/2}}+\frac {89 \tan ^2(c+d x)}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {89 \sqrt {a+i a \tan (c+d x)}}{5 a^3 d}+\frac {361 (a+i a \tan (c+d x))^{3/2}}{60 a^4 d} \] Output:
-1/8*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/a^(5/2) /d-1/5*tan(d*x+c)^4/d/(a+I*a*tan(d*x+c))^(5/2)+7/10*I*tan(d*x+c)^3/a/d/(a+ I*a*tan(d*x+c))^(3/2)+89/20*tan(d*x+c)^2/a^2/d/(a+I*a*tan(d*x+c))^(1/2)-89 /5*(a+I*a*tan(d*x+c))^(1/2)/a^3/d+361/60*(a+I*a*tan(d*x+c))^(3/2)/a^4/d
Time = 1.66 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.63 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {-15 \sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )+\frac {2 \sqrt {a+i a \tan (c+d x)} \left (-707 i+1760 \tan (c+d x)+1305 i \tan ^2(c+d x)-200 \tan ^3(c+d x)+40 i \tan ^4(c+d x)\right )}{(-i+\tan (c+d x))^3}}{120 a^3 d} \] Input:
Integrate[Tan[c + d*x]^5/(a + I*a*Tan[c + d*x])^(5/2),x]
Output:
(-15*Sqrt[2]*Sqrt[a]*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])] + (2*Sqrt[a + I*a*Tan[c + d*x]]*(-707*I + 1760*Tan[c + d*x] + (1305*I)*Ta n[c + d*x]^2 - 200*Tan[c + d*x]^3 + (40*I)*Tan[c + d*x]^4))/(-I + Tan[c + d*x])^3)/(120*a^3*d)
Time = 1.22 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.07, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {3042, 4041, 27, 3042, 4078, 27, 3042, 4078, 27, 3042, 4075, 3042, 4010, 3042, 3961, 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.
| \(\displaystyle \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^5}{(a+i a \tan (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4041 |
| \(\displaystyle -\frac {\int -\frac {\tan ^3(c+d x) (8 a-13 i a \tan (c+d x))}{2 (i \tan (c+d x) a+a)^{3/2}}dx}{5 a^2}-\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\tan ^3(c+d x) (8 a-13 i a \tan (c+d x))}{(i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}-\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\tan (c+d x)^3 (8 a-13 i a \tan (c+d x))}{(i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}-\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \frac {\frac {7 i a \tan ^3(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {\int \frac {3 \tan ^2(c+d x) \left (47 \tan (c+d x) a^2+42 i a^2\right )}{2 \sqrt {i \tan (c+d x) a+a}}dx}{3 a^2}}{10 a^2}-\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {7 i a \tan ^3(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {\int \frac {\tan ^2(c+d x) \left (47 \tan (c+d x) a^2+42 i a^2\right )}{\sqrt {i \tan (c+d x) a+a}}dx}{2 a^2}}{10 a^2}-\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {7 i a \tan ^3(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {\int \frac {\tan (c+d x)^2 \left (47 \tan (c+d x) a^2+42 i a^2\right )}{\sqrt {i \tan (c+d x) a+a}}dx}{2 a^2}}{10 a^2}-\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \frac {\frac {7 i a \tan ^3(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {-\frac {\int -\frac {1}{2} \tan (c+d x) \sqrt {i \tan (c+d x) a+a} \left (356 a^3-361 i a^3 \tan (c+d x)\right )dx}{a^2}-\frac {89 a^2 \tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}}{10 a^2}-\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {7 i a \tan ^3(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\int \tan (c+d x) \sqrt {i \tan (c+d x) a+a} \left (356 a^3-361 i a^3 \tan (c+d x)\right )dx}{2 a^2}-\frac {89 a^2 \tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}}{10 a^2}-\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {7 i a \tan ^3(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\int \tan (c+d x) \sqrt {i \tan (c+d x) a+a} \left (356 a^3-361 i a^3 \tan (c+d x)\right )dx}{2 a^2}-\frac {89 a^2 \tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}}{10 a^2}-\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4075 |
| \(\displaystyle \frac {\frac {7 i a \tan ^3(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\int \sqrt {i \tan (c+d x) a+a} \left (356 \tan (c+d x) a^3+361 i a^3\right )dx-\frac {722 a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}}{2 a^2}-\frac {89 a^2 \tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}}{10 a^2}-\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {7 i a \tan ^3(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\int \sqrt {i \tan (c+d x) a+a} \left (356 \tan (c+d x) a^3+361 i a^3\right )dx-\frac {722 a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}}{2 a^2}-\frac {89 a^2 \tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}}{10 a^2}-\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4010 |
