\(\int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\) [125]

Optimal result

Integrand size = 36, antiderivative size = 127 \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\frac {4 \sqrt [4]{-1} a^2 (i A+B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a^2 (7 i A+5 B)}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (A-i B)}{d \sqrt {\tan (c+d x)}}-\frac {2 A \left (a^2+i a^2 \tan (c+d x)\right )}{5 d \tan ^{\frac {5}{2}}(c+d x)} \] Output:

4*(-1)^(1/4)*a^2*(I*A+B)*arctan((-1)^(3/4)*tan(d*x+c)^(1/2))/d-2/15*a^2*(7 
*I*A+5*B)/d/tan(d*x+c)^(3/2)+4*a^2*(A-I*B)/d/tan(d*x+c)^(1/2)-2/5*A*(a^2+I 
*a^2*tan(d*x+c))/d/tan(d*x+c)^(5/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.70 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.80 \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {2 a^2 \left (3 A+5 (2 i A+B) \tan (c+d x)-30 (A-i B) \tan ^2(c+d x)+(15+15 i) \sqrt {2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {\tan (c+d x)}}{\sqrt {2}}\right ) \tan ^{\frac {5}{2}}(c+d x)\right )}{15 d \tan ^{\frac {5}{2}}(c+d x)} \] Input:

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

Output:

(-2*a^2*(3*A + 5*((2*I)*A + B)*Tan[c + d*x] - 30*(A - I*B)*Tan[c + d*x]^2 
+ (15 + 15*I)*Sqrt[2]*(A - I*B)*ArcTanh[((1 + I)*Sqrt[Tan[c + d*x]])/Sqrt[ 
2]]*Tan[c + d*x]^(5/2)))/(15*d*Tan[c + d*x]^(5/2))
 

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.06, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.306, Rules used = {3042, 4076, 27, 3042, 4074, 27, 3042, 4012, 3042, 4016, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4076

\(\displaystyle \frac {2}{5} \int \frac {(i \tan (c+d x) a+a) (a (7 i A+5 B)-a (3 A-5 i B) \tan (c+d x))}{2 \tan ^{\frac {5}{2}}(c+d x)}dx-\frac {2 A \left (a^2+i a^2 \tan (c+d x)\right )}{5 d \tan ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \int \frac {(i \tan (c+d x) a+a) (a (7 i A+5 B)-a (3 A-5 i B) \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)}dx-\frac {2 A \left (a^2+i a^2 \tan (c+d x)\right )}{5 d \tan ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \int \frac {(i \tan (c+d x) a+a) (a (7 i A+5 B)-a (3 A-5 i B) \tan (c+d x))}{\tan (c+d x)^{5/2}}dx-\frac {2 A \left (a^2+i a^2 \tan (c+d x)\right )}{5 d \tan ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 4074

\(\displaystyle \frac {1}{5} \left (\int -\frac {10 \left ((A-i B) a^2+(i A+B) \tan (c+d x) a^2\right )}{\tan ^{\frac {3}{2}}(c+d x)}dx-\frac {2 a^2 (5 B+7 i A)}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 A \left (a^2+i a^2 \tan (c+d x)\right )}{5 d \tan ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (-10 \int \frac {(A-i B) a^2+(i A+B) \tan (c+d x) a^2}{\tan ^{\frac {3}{2}}(c+d x)}dx-\frac {2 a^2 (5 B+7 i A)}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 A \left (a^2+i a^2 \tan (c+d x)\right )}{5 d \tan ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (-10 \int \frac {(A-i B) a^2+(i A+B) \tan (c+d x) a^2}{\tan (c+d x)^{3/2}}dx-\frac {2 a^2 (5 B+7 i A)}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 A \left (a^2+i a^2 \tan (c+d x)\right )}{5 d \tan ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 4012

\(\displaystyle \frac {1}{5} \left (-10 \left (\int \frac {a^2 (i A+B)-a^2 (A-i B) \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx-\frac {2 a^2 (A-i B)}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 a^2 (5 B+7 i A)}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 A \left (a^2+i a^2 \tan (c+d x)\right )}{5 d \tan ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (-10 \left (\int \frac {a^2 (i A+B)-a^2 (A-i B) \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx-\frac {2 a^2 (A-i B)}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 a^2 (5 B+7 i A)}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 A \left (a^2+i a^2 \tan (c+d x)\right )}{5 d \tan ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 4016

