\(\int \frac {\sqrt {\tan (c+d x)}}{\sqrt {2+3 \tan (c+d x)}} \, dx\) [663]

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

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

\(\Big \downarrow \) 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*T 
an[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 = 0.30 (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)*(13 
^(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*1 
3^(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)))-3*(2*13^(1/2)+6)^(1/2)*(-6+2*13^(1/2))^(1/2)*ar 
ctan(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)))-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*arc 
tanh(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*ta 
n(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) 
 + ((33*d^2*tan(d*x + c) - 56*d^2)*sqrt(-1/d^4) - 56*tan(d*x + c) - 33)*sq 
rt(3*tan(d*x + c) + 2)*sqrt(tan(d*x + c)))/(tan(d*x + c)^2 + 1)) - 1/8*sqr 
t(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) - ((33 
*d^2*tan(d*x + c) - 56*d^2)*sqrt(-1/d^4) - 56*tan(d*x + c) - 33)*sqrt(3*ta 
n(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 = 2.74 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.18 \[ \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {2+3 \tan (c+d x)}} \, dx=\mathrm {atan}\left (\frac {\sqrt {2}\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {-\frac {3}{52}-\frac {1}{26}{}\mathrm {i}}{d^2}}\,\left (6-4{}\mathrm {i}\right )+d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {-\frac {3}{52}-\frac {1}{26}{}\mathrm {i}}{d^2}}\,\sqrt {3\,\mathrm {tan}\left (c+d\,x\right )+2}\,\left (-6+4{}\mathrm {i}\right )}{3\,\mathrm {tan}\left (c+d\,x\right )-\sqrt {2}\,\sqrt {3\,\mathrm {tan}\left (c+d\,x\right )+2}+2}\right )\,\sqrt {\frac {-\frac {3}{52}-\frac {1}{26}{}\mathrm {i}}{d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {\sqrt {2}\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {-\frac {3}{52}+\frac {1}{26}{}\mathrm {i}}{d^2}}\,\left (6+4{}\mathrm {i}\right )+d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {-\frac {3}{52}+\frac {1}{26}{}\mathrm {i}}{d^2}}\,\sqrt {3\,\mathrm {tan}\left (c+d\,x\right )+2}\,\left (-6-4{}\mathrm {i}\right )}{3\,\mathrm {tan}\left (c+d\,x\right )-\sqrt {2}\,\sqrt {3\,\mathrm {tan}\left (c+d\,x\right )+2}+2}\right )\,\sqrt {\frac {-\frac {3}{52}+\frac {1}{26}{}\mathrm {i}}{d^2}}\,2{}\mathrm {i} \] Input:

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

Output:

atan((2^(1/2)*d*tan(c + d*x)^(1/2)*((- 3/52 - 1i/26)/d^2)^(1/2)*(6 - 4i) - 
 d*tan(c + d*x)^(1/2)*((- 3/52 - 1i/26)/d^2)^(1/2)*(3*tan(c + d*x) + 2)^(1 
/2)*(6 - 4i))/(3*tan(c + d*x) - 2^(1/2)*(3*tan(c + d*x) + 2)^(1/2) + 2))*( 
(- 3/52 - 1i/26)/d^2)^(1/2)*2i - atan((2^(1/2)*d*tan(c + d*x)^(1/2)*((- 3/ 
52 + 1i/26)/d^2)^(1/2)*(6 + 4i) - d*tan(c + d*x)^(1/2)*((- 3/52 + 1i/26)/d 
^2)^(1/2)*(3*tan(c + d*x) + 2)^(1/2)*(6 + 4i))/(3*tan(c + d*x) - 2^(1/2)*( 
3*tan(c + d*x) + 2)^(1/2) + 2))*((- 3/52 + 1i/26)/d^2)^(1/2)*2i
 

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)