\(\int \frac {\sqrt {\tan (c+d x)}}{\sqrt {-2+3 \tan (c+d x)}} \, dx\) [664]

Optimal result

Integrand size = 25, antiderivative size = 95 \[ \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {-2+3 \tan (c+d x)}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {3-2 i} \sqrt {\tan (c+d x)}}{\sqrt {-2+3 \tan (c+d x)}}\right )}{\sqrt {3-2 i} d}+\frac {i \text {arctanh}\left (\frac {\sqrt {3+2 i} \sqrt {\tan (c+d x)}}{\sqrt {-2+3 \tan (c+d x)}}\right )}{\sqrt {3+2 i} d} \] Output:

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {-2+3 \tan (c+d x)}} \, dx=-\frac {i \arctan \left (\frac {\sqrt {-3+2 i} \sqrt {\tan (c+d x)}}{\sqrt {-2+3 \tan (c+d x)}}\right )}{\sqrt {-3+2 i} d}+\frac {i \text {arctanh}\left (\frac {\sqrt {3+2 i} \sqrt {\tan (c+d x)}}{\sqrt {-2+3 \tan (c+d x)}}\right )}{\sqrt {3+2 i} d} \] Input:

Integrate[Sqrt[Tan[c + d*x]]/Sqrt[-2 + 3*Tan[c + d*x]],x]
 

Output:

((-I)*ArcTan[(Sqrt[-3 + 2*I]*Sqrt[Tan[c + d*x]])/Sqrt[-2 + 3*Tan[c + d*x]] 
])/(Sqrt[-3 + 2*I]*d) + (I*ArcTanh[(Sqrt[3 + 2*I]*Sqrt[Tan[c + d*x]])/Sqrt 
[-2 + 3*Tan[c + d*x]]])/(Sqrt[3 + 2*I]*d)
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4058, 613, 104, 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.

\(\displaystyle \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {3 \tan (c+d x)-2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {3 \tan (c+d x)-2}}dx\)

\(\Big \downarrow \) 4058

\(\displaystyle \frac {\int \frac {\sqrt {\tan (c+d x)}}{\sqrt {3 \tan (c+d x)-2} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}\)

\(\Big \downarrow \) 613

\(\displaystyle \frac {\frac {1}{2} \int \frac {1}{\sqrt {\tan (c+d x)} (\tan (c+d x)+i) \sqrt {3 \tan (c+d x)-2}}d\tan (c+d x)-\frac {1}{2} \int \frac {1}{(i-\tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {3 \tan (c+d x)-2}}d\tan (c+d x)}{d}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {\int \frac {1}{i-\frac {(2+3 i) \tan (c+d x)}{3 \tan (c+d x)-2}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {3 \tan (c+d x)-2}}-\int \frac {1}{\frac {(2-3 i) \tan (c+d x)}{3 \tan (c+d x)-2}+i}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {3 \tan (c+d x)-2}}}{d}\)

\(\Big \downarrow \) 221

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

Input:

Int[Sqrt[Tan[c + d*x]]/Sqrt[-2 + 3*Tan[c + d*x]],x]
 

Output:

(((-I)*ArcTanh[(Sqrt[3 - 2*I]*Sqrt[Tan[c + d*x]])/Sqrt[-2 + 3*Tan[c + d*x] 
]])/Sqrt[3 - 2*I] + (I*ArcTanh[(Sqrt[3 + 2*I]*Sqrt[Tan[c + d*x]])/Sqrt[-2 
+ 3*Tan[c + d*x]]])/Sqrt[3 + 2*I])/d
 

Defintions of rubi rules used

rule 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*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 613

Int[Sqrt[(e_.)*(x_)]/(Sqrt[(c_) + (d_.)*(x_)]*((a_) + (b_.)*(x_)^2)), x_Sym 
bol] :> Simp[e/(2*b)   Int[1/(Sqrt[e*x]*Sqrt[c + d*x]*(Rt[-a/b, 2] + x)), x 
], x] - Simp[e/(2*b)   Int[1/(Sqrt[e*x]*Sqrt[c + d*x]*(Rt[-a/b, 2] - x)), x 
], x] /; FreeQ[{a, b, c, d, e}, x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4058

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, S 
imp[ff/f   Subst[Int[(a + b*ff*x)^m*((c + d*ff*x)^n/(1 + ff^2*x^2)), x], x, 
 Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a 
*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(478\) vs. \(2(77)=154\).