\(\displaystyle \frac {1}{5} \left (-10 \left (\frac {2 a^4 (B+i A)^2 \int \frac {1}{(i A+B) a^2+(A-i B) \tan (c+d x) a^2}d\sqrt {\tan (c+d x)}}{d}-\frac {2 a^2 (A-i B)}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 a^2 (5 B+7 i A)}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 A \left (a^2+i a^2 \tan (c+d x)\right )}{5 d \tan ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {1}{5} \left (-10 \left (-\frac {2 \sqrt [4]{-1} a^2 (B+i A) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a^2 (A-i B)}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 a^2 (5 B+7 i A)}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 A \left (a^2+i a^2 \tan (c+d x)\right )}{5 d \tan ^{\frac {5}{2}}(c+d x)}\)

Input:

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

Output:

(-10*((-2*(-1)^(1/4)*a^2*(I*A + B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/ 
d - (2*a^2*(A - I*B))/(d*Sqrt[Tan[c + d*x]])) - (2*a^2*((7*I)*A + 5*B))/(3 
*d*Tan[c + d*x]^(3/2)))/5 - (2*A*(a^2 + I*a^2*Tan[c + d*x]))/(5*d*Tan[c + 
d*x]^(5/2))
 

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 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 4012

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

rule 4016

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

rule 4074

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

rule 4076

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^2)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n 
+ 1)/(d*f*(b*c + a*d)*(n + 1))), x] - Simp[a/(d*(b*c + a*d)*(n + 1))   Int[ 
(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*b*d*(m - n 
 - 2) - B*(b*c*(m - 1) + a*d*(n + 1)) + (a*A*d*(m + n) - B*(a*c*(m - 1) + b 
*d*(n + 1)))*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] && GtQ[m, 1] && 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. 239 vs. \(2 (107 ) = 214\).

Time = 0.10 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.89

method	result	size
derivativedivides	\(\frac {a^{2} \left (\frac {\left (-2 i A -2 B \right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {\left (-2 i B +2 A \right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}-\frac {2 A}{5 \tan \left (d x +c \right )^{\frac {5}{2}}}-\frac {2 \left (2 i A +B \right )}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 \left (2 i B -2 A \right )}{\sqrt {\tan \left (d x +c \right )}}\right )}{d}\)	\(240\)
default	\(\frac {a^{2} \left (\frac {\left (-2 i A -2 B \right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {\left (-2 i B +2 A \right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}-\frac {2 A}{5 \tan \left (d x +c \right )^{\frac {5}{2}}}-\frac {2 \left (2 i A +B \right )}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 \left (2 i B -2 A \right )}{\sqrt {\tan \left (d x +c \right )}}\right )}{d}\)	\(240\)
parts	\(\frac {\left (2 i A \,a^{2}+B \,a^{2}\right ) \left (-\frac {2}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}\right )}{d}+\frac {\left (2 i B \,a^{2}-A \,a^{2}\right ) \left (-\frac {\sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}-\frac {2}{\sqrt {\tan \left (d x +c \right )}}\right )}{d}+\frac {A \,a^{2} \left (\frac {\sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}-\frac {2}{5 \tan \left (d x +c \right )^{\frac {5}{2}}}+\frac {2}{\sqrt {\tan \left (d x +c \right )}}\right )}{d}-\frac {B \,a^{2} \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4 d}\)	\(435\)

Input:

int((a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x,method=_RETUR 
NVERBOSE)
 

Output:

1/d*a^2*(1/4*(-2*B-2*I*A)*2^(1/2)*(ln((tan(d*x+c)+2^(1/2)*tan(d*x+c)^(1/2) 
+1)/(tan(d*x+c)-2^(1/2)*tan(d*x+c)^(1/2)+1))+2*arctan(1+2^(1/2)*tan(d*x+c) 
^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)))+1/4*(-2*I*B+2*A)*2^(1/2)*(l 
n((tan(d*x+c)-2^(1/2)*tan(d*x+c)^(1/2)+1)/(tan(d*x+c)+2^(1/2)*tan(d*x+c)^( 
1/2)+1))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+ 
c)^(1/2)))-2/5*A/tan(d*x+c)^(5/2)-2/3*(2*I*A+B)/tan(d*x+c)^(3/2)-2*(2*I*B- 
2*A)/tan(d*x+c)^(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. 502 vs. \(2 (103) = 206\).

