\(\int \frac {1}{\sqrt {3-2 \tan (c+d x)} \sqrt {\tan (c+d x)}} \, dx\) [660]

Optimal result

Integrand size = 25, antiderivative size = 89 \[ \int \frac {1}{\sqrt {3-2 \tan (c+d x)} \sqrt {\tan (c+d x)}} \, dx=\frac {\arctan \left (\frac {\sqrt {2-3 i} \sqrt {\tan (c+d x)}}{\sqrt {3-2 \tan (c+d x)}}\right )}{\sqrt {2-3 i} d}+\frac {\arctan \left (\frac {\sqrt {2+3 i} \sqrt {\tan (c+d x)}}{\sqrt {3-2 \tan (c+d x)}}\right )}{\sqrt {2+3 i} d} \] Output:

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.13 \[ \int \frac {1}{\sqrt {3-2 \tan (c+d x)} \sqrt {\tan (c+d x)}} \, dx=\frac {-\sqrt {2+3 i} \arctan \left (\frac {\sqrt {\frac {2}{13}+\frac {3 i}{13}} \sqrt {3-2 \tan (c+d x)}}{\sqrt {\tan (c+d x)}}\right )+\sqrt {-2+3 i} \text {arctanh}\left (\frac {\sqrt {-\frac {2}{13}+\frac {3 i}{13}} \sqrt {3-2 \tan (c+d x)}}{\sqrt {\tan (c+d x)}}\right )}{\sqrt {13} d} \] Input:

Integrate[1/(Sqrt[3 - 2*Tan[c + d*x]]*Sqrt[Tan[c + d*x]]),x]
 

Output:

(-(Sqrt[2 + 3*I]*ArcTan[(Sqrt[2/13 + (3*I)/13]*Sqrt[3 - 2*Tan[c + d*x]])/S 
qrt[Tan[c + d*x]]]) + Sqrt[-2 + 3*I]*ArcTanh[(Sqrt[-2/13 + (3*I)/13]*Sqrt[ 
3 - 2*Tan[c + d*x]])/Sqrt[Tan[c + d*x]]])/(Sqrt[13]*d)
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3042, 4058, 615, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4058

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

\(\Big \downarrow \) 615

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

\(\Big \downarrow \) 2009

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

Input:

Int[1/(Sqrt[3 - 2*Tan[c + d*x]]*Sqrt[Tan[c + d*x]]),x]
 

Output:

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

Defintions of rubi rules used

rule 615

Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), 
 x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] 
 /; FreeQ[{a, b, c, d, e, m, n}, x] && ILtQ[p, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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. \(426\) vs. \(2(73)=146\).

Time = 2.25 (sec) , antiderivative size = 427, normalized size of antiderivative = 4.80