Time = 0.45 (sec) , antiderivative size = 479, normalized size of antiderivative = 5.04

method	result	size
derivativedivides	\(-\frac {\sqrt {\frac {\tan \left (d x +c \right ) \left (-2+3 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-3-2 \tan \left (d x +c \right )\right )^{2}}}\, \left (\sqrt {13}-3-2 \tan \left (d x +c \right )\right ) \left (\sqrt {13}\, \sqrt {2 \sqrt {13}+6}\, \sqrt {-6+2 \sqrt {13}}\, \arctan \left (\frac {\sqrt {-6+2 \sqrt {13}}\, \sqrt {\frac {\left (11 \sqrt {13}-39\right ) \tan \left (d x +c \right ) \left (39+11 \sqrt {13}\right ) \left (-2+3 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-3-2 \tan \left (d x +c \right )\right )^{2}}}\, \left (3 \sqrt {13}+11\right ) \left (\sqrt {13}+3+2 \tan \left (d x +c \right )\right ) \left (11 \sqrt {13}-39\right ) \left (\sqrt {13}-3-2 \tan \left (d x +c \right )\right )}{416 \tan \left (d x +c \right ) \left (-2+3 \tan \left (d x +c \right )\right )}\right )-3 \sqrt {2 \sqrt {13}+6}\, \sqrt {-6+2 \sqrt {13}}\, \arctan \left (\frac {\sqrt {-6+2 \sqrt {13}}\, \sqrt {\frac {\left (11 \sqrt {13}-39\right ) \tan \left (d x +c \right ) \left (39+11 \sqrt {13}\right ) \left (-2+3 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-3-2 \tan \left (d x +c \right )\right )^{2}}}\, \left (3 \sqrt {13}+11\right ) \left (\sqrt {13}+3+2 \tan \left (d x +c \right )\right ) \left (11 \sqrt {13}-39\right ) \left (\sqrt {13}-3-2 \tan \left (d x +c \right )\right )}{416 \tan \left (d x +c \right ) \left (-2+3 \tan \left (d x +c \right )\right )}\right )-12 \,\operatorname {arctanh}\left (\frac {4 \sqrt {13}\, \sqrt {\frac {\tan \left (d x +c \right ) \left (-2+3 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-3-2 \tan \left (d x +c \right )\right )^{2}}}}{\sqrt {26 \sqrt {13}+78}}\right ) \sqrt {13}+44 \,\operatorname {arctanh}\left (\frac {4 \sqrt {13}\, \sqrt {\frac {\tan \left (d x +c \right ) \left (-2+3 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-3-2 \tan \left (d x +c \right )\right )^{2}}}}{\sqrt {26 \sqrt {13}+78}}\right )\right )}{2 d \sqrt {\tan \left (d x +c \right )}\, \sqrt {-2+3 \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {13}+6}\, \left (11 \sqrt {13}-39\right )}\)	\(479\)
default	\(-\frac {\sqrt {\frac {\tan \left (d x +c \right ) \left (-2+3 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-3-2 \tan \left (d x +c \right )\right )^{2}}}\, \left (\sqrt {13}-3-2 \tan \left (d x +c \right )\right ) \left (\sqrt {13}\, \sqrt {2 \sqrt {13}+6}\, \sqrt {-6+2 \sqrt {13}}\, \arctan \left (\frac {\sqrt {-6+2 \sqrt {13}}\, \sqrt {\frac {\left (11 \sqrt {13}-39\right ) \tan \left (d x +c \right ) \left (39+11 \sqrt {13}\right ) \left (-2+3 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-3-2 \tan \left (d x +c \right )\right )^{2}}}\, \left (3 \sqrt {13}+11\right ) \left (\sqrt {13}+3+2 \tan \left (d x +c \right )\right ) \left (11 \sqrt {13}-39\right ) \left (\sqrt {13}-3-2 \tan \left (d x +c \right )\right )}{416 \tan \left (d x +c \right ) \left (-2+3 \tan \left (d x +c \right )\right )}\right )-3 \sqrt {2 \sqrt {13}+6}\, \sqrt {-6+2 \sqrt {13}}\, \arctan \left (\frac {\sqrt {-6+2 \sqrt {13}}\, \sqrt {\frac {\left (11 \sqrt {13}-39\right ) \tan \left (d x +c \right ) \left (39+11 \sqrt {13}\right ) \left (-2+3 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-3-2 \tan \left (d x +c \right )\right )^{2}}}\, \left (3 \sqrt {13}+11\right ) \left (\sqrt {13}+3+2 \tan \left (d x +c \right )\right ) \left (11 \sqrt {13}-39\right ) \left (\sqrt {13}-3-2 \tan \left (d x +c \right )\right )}{416 \tan \left (d x +c \right ) \left (-2+3 \tan \left (d x +c \right )\right )}\right )-12 \,\operatorname {arctanh}\left (\frac {4 \sqrt {13}\, \sqrt {\frac {\tan \left (d x +c \right ) \left (-2+3 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-3-2 \tan \left (d x +c \right )\right )^{2}}}}{\sqrt {26 \sqrt {13}+78}}\right ) \sqrt {13}+44 \,\operatorname {arctanh}\left (\frac {4 \sqrt {13}\, \sqrt {\frac {\tan \left (d x +c \right ) \left (-2+3 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-3-2 \tan \left (d x +c \right )\right )^{2}}}}{\sqrt {26 \sqrt {13}+78}}\right )\right )}{2 d \sqrt {\tan \left (d x +c \right )}\, \sqrt {-2+3 \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {13}+6}\, \left (11 \sqrt {13}-39\right )}\)	\(479\)