Time = 0.11 (sec) , antiderivative size = 502, normalized size of antiderivative = 3.95 \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {15 \, \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{4}}{d^{2}}} {\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 )} \log \left (-\frac {2 \, {\left ({\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \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}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) - 15 \, \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{4}}{d^{2}}} {\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 )} \log \left (-\frac {2 \, {\left ({\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \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}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) + 2 \, {\left ({\left (-43 i \, A - 35 \, B\right )} a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (11 i \, A + 25 \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (31 i \, A + 35 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-23 i \, A - 25 \, B\right )} a^{2}\right )} \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}}}{15 \, {\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))^2*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x, algori 
thm="fricas")
 

Output:

-1/15*(15*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^4/d^2)*(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)*log(-2*((A - I*B)* 
a^2*e^(2*I*d*x + 2*I*c) + sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^4/d^2)*(d*e^(2* 
I*d*x + 2*I*c) + d)*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)/((-I*A - B)*a^2)) - 15*sqrt(-(-I*A^2 - 2*A*B 
+ I*B^2)*a^4/d^2)*(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)*log(-2*((A - I*B)*a^2*e^(2*I*d*x + 2*I*c) - sqrt(- 
(-I*A^2 - 2*A*B + I*B^2)*a^4/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*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)/((- 
I*A - B)*a^2)) + 2*((-43*I*A - 35*B)*a^2*e^(6*I*d*x + 6*I*c) + (11*I*A + 2 
5*B)*a^2*e^(4*I*d*x + 4*I*c) + (31*I*A + 35*B)*a^2*e^(2*I*d*x + 2*I*c) + ( 
-23*I*A - 25*B)*a^2)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c 
) + 1)))/(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]

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

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

Output:

-a**2*(Integral(-A/tan(c + d*x)**(7/2), x) + Integral(A/tan(c + d*x)**(3/2 
), x) + Integral(-B/tan(c + d*x)**(5/2), x) + Integral(B/sqrt(tan(c + d*x) 
), x) + Integral(-2*I*A/tan(c + d*x)**(5/2), x) + Integral(-2*I*B/tan(c + 
d*x)**(3/2), x))
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.54 \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {15 \, {\left (2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a^{2} - \frac {4 \, {\left (30 \, {\left (A - i \, B\right )} a^{2} \tan \left (d x + c\right )^{2} + 5 \, {\left (-2 i \, A - B\right )} a^{2} \tan \left (d x + c\right ) - 3 \, A a^{2}\right )}}{\tan \left (d x + c\right )^{\frac {5}{2}}}}{30 \, d} \] Input:

integrate((a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x, algori 
thm="maxima")
 

Output:

-1/30*(15*(2*sqrt(2)*((I - 1)*A + (I + 1)*B)*arctan(1/2*sqrt(2)*(sqrt(2) + 
 2*sqrt(tan(d*x + c)))) + 2*sqrt(2)*((I - 1)*A + (I + 1)*B)*arctan(-1/2*sq 
rt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) - sqrt(2)*(-(I + 1)*A + (I - 1)*B) 
*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + sqrt(2)*(-(I + 1)*A 
+ (I - 1)*B)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1))*a^2 - 4* 
(30*(A - I*B)*a^2*tan(d*x + c)^2 + 5*(-2*I*A - B)*a^2*tan(d*x + c) - 3*A*a 
^2)/tan(d*x + c)^(5/2))/d
 

Giac [A] (verification not implemented)