| \(\displaystyle \frac {\frac {7 i a \tan ^3(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {5 i a^3 \int \sqrt {i \tan (c+d x) a+a}dx+\frac {712 a^3 \sqrt {a+i a \tan (c+d x)}}{d}-\frac {722 a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}}{2 a^2}-\frac {89 a^2 \tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}}{10 a^2}-\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {7 i a \tan ^3(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {5 i a^3 \int \sqrt {i \tan (c+d x) a+a}dx+\frac {712 a^3 \sqrt {a+i a \tan (c+d x)}}{d}-\frac {722 a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}}{2 a^2}-\frac {89 a^2 \tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}}{10 a^2}-\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3961 |
| \(\displaystyle \frac {\frac {7 i a \tan ^3(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\frac {10 a^4 \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}+\frac {712 a^3 \sqrt {a+i a \tan (c+d x)}}{d}-\frac {722 a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}}{2 a^2}-\frac {89 a^2 \tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}}{10 a^2}-\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {7 i a \tan ^3(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\frac {5 \sqrt {2} a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {712 a^3 \sqrt {a+i a \tan (c+d x)}}{d}-\frac {722 a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}}{2 a^2}-\frac {89 a^2 \tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}}{10 a^2}-\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
Input:
Int[Tan[c + d*x]^5/(a + I*a*Tan[c + d*x])^(5/2),x]
Output:
-1/5*Tan[c + d*x]^4/(d*(a + I*a*Tan[c + d*x])^(5/2)) + (((7*I)*a*Tan[c + d *x]^3)/(d*(a + I*a*Tan[c + d*x])^(3/2)) - ((-89*a^2*Tan[c + d*x]^2)/(d*Sqr t[a + I*a*Tan[c + d*x]]) + ((5*Sqrt[2]*a^(7/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + (712*a^3*Sqrt[a + I*a*Tan[c + d*x]])/d - ( 722*a^2*(a + I*a*Tan[c + d*x])^(3/2))/(3*d))/(2*a^2))/(2*a^2))/(10*a^2)
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_)^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[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*(b/d) Subst[Int[1/(2*a - x^2), x], x, Sqrt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a , b, c, d}, x] && EqQ[a^2 + b^2, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Simp [(b*c + a*d)/b Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e , f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && !LtQ[m, 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_)]), x_Symbol] :> Simp[B *d*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f* x])^m*Simp[A*c - B*d + (B*c + A*d)*Tan[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && !LeQ[m, -1]
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]
Time = 1.38 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.64
| method | result | size |
| derivativedivides | \(\frac {\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-6 a \sqrt {a +i a \tan \left (d x +c \right )}-\frac {a^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8}-\frac {31 a^{2}}{4 \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {3 a^{3}}{2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {a^{4}}{5 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}}{a^{4} d}\) | \(131\) |
| default | \(\frac {\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-6 a \sqrt {a +i a \tan \left (d x +c \right )}-\frac {a^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8}-\frac {31 a^{2}}{4 \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {3 a^{3}}{2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {a^{4}}{5 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}}{a^{4} d}\) | \(131\) |
Input:
int(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
2/d/a^4*(1/3*(a+I*a*tan(d*x+c))^(3/2)-3*a*(a+I*a*tan(d*x+c))^(1/2)-1/16*a^ (3/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))-31/8*a ^2/(a+I*a*tan(d*x+c))^(1/2)+3/4*a^3/(a+I*a*tan(d*x+c))^(3/2)-1/10*a^4/(a+I *a*tan(d*x+c))^(5/2))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 342 vs. \(2 (160) = 320\).
