\(\int \frac {\sqrt {\tan (c+d x)}}{\sqrt {-3+2 \tan (c+d x)}} \, dx\) [669]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3656, 924, 95, 214} \[ \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {-3+2 \tan (c+d x)}} \, dx=\frac {i \text {arctanh}\left (\frac {\sqrt {2+3 i} \sqrt {\tan (c+d x)}}{\sqrt {2 \tan (c+d x)-3}}\right )}{\sqrt {2+3 i} d}-\frac {i \text {arctanh}\left (\frac {\sqrt {2-3 i} \sqrt {\tan (c+d x)}}{\sqrt {2 \tan (c+d x)-3}}\right )}{\sqrt {2-3 i} d} \]

[In]

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

[Out]

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

Rule 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[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] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 214

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 924

Int[((d_.) + (e_.)*(x_))^(m_)/(Sqrt[(f_.) + (g_.)*(x_)]*((a_.) + (c_.)*(x_)^2)), x_Symbol] :> Int[ExpandIntegr
and[1/(Sqrt[d + e*x]*Sqrt[f + g*x]), (d + e*x)^(m + 1/2)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] &
& NeQ[c*d^2 + a*e^2, 0] && IGtQ[m + 1/2, 0]

Rule 3656

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Wit
h[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\sqrt {x}}{\sqrt {-3+2 x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \left (-\frac {1}{2 (i-x) \sqrt {x} \sqrt {-3+2 x}}+\frac {1}{2 \sqrt {x} (i+x) \sqrt {-3+2 x}}\right ) \, dx,x,\tan (c+d x)\right )}{d} \\ & = -\frac {\text {Subst}\left (\int \frac {1}{(i-x) \sqrt {x} \sqrt {-3+2 x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {\text {Subst}\left (\int \frac {1}{\sqrt {x} (i+x) \sqrt {-3+2 x}} \, dx,x,\tan (c+d x)\right )}{2 d} \\ & = \frac {\text {Subst}\left (\int \frac {1}{i-(3+2 i) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {-3+2 \tan (c+d x)}}\right )}{d}-\frac {\text {Subst}\left (\int \frac {1}{i+(3-2 i) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {-3+2 \tan (c+d x)}}\right )}{d} \\ & = -\frac {i \text {arctanh}\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 {i \text {arctanh}\left (\frac {\sqrt {2+3 i} \sqrt {\tan (c+d x)}}{\sqrt {-3+2 \tan (c+d x)}}\right )}{\sqrt {2+3 i} d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {-3+2 \tan (c+d x)}} \, dx=-\frac {i \arctan \left (\frac {\sqrt {-2+3 i} \sqrt {\tan (c+d x)}}{\sqrt {-3+2 \tan (c+d x)}}\right )}{\sqrt {-2+3 i} d}+\frac {i \text {arctanh}\left (\frac {\sqrt {2+3 i} \sqrt {\tan (c+d x)}}{\sqrt {-3+2 \tan (c+d x)}}\right )}{\sqrt {2+3 i} d} \]

[In]

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

[Out]

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

Maple [B] (verified)

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

Time = 4.15 (sec) , antiderivative size = 479, normalized size of antiderivative = 5.04