Input:

int(tan(d*x+c)^(1/2)/(-2+3*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/2/d*(tan(d*x+c)*(-2+3*tan(d*x+c))/(13^(1/2)-3-2*tan(d*x+c))^2)^(1/2)*(1 
3^(1/2)-3-2*tan(d*x+c))*(13^(1/2)*(2*13^(1/2)+6)^(1/2)*(-6+2*13^(1/2))^(1/ 
2)*arctan(1/416*(-6+2*13^(1/2))^(1/2)*((11*13^(1/2)-39)*tan(d*x+c)*(39+11* 
13^(1/2))*(-2+3*tan(d*x+c))/(13^(1/2)-3-2*tan(d*x+c))^2)^(1/2)*(3*13^(1/2) 
+11)*(13^(1/2)+3+2*tan(d*x+c))*(11*13^(1/2)-39)*(13^(1/2)-3-2*tan(d*x+c))/ 
tan(d*x+c)/(-2+3*tan(d*x+c)))-3*(2*13^(1/2)+6)^(1/2)*(-6+2*13^(1/2))^(1/2) 
*arctan(1/416*(-6+2*13^(1/2))^(1/2)*((11*13^(1/2)-39)*tan(d*x+c)*(39+11*13 
^(1/2))*(-2+3*tan(d*x+c))/(13^(1/2)-3-2*tan(d*x+c))^2)^(1/2)*(3*13^(1/2)+1 
1)*(13^(1/2)+3+2*tan(d*x+c))*(11*13^(1/2)-39)*(13^(1/2)-3-2*tan(d*x+c))/ta 
n(d*x+c)/(-2+3*tan(d*x+c)))-12*arctanh(4*13^(1/2)*(tan(d*x+c)*(-2+3*tan(d* 
x+c))/(13^(1/2)-3-2*tan(d*x+c))^2)^(1/2)/(26*13^(1/2)+78)^(1/2))*13^(1/2)+ 
44*arctanh(4*13^(1/2)*(tan(d*x+c)*(-2+3*tan(d*x+c))/(13^(1/2)-3-2*tan(d*x+ 
c))^2)^(1/2)/(26*13^(1/2)+78)^(1/2)))/tan(d*x+c)^(1/2)/(-2+3*tan(d*x+c))^( 
1/2)/(2*13^(1/2)+6)^(1/2)/(11*13^(1/2)-39)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1477 vs. \(2 (67) = 134\).