Time = 0.09 (sec) , antiderivative size = 342, normalized size of antiderivative = 1.67 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {15 \, \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} + a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )} \sqrt {\frac {1}{a^{5} d^{2}}} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 15 \, \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} + a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )} \sqrt {\frac {1}{a^{5} d^{2}}} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (983 \, e^{\left (8 i \, d x + 8 i \, c\right )} + 1527 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 348 \, e^{\left (4 i \, d x + 4 i \, c\right )} - 33 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3\right )}}{120 \, {\left (a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} + a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )}} \] Input:
integrate(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")
Output:
-1/120*(15*sqrt(1/2)*(a^3*d*e^(7*I*d*x + 7*I*c) + a^3*d*e^(5*I*d*x + 5*I*c ))*sqrt(1/(a^5*d^2))*log(4*(sqrt(2)*sqrt(1/2)*(a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^5*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 15*sqrt(1/2)*(a^3*d*e^(7*I*d*x + 7*I*c) + a^3*d *e^(5*I*d*x + 5*I*c))*sqrt(1/(a^5*d^2))*log(-4*(sqrt(2)*sqrt(1/2)*(a^3*d*e ^(2*I*d*x + 2*I*c) + a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^5* d^2)) - a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(983*e^(8*I*d*x + 8*I*c) + 1527*e^(6*I*d*x + 6*I*c) + 348*e ^(4*I*d*x + 4*I*c) - 33*e^(2*I*d*x + 2*I*c) + 3))/(a^3*d*e^(7*I*d*x + 7*I* c) + a^3*d*e^(5*I*d*x + 5*I*c))
\[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {\tan ^{5}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(tan(d*x+c)**5/(a+I*a*tan(d*x+c))**(5/2),x)
Output:
Integral(tan(c + d*x)**5/(I*a*(tan(c + d*x) - I))**(5/2), x)
Time = 0.14 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.77 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {15 \, \sqrt {2} a^{\frac {7}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + 160 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{2} - 1440 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{3} - \frac {12 \, {\left (155 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{4} - 30 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{5} + 4 \, a^{6}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}}{240 \, a^{6} d} \] Input:
integrate(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")
Output:
1/240*(15*sqrt(2)*a^(7/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(d*x + c) + a))) + 160*(I*a*tan(d*x + c ) + a)^(3/2)*a^2 - 1440*sqrt(I*a*tan(d*x + c) + a)*a^3 - 12*(155*(I*a*tan( d*x + c) + a)^2*a^4 - 30*(I*a*tan(d*x + c) + a)*a^5 + 4*a^6)/(I*a*tan(d*x + c) + a)^(5/2))/(a^6*d)
Exception generated. \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad Argument TypeError: Bad Argument TypeError: Bad Argument TypeDone
Time = 1.09 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.68 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {6\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a^3\,d}+\frac {2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,a^4\,d}-\frac {\frac {31\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2}{4}-\frac {3\,a\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{2}+\frac {a^2}{5}}{a^2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {a}}\right )\,1{}\mathrm {i}}{8\,a^{5/2}\,d} \] Input:
int(tan(c + d*x)^5/(a + a*tan(c + d*x)*1i)^(5/2),x)
Output:
(2*(a + a*tan(c + d*x)*1i)^(3/2))/(3*a^4*d) - (6*(a + a*tan(c + d*x)*1i)^( 1/2))/(a^3*d) - ((31*(a + a*tan(c + d*x)*1i)^2)/4 - (3*a*(a + a*tan(c + d* x)*1i))/2 + a^2/5)/(a^2*d*(a + a*tan(c + d*x)*1i)^(5/2)) + (2^(1/2)*atan(( 2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*1i)/(2*a^(1/2)))*1i)/(8*a^(5/2)*d)
\[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {\int \frac {\tan \left (d x +c \right )^{5}}{\sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}-2 \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) i -\sqrt {\tan \left (d x +c \right ) i +1}}d x}{\sqrt {a}\, a^{2}} \] Input:
int(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^(5/2),x)
Output:
( - int(tan(c + d*x)**5/(sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2 - 2*sqrt (tan(c + d*x)*i + 1)*tan(c + d*x)*i - sqrt(tan(c + d*x)*i + 1)),x))/(sqrt( a)*a**2)