method	result	size
derivativedivides	\(-\frac {\left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right ) \left (4 \sqrt {-4+2 \sqrt {13}}\, \sqrt {13}\, \sqrt {2 \sqrt {13}+4}\, \operatorname {arctanh}\left (\frac {\sqrt {-4+2 \sqrt {13}}\, \left (4 \sqrt {13}+17\right ) \left (\sqrt {13}+2+3 \tan \left (d x +c \right )\right ) \left (17 \sqrt {13}-52\right ) \sqrt {13}}{6318 \left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right ) \sqrt {-\frac {\tan \left (d x +c \right ) \left (-3+2 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right )^{2}}}}\right )-17 \sqrt {-4+2 \sqrt {13}}\, \sqrt {2 \sqrt {13}+4}\, \operatorname {arctanh}\left (\frac {\sqrt {-4+2 \sqrt {13}}\, \left (4 \sqrt {13}+17\right ) \left (\sqrt {13}+2+3 \tan \left (d x +c \right )\right ) \left (17 \sqrt {13}-52\right ) \sqrt {13}}{6318 \left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right ) \sqrt {-\frac {\tan \left (d x +c \right ) \left (-3+2 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right )^{2}}}}\right )-18 \arctan \left (\frac {6 \sqrt {13}\, \sqrt {-\frac {\tan \left (d x +c \right ) \left (-3+2 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right )^{2}}}}{\sqrt {26 \sqrt {13}+52}}\right ) \sqrt {13}+36 \arctan \left (\frac {6 \sqrt {13}\, \sqrt {-\frac {\tan \left (d x +c \right ) \left (-3+2 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right )^{2}}}}{\sqrt {26 \sqrt {13}+52}}\right )\right ) \sqrt {-\frac {\tan \left (d x +c \right ) \left (-3+2 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right )^{2}}}}{2 d \sqrt {2 \sqrt {13}+4}\, \sqrt {3-2 \tan \left (d x +c \right )}\, \left (17 \sqrt {13}-52\right ) \sqrt {\tan \left (d x +c \right )}}\)	\(427\)
default	\(-\frac {\left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right ) \left (4 \sqrt {-4+2 \sqrt {13}}\, \sqrt {13}\, \sqrt {2 \sqrt {13}+4}\, \operatorname {arctanh}\left (\frac {\sqrt {-4+2 \sqrt {13}}\, \left (4 \sqrt {13}+17\right ) \left (\sqrt {13}+2+3 \tan \left (d x +c \right )\right ) \left (17 \sqrt {13}-52\right ) \sqrt {13}}{6318 \left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right ) \sqrt {-\frac {\tan \left (d x +c \right ) \left (-3+2 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right )^{2}}}}\right )-17 \sqrt {-4+2 \sqrt {13}}\, \sqrt {2 \sqrt {13}+4}\, \operatorname {arctanh}\left (\frac {\sqrt {-4+2 \sqrt {13}}\, \left (4 \sqrt {13}+17\right ) \left (\sqrt {13}+2+3 \tan \left (d x +c \right )\right ) \left (17 \sqrt {13}-52\right ) \sqrt {13}}{6318 \left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right ) \sqrt {-\frac {\tan \left (d x +c \right ) \left (-3+2 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right )^{2}}}}\right )-18 \arctan \left (\frac {6 \sqrt {13}\, \sqrt {-\frac {\tan \left (d x +c \right ) \left (-3+2 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right )^{2}}}}{\sqrt {26 \sqrt {13}+52}}\right ) \sqrt {13}+36 \arctan \left (\frac {6 \sqrt {13}\, \sqrt {-\frac {\tan \left (d x +c \right ) \left (-3+2 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right )^{2}}}}{\sqrt {26 \sqrt {13}+52}}\right )\right ) \sqrt {-\frac {\tan \left (d x +c \right ) \left (-3+2 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right )^{2}}}}{2 d \sqrt {2 \sqrt {13}+4}\, \sqrt {3-2 \tan \left (d x +c \right )}\, \left (17 \sqrt {13}-52\right ) \sqrt {\tan \left (d x +c \right )}}\)	\(427\)

Input:

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

Output:

-1/2/d*(13^(1/2)-2-3*tan(d*x+c))*(4*(-4+2*13^(1/2))^(1/2)*13^(1/2)*(2*13^( 
1/2)+4)^(1/2)*arctanh(1/6318*(-4+2*13^(1/2))^(1/2)*(4*13^(1/2)+17)*(13^(1/ 
2)+2+3*tan(d*x+c))*(17*13^(1/2)-52)/(13^(1/2)-2-3*tan(d*x+c))*13^(1/2)/(-t 
an(d*x+c)*(-3+2*tan(d*x+c))/(13^(1/2)-2-3*tan(d*x+c))^2)^(1/2))-17*(-4+2*1 
3^(1/2))^(1/2)*(2*13^(1/2)+4)^(1/2)*arctanh(1/6318*(-4+2*13^(1/2))^(1/2)*( 
4*13^(1/2)+17)*(13^(1/2)+2+3*tan(d*x+c))*(17*13^(1/2)-52)/(13^(1/2)-2-3*ta 
n(d*x+c))*13^(1/2)/(-tan(d*x+c)*(-3+2*tan(d*x+c))/(13^(1/2)-2-3*tan(d*x+c) 
)^2)^(1/2))-18*arctan(6*13^(1/2)*(-tan(d*x+c)*(-3+2*tan(d*x+c))/(13^(1/2)- 
2-3*tan(d*x+c))^2)^(1/2)/(26*13^(1/2)+52)^(1/2))*13^(1/2)+36*arctan(6*13^( 
1/2)*(-tan(d*x+c)*(-3+2*tan(d*x+c))/(13^(1/2)-2-3*tan(d*x+c))^2)^(1/2)/(26 
*13^(1/2)+52)^(1/2)))/(2*13^(1/2)+4)^(1/2)/(3-2*tan(d*x+c))^(1/2)/(17*13^( 
1/2)-52)*(-tan(d*x+c)*(-3+2*tan(d*x+c))/(13^(1/2)-2-3*tan(d*x+c))^2)^(1/2) 
/tan(d*x+c)^(1/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1485 vs. \(2 (65) = 130\).

