\(\int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\) [119]

Optimal result

Integrand size = 26, antiderivative size = 133 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {\tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {11}{6 a d \sqrt {a+i a \tan (c+d x)}}-\frac {7 \sqrt {a+i a \tan (c+d x)}}{3 a^2 d} \] 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)^2/d/(a+I*a*tan(d*x+c))^(3/2)-11/6/a/d/(a+I*a*tan(d*x+c))^ 
(1/2)-7/3*(a+I*a*tan(d*x+c))^(1/2)/a^2/d
 

Mathematica [A] (verified)

Time = 0.83 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.79 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {3 \sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )+\frac {2 \sqrt {a+i a \tan (c+d x)} \left (25+39 i \tan (c+d x)-12 \tan ^2(c+d x)\right )}{(-i+\tan (c+d x))^2}}{12 a^2 d} \] Input:

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

Output:

(3*Sqrt[2]*Sqrt[a]*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])] + 
 (2*Sqrt[a + I*a*Tan[c + d*x]]*(25 + (39*I)*Tan[c + d*x] - 12*Tan[c + d*x] 
^2))/(-I + Tan[c + d*x])^2)/(12*a^2*d)
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.98, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3042, 4041, 27, 3042, 4075, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4041

\(\displaystyle -\frac {\int -\frac {\tan (c+d x) (4 a-7 i a \tan (c+d x))}{2 \sqrt {i \tan (c+d x) a+a}}dx}{3 a^2}-\frac {\tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\tan (c+d x) (4 a-7 i a \tan (c+d x))}{\sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}-\frac {\tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\tan (c+d x) (4 a-7 i a \tan (c+d x))}{\sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}-\frac {\tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4075

\(\displaystyle \frac {\int \frac {4 \tan (c+d x) a+7 i a}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {14 \sqrt {a+i a \tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {4 \tan (c+d x) a+7 i a}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {14 \sqrt {a+i a \tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4009

\(\displaystyle \frac {\frac {3}{2} i \int \sqrt {i \tan (c+d x) a+a}dx-\frac {11 a}{d \sqrt {a+i a \tan (c+d x)}}-\frac {14 \sqrt {a+i a \tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{2} i \int \sqrt {i \tan (c+d x) a+a}dx-\frac {11 a}{d \sqrt {a+i a \tan (c+d x)}}-\frac {14 \sqrt {a+i a \tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3961

\(\displaystyle \frac {\frac {3 a \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}-\frac {11 a}{d \sqrt {a+i a \tan (c+d x)}}-\frac {14 \sqrt {a+i a \tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {3 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} d}-\frac {11 a}{d \sqrt {a+i a \tan (c+d x)}}-\frac {14 \sqrt {a+i a \tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

Input:

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

Output:

-1/3*Tan[c + d*x]^2/(d*(a + I*a*Tan[c + d*x])^(3/2)) + ((3*Sqrt[a]*ArcTanh 
[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*d) - (11*a)/(d*Sq 
rt[a + I*a*Tan[c + d*x]]) - (14*Sqrt[a + I*a*Tan[c + d*x]])/d)/(6*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 4041

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])
 

rule 4075

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]
 

Maple [A] (verified)

Time = 1.36 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.70

method	result	size
derivativedivides	\(\frac {-2 \sqrt {a +i a \tan \left (d x +c \right )}+\frac {\sqrt {a}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4}-\frac {5 a}{2 \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {a^{2}}{3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}}{a^{2} d}\)	\(93\)
default	\(\frac {-2 \sqrt {a +i a \tan \left (d x +c \right )}+\frac {\sqrt {a}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4}-\frac {5 a}{2 \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {a^{2}}{3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}}{a^{2} d}\)	\(93\)

Input:

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

Output:

2/d/a^2*(-(a+I*a*tan(d*x+c))^(1/2)+1/8*a^(1/2)*2^(1/2)*arctanh(1/2*(a+I*a* 
tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))-5/4*a/(a+I*a*tan(d*x+c))^(1/2)+1/6*a^2/ 
(a+I*a*tan(d*x+c))^(3/2))
 

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. 273 vs. \(2 (102) = 204\).

Time = 0.11 (sec) , antiderivative size = 273, normalized size of antiderivative = 2.05 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {1}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {1}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (38 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 13 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{12 \, a^{2} d} \] Input:

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

Output:

1/12*(3*sqrt(1/2)*a^2*d*sqrt(1/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*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)) - 3*sqrt 
(1/2)*a^2*d*sqrt(1/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*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))*sq 
rt(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))*(38*e^(4*I*d*x + 4*I*c) + 13*e^(2*I*d*x + 2*I*c) 
- 1))*e^(-3*I*d*x - 3*I*c)/(a^2*d)
 

Sympy [F]

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

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

Output:

Integral(tan(c + d*x)**3/(I*a*(tan(c + d*x) - I))**(3/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.91 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {3 \, \sqrt {2} a^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + 48 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{2} + \frac {4 \, {\left (15 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{3} - 2 \, a^{4}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}}{24 \, a^{4} d} \] Input:

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

Output:

-1/24*(3*sqrt(2)*a^(5/2)*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*sqrt(I*a*tan(d*x + 
 c) + a)*a^2 + 4*(15*(I*a*tan(d*x + c) + a)*a^3 - 2*a^4)/(I*a*tan(d*x + c) 
 + a)^(3/2))/(a^4*d)
 

Giac [F(-2)]

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

integrate(tan(d*x+c)^3/(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 TypeDone
 

Mupad [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.70 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {2\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a^2\,d}+\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}}\right )}{4\,a^{3/2}\,d}-\frac {\frac {13\,a}{6}+\frac {a\,\mathrm {tan}\left (c+d\,x\right )\,5{}\mathrm {i}}{2}}{a\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \] Input:

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

Output:

(2^(1/2)*atanh((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(2*a^(1/2))))/(4*a^ 
(3/2)*d) - (2*(a + a*tan(c + d*x)*1i)^(1/2))/(a^2*d) - ((13*a)/6 + (a*tan( 
c + d*x)*5i)/2)/(a*d*(a + a*tan(c + d*x)*1i)^(3/2))
 

Reduce [F]

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

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

Output:

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