Time = 0.16 (sec) , antiderivative size = 1477, normalized size of antiderivative = 15.55 \[ \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {-2+3 \tan (c+d x)}} \, dx=\text {Too large to display} \] Input:

integrate(tan(d*x+c)^(1/2)/(-2+3*tan(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

-1/8*sqrt(1/13)*sqrt(-(2*d^2*sqrt(-1/d^4) + 3)/d^2)*log((sqrt(1/13)*(155*d 
*tan(d*x + c)^2 + 102*d*tan(d*x + c) + (135*d^3*tan(d*x + c)^2 - 211*d^3*t 
an(d*x + c) + 33*d^3)*sqrt(-1/d^4) - 56*d)*sqrt(-(2*d^2*sqrt(-1/d^4) + 3)/ 
d^2) + ((33*d^2*tan(d*x + c) + 56*d^2)*sqrt(-1/d^4) - 56*tan(d*x + c) + 33 
)*sqrt(3*tan(d*x + c) - 2)*sqrt(tan(d*x + c)))/(tan(d*x + c)^2 + 1)) - 1/8 
*sqrt(1/13)*sqrt(-(2*d^2*sqrt(-1/d^4) + 3)/d^2)*log(-(sqrt(1/13)*(155*d*ta 
n(d*x + c)^2 + 102*d*tan(d*x + c) + (135*d^3*tan(d*x + c)^2 - 211*d^3*tan( 
d*x + c) + 33*d^3)*sqrt(-1/d^4) - 56*d)*sqrt(-(2*d^2*sqrt(-1/d^4) + 3)/d^2 
) + ((33*d^2*tan(d*x + c) + 56*d^2)*sqrt(-1/d^4) - 56*tan(d*x + c) + 33)*s 
qrt(3*tan(d*x + c) - 2)*sqrt(tan(d*x + c)))/(tan(d*x + c)^2 + 1)) + 1/8*sq 
rt(1/13)*sqrt(-(2*d^2*sqrt(-1/d^4) + 3)/d^2)*log((sqrt(1/13)*(155*d*tan(d* 
x + c)^2 + 102*d*tan(d*x + c) + (135*d^3*tan(d*x + c)^2 - 211*d^3*tan(d*x 
+ c) + 33*d^3)*sqrt(-1/d^4) - 56*d)*sqrt(-(2*d^2*sqrt(-1/d^4) + 3)/d^2) - 
((33*d^2*tan(d*x + c) + 56*d^2)*sqrt(-1/d^4) - 56*tan(d*x + c) + 33)*sqrt( 
3*tan(d*x + c) - 2)*sqrt(tan(d*x + c)))/(tan(d*x + c)^2 + 1)) + 1/8*sqrt(1 
/13)*sqrt(-(2*d^2*sqrt(-1/d^4) + 3)/d^2)*log(-(sqrt(1/13)*(155*d*tan(d*x + 
 c)^2 + 102*d*tan(d*x + c) + (135*d^3*tan(d*x + c)^2 - 211*d^3*tan(d*x + c 
) + 33*d^3)*sqrt(-1/d^4) - 56*d)*sqrt(-(2*d^2*sqrt(-1/d^4) + 3)/d^2) - ((3 
3*d^2*tan(d*x + c) + 56*d^2)*sqrt(-1/d^4) - 56*tan(d*x + c) + 33)*sqrt(3*t 
an(d*x + c) - 2)*sqrt(tan(d*x + c)))/(tan(d*x + c)^2 + 1)) + 1/8*sqrt(1...
 