Time = 0.16 (sec) , antiderivative size = 1485, normalized size of antiderivative = 16.69 \[ \int \frac {1}{\sqrt {3-2 \tan (c+d x)} \sqrt {\tan (c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/8*sqrt(1/13)*sqrt(-(3*d^2*sqrt(-1/d^4) + 2)/d^2)*log(1/2*(sqrt(1/13)*(40 
0*d*tan(d*x + c)^2 - 2334*d*tan(d*x + c) - (1575*d^3*tan(d*x + c)^2 - 212* 
d^3*tan(d*x + c) - 759*d^3)*sqrt(-1/d^4) + 612*d)*sqrt(-(3*d^2*sqrt(-1/d^4 
) + 2)/d^2) + 2*((204*d^2*tan(d*x + c) + 253*d^2)*sqrt(-1/d^4) - 253*tan(d 
*x + c) + 204)*sqrt(-2*tan(d*x + c) + 3)*sqrt(tan(d*x + c)))/(tan(d*x + c) 
^2 + 1)) + 1/8*sqrt(1/13)*sqrt(-(3*d^2*sqrt(-1/d^4) + 2)/d^2)*log(-1/2*(sq 
rt(1/13)*(400*d*tan(d*x + c)^2 - 2334*d*tan(d*x + c) - (1575*d^3*tan(d*x + 
 c)^2 - 212*d^3*tan(d*x + c) - 759*d^3)*sqrt(-1/d^4) + 612*d)*sqrt(-(3*d^2 
*sqrt(-1/d^4) + 2)/d^2) + 2*((204*d^2*tan(d*x + c) + 253*d^2)*sqrt(-1/d^4) 
 - 253*tan(d*x + c) + 204)*sqrt(-2*tan(d*x + c) + 3)*sqrt(tan(d*x + c)))/( 
tan(d*x + c)^2 + 1)) - 1/8*sqrt(1/13)*sqrt(-(3*d^2*sqrt(-1/d^4) + 2)/d^2)* 
log(1/2*(sqrt(1/13)*(400*d*tan(d*x + c)^2 - 2334*d*tan(d*x + c) - (1575*d^ 
3*tan(d*x + c)^2 - 212*d^3*tan(d*x + c) - 759*d^3)*sqrt(-1/d^4) + 612*d)*s 
qrt(-(3*d^2*sqrt(-1/d^4) + 2)/d^2) - 2*((204*d^2*tan(d*x + c) + 253*d^2)*s 
qrt(-1/d^4) - 253*tan(d*x + c) + 204)*sqrt(-2*tan(d*x + c) + 3)*sqrt(tan(d 
*x + c)))/(tan(d*x + c)^2 + 1)) - 1/8*sqrt(1/13)*sqrt(-(3*d^2*sqrt(-1/d^4) 
 + 2)/d^2)*log(-1/2*(sqrt(1/13)*(400*d*tan(d*x + c)^2 - 2334*d*tan(d*x + c 
) - (1575*d^3*tan(d*x + c)^2 - 212*d^3*tan(d*x + c) - 759*d^3)*sqrt(-1/d^4 
) + 612*d)*sqrt(-(3*d^2*sqrt(-1/d^4) + 2)/d^2) - 2*((204*d^2*tan(d*x + c) 
+ 253*d^2)*sqrt(-1/d^4) - 253*tan(d*x + c) + 204)*sqrt(-2*tan(d*x + c) ...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\sqrt {3-2 \tan (c+d x)} \sqrt {\tan (c+d x)}} \, dx=\text {Exception raised: RuntimeError} \] Input:

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

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1061 vs. \(2 (65) = 130\).