method	result	size
derivativedivides	\(-\frac {3 \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}}}\, \left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right ) \left (\sqrt {-4+2 \sqrt {13}}\, \sqrt {13}\, \sqrt {2 \sqrt {13}+4}\, \arctan \left (\frac {\sqrt {-4+2 \sqrt {13}}\, \sqrt {\frac {\left (17 \sqrt {13}-52\right ) \tan \left (d x +c \right ) \left (52+17 \sqrt {13}\right ) \left (-3+2 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right )^{2}}}\, \left (4 \sqrt {13}+17\right ) \left (\sqrt {13}+2+3 \tan \left (d x +c \right )\right ) \left (17 \sqrt {13}-52\right ) \left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right )}{56862 \tan \left (d x +c \right ) \left (-3+2 \tan \left (d x +c \right )\right )}\right )-2 \sqrt {-4+2 \sqrt {13}}\, \sqrt {2 \sqrt {13}+4}\, \arctan \left (\frac {\sqrt {-4+2 \sqrt {13}}\, \sqrt {\frac {\left (17 \sqrt {13}-52\right ) \tan \left (d x +c \right ) \left (52+17 \sqrt {13}\right ) \left (-3+2 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right )^{2}}}\, \left (4 \sqrt {13}+17\right ) \left (\sqrt {13}+2+3 \tan \left (d x +c \right )\right ) \left (17 \sqrt {13}-52\right ) \left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right )}{56862 \tan \left (d x +c \right ) \left (-3+2 \tan \left (d x +c \right )\right )}\right )-8 \,\operatorname {arctanh}\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}+34 \,\operatorname {arctanh}\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 )}{2 d \sqrt {\tan \left (d x +c \right )}\, \sqrt {-3+2 \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {13}+4}\, \left (17 \sqrt {13}-52\right )}\)	\(479\)
default	\(-\frac {3 \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}}}\, \left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right ) \left (\sqrt {-4+2 \sqrt {13}}\, \sqrt {13}\, \sqrt {2 \sqrt {13}+4}\, \arctan \left (\frac {\sqrt {-4+2 \sqrt {13}}\, \sqrt {\frac {\left (17 \sqrt {13}-52\right ) \tan \left (d x +c \right ) \left (52+17 \sqrt {13}\right ) \left (-3+2 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right )^{2}}}\, \left (4 \sqrt {13}+17\right ) \left (\sqrt {13}+2+3 \tan \left (d x +c \right )\right ) \left (17 \sqrt {13}-52\right ) \left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right )}{56862 \tan \left (d x +c \right ) \left (-3+2 \tan \left (d x +c \right )\right )}\right )-2 \sqrt {-4+2 \sqrt {13}}\, \sqrt {2 \sqrt {13}+4}\, \arctan \left (\frac {\sqrt {-4+2 \sqrt {13}}\, \sqrt {\frac {\left (17 \sqrt {13}-52\right ) \tan \left (d x +c \right ) \left (52+17 \sqrt {13}\right ) \left (-3+2 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right )^{2}}}\, \left (4 \sqrt {13}+17\right ) \left (\sqrt {13}+2+3 \tan \left (d x +c \right )\right ) \left (17 \sqrt {13}-52\right ) \left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right )}{56862 \tan \left (d x +c \right ) \left (-3+2 \tan \left (d x +c \right )\right )}\right )-8 \,\operatorname {arctanh}\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}+34 \,\operatorname {arctanh}\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 )}{2 d \sqrt {\tan \left (d x +c \right )}\, \sqrt {-3+2 \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {13}+4}\, \left (17 \sqrt {13}-52\right )}\)	\(479\)

[In]

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

[Out]

-3/2/d*(tan(d*x+c)*(-3+2*tan(d*x+c))/(13^(1/2)-2-3*tan(d*x+c))^2)^(1/2)*(13^(1/2)-2-3*tan(d*x+c))*((-4+2*13^(1
/2))^(1/2)*13^(1/2)*(2*13^(1/2)+4)^(1/2)*arctan(1/56862*(-4+2*13^(1/2))^(1/2)*((17*13^(1/2)-52)*tan(d*x+c)*(52
+17*13^(1/2))*(-3+2*tan(d*x+c))/(13^(1/2)-2-3*tan(d*x+c))^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))/tan(d*x+c)/(-3+2*tan(d*x+c)))-2*(-4+2*13^(1/2))^(1/2)*(2*13^(1/2)+4)
^(1/2)*arctan(1/56862*(-4+2*13^(1/2))^(1/2)*((17*13^(1/2)-52)*tan(d*x+c)*(52+17*13^(1/2))*(-3+2*tan(d*x+c))/(1
3^(1/2)-2-3*tan(d*x+c))^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))/tan(d*x+c)/(-3+2*tan(d*x+c)))-8*arctanh(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)+34*arctanh(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)))/tan(d*x+c)^(1/2)/(-3+2*tan(d*x+c))^(1/2)/(2*13^(1/2)+4)^(1/2)
/(17*13^(1/2)-52)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1485 vs. \(2 (67) = 134\).

