\(\int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^{3/2}} \, dx\) [1126]

Optimal result

Integrand size = 30, antiderivative size = 92 \[ \int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^{3/2}} \, dx=-\frac {4 i a^2 \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{3/2} f}+\frac {2 a^2 (i c-d)}{d (i c+d) f \sqrt {c+d \tan (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.86 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.07 \[ \int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^{3/2}} \, dx=\frac {2 a^2 \left (c^2+d^2-2 i \sqrt {c-i d} d \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right ) \sqrt {c+d \tan (e+f x)}\right )}{(c-i d)^2 d f \sqrt {c+d \tan (e+f x)}} \] Input:

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

Output:

(2*a^2*(c^2 + d^2 - (2*I)*Sqrt[c - I*d]*d*ArcTanh[Sqrt[c + d*Tan[e + f*x]] 
/Sqrt[c - I*d]]*Sqrt[c + d*Tan[e + f*x]]))/((c - I*d)^2*d*f*Sqrt[c + d*Tan 
[e + f*x]])
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.25, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3042, 4025, 27, 3042, 4020, 25, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4025

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 a^2 (-d+i c)}{d f (d+i c) \sqrt {c+d \tan (e+f x)}}-\frac {4 a^2 (-d+i c) \sqrt {c-i d} (c+i d) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f \left (c^2+d^2\right )^2}\)

Input:

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

Output:

(-4*a^2*(I*c - d)*Sqrt[c - I*d]*(c + I*d)*ArcTanh[Sqrt[c + d*Tan[e + f*x]] 
/Sqrt[c - I*d]])/((c^2 + d^2)^2*f) + (2*a^2*(I*c - d))/(d*(I*c + d)*f*Sqrt 
[c + d*Tan[e + f*x]])
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 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 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 4020

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

rule 4025

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2527 vs. \(2 (79 ) = 158\).

Time = 0.41 (sec) , antiderivative size = 2528, normalized size of antiderivative = 27.48

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(2528\)
default	\(\text {Expression too large to display}\)	\(2528\)
parts	\(\text {Expression too large to display}\)	\(5622\)

Input:

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

Output:

2*I/f*a^2*d^2/(c^2+d^2)^(3/2)/((c^2+d^2)^(1/2)+c)/(2*(c^2+d^2)^(1/2)-2*c)^ 
(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)-(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*( 
c^2+d^2)^(1/2)-2*c)^(1/2))*c+2*I/f*a^2*d^2/(c^2+d^2)^(3/2)/((c^2+d^2)^(1/2 
)+c)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^ 
2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c+2/f*a^2/d/(c^2+d 
^2)/(c+d*tan(f*x+e))^(1/2)*c^2+4*I/f*a^2/(c^2+d^2)/(c+d*tan(f*x+e))^(1/2)* 
c+2/f*a^2*d/(c^2+d^2)/((c^2+d^2)^(1/2)+c)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*ar 
ctan((2*(c+d*tan(f*x+e))^(1/2)-(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2) 
^(1/2)-2*c)^(1/2))*c+2/f*a^2*d/(c^2+d^2)/((c^2+d^2)^(1/2)+c)/(2*(c^2+d^2)^ 
(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^ 
(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c+4/f*a^2*d/(c^2+d^2)^(3/2)/((c^2+d^ 
2)^(1/2)+c)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2) 
-(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c^2+1/f*a^2 
*d/(c^2+d^2)^(3/2)/((c^2+d^2)^(1/2)+c)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^ 
(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2* 
c)^(1/2)*c+2*I/f*a^2*d^2/(c^2+d^2)/((c^2+d^2)^(1/2)+c)/(2*(c^2+d^2)^(1/2)- 
2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)-(2*(c^2+d^2)^(1/2)+2*c)^(1/2)) 
/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))+1/2*I/f*a^2*d^2/(c^2+d^2)^(3/2)/((c^2+d^2) 
^(1/2)+c)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c) 
^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-1/2*I/f*a^2*d^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. 584 vs. \(2 (74) = 148\).

Time = 0.11 (sec) , antiderivative size = 584, normalized size of antiderivative = 6.35 \[ \int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

1/4*(((c^2*d - 2*I*c*d^2 - d^3)*f*e^(2*I*f*x + 2*I*e) + (c^2*d + d^3)*f)*s 
qrt(16*I*a^4/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*log(1/2*(4*a^2*c 
+ ((I*c^2 + 2*c*d - I*d^2)*f*e^(2*I*f*x + 2*I*e) + (I*c^2 + 2*c*d - I*d^2) 
*f)*sqrt(16*I*a^4/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*sqrt(((c - I 
*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1)) + 4*(a^2*c - 
 I*a^2*d)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a^2) - ((c^2*d - 2*I*c 
*d^2 - d^3)*f*e^(2*I*f*x + 2*I*e) + (c^2*d + d^3)*f)*sqrt(16*I*a^4/((-I*c^ 
3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*log(1/2*(4*a^2*c + ((-I*c^2 - 2*c*d + 
 I*d^2)*f*e^(2*I*f*x + 2*I*e) + (-I*c^2 - 2*c*d + I*d^2)*f)*sqrt(16*I*a^4/ 
((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*sqrt(((c - I*d)*e^(2*I*f*x + 2 
*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1)) + 4*(a^2*c - I*a^2*d)*e^(2*I*f 
*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a^2) + 8*(a^2*c + I*a^2*d + (a^2*c + I*a 
^2*d)*e^(2*I*f*x + 2*I*e))*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/ 
(e^(2*I*f*x + 2*I*e) + 1)))/((c^2*d - 2*I*c*d^2 - d^3)*f*e^(2*I*f*x + 2*I* 
e) + (c^2*d + d^3)*f)
 

