\(\int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\) [199]

Optimal result

Integrand size = 28, antiderivative size = 198 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {(2+2 i) a^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {2 i a^2}{5 d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {4 i a \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {12 a \sqrt {a+i a \tan (c+d x)}}{5 d \sqrt {\tan (c+d x)}} \] Output:

(-2-2*I)*a^(3/2)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c)) 
^(1/2))/d-2/5*a^2/d/tan(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^(1/2)-2/5*I*a^2/d/ 
tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(1/2)-4/5*I*a*(a+I*a*tan(d*x+c))^(1/2) 
/d/tan(d*x+c)^(3/2)+12/5*a*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 6.35 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.69 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {2 \sqrt {2} a \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}+\frac {2 a^{3/2} \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+i \tan (c+d x)} \sqrt {i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {2 \sqrt [4]{-1} a \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {1+i \tan (c+d x)}}-\frac {2 a \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {4 i a \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {12 a \sqrt {a+i a \tan (c+d x)}}{5 d \sqrt {\tan (c+d x)}} \] Input:

Integrate[(a + I*a*Tan[c + d*x])^(3/2)/Tan[c + d*x]^(7/2),x]
 

Output:

(-2*Sqrt[2]*a*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d*x]])/Sqrt[a + I*a*Tan[c 
+ d*x]]]*Sqrt[I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]) + (2*a^(3/2)*ArcSi 
nh[Sqrt[I*a*Tan[c + d*x]]/Sqrt[a]]*Sqrt[1 + I*Tan[c + d*x]]*Sqrt[I*a*Tan[c 
 + d*x]])/(d*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (2*(-1)^(1/4 
)*a*ArcSinh[(-1)^(1/4)*Sqrt[Tan[c + d*x]]]*Sqrt[a + I*a*Tan[c + d*x]])/(d* 
Sqrt[1 + I*Tan[c + d*x]]) - (2*a*Sqrt[a + I*a*Tan[c + d*x]])/(5*d*Tan[c + 
d*x]^(5/2)) - (((4*I)/5)*a*Sqrt[a + I*a*Tan[c + d*x]])/(d*Tan[c + d*x]^(3/ 
2)) + (12*a*Sqrt[a + I*a*Tan[c + d*x]])/(5*d*Sqrt[Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 1.17 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.08, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.536, Rules used = {3042, 4036, 27, 3042, 4079, 27, 3042, 4081, 27, 3042, 4081, 27, 3042, 4027, 218}

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.

\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan (c+d x)^{7/2}}dx\)

\(\Big \downarrow \) 4036

\(\displaystyle -\frac {2}{5} \int -\frac {9 i a^2-11 a^2 \tan (c+d x)}{2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {i \tan (c+d x) a+a}}dx-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \int \frac {9 i a^2-11 a^2 \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {i \tan (c+d x) a+a}}dx-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \int \frac {9 i a^2-11 a^2 \tan (c+d x)}{\tan (c+d x)^{5/2} \sqrt {i \tan (c+d x) a+a}}dx-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {1}{5} \left (\frac {\int \frac {2 \sqrt {i \tan (c+d x) a+a} \left (3 i a^3-2 a^3 \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)}dx}{a^2}-\frac {2 i a^2}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (\frac {2 \int \frac {\sqrt {i \tan (c+d x) a+a} \left (3 i a^3-2 a^3 \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)}dx}{a^2}-\frac {2 i a^2}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {2 \int \frac {\sqrt {i \tan (c+d x) a+a} \left (3 i a^3-2 a^3 \tan (c+d x)\right )}{\tan (c+d x)^{5/2}}dx}{a^2}-\frac {2 i a^2}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 4081

\(\displaystyle \frac {1}{5} \left (\frac {2 \left (\frac {2 \int -\frac {3 \sqrt {i \tan (c+d x) a+a} \left (2 i \tan (c+d x) a^4+3 a^4\right )}{2 \tan ^{\frac {3}{2}}(c+d x)}dx}{3 a}-\frac {2 i a^3 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {3}{2}}(c+d x)}\right )}{a^2}-\frac {2 i a^2}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (\frac {2 \left (-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (2 i \tan (c+d x) a^4+3 a^4\right )}{\tan ^{\frac {3}{2}}(c+d x)}dx}{a}-\frac {2 i a^3 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {3}{2}}(c+d x)}\right )}{a^2}-\frac {2 i a^2}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {2 \left (-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (2 i \tan (c+d x) a^4+3 a^4\right )}{\tan (c+d x)^{3/2}}dx}{a}-\frac {2 i a^3 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {3}{2}}(c+d x)}\right )}{a^2}-\frac {2 i a^2}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 4081

