\(\int \frac {(a+i a \tan (e+f x))^{3/2}}{(c+d \tan (e+f x))^{3/2}} \, dx\) [1162]

Optimal result

Integrand size = 32, antiderivative size = 129 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c+d \tan (e+f x))^{3/2}} \, dx=-\frac {2 i \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{(c-i d)^{3/2} f}-\frac {2 a \sqrt {a+i a \tan (e+f x)}}{(i c+d) f \sqrt {c+d \tan (e+f x)}} \] Output:

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

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(981\) vs. \(2(129)=258\).

Time = 6.56 (sec) , antiderivative size = 981, normalized size of antiderivative = 7.60 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c+d \tan (e+f x))^{3/2}} \, dx=-\frac {2 d (a+i a \tan (e+f x))^{3/2}}{\left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}+\frac {2 \left (\frac {a \left (\frac {1}{2} i a^2 (c+3 i d)+a^2 d\right ) \left (-\frac {2 \sqrt {2} \arctan \left (\frac {\sqrt {-a c+i a d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {2} a \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-a c+i a d}}-\frac {2 (-1)^{3/4} \sqrt {c+i d} \sqrt {\frac {1}{\frac {c}{c+i d}+\frac {i d}{c+i d}}} \sqrt {\frac {c}{c+i d}+\frac {i d}{c+i d}} \arcsin \left (\frac {\sqrt [4]{-1} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+i d} \sqrt {\frac {c}{c+i d}+\frac {i d}{c+i d}}}\right ) \sqrt {\frac {c+d \tan (e+f x)}{c+i d}}}{\sqrt {a} \sqrt {d} \sqrt {c+d \tan (e+f x)}}\right )}{f}-\frac {(i a c-a d)^2 \sqrt {\frac {i a}{-\frac {a^2 c}{i a c-a d}-\frac {i a^2 d}{i a c-a d}}} \left (-\frac {a^2 c}{i a c-a d}-\frac {i a^2 d}{i a c-a d}\right )^2 \sqrt {\frac {i a (c+d \tan (e+f x))}{i a c-a d}} \sqrt {1+\frac {i a d (a+i a \tan (e+f x))}{(i a c-a d) \left (-\frac {a^2 c}{i a c-a d}-\frac {i a^2 d}{i a c-a d}\right )}} \left (\frac {2 i a d (a+i a \tan (e+f x))}{(i a c-a d) \left (-\frac {a^2 c}{i a c-a d}-\frac {i a^2 d}{i a c-a d}\right )}-\frac {2 \sqrt [4]{-1} \sqrt {a} \sqrt {d} \text {arcsinh}\left (\frac {\sqrt [4]{-1} \sqrt {a} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {i a c-a d} \sqrt {-\frac {a^2 c}{i a c-a d}-\frac {i a^2 d}{i a c-a d}}}\right ) \sqrt {a+i a \tan (e+f x)}}{\sqrt {i a c-a d} \sqrt {-\frac {a^2 c}{i a c-a d}-\frac {i a^2 d}{i a c-a d}} \sqrt {1+\frac {i a d (a+i a \tan (e+f x))}{(i a c-a d) \left (-\frac {a^2 c}{i a c-a d}-\frac {i a^2 d}{i a c-a d}\right )}}}\right )}{2 a d f \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}\right )}{a \left (c^2+d^2\right )} \] Input:

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

Output:

(-2*d*(a + I*a*Tan[e + f*x])^(3/2))/((c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x] 
]) + (2*((a*((I/2)*a^2*(c + (3*I)*d) + a^2*d)*((-2*Sqrt[2]*ArcTan[(Sqrt[-( 
a*c) + I*a*d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[2]*a*Sqrt[c + d*Tan[e + f* 
x]])])/Sqrt[-(a*c) + I*a*d] - (2*(-1)^(3/4)*Sqrt[c + I*d]*Sqrt[(c/(c + I*d 
) + (I*d)/(c + I*d))^(-1)]*Sqrt[c/(c + I*d) + (I*d)/(c + I*d)]*ArcSin[((-1 
)^(1/4)*Sqrt[d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c + I*d]*Sqrt[c/ 
(c + I*d) + (I*d)/(c + I*d)])]*Sqrt[(c + d*Tan[e + f*x])/(c + I*d)])/(Sqrt 
[a]*Sqrt[d]*Sqrt[c + d*Tan[e + f*x]])))/f - ((I*a*c - a*d)^2*Sqrt[(I*a)/(- 
((a^2*c)/(I*a*c - a*d)) - (I*a^2*d)/(I*a*c - a*d))]*(-((a^2*c)/(I*a*c - a* 
d)) - (I*a^2*d)/(I*a*c - a*d))^2*Sqrt[(I*a*(c + d*Tan[e + f*x]))/(I*a*c - 
a*d)]*Sqrt[1 + (I*a*d*(a + I*a*Tan[e + f*x]))/((I*a*c - a*d)*(-((a^2*c)/(I 
*a*c - a*d)) - (I*a^2*d)/(I*a*c - a*d)))]*(((2*I)*a*d*(a + I*a*Tan[e + f*x 
]))/((I*a*c - a*d)*(-((a^2*c)/(I*a*c - a*d)) - (I*a^2*d)/(I*a*c - a*d))) - 
 (2*(-1)^(1/4)*Sqrt[a]*Sqrt[d]*ArcSinh[((-1)^(1/4)*Sqrt[a]*Sqrt[d]*Sqrt[a 
+ I*a*Tan[e + f*x]])/(Sqrt[I*a*c - a*d]*Sqrt[-((a^2*c)/(I*a*c - a*d)) - (I 
*a^2*d)/(I*a*c - a*d)])]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[I*a*c - a*d]*Sq 
rt[-((a^2*c)/(I*a*c - a*d)) - (I*a^2*d)/(I*a*c - a*d)]*Sqrt[1 + (I*a*d*(a 
+ I*a*Tan[e + f*x]))/((I*a*c - a*d)*(-((a^2*c)/(I*a*c - a*d)) - (I*a^2*d)/ 
(I*a*c - a*d)))])))/(2*a*d*f*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + 
 f*x]])))/(a*(c^2 + d^2))
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {3042, 4028, 3042, 4027, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4028

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4027

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

((-2*I)*Sqrt[2]*a^(3/2)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]]) 
/(Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e + f*x]])])/((c - I*d)^(3/2)*f) - (2*a*S 
qrt[a + I*a*Tan[e + f*x]])/((I*c + d)*f*Sqrt[c + d*Tan[e + f*x]])
 

Defintions of rubi rules used

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 4027

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

rule 4028

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2563 vs. \(2 (103 ) = 206\).

Time = 0.60 (sec) , antiderivative size = 2564, normalized size of antiderivative = 19.88

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(2564\)
default	\(\text {Expression too large to display}\)	\(2564\)

Input:

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

Output:

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

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. 551 vs. \(2 (97) = 194\).

Time = 0.10 (sec) , antiderivative size = 551, normalized size of antiderivative = 4.27 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c+d \tan (e+f x))^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeRecursive assumption sageVARc>=(-sageVARd*t_nostep) ignoredDo 
ne
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(2*sqrt(a)*a*(sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*c*i + sqrt 
(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*d + int((sqrt(tan(e + f*x)*i 
 + 1)*sqrt(tan(e + f*x)*d + c))/(tan(e + f*x)**2*d**2 + 2*tan(e + f*x)*c*d 
 + c**2),x)*tan(e + f*x)*c**2*d*f + int((sqrt(tan(e + f*x)*i + 1)*sqrt(tan 
(e + f*x)*d + c))/(tan(e + f*x)**2*d**2 + 2*tan(e + f*x)*c*d + c**2),x)*ta 
n(e + f*x)*d**3*f + int((sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c) 
)/(tan(e + f*x)**2*d**2 + 2*tan(e + f*x)*c*d + c**2),x)*c**3*f + int((sqrt 
(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c))/(tan(e + f*x)**2*d**2 + 2*t 
an(e + f*x)*c*d + c**2),x)*c*d**2*f))/(f*(tan(e + f*x)*c**2*d + tan(e + f* 
x)*d**3 + c**3 + c*d**2))