3.1.71 \(\int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx\) [71]

3.1.71.1 Optimal result

Integrand size = 34, antiderivative size = 86 \[ \int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=-\frac {2 \sqrt {a} A \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {\sqrt {2} \sqrt {a} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d} \]

output

-2*A*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))*a^(1/2)/d+(A-I*B)*arctanh(1 
/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*a^(1/2)/d
 

3.1.71.2 Mathematica [A] (verified)

Time = 1.11 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.92 \[ \int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {\sqrt {a} \left (-2 A \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )+\sqrt {2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )\right )}{d} \]

input

Integrate[Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]]*(A + B*Tan[c + d*x]),x]
 

output

(Sqrt[a]*(-2*A*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]] + Sqrt[2]*(A - 
I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])]))/d
 

3.1.71.3 Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {3042, 4083, 3042, 3961, 219, 4082, 73, 221}

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.

\(\displaystyle \int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x))}{\tan (c+d x)}dx\)

\(\Big \downarrow \) 4083

\(\displaystyle (B+i A) \int \sqrt {i \tan (c+d x) a+a}dx+\frac {A \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle (B+i A) \int \sqrt {i \tan (c+d x) a+a}dx+\frac {A \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx}{a}\)

\(\Big \downarrow \) 3961

\(\displaystyle \frac {A \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx}{a}-\frac {2 i a (B+i A) \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {A \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx}{a}-\frac {i \sqrt {2} \sqrt {a} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}\)

\(\Big \downarrow \) 4082

\(\displaystyle \frac {a A \int \frac {\cot (c+d x)}{\sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}-\frac {i \sqrt {2} \sqrt {a} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {2 i A \int \frac {1}{i-\frac {i (i \tan (c+d x) a+a)}{a}}d\sqrt {i \tan (c+d x) a+a}}{d}-\frac {i \sqrt {2} \sqrt {a} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {i \sqrt {2} \sqrt {a} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {2 \sqrt {a} A \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}\)

input

Int[Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]]*(A + B*Tan[c + d*x]),x]
 

output

(-2*Sqrt[a]*A*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/d - (I*Sqrt[2]* 
Sqrt[a]*(I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d
 

3.1.71.3.1 Defintions of rubi rules used

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, 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 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

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 4082

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

rule 4083

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)/(b*c + a*d)   Int[(a + b*Tan[e + f*x])^m, x], x] - Simp[(B*c - A 
*d)/(b*c + a*d)   Int[(a + b*Tan[e + f*x])^m*((a - b*Tan[e + f*x])/(c + d*T 
an[e + f*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] && NeQ[A*b + a*B, 0]
 

3.1.71.4 Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(\frac {2 a \left (-\frac {\left (i B -A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2 \sqrt {a}}-\frac {A \,\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{\sqrt {a}}\right )}{d}\)	\(72\)
default	\(\frac {2 a \left (-\frac {\left (i B -A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2 \sqrt {a}}-\frac {A \,\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{\sqrt {a}}\right )}{d}\)	\(72\)

input

int(cot(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x,method=_RETURNV 
ERBOSE)
 

output

2/d*a*(-1/2*(-A+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))-A/a^(1/2)*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2)))
 

3.1.71.5 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. 447 vs. \(2 (65) = 130\).

Time = 0.26 (sec) , antiderivative size = 447, normalized size of antiderivative = 5.20 \[ \int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {1}{2} \, \sqrt {2} \sqrt {\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a}{d^{2}}} \log \left (-\frac {4 \, {\left ({\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )} - {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - \frac {1}{2} \, \sqrt {2} \sqrt {\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a}{d^{2}}} \log \left (-\frac {4 \, {\left ({\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )} - {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - \frac {1}{2} \, \sqrt {\frac {A^{2} a}{d^{2}}} \log \left (\frac {16 \, {\left (3 \, A a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + A a^{2} + 2 \, \sqrt {2} {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {A^{2} a}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{A}\right ) + \frac {1}{2} \, \sqrt {\frac {A^{2} a}{d^{2}}} \log \left (\frac {16 \, {\left (3 \, A a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + A a^{2} - 2 \, \sqrt {2} {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {A^{2} a}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{A}\right ) \]

input

integrate(cot(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorith 
m="fricas")
 

output

1/2*sqrt(2)*sqrt((A^2 - 2*I*A*B - B^2)*a/d^2)*log(-4*((-I*A - B)*a*e^(I*d* 
x + I*c) - (I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt((A^2 - 2*I*A*B - B^2)*a/d^ 
2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/(I*A + B)) - 1/2*sq 
rt(2)*sqrt((A^2 - 2*I*A*B - B^2)*a/d^2)*log(-4*((-I*A - B)*a*e^(I*d*x + I* 
c) - (-I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt((A^2 - 2*I*A*B - B^2)*a/d^2)*sq 
rt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/(I*A + B)) - 1/2*sqrt(A^ 
2*a/d^2)*log(16*(3*A*a^2*e^(2*I*d*x + 2*I*c) + A*a^2 + 2*sqrt(2)*(a*d*e^(3 
*I*d*x + 3*I*c) + a*d*e^(I*d*x + I*c))*sqrt(A^2*a/d^2)*sqrt(a/(e^(2*I*d*x 
+ 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/A) + 1/2*sqrt(A^2*a/d^2)*log(16*(3*A* 
a^2*e^(2*I*d*x + 2*I*c) + A*a^2 - 2*sqrt(2)*(a*d*e^(3*I*d*x + 3*I*c) + a*d 
*e^(I*d*x + I*c))*sqrt(A^2*a/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2 
*I*d*x - 2*I*c)/A)
 