\(\displaystyle \frac {1}{5} \left (\frac {2 \left (-\frac {\frac {2 \int \frac {5 i a^5 \sqrt {i \tan (c+d x) a+a}}{2 \sqrt {\tan (c+d x)}}dx}{a}-\frac {6 a^4 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{a}-\frac {2 i a^3 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {3}{2}}(c+d x)}\right )}{a^2}-\frac {2 i a^2}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (\frac {2 \left (-\frac {5 i a^4 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {6 a^4 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{a}-\frac {2 i a^3 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {3}{2}}(c+d x)}\right )}{a^2}-\frac {2 i a^2}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {2 \left (-\frac {5 i a^4 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {6 a^4 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{a}-\frac {2 i a^3 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {3}{2}}(c+d x)}\right )}{a^2}-\frac {2 i a^2}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 4027

\(\displaystyle \frac {1}{5} \left (\frac {2 \left (-\frac {\frac {10 a^6 \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {6 a^4 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{a}-\frac {2 i a^3 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {3}{2}}(c+d x)}\right )}{a^2}-\frac {2 i a^2}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {1}{5} \left (\frac {2 \left (-\frac {2 i a^3 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {3}{2}}(c+d x)}-\frac {\frac {(5+5 i) a^{9/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {6 a^4 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{a}\right )}{a^2}-\frac {2 i a^2}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\)

Input:

Int[(a + I*a*Tan[c + d*x])^(3/2)/Tan[c + d*x]^(7/2),x]
 

Output:

(-2*a^2)/(5*d*Tan[c + d*x]^(5/2)*Sqrt[a + I*a*Tan[c + d*x]]) + (((-2*I)*a^ 
2)/(d*Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]) + (2*(((-2*I)*a^3*Sqr 
t[a + I*a*Tan[c + d*x]])/(d*Tan[c + d*x]^(3/2)) - (((5 + 5*I)*a^(9/2)*ArcT 
anh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - 
(6*a^4*Sqrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]))/a))/a^2)/5
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 218

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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4027

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]
 

rule 4036

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-a^2)*(b*c - a*d)*(a + b*Tan[e + f*x] 
)^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(b*c + a*d)*(n + 1))), x] + Si 
mp[a/(d*(b*c + a*d)*(n + 1))   Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[ 
e + f*x])^(n + 1)*Simp[b*(b*c*(m - 2) - a*d*(m - 2*n - 4)) + (a*b*c*(m - 2) 
 + b^2*d*(n + 1) - a^2*d*(m + n - 1))*Tan[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] && GtQ[m, 1] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
 

rule 4079

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*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*( 
b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d))   Int[(a + b*Tan[e + f*x])^(m 
 + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m 
- b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Free 
Q[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] 
 && LtQ[m, 0] &&  !GtQ[n, 0]
 

rule 4081

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*d - B*c)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*(n + 
1)*(c^2 + d^2))), x] - Simp[1/(a*(n + 1)*(c^2 + d^2))   Int[(a + b*Tan[e + 
f*x])^m*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c* 
m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Fr 
eeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 
0] && LtQ[n, -1]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 409 vs. \(2 (159 ) = 318\).

Time = 1.63 (sec) , antiderivative size = 410, normalized size of antiderivative = 2.07

method	result	size
derivativedivides	\(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (5 i \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{3}+5 \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{3}+20 \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \tan \left (d x +c \right )^{3}-8 i \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+24 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )^{2}-4 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{10 d \tan \left (d x +c \right )^{\frac {5}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(410\)
default	\(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (5 i \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{3}+5 \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{3}+20 \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \tan \left (d x +c \right )^{3}-8 i \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+24 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )^{2}-4 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{10 d \tan \left (d x +c \right )^{\frac {5}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(410\)

Input:

int((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(7/2),x,method=_RETURNVERBOSE)
 

Output:

1/10/d*(a*(1+I*tan(d*x+c)))^(1/2)*a*(5*I*(I*a)^(1/2)*2^(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))/(t 
an(d*x+c)+I))*a*tan(d*x+c)^3+5*(I*a)^(1/2)*2^(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)^3+20*(-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)^3-8*I*t 
an(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)+2 
4*(-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 
)^2-4*(-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)^(5/2)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(I*a)^(1/2)/(-I*a)^(1/2 
)
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 450 vs. \(2 (148) = 296\).