Time = 0.56 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.84 \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (15 \, \sqrt {2} {\left (\left (i - 1\right ) \, A a^{2} + \left (i + 1\right ) \, B a^{2}\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right ) - \frac {30 \, A a^{2} \tan \left (d x + c\right )^{2} - 30 i \, B a^{2} \tan \left (d x + c\right )^{2} - 10 i \, A a^{2} \tan \left (d x + c\right ) - 5 \, B a^{2} \tan \left (d x + c\right ) - 3 \, A a^{2}}{\tan \left (d x + c\right )^{\frac {5}{2}}}\right )}}{15 \, d} \] Input:

integrate((a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x, algori 
thm="giac")
 

Output:

-2/15*(15*sqrt(2)*((I - 1)*A*a^2 + (I + 1)*B*a^2)*arctan(-(1/2*I - 1/2)*sq 
rt(2)*sqrt(tan(d*x + c))) - (30*A*a^2*tan(d*x + c)^2 - 30*I*B*a^2*tan(d*x 
+ c)^2 - 10*I*A*a^2*tan(d*x + c) - 5*B*a^2*tan(d*x + c) - 3*A*a^2)/tan(d*x 
 + c)^(5/2))/d
 

Mupad [B] (verification not implemented)

Time = 5.21 (sec) , antiderivative size = 258, normalized size of antiderivative = 2.03 \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {\frac {2\,A\,a^2}{5\,d}-\frac {4\,A\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^2}{d}+\frac {A\,a^2\,\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}}{3\,d}}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}-\frac {\frac {2\,B\,a^2}{3\,d}+\frac {B\,a^2\,\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}}{d}}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}+\frac {\sqrt {2}\,A\,a^2\,\ln \left (4\,A\,a^2\,d+\sqrt {2}\,A\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-2-2{}\mathrm {i}\right )\right )\,\left (1+1{}\mathrm {i}\right )}{d}-\frac {\sqrt {4{}\mathrm {i}}\,A\,a^2\,\ln \left (4\,A\,a^2\,d+2\,\sqrt {4{}\mathrm {i}}\,A\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d}+\frac {\sqrt {2}\,B\,a^2\,\ln \left (-B\,a^2\,d\,4{}\mathrm {i}+\sqrt {2}\,B\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-2+2{}\mathrm {i}\right )\right )\,\left (1-\mathrm {i}\right )}{d}-\frac {\sqrt {-4{}\mathrm {i}}\,B\,a^2\,\ln \left (-B\,a^2\,d\,4{}\mathrm {i}+2\,\sqrt {-4{}\mathrm {i}}\,B\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d} \] Input:

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

Output:

(2^(1/2)*A*a^2*log(4*A*a^2*d - 2^(1/2)*A*a^2*d*tan(c + d*x)^(1/2)*(2 + 2i) 
)*(1 + 1i))/d - ((2*B*a^2)/(3*d) + (B*a^2*tan(c + d*x)*4i)/d)/tan(c + d*x) 
^(3/2) - ((2*A*a^2)/(5*d) + (A*a^2*tan(c + d*x)*4i)/(3*d) - (4*A*a^2*tan(c 
 + d*x)^2)/d)/tan(c + d*x)^(5/2) - (4i^(1/2)*A*a^2*log(4*A*a^2*d + 2*4i^(1 
/2)*A*a^2*d*tan(c + d*x)^(1/2)))/d + (2^(1/2)*B*a^2*log(- B*a^2*d*4i - 2^( 
1/2)*B*a^2*d*tan(c + d*x)^(1/2)*(2 - 2i))*(1 - 1i))/d - ((-4i)^(1/2)*B*a^2 
*log(2*(-4i)^(1/2)*B*a^2*d*tan(c + d*x)^(1/2) - B*a^2*d*4i))/d
 

Reduce [F]

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

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

Output:

a**2*(int(sqrt(tan(c + d*x))/tan(c + d*x)**4,x)*a + 2*int(sqrt(tan(c + d*x 
))/tan(c + d*x)**3,x)*a*i + int(sqrt(tan(c + d*x))/tan(c + d*x)**3,x)*b - 
int(sqrt(tan(c + d*x))/tan(c + d*x)**2,x)*a + 2*int(sqrt(tan(c + d*x))/tan 
(c + d*x)**2,x)*b*i - int(sqrt(tan(c + d*x))/tan(c + d*x),x)*b)