Integrand size = 26, antiderivative size = 205 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {19 i \tan ^3(c+d x)}{6 a d \sqrt {a+i a \tan (c+d x)}}+\frac {78 \sqrt {a+i a \tan (c+d x)}}{5 a^2 d}-\frac {39 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{10 a^2 d}-\frac {151 (a+i a \tan (c+d x))^{3/2}}{30 a^3 d} \] Output:
-1/4*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/a^(3/2) /d-1/3*tan(d*x+c)^4/d/(a+I*a*tan(d*x+c))^(3/2)+19/6*I*tan(d*x+c)^3/a/d/(a+ I*a*tan(d*x+c))^(1/2)+78/5*(a+I*a*tan(d*x+c))^(1/2)/a^2/d-39/10*tan(d*x+c) ^2*(a+I*a*tan(d*x+c))^(1/2)/a^2/d-151/30*(a+I*a*tan(d*x+c))^(3/2)/a^3/d
Time = 2.46 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.73 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {e^{-2 i (c+d x)} \left (-5+115 e^{2 i (c+d x)}+855 e^{4 i (c+d x)}+1105 e^{6 i (c+d x)}+466 e^{8 i (c+d x)}-15 e^{3 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{5/2} \text {arcsinh}\left (e^{i (c+d x)}\right )\right )}{30 a d \left (1+e^{2 i (c+d x)}\right )^3 \sqrt {a+i a \tan (c+d x)}} \] Input:
Integrate[Tan[c + d*x]^5/(a + I*a*Tan[c + d*x])^(3/2),x]
Output:
(-5 + 115*E^((2*I)*(c + d*x)) + 855*E^((4*I)*(c + d*x)) + 1105*E^((6*I)*(c + d*x)) + 466*E^((8*I)*(c + d*x)) - 15*E^((3*I)*(c + d*x))*(1 + E^((2*I)* (c + d*x)))^(5/2)*ArcSinh[E^(I*(c + d*x))])/(30*a*d*E^((2*I)*(c + d*x))*(1 + E^((2*I)*(c + d*x)))^3*Sqrt[a + I*a*Tan[c + d*x]])
Time = 1.21 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.08, 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, 4080, 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.
\(\displaystyle \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)^5}{(a+i a \tan (c+d x))^{3/2}}dx\) |
\(\Big \downarrow \) 4041 |
\(\displaystyle -\frac {\int -\frac {\tan ^3(c+d x) (8 a-11 i a \tan (c+d x))}{2 \sqrt {i \tan (c+d x) a+a}}dx}{3 a^2}-\frac {\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\tan ^3(c+d x) (8 a-11 i a \tan (c+d x))}{\sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}-\frac {\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\tan (c+d x)^3 (8 a-11 i a \tan (c+d x))}{\sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}-\frac {\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 4078 |
\(\displaystyle \frac {\frac {19 i a \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \frac {3}{2} \tan ^2(c+d x) \sqrt {i \tan (c+d x) a+a} \left (39 \tan (c+d x) a^2+38 i a^2\right )dx}{a^2}}{6 a^2}-\frac {\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {19 i a \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {3 \int \tan ^2(c+d x) \sqrt {i \tan (c+d x) a+a} \left (39 \tan (c+d x) a^2+38 i a^2\right )dx}{2 a^2}}{6 a^2}-\frac {\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {19 i a \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {3 \int \tan (c+d x)^2 \sqrt {i \tan (c+d x) a+a} \left (39 \tan (c+d x) a^2+38 i a^2\right )dx}{2 a^2}}{6 a^2}-\frac {\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 4080 |
\(\displaystyle \frac {\frac {19 i a \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {3 \left (\frac {2 \int -\frac {1}{2} \tan (c+d x) \sqrt {i \tan (c+d x) a+a} \left (156 a^3-151 i a^3 \tan (c+d x)\right )dx}{5 a}+\frac {78 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}\right )}{2 a^2}}{6 a^2}-\frac {\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {19 i a \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {3 \left (\frac {78 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {\int \tan (c+d x) \sqrt {i \tan (c+d x) a+a} \left (156 a^3-151 i a^3 \tan (c+d x)\right )dx}{5 a}\right )}{2 a^2}}{6 a^2}-\frac {\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {19 i a \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {3 \left (\frac {78 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {\int \tan (c+d x) \sqrt {i \tan (c+d x) a+a} \left (156 a^3-151 i a^3 \tan (c+d x)\right )dx}{5 a}\right )}{2 a^2}}{6 a^2}-\frac {\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 4075 |