Sympy [F]

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

integrate(tan(d*x+c)**(1/2)/(-2+3*tan(d*x+c))**(1/2),x)
 

Output:

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

Maxima [F]

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

integrate(tan(d*x+c)^(1/2)/(-2+3*tan(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(tan(d*x + c))/sqrt(3*tan(d*x + c) - 2), x)
 

Giac [F(-2)]

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

integrate(tan(d*x+c)^(1/2)/(-2+3*tan(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to find common minimal polyn 
omial Error: Bad Argument ValueDone
 

Mupad [B] (verification not implemented)

Time = 2.94 (sec) , antiderivative size = 191, normalized size of antiderivative = 2.01 \[ \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {-2+3 \tan (c+d x)}} \, dx=-\mathrm {atan}\left (\frac {12\,d\,\sqrt {\frac {-\frac {3}{52}-\frac {1}{26}{}\mathrm {i}}{d^2}}\,\left (\frac {\sqrt {2}\,\sqrt {3}}{3}-\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{\sqrt {3\,\mathrm {tan}\left (c+d\,x\right )-2}\,\left (\frac {3\,{\left (\frac {\sqrt {2}\,\sqrt {3}}{3}-\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}^2}{3\,\mathrm {tan}\left (c+d\,x\right )-2}+1\right )}\right )\,\sqrt {\frac {-\frac {3}{52}-\frac {1}{26}{}\mathrm {i}}{d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {12\,d\,\sqrt {\frac {-\frac {3}{52}+\frac {1}{26}{}\mathrm {i}}{d^2}}\,\left (\frac {\sqrt {2}\,\sqrt {3}}{3}-\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{\sqrt {3\,\mathrm {tan}\left (c+d\,x\right )-2}\,\left (\frac {3\,{\left (\frac {\sqrt {2}\,\sqrt {3}}{3}-\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}^2}{3\,\mathrm {tan}\left (c+d\,x\right )-2}+1\right )}\right )\,\sqrt {\frac {-\frac {3}{52}+\frac {1}{26}{}\mathrm {i}}{d^2}}\,2{}\mathrm {i} \] Input:

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

Output:

atan((12*d*((- 3/52 + 1i/26)/d^2)^(1/2)*((2^(1/2)*3^(1/2))/3 - tan(c + d*x 
)^(1/2)))/((3*tan(c + d*x) - 2)^(1/2)*((3*((2^(1/2)*3^(1/2))/3 - tan(c + d 
*x)^(1/2))^2)/(3*tan(c + d*x) - 2) + 1)))*((- 3/52 + 1i/26)/d^2)^(1/2)*2i 
- atan((12*d*((- 3/52 - 1i/26)/d^2)^(1/2)*((2^(1/2)*3^(1/2))/3 - tan(c + d 
*x)^(1/2)))/((3*tan(c + d*x) - 2)^(1/2)*((3*((2^(1/2)*3^(1/2))/3 - tan(c + 
 d*x)^(1/2))^2)/(3*tan(c + d*x) - 2) + 1)))*((- 3/52 - 1i/26)/d^2)^(1/2)*2 
i
 

Reduce [F]

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

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

Output:

(sqrt(tan(c + d*x))*sqrt(3*tan(c + d*x) - 2) - 3*int((sqrt(tan(c + d*x))*s 
qrt(3*tan(c + d*x) - 2)*tan(c + d*x)**2)/(3*tan(c + d*x) - 2),x)*d + int(( 
sqrt(tan(c + d*x))*sqrt(3*tan(c + d*x) - 2)*tan(c + d*x))/(3*tan(c + d*x) 
- 2),x)*d + int((sqrt(tan(c + d*x))*sqrt(3*tan(c + d*x) - 2))/(3*tan(c + d 
*x)**2 - 2*tan(c + d*x)),x)*d)/(3*d)