3.1.71.6 Sympy [F]

\[ \int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \left (A + B \tan {\left (c + d x \right )}\right ) \cot {\left (c + d x \right )}\, dx \]

input

integrate(cot(d*x+c)*(a+I*a*tan(d*x+c))**(1/2)*(A+B*tan(d*x+c)),x)
 

output

Integral(sqrt(I*a*(tan(c + d*x) - I))*(A + B*tan(c + d*x))*cot(c + d*x), x 
)
 

3.1.71.7 Maxima [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.31 \[ \int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=-\frac {\sqrt {2} {\left (A - i \, B\right )} \sqrt {a} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - 2 \, A \sqrt {a} \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{2 \, d} \]

input

integrate(cot(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorith 
m="maxima")
 

output

-1/2*(sqrt(2)*(A - I*B)*sqrt(a)*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))) - 2*A*sqrt(a)*lo 
g((sqrt(I*a*tan(d*x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + c) + a) + sqr 
t(a))))/d
 

3.1.71.8 Giac [F]

\[ \int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {i \, a \tan \left (d x + c\right ) + a} \cot \left (d x + c\right ) \,d x } \]

input

integrate(cot(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorith 
m="giac")
 

output

integrate((B*tan(d*x + c) + A)*sqrt(I*a*tan(d*x + c) + a)*cot(d*x + c), x)
 

3.1.71.9 Mupad [B] (verification not implemented)

Time = 8.16 (sec) , antiderivative size = 493, normalized size of antiderivative = 5.73 \[ \int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=-\frac {2\,A\,\sqrt {a}\,\mathrm {atanh}\left (\frac {16\,A^3\,a^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{16\,d\,A^3\,a^5+32{}\mathrm {i}\,d\,A^2\,B\,a^5+16\,d\,A\,B^2\,a^5}+\frac {16\,A\,B^2\,a^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{16\,d\,A^3\,a^5+32{}\mathrm {i}\,d\,A^2\,B\,a^5+16\,d\,A\,B^2\,a^5}+\frac {A^2\,B\,a^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,32{}\mathrm {i}}{16\,d\,A^3\,a^5+32{}\mathrm {i}\,d\,A^2\,B\,a^5+16\,d\,A\,B^2\,a^5}\right )}{d}+\frac {\sqrt {2}\,\sqrt {-a}\,\mathrm {atan}\left (\frac {4\,\sqrt {2}\,A^3\,{\left (-a\right )}^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{8\,d\,A^3\,a^5+8{}\mathrm {i}\,d\,A^2\,B\,a^5+24\,d\,A\,B^2\,a^5-8{}\mathrm {i}\,d\,B^3\,a^5}-\frac {\sqrt {2}\,B^3\,{\left (-a\right )}^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,4{}\mathrm {i}}{8\,d\,A^3\,a^5+8{}\mathrm {i}\,d\,A^2\,B\,a^5+24\,d\,A\,B^2\,a^5-8{}\mathrm {i}\,d\,B^3\,a^5}+\frac {12\,\sqrt {2}\,A\,B^2\,{\left (-a\right )}^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{8\,d\,A^3\,a^5+8{}\mathrm {i}\,d\,A^2\,B\,a^5+24\,d\,A\,B^2\,a^5-8{}\mathrm {i}\,d\,B^3\,a^5}+\frac {\sqrt {2}\,A^2\,B\,{\left (-a\right )}^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,4{}\mathrm {i}}{8\,d\,A^3\,a^5+8{}\mathrm {i}\,d\,A^2\,B\,a^5+24\,d\,A\,B^2\,a^5-8{}\mathrm {i}\,d\,B^3\,a^5}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{d} \]

input

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

output

(2^(1/2)*(-a)^(1/2)*atan((4*2^(1/2)*A^3*(-a)^(9/2)*d*(a + a*tan(c + d*x)*1 
i)^(1/2))/(8*A^3*a^5*d - B^3*a^5*d*8i + 24*A*B^2*a^5*d + A^2*B*a^5*d*8i) - 
 (2^(1/2)*B^3*(-a)^(9/2)*d*(a + a*tan(c + d*x)*1i)^(1/2)*4i)/(8*A^3*a^5*d 
- B^3*a^5*d*8i + 24*A*B^2*a^5*d + A^2*B*a^5*d*8i) + (12*2^(1/2)*A*B^2*(-a) 
^(9/2)*d*(a + a*tan(c + d*x)*1i)^(1/2))/(8*A^3*a^5*d - B^3*a^5*d*8i + 24*A 
*B^2*a^5*d + A^2*B*a^5*d*8i) + (2^(1/2)*A^2*B*(-a)^(9/2)*d*(a + a*tan(c + 
d*x)*1i)^(1/2)*4i)/(8*A^3*a^5*d - B^3*a^5*d*8i + 24*A*B^2*a^5*d + A^2*B*a^ 
5*d*8i))*(A*1i + B)*1i)/d - (2*A*a^(1/2)*atanh((16*A^3*a^(9/2)*d*(a + a*ta 
n(c + d*x)*1i)^(1/2))/(16*A^3*a^5*d + 16*A*B^2*a^5*d + A^2*B*a^5*d*32i) + 
(16*A*B^2*a^(9/2)*d*(a + a*tan(c + d*x)*1i)^(1/2))/(16*A^3*a^5*d + 16*A*B^ 
2*a^5*d + A^2*B*a^5*d*32i) + (A^2*B*a^(9/2)*d*(a + a*tan(c + d*x)*1i)^(1/2 
)*32i)/(16*A^3*a^5*d + 16*A*B^2*a^5*d + A^2*B*a^5*d*32i)))/d