\(\int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [106]

Optimal result

Integrand size = 36, antiderivative size = 167 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {(i A-B) \tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {3 i A-13 B}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac {i A-31 B}{20 a^2 d \sqrt {a+i a \tan (c+d x)}} \] Output:

1/8*(I*A+B)*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*(I*A-B)*tan(d*x+c)^2/d/(a+I*a*tan(d*x+c))^(5/2)+1/30*(3*I*A- 
13*B)/a/d/(a+I*a*tan(d*x+c))^(3/2)-1/20*(I*A-31*B)/a^2/d/(a+I*a*tan(d*x+c) 
)^(1/2)
 

Mathematica [A] (verified)

Time = 1.43 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.06 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {2 B \tan ^2(c+d x)}{d (a+i a \tan (c+d x))^{5/2}}+\frac {i \left (\frac {15 \sqrt {2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{a^{3/2}}-\frac {24 a (A-9 i B)}{(a+i a \tan (c+d x))^{5/2}}+\frac {20 (3 A-19 i B)}{(a+i a \tan (c+d x))^{3/2}}-\frac {30 (A-i B)}{a \sqrt {a+i a \tan (c+d x)}}\right )}{120 a d} \] Input:

Integrate[(Tan[c + d*x]^2*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^(5/ 
2),x]
 

Output:

(-2*B*Tan[c + d*x]^2)/(d*(a + I*a*Tan[c + d*x])^(5/2)) + ((I/120)*((15*Sqr 
t[2]*(A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/a^(3 
/2) - (24*a*(A - (9*I)*B))/(a + I*a*Tan[c + d*x])^(5/2) + (20*(3*A - (19*I 
)*B))/(a + I*a*Tan[c + d*x])^(3/2) - (30*(A - I*B))/(a*Sqrt[a + I*a*Tan[c 
+ d*x]])))/(a*d)
 

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {3042, 4078, 27, 3042, 4073, 3042, 4009, 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.

\(\displaystyle \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)^2 (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}}dx\)

\(\Big \downarrow \) 4078

\(\displaystyle \frac {(-B+i A) \tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\int \frac {\tan (c+d x) (4 a (i A-B)-a (A-9 i B) \tan (c+d x))}{2 (i \tan (c+d x) a+a)^{3/2}}dx}{5 a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(-B+i A) \tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\int \frac {\tan (c+d x) (4 a (i A-B)-a (A-9 i B) \tan (c+d x))}{(i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(-B+i A) \tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\int \frac {\tan (c+d x) (4 a (i A-B)-a (A-9 i B) \tan (c+d x))}{(i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}\)

\(\Big \downarrow \) 4073

\(\displaystyle \frac {(-B+i A) \tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {-\frac {i \int \frac {a^2 (3 i A-13 B)-2 a^2 (A-9 i B) \tan (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx}{2 a^2}-\frac {a (-13 B+3 i A)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(-B+i A) \tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {-\frac {i \int \frac {a^2 (3 i A-13 B)-2 a^2 (A-9 i B) \tan (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx}{2 a^2}-\frac {a (-13 B+3 i A)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}\)

\(\Big \downarrow \) 4009

\(\displaystyle \frac {(-B+i A) \tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {-\frac {i \left (\frac {5}{2} a (B+i A) \int \sqrt {i \tan (c+d x) a+a}dx-\frac {a^2 (A+31 i B)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{2 a^2}-\frac {a (-13 B+3 i A)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(-B+i A) \tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {-\frac {i \left (\frac {5}{2} a (B+i A) \int \sqrt {i \tan (c+d x) a+a}dx-\frac {a^2 (A+31 i B)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{2 a^2}-\frac {a (-13 B+3 i A)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}\)

\(\Big \downarrow \) 3961

\(\displaystyle \frac {(-B+i A) \tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {-\frac {i \left (-\frac {5 i a^2 (B+i A) \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}-\frac {a^2 (A+31 i B)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{2 a^2}-\frac {a (-13 B+3 i A)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {(-B+i A) \tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {-\frac {i \left (-\frac {5 i a^{3/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} d}-\frac {a^2 (A+31 i B)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{2 a^2}-\frac {a (-13 B+3 i A)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}\)