Time = 0.09 (sec) , antiderivative size = 450, normalized size of antiderivative = 2.27 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {4 \, \sqrt {2} {\left (-9 i \, a e^{\left (7 i \, d x + 7 i \, c\right )} + i \, a e^{\left (5 i \, d x + 5 i \, c\right )} + 5 i \, a e^{\left (3 i \, d x + 3 i \, c\right )} - 5 i \, a e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - 5 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {8 i \, a^{3}}{d^{2}}} \log \left (\frac {{\left (2 \, \sqrt {2} {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + i \, \sqrt {\frac {8 i \, a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, a}\right ) + 5 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {8 i \, a^{3}}{d^{2}}} \log \left (\frac {{\left (2 \, \sqrt {2} {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - i \, \sqrt {\frac {8 i \, a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, a}\right )}{10 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \] Input:

integrate((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(7/2),x, algorithm="fricas")
 

Output:

-1/10*(4*sqrt(2)*(-9*I*a*e^(7*I*d*x + 7*I*c) + I*a*e^(5*I*d*x + 5*I*c) + 5 
*I*a*e^(3*I*d*x + 3*I*c) - 5*I*a*e^(I*d*x + I*c))*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)) - 5 
*(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c 
) - d)*sqrt(8*I*a^3/d^2)*log(1/2*(2*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*sq 
rt(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)) + I*sqrt(8*I*a^3/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I* 
c)/a) + 5*(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d* 
x + 2*I*c) - d)*sqrt(8*I*a^3/d^2)*log(1/2*(2*sqrt(2)*(a*e^(2*I*d*x + 2*I*c 
) + a)*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)) - I*sqrt(8*I*a^3/d^2)*d*e^(I*d*x + I*c))*e^(-I 
*d*x - I*c)/a))/(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^( 
2*I*d*x + 2*I*c) - d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:

integrate((a+I*a*tan(d*x+c))**(3/2)/tan(d*x+c)**(7/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1198 vs. \(2 (148) = 296\).

Time = 0.32 (sec) , antiderivative size = 1198, normalized size of antiderivative = 6.05 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:

integrate((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(7/2),x, algorithm="maxima")
 

Output:

1/15*(2*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) 
+ 1)*(((15*I - 15)*a*cos(3*d*x + 3*c) - (16*I - 16)*a*cos(d*x + c) - (15*I 
 + 15)*a*sin(3*d*x + 3*c) + (16*I + 16)*a*sin(d*x + c))*cos(3/2*arctan2(si 
n(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + (-(15*I + 15)*a*cos(3*d*x + 3*c) 
 + (16*I + 16)*a*cos(d*x + c) - (15*I - 15)*a*sin(3*d*x + 3*c) + (16*I - 1 
6)*a*sin(d*x + c))*sin(3/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1 
)))*sqrt(a) + 15*(2*(-(I - 1)*a*cos(2*d*x + 2*c)^2 - (I - 1)*a*sin(2*d*x + 
 2*c)^2 + (2*I - 2)*a*cos(2*d*x + 2*c) - (I - 1)*a)*arctan2((cos(2*d*x + 2 
*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2 
(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) - cos(d*x + c), (cos(2*d*x + 2* 
c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2( 
sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) - sin(d*x + c)) + ((I + 1)*a*cos 
(2*d*x + 2*c)^2 + (I + 1)*a*sin(2*d*x + 2*c)^2 - (2*I + 2)*a*cos(2*d*x + 2 
*c) + (I + 1)*a)*log(cos(d*x + c)^2 + sin(d*x + c)^2 + sqrt(cos(2*d*x + 2* 
c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*(cos(1/2*arctan2(sin(2 
*d*x + 2*c), -cos(2*d*x + 2*c) + 1))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), 
 -cos(2*d*x + 2*c) + 1))^2) - 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 
 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d 
*x + 2*c) + 1))*sin(d*x + c) + cos(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2* 
c), -cos(2*d*x + 2*c) + 1)))))*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2...
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(7/2),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 TypeDone
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{7/2}} \,d x \] Input:

int((a + a*tan(c + d*x)*1i)^(3/2)/tan(c + d*x)^(7/2),x)
 

Output:

int((a + a*tan(c + d*x)*1i)^(3/2)/tan(c + d*x)^(7/2), x)
 

Reduce [F]

\[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 \sqrt {a}\, a \left (-4 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}-2 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) i -\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}-5 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}}{\tan \left (d x +c \right )^{2}}d x \right ) \tan \left (d x +c \right )^{3} d \right )}{5 \tan \left (d x +c \right )^{3} d} \] Input:

int((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(7/2),x)
 

Output:

(2*sqrt(a)*a*( - 4*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x 
)**2 - 2*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)*i - sqrt 
(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1) - 5*int((sqrt(tan(c + d*x))*sqrt(t 
an(c + d*x)*i + 1))/tan(c + d*x)**2,x)*tan(c + d*x)**3*d))/(5*tan(c + d*x) 
**3*d)