Time = 0.34 (sec) , antiderivative size = 1485, normalized size of antiderivative = 15.63 \[ \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {-3+2 \tan (c+d x)}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/8*sqrt(1/13)*sqrt(-(3*d^2*sqrt(-1/d^4) + 2)/d^2)*log((sqrt(1/13)*(1575*d*tan(d*x + c)^2 - 212*d*tan(d*x + c
) + 2*(200*d^3*tan(d*x + c)^2 - 1167*d^3*tan(d*x + c) + 306*d^3)*sqrt(-1/d^4) - 759*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(-(sqrt
(1/13)*(1575*d*tan(d*x + c)^2 - 212*d*tan(d*x + c) + 2*(200*d^3*tan(d*x + c)^2 - 1167*d^3*tan(d*x + c) + 306*d
^3)*sqrt(-1/d^4) - 759*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(1/1
3)*sqrt(-(3*d^2*sqrt(-1/d^4) + 2)/d^2)*log((sqrt(1/13)*(1575*d*tan(d*x + c)^2 - 212*d*tan(d*x + c) + 2*(200*d^
3*tan(d*x + c)^2 - 1167*d^3*tan(d*x + c) + 306*d^3)*sqrt(-1/d^4) - 759*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(-(sqrt(1/13)*(1575*
d*tan(d*x + c)^2 - 212*d*tan(d*x + c) + 2*(200*d^3*tan(d*x + c)^2 - 1167*d^3*tan(d*x + c) + 306*d^3)*sqrt(-1/d
^4) - 759*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((sqrt(1/13)*(1575*d*tan(d*x + c)^2 - 212*d*tan(d*x + c) - 2*(200*d^3*tan(d*x + c)
^2 - 1167*d^3*tan(d*x + c) + 306*d^3)*sqrt(-1/d^4) - 759*d)*sqrt((3*d^2*sqrt(-1/d^4) - 2)/d^2) + 2*((204*d^2*t
an(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)))/(ta
n(d*x + c)^2 + 1)) + 1/8*sqrt(1/13)*sqrt((3*d^2*sqrt(-1/d^4) - 2)/d^2)*log(-(sqrt(1/13)*(1575*d*tan(d*x + c)^2
- 212*d*tan(d*x + c) - 2*(200*d^3*tan(d*x + c)^2 - 1167*d^3*tan(d*x + c) + 306*d^3)*sqrt(-1/d^4) - 759*d)*sqr
t((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)*s
qrt(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((sqrt(1/13)*(1575*d*tan(d*x + c)^2 - 212*d*tan(d*x + c) - 2*(200*d^3*tan(d*x + c)^2 - 1167*d^3*tan
(d*x + c) + 306*d^3)*sqrt(-1/d^4) - 759*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(-(sqrt(1/13)*(1575*d*tan(d*x + c)^2 - 212*d*tan(d*x
+ c) - 2*(200*d^3*tan(d*x + c)^2 - 1167*d^3*tan(d*x + c) + 306*d^3)*sqrt(-1/d^4) - 759*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))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 498 vs. \(2 (67) = 134\).

Time = 1.32 (sec) , antiderivative size = 498, normalized size of antiderivative = 5.24 \[ \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {-3+2 \tan (c+d x)}} \, dx=\frac {1}{676} \, \sqrt {2} {\left (\frac {3 \, {\left (2 \, \sqrt {169 \, \sqrt {13} + 598} \arctan \left (\frac {13 \, \left (\frac {4}{13}\right )^{\frac {3}{4}} {\left (\left (\frac {4}{13}\right )^{\frac {1}{4}} \sqrt {\frac {1}{13} \, \sqrt {13} + \frac {1}{2}} + 1\right )}}{4 \, \sqrt {-\frac {1}{13} \, \sqrt {13} + \frac {1}{2}}}\right ) + 2 \, \sqrt {169 \, \sqrt {13} + 598} \arctan \left (-\frac {13 \, \left (\frac {4}{13}\right )^{\frac {3}{4}} {\left (\left (\frac {4}{13}\right )^{\frac {1}{4}} \sqrt {\frac {1}{13} \, \sqrt {13} + \frac {1}{2}} - 1\right )}}{4 \, \sqrt {-\frac {1}{13} \, \sqrt {13} + \frac {1}{2}}}\right ) + \sqrt {169 \, \sqrt {13} - 598} \log \left (2 \, \left (\frac {4}{13}\right )^{\frac {1}{4}} \sqrt {\frac {1}{13} \, \sqrt {13} + \frac {1}{2}} + 2 \, \sqrt {\frac {1}{13}} + 1\right ) - \sqrt {169 \, \sqrt {13} - 598} \log \left (-2 \, \left (\frac {4}{13}\right )^{\frac {1}{4}} \sqrt {\frac {1}{13} \, \sqrt {13} + \frac {1}{2}} + 2 \, \sqrt {\frac {1}{13}} + 1\right )\right )}}{d} - \frac {2 \, {\left (3 \, d^{2} \sqrt {169 \, \sqrt {13} + 598} + 2 \, d \sqrt {169 \, \sqrt {13} - 598} {\left | d \right |}\right )} \arctan \left (\frac {13 \, \left (\frac {4}{13}\right )^{\frac {3}{4}} {\left (\left (\frac {4}{13}\right )^{\frac {1}{4}} \sqrt {\frac {1}{13} \, \sqrt {13} + \frac {1}{2}} + \sqrt {\frac {3}{2 \, \tan \left (d x + c\right ) - 3} + 1}\right )}}{4 \, \sqrt {-\frac {1}{13} \, \sqrt {13} + \frac {1}{2}}}\right )}{d^{3}} - \frac {2 \, {\left (3 \, d^{2} \sqrt {169 \, \sqrt {13} + 598} + 2 \, d \sqrt {169 \, \sqrt {13} - 598} {\left | d \right |}\right )} \arctan \left (-\frac {13 \, \left (\frac {4}{13}\right )^{\frac {3}{4}} {\left (\left (\frac {4}{13}\right )^{\frac {1}{4}} \sqrt {\frac {1}{13} \, \sqrt {13} + \frac {1}{2}} - \sqrt {\frac {3}{2 \, \tan \left (d x + c\right ) - 3} + 1}\right )}}{4 \, \sqrt {-\frac {1}{13} \, \sqrt {13} + \frac {1}{2}}}\right )}{d^{3}} - \frac {{\left (3 \, d^{2} \sqrt {169 \, \sqrt {13} - 598} - 2 \, d \sqrt {169 \, \sqrt {13} + 598} {\left | d \right |}\right )} \log \left (2 \, \left (\frac {4}{13}\right )^{\frac {1}{4}} \sqrt {\frac {1}{13} \, \sqrt {13} + \frac {1}{2}} \sqrt {\frac {3}{2 \, \tan \left (d x + c\right ) - 3} + 1} + 2 \, \sqrt {\frac {1}{13}} + \frac {3}{2 \, \tan \left (d x + c\right ) - 3} + 1\right )}{d^{3}} + \frac {{\left (3 \, d^{2} \sqrt {169 \, \sqrt {13} - 598} - 2 \, d \sqrt {169 \, \sqrt {13} + 598} {\left | d \right |}\right )} \log \left (-2 \, \left (\frac {4}{13}\right )^{\frac {1}{4}} \sqrt {\frac {1}{13} \, \sqrt {13} + \frac {1}{2}} \sqrt {\frac {3}{2 \, \tan \left (d x + c\right ) - 3} + 1} + 2 \, \sqrt {\frac {1}{13}} + \frac {3}{2 \, \tan \left (d x + c\right ) - 3} + 1\right )}{d^{3}}\right )} \]