\(\displaystyle \frac {\frac {19 i a \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {3 \left (\frac {78 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {\int \sqrt {i \tan (c+d x) a+a} \left (156 \tan (c+d x) a^3+151 i a^3\right )dx-\frac {302 a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}\right )}{2 a^2}}{6 a^2}-\frac {\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {19 i a \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {3 \left (\frac {78 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {\int \sqrt {i \tan (c+d x) a+a} \left (156 \tan (c+d x) a^3+151 i a^3\right )dx-\frac {302 a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}\right )}{2 a^2}}{6 a^2}-\frac {\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 4010 |
\(\displaystyle \frac {\frac {19 i a \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {3 \left (\frac {78 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {-5 i a^3 \int \sqrt {i \tan (c+d x) a+a}dx+\frac {312 a^3 \sqrt {a+i a \tan (c+d x)}}{d}-\frac {302 a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}\right )}{2 a^2}}{6 a^2}-\frac {\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {19 i a \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {3 \left (\frac {78 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {-5 i a^3 \int \sqrt {i \tan (c+d x) a+a}dx+\frac {312 a^3 \sqrt {a+i a \tan (c+d x)}}{d}-\frac {302 a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}\right )}{2 a^2}}{6 a^2}-\frac {\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3961 |
\(\displaystyle \frac {\frac {19 i a \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {3 \left (\frac {78 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {-\frac {10 a^4 \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}+\frac {312 a^3 \sqrt {a+i a \tan (c+d x)}}{d}-\frac {302 a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}\right )}{2 a^2}}{6 a^2}-\frac {\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {19 i a \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {3 \left (\frac {78 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {-\frac {5 \sqrt {2} a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {312 a^3 \sqrt {a+i a \tan (c+d x)}}{d}-\frac {302 a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}\right )}{2 a^2}}{6 a^2}-\frac {\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\) |
Input:
Int[Tan[c + d*x]^5/(a + I*a*Tan[c + d*x])^(3/2),x]
Output:
-1/3*Tan[c + d*x]^4/(d*(a + I*a*Tan[c + d*x])^(3/2)) + (((19*I)*a*Tan[c + d*x]^3)/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (3*((78*a^2*Tan[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(5*d) - ((-5*Sqrt[2]*a^(7/2)*ArcTanh[Sqrt[a + I*a*Ta n[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + (312*a^3*Sqrt[a + I*a*Tan[c + d*x]])/d - (302*a^2*(a + I*a*Tan[c + d*x])^(3/2))/(3*d))/(5*a)))/(2*a^2))/(6*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]
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]
Time = 1.40 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.64
method | result | size |
derivativedivides | \(\frac {\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}+8 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}+\frac {9 a^{3}}{2 \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {a^{4}}{3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {a^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4}}{a^{4} d}\) | \(131\) |
default | \(\frac {\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}+8 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}+\frac {9 a^{3}}{2 \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {a^{4}}{3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {a^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4}}{a^{4} d}\) | \(131\) |
Input:
int(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
Output:
2/d/a^4*(1/5*(a+I*a*tan(d*x+c))^(5/2)-a*(a+I*a*tan(d*x+c))^(3/2)+4*a^2*(a+ I*a*tan(d*x+c))^(1/2)+9/4*a^3/(a+I*a*tan(d*x+c))^(1/2)-1/6*a^4/(a+I*a*tan( d*x+c))^(3/2)-1/8*a^(5/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^( 1/2)/a^(1/2)))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 388 vs. \(2 (160) = 320\).