Input:

Int[(Tan[c + d*x]^2*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^(5/2),x]
 

Output:

((I*A - B)*Tan[c + d*x]^2)/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) - (-1/3*(a*( 
(3*I)*A - 13*B))/(d*(a + I*a*Tan[c + d*x])^(3/2)) - ((I/2)*(((-5*I)*a^(3/2 
)*(I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2 
]*d) - (a^2*(A + (31*I)*B))/(d*Sqrt[a + I*a*Tan[c + d*x]])))/a^2)/(10*a^2)
 

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 219

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])
 

rule 3042

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

rule 3961

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]
 

rule 4009

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*((a + b*Tan[e + f*x])^m/(2*a 
*f*m)), x] + Simp[(b*c + a*d)/(2*a*b)   Int[(a + b*Tan[e + f*x])^(m + 1), x 
], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2 
, 0] && LtQ[m, 0]
 

rule 4073

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-( 
A*b - a*B))*(a*c + b*d)*((a + b*Tan[e + f*x])^m/(2*a^2*f*m)), x] + Simp[1/( 
2*a*b)   Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[A*b*c + a*B*c + a*A*d + b*B* 
d + 2*a*B*d*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] 
&& NeQ[b*c - a*d, 0] && LtQ[m, -1] && EqQ[a^2 + b^2, 0]
 

rule 4078

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]
 

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(\frac {2 i \left (-\frac {\left (-\frac {A}{8}+\frac {i B}{8}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2 \sqrt {a}}-\frac {\frac {A}{8}+\frac {7 i B}{8}}{\sqrt {a +i a \tan \left (d x +c \right )}}+\frac {a \left (5 i B +3 A \right )}{12 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {a^{2} \left (i B +A \right )}{10 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}\right )}{d \,a^{2}}\)	\(124\)
default	\(\frac {2 i \left (-\frac {\left (-\frac {A}{8}+\frac {i B}{8}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2 \sqrt {a}}-\frac {\frac {A}{8}+\frac {7 i B}{8}}{\sqrt {a +i a \tan \left (d x +c \right )}}+\frac {a \left (5 i B +3 A \right )}{12 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {a^{2} \left (i B +A \right )}{10 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}\right )}{d \,a^{2}}\)	\(124\)
parts	\(\frac {2 i A \left (\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 a^{\frac {3}{2}}}+\frac {1}{4 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {1}{8 a \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {a}{10 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}\right )}{d a}+\frac {2 B \left (\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 \sqrt {a}}+\frac {7}{8 \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {5 a}{12 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {a^{2}}{10 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}\right )}{d \,a^{2}}\)	\(189\)

Input:

int(tan(d*x+c)^2*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETUR 
NVERBOSE)
 

Output:

2*I/d/a^2*(-1/2*(-1/8*A+1/8*I*B)*2^(1/2)/a^(1/2)*arctanh(1/2*(a+I*a*tan(d* 
x+c))^(1/2)*2^(1/2)/a^(1/2))-(1/8*A+7/8*I*B)/(a+I*a*tan(d*x+c))^(1/2)+1/12 
*a*(3*A+5*I*B)/(a+I*a*tan(d*x+c))^(3/2)-1/10*a^2*(A+I*B)/(a+I*a*tan(d*x+c) 
)^(5/2))
 

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. 393 vs. \(2 (128) = 256\).