[In]

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

[Out]

1/676*sqrt(2)*(3*(2*sqrt(169*sqrt(13) + 598)*arctan(13/4*(4/13)^(3/4)*((4/13)^(1/4)*sqrt(1/13*sqrt(13) + 1/2)
+ 1)/sqrt(-1/13*sqrt(13) + 1/2)) + 2*sqrt(169*sqrt(13) + 598)*arctan(-13/4*(4/13)^(3/4)*((4/13)^(1/4)*sqrt(1/1
3*sqrt(13) + 1/2) - 1)/sqrt(-1/13*sqrt(13) + 1/2)) + sqrt(169*sqrt(13) - 598)*log(2*(4/13)^(1/4)*sqrt(1/13*sqr
t(13) + 1/2) + 2*sqrt(1/13) + 1) - sqrt(169*sqrt(13) - 598)*log(-2*(4/13)^(1/4)*sqrt(1/13*sqrt(13) + 1/2) + 2*
sqrt(1/13) + 1))/d - 2*(3*d^2*sqrt(169*sqrt(13) + 598) + 2*d*sqrt(169*sqrt(13) - 598)*abs(d))*arctan(13/4*(4/1
3)^(3/4)*((4/13)^(1/4)*sqrt(1/13*sqrt(13) + 1/2) + sqrt(3/(2*tan(d*x + c) - 3) + 1))/sqrt(-1/13*sqrt(13) + 1/2
))/d^3 - 2*(3*d^2*sqrt(169*sqrt(13) + 598) + 2*d*sqrt(169*sqrt(13) - 598)*abs(d))*arctan(-13/4*(4/13)^(3/4)*((
4/13)^(1/4)*sqrt(1/13*sqrt(13) + 1/2) - sqrt(3/(2*tan(d*x + c) - 3) + 1))/sqrt(-1/13*sqrt(13) + 1/2))/d^3 - (3
*d^2*sqrt(169*sqrt(13) - 598) - 2*d*sqrt(169*sqrt(13) + 598)*abs(d))*log(2*(4/13)^(1/4)*sqrt(1/13*sqrt(13) + 1
/2)*sqrt(3/(2*tan(d*x + c) - 3) + 1) + 2*sqrt(1/13) + 3/(2*tan(d*x + c) - 3) + 1)/d^3 + (3*d^2*sqrt(169*sqrt(1
3) - 598) - 2*d*sqrt(169*sqrt(13) + 598)*abs(d))*log(-2*(4/13)^(1/4)*sqrt(1/13*sqrt(13) + 1/2)*sqrt(3/(2*tan(d
*x + c) - 3) + 1) + 2*sqrt(1/13) + 3/(2*tan(d*x + c) - 3) + 1)/d^3)

Mupad [B] (verification not implemented)

Time = 7.31 (sec) , antiderivative size = 191, normalized size of antiderivative = 2.01 \[ \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {-3+2 \tan (c+d x)}} \, dx=-\mathrm {atan}\left (\frac {8\,d\,\sqrt {\frac {-\frac {1}{26}-\frac {3}{52}{}\mathrm {i}}{d^2}}\,\left (\frac {\sqrt {