Time = 0.09 (sec) , antiderivative size = 388, normalized size of antiderivative = 1.89 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {15 \, \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (7 i \, d x + 7 i \, c\right )} + 2 \, a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )} \sqrt {\frac {1}{a^{3} d^{2}}} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 15 \, \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (7 i \, d x + 7 i \, c\right )} + 2 \, a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )} \sqrt {\frac {1}{a^{3} d^{2}}} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (466 \, e^{\left (8 i \, d x + 8 i \, c\right )} + 1105 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 855 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 115 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 5\right )}}{60 \, {\left (a^{2} d e^{\left (7 i \, d x + 7 i \, c\right )} + 2 \, a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )}} \] Input:
integrate(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")
Output:
-1/60*(15*sqrt(1/2)*(a^2*d*e^(7*I*d*x + 7*I*c) + 2*a^2*d*e^(5*I*d*x + 5*I* c) + a^2*d*e^(3*I*d*x + 3*I*c))*sqrt(1/(a^3*d^2))*log(4*(sqrt(2)*sqrt(1/2) *(a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqr t(1/(a^3*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 15*sqrt(1/2)*(a^2* d*e^(7*I*d*x + 7*I*c) + 2*a^2*d*e^(5*I*d*x + 5*I*c) + a^2*d*e^(3*I*d*x + 3 *I*c))*sqrt(1/(a^3*d^2))*log(-4*(sqrt(2)*sqrt(1/2)*(a^2*d*e^(2*I*d*x + 2*I *c) + a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^3*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))* (466*e^(8*I*d*x + 8*I*c) + 1105*e^(6*I*d*x + 6*I*c) + 855*e^(4*I*d*x + 4*I *c) + 115*e^(2*I*d*x + 2*I*c) - 5))/(a^2*d*e^(7*I*d*x + 7*I*c) + 2*a^2*d*e ^(5*I*d*x + 5*I*c) + a^2*d*e^(3*I*d*x + 3*I*c))
\[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {\tan ^{5}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(tan(d*x+c)**5/(a+I*a*tan(d*x+c))**(3/2),x)
Output:
Integral(tan(c + d*x)**5/(I*a*(tan(c + d*x) - I))**(3/2), x)
Time = 0.13 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.77 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {15 \, \sqrt {2} a^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + 48 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2} - 240 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{3} + 960 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{4} + \frac {20 \, {\left (27 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{5} - 2 \, a^{6}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}}{120 \, a^{6} d} \] Input:
integrate(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")
Output:
1/120*(15*sqrt(2)*a^(9/2)*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))) + 48*(I*a*tan(d*x + c) + a)^(5/2)*a^2 - 240*(I*a*tan(d*x + c) + a)^(3/2)*a^3 + 960*sqrt(I*a*tan( d*x + c) + a)*a^4 + 20*(27*(I*a*tan(d*x + c) + a)*a^5 - 2*a^6)/(I*a*tan(d* x + c) + a)^(3/2))/(a^6*d)
Exception generated. \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^(3/2),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 = 1.16 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.67 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {8\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a^2\,d}-\frac {2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{a^3\,d}+\frac {2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{5\,a^4\,d}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {a}}\right )\,1{}\mathrm {i}}{4\,a^{3/2}\,d}+\frac {\frac {25\,a}{6}+\frac {a\,\mathrm {tan}\left (c+d\,x\right )\,9{}\mathrm {i}}{2}}{a\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \] Input:
int(tan(c + d*x)^5/(a + a*tan(c + d*x)*1i)^(3/2),x)
Output:
(8*(a + a*tan(c + d*x)*1i)^(1/2))/(a^2*d) - (2*(a + a*tan(c + d*x)*1i)^(3/ 2))/(a^3*d) + (2*(a + a*tan(c + d*x)*1i)^(5/2))/(5*a^4*d) + (2^(1/2)*atan( (2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*1i)/(2*a^(1/2)))*1i)/(4*a^(3/2)*d) + ((25*a)/6 + (a*tan(c + d*x)*9i)/2)/(a*d*(a + a*tan(c + d*x)*1i)^(3/2))
\[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (-\left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{6}}{\tan \left (d x +c \right )^{3} i +\tan \left (d x +c \right )^{2}+\tan \left (d x +c \right ) i +1}d x \right ) i +\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{5}}{\tan \left (d x +c \right )^{3} i +\tan \left (d x +c \right )^{2}+\tan \left (d x +c \right ) i +1}d x \right )}{a^{2}} \] Input:
int(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^(3/2),x)
Output:
(sqrt(a)*( - int((sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**6)/(tan(c + d*x)* *3*i + tan(c + d*x)**2 + tan(c + d*x)*i + 1),x)*i + int((sqrt(tan(c + d*x) *i + 1)*tan(c + d*x)**5)/(tan(c + d*x)**3*i + tan(c + d*x)**2 + tan(c + d* x)*i + 1),x)))/a**2