Time = 0.09 (sec) , antiderivative size = 393, normalized size of antiderivative = 2.35 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {{\left (15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} + {\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} - {\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - \sqrt {2} {\left ({\left (-3 i \, A + 83 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - 2 \, {\left (-3 i \, A - 32 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, {\left (-3 i \, A + 8 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 3 i \, A + 3 \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{120 \, a^{3} d} \] Input:

integrate(tan(d*x+c)^2*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, algori 
thm="fricas")
 

Output:

-1/120*(15*sqrt(1/2)*a^3*d*sqrt(-(A^2 - 2*I*A*B - B^2)/(a^5*d^2))*e^(5*I*d 
*x + 5*I*c)*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(-(A^2 - 2*I*A*B - B^2)/(a^5*d^2)) + 
 (-I*A - B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(I*A + B)) - 15*sqrt(1/2)* 
a^3*d*sqrt(-(A^2 - 2*I*A*B - B^2)/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(4*(sq 
rt(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(-(A^2 - 2*I*A*B - B^2)/(a^5*d^2)) - (-I*A - B)*a*e^(I*d*x 
 + I*c))*e^(-I*d*x - I*c)/(I*A + B)) - sqrt(2)*((-3*I*A + 83*B)*e^(6*I*d*x 
 + 6*I*c) - 2*(-3*I*A - 32*B)*e^(4*I*d*x + 4*I*c) - 2*(-3*I*A + 8*B)*e^(2* 
I*d*x + 2*I*c) - 3*I*A + 3*B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-5*I*d 
*x - 5*I*c)/(a^3*d)
 

Sympy [F]

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

integrate(tan(d*x+c)**2*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**(5/2),x)
 

Output:

Integral((A + B*tan(c + d*x))*tan(c + d*x)**2/(I*a*(tan(c + d*x) - I))**(5 
/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.84 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {i \, {\left (15 \, \sqrt {2} {\left (A - i \, B\right )} \sqrt {a} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (15 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} {\left (A + 7 i \, B\right )} a - 10 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} {\left (3 \, A + 5 i \, B\right )} a^{2} + 12 \, {\left (A + i \, B\right )} a^{3}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\right )}}{240 \, a^{3} d} \] Input:

integrate(tan(d*x+c)^2*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, algori 
thm="maxima")
 

Output:

-1/240*I*(15*sqrt(2)*(A - I*B)*sqrt(a)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*ta 
n(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(d*x + c) + a))) + 4*(15*( 
I*a*tan(d*x + c) + a)^2*(A + 7*I*B)*a - 10*(I*a*tan(d*x + c) + a)*(3*A + 5 
*I*B)*a^2 + 12*(A + I*B)*a^3)/(I*a*tan(d*x + c) + a)^(5/2))/(a^3*d)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(tan(d*x+c)^2*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, algori 
thm="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 [B] (verification not implemented)

Time = 3.64 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.12 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {A\,1{}\mathrm {i}}{20\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{4\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {\frac {B\,a^2}{5}+\frac {7\,B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2}{4}-\frac {5\,B\,a\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{6}}{a^2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}-\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{8\,{\left (-a\right )}^{5/2}\,d}+\frac {\sqrt {2}\,B\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}}\right )}{8\,a^{5/2}\,d} \] Input:

int((tan(c + d*x)^2*(A + B*tan(c + d*x)))/(a + a*tan(c + d*x)*1i)^(5/2),x)
 

Output:

(A*1i)/(20*d*(a + a*tan(c + d*x)*1i)^(5/2)) + (A*tan(c + d*x)^2*1i)/(4*d*( 
a + a*tan(c + d*x)*1i)^(5/2)) + ((B*a^2)/5 + (7*B*(a + a*tan(c + d*x)*1i)^ 
2)/4 - (5*B*a*(a + a*tan(c + d*x)*1i))/6)/(a^2*d*(a + a*tan(c + d*x)*1i)^( 
5/2)) - (2^(1/2)*A*atan((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(2*(-a)^(1 
/2)))*1i)/(8*(-a)^(5/2)*d) + (2^(1/2)*B*atanh((2^(1/2)*(a + a*tan(c + d*x) 
*1i)^(1/2))/(2*a^(1/2))))/(8*a^(5/2)*d)
 

Reduce [F]

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

int(tan(d*x+c)^2*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x)
 

Output:

( - (int(tan(c + d*x)**3/(sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2 - 2*sqr 
t(tan(c + d*x)*i + 1)*tan(c + d*x)*i - sqrt(tan(c + d*x)*i + 1)),x)*b + in 
t(tan(c + d*x)**2/(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)*a))/(sqrt(a)* 
a**2)