Time = 0.39 (sec) , antiderivative size = 1061, normalized size of antiderivative = 11.92 \[ \int \frac {1}{\sqrt {3-2 \tan (c+d x)} \sqrt {\tan (c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/105456*(3*sqrt(2)*(4/13)^(3/4)*d^2*(52*sqrt(13) + 338)^(3/2) + 234*sqrt 
(2)*(4/13)^(3/4)*d^2*sqrt(52*sqrt(13) + 338)*(2*sqrt(13) - 13) - 156*sqrt( 
2)*(4/13)^(3/4)*d*(2*sqrt(13) + 13)*sqrt(-52*sqrt(13) + 338)*abs(d) + 2*sq 
rt(2)*(4/13)^(3/4)*d*(-52*sqrt(13) + 338)^(3/2)*abs(d) - 2028*sqrt(2)*(4/1 
3)^(1/4)*d^2*sqrt(52*sqrt(13) + 338) + 1352*sqrt(2)*(4/13)^(1/4)*d*sqrt(-5 
2*sqrt(13) + 338)*abs(d))*arctan(13/8*(4/13)^(3/4)*(2*(4/13)^(1/4)*sqrt(-1 
/13*sqrt(13) + 1/2) + (sqrt(2)*sqrt(tan(d*x + c)) - sqrt(3))/sqrt(-2*tan(d 
*x + c) + 3) - sqrt(-2*tan(d*x + c) + 3)/(sqrt(2)*sqrt(tan(d*x + c)) - sqr 
t(3)))/sqrt(1/13*sqrt(13) + 1/2))/d^3 - 1/105456*(3*sqrt(2)*(4/13)^(3/4)*d 
^2*(52*sqrt(13) + 338)^(3/2) + 234*sqrt(2)*(4/13)^(3/4)*d^2*sqrt(52*sqrt(1 
3) + 338)*(2*sqrt(13) - 13) - 156*sqrt(2)*(4/13)^(3/4)*d*(2*sqrt(13) + 13) 
*sqrt(-52*sqrt(13) + 338)*abs(d) + 2*sqrt(2)*(4/13)^(3/4)*d*(-52*sqrt(13) 
+ 338)^(3/2)*abs(d) - 2028*sqrt(2)*(4/13)^(1/4)*d^2*sqrt(52*sqrt(13) + 338 
) + 1352*sqrt(2)*(4/13)^(1/4)*d*sqrt(-52*sqrt(13) + 338)*abs(d))*arctan(-1 
3/8*(4/13)^(3/4)*(2*(4/13)^(1/4)*sqrt(-1/13*sqrt(13) + 1/2) - (sqrt(2)*sqr 
t(tan(d*x + c)) - sqrt(3))/sqrt(-2*tan(d*x + c) + 3) + sqrt(-2*tan(d*x + c 
) + 3)/(sqrt(2)*sqrt(tan(d*x + c)) - sqrt(3)))/sqrt(1/13*sqrt(13) + 1/2))/ 
d^3 - 1/210912*(234*sqrt(2)*(4/13)^(3/4)*d^2*(2*sqrt(13) + 13)*sqrt(-52*sq 
rt(13) + 338) - 3*sqrt(2)*(4/13)^(3/4)*d^2*(-52*sqrt(13) + 338)^(3/2) + 2* 
sqrt(2)*(4/13)^(3/4)*d*(52*sqrt(13) + 338)^(3/2)*abs(d) + 156*sqrt(2)*(...
 

Mupad [B] (verification not implemented)

Time = 3.54 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.26 \[ \int \frac {1}{\sqrt {3-2 \tan (c+d x)} \sqrt {\tan (c+d x)}} \, dx=-\mathrm {atan}\left (\frac {\sqrt {3}\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {-\frac {1}{26}+\frac {3}{52}{}\mathrm {i}}{d^2}}\,\left (6-4{}\mathrm {i}\right )+d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {-\frac {1}{26}+\frac {3}{52}{}\mathrm {i}}{d^2}}\,\sqrt {3-2\,\mathrm {tan}\left (c+d\,x\right )}\,\left (-6+4{}\mathrm {i}\right )}{2\,\mathrm {tan}\left (c+d\,x\right )+\sqrt {3}\,\sqrt {3-2\,\mathrm {tan}\left (c+d\,x\right )}-3}\right )\,\sqrt {\frac {-\frac {1}{26}+\frac {3}{52}{}\mathrm {i}}{d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {\sqrt {3}\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {-\frac {1}{26}-\frac {3}{52}{}\mathrm {i}}{d^2}}\,\left (6+4{}\mathrm {i}\right )+d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {-\frac {1}{26}-\frac {3}{52}{}\mathrm {i}}{d^2}}\,\sqrt {3-2\,\mathrm {tan}\left (c+d\,x\right )}\,\left (-6-4{}\mathrm {i}\right )}{2\,\mathrm {tan}\left (c+d\,x\right )+\sqrt {9-6\,\mathrm {tan}\left (c+d\,x\right )}-3}\right )\,\sqrt {\frac {-\frac {1}{26}-\frac {3}{52}{}\mathrm {i}}{d^2}}\,2{}\mathrm {i} \] Input:

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

Output:

atan((3^(1/2)*d*tan(c + d*x)^(1/2)*((- 1/26 - 3i/52)/d^2)^(1/2)*(6 + 4i) - 
 d*tan(c + d*x)^(1/2)*((- 1/26 - 3i/52)/d^2)^(1/2)*(3 - 2*tan(c + d*x))^(1 
/2)*(6 + 4i))/(2*tan(c + d*x) + (9 - 6*tan(c + d*x))^(1/2) - 3))*((- 1/26 
- 3i/52)/d^2)^(1/2)*2i - atan((3^(1/2)*d*tan(c + d*x)^(1/2)*((- 1/26 + 3i/ 
52)/d^2)^(1/2)*(6 - 4i) - d*tan(c + d*x)^(1/2)*((- 1/26 + 3i/52)/d^2)^(1/2 
)*(3 - 2*tan(c + d*x))^(1/2)*(6 - 4i))/(2*tan(c + d*x) + 3^(1/2)*(3 - 2*ta 
n(c + d*x))^(1/2) - 3))*((- 1/26 + 3i/52)/d^2)^(1/2)*2i
 

Reduce [F]

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

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

Output:

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