Sympy [F]

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

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

Output:

-a**2*(Integral(tan(e + f*x)**2/(c*sqrt(c + d*tan(e + f*x)) + d*sqrt(c + d 
*tan(e + f*x))*tan(e + f*x)), x) + Integral(-2*I*tan(e + f*x)/(c*sqrt(c + 
d*tan(e + f*x)) + d*sqrt(c + d*tan(e + f*x))*tan(e + f*x)), x) + Integral( 
-1/(c*sqrt(c + d*tan(e + f*x)) + d*sqrt(c + d*tan(e + f*x))*tan(e + f*x)), 
 x))
 

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 203 vs. \(2 (74) = 148\).

Time = 0.57 (sec) , antiderivative size = 203, normalized size of antiderivative = 2.21 \[ \int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^{3/2}} \, dx=\frac {2 \, a^{2} {\left (\frac {c + i \, d}{{\left (c d - i \, d^{2}\right )} \sqrt {d \tan \left (f x + e\right ) + c}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{\sqrt {2} c \sqrt {-c + \sqrt {c^{2} + d^{2}}} - i \, \sqrt {2} \sqrt {-c + \sqrt {c^{2} + d^{2}}} d - \sqrt {2} \sqrt {c^{2} + d^{2}} \sqrt {-c + \sqrt {c^{2} + d^{2}}}}\right )}{{\left (-i \, c - d\right )} \sqrt {-c + \sqrt {c^{2} + d^{2}}} {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}}\right )}}{f} \] Input:

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

Output:

2*a^2*((c + I*d)/((c*d - I*d^2)*sqrt(d*tan(f*x + e) + c)) + 2*sqrt(2)*arct 
an(2*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c 
))/(sqrt(2)*c*sqrt(-c + sqrt(c^2 + d^2)) - I*sqrt(2)*sqrt(-c + sqrt(c^2 + 
d^2))*d - sqrt(2)*sqrt(c^2 + d^2)*sqrt(-c + sqrt(c^2 + d^2))))/((-I*c - d) 
*sqrt(-c + sqrt(c^2 + d^2))*(-I*d/(c - sqrt(c^2 + d^2)) + 1)))/f
 

Mupad [B] (verification not implemented)

Time = 3.54 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.54 \[ \int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^{3/2}} \, dx=\frac {a^2\,\mathrm {atan}\left (\frac {\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,\left (2\,c^4\,f^2+4\,c^2\,d^2\,f^2+2\,d^4\,f^2\right )}{2\,f\,{\left (-c+d\,1{}\mathrm {i}\right )}^{3/2}\,\left (f\,c^3+1{}\mathrm {i}\,f\,c^2\,d+f\,c\,d^2+1{}\mathrm {i}\,f\,d^3\right )}\right )\,4{}\mathrm {i}}{f\,{\left (-c+d\,1{}\mathrm {i}\right )}^{3/2}}+\frac {2\,a^2\,\left (c+d\,1{}\mathrm {i}\right )}{d\,f\,\left (c-d\,1{}\mathrm {i}\right )\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}} \] Input:

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

Output:

(a^2*atan(((c + d*tan(e + f*x))^(1/2)*(2*c^4*f^2 + 2*d^4*f^2 + 4*c^2*d^2*f 
^2))/(2*f*(d*1i - c)^(3/2)*(c^3*f + d^3*f*1i + c*d^2*f + c^2*d*f*1i)))*4i) 
/(f*(d*1i - c)^(3/2)) + (2*a^2*(c + d*1i))/(d*f*(c - d*1i)*(c + d*tan(e + 
f*x))^(1/2))
 

Reduce [F]

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

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

Output:

(2*a**2*( - sqrt(tan(e + f*x)*d + c) - int((sqrt(tan(e + f*x)*d + c)*tan(e 
 + f*x)**2)/(tan(e + f*x)**2*d**2 + 2*tan(e + f*x)*c*d + c**2),x)*tan(e + 
f*x)*d**2*f - int((sqrt(tan(e + f*x)*d + c)*tan(e + f*x)**2)/(tan(e + f*x) 
**2*d**2 + 2*tan(e + f*x)*c*d + c**2),x)*c*d*f + int((sqrt(tan(e + f*x)*d 
+ c)*tan(e + f*x))/(tan(e + f*x)**2*d**2 + 2*tan(e + f*x)*c*d + c**2),x)*t 
an(e + f*x)*d**2*f*i + int((sqrt(tan(e + f*x)*d + c)*tan(e + f*x))/(tan(e 
+ f*x)**2*d**2 + 2*tan(e + f*x)*c*d + c**2),x)*c*d*f*i))/(d*f*(tan(e + f*x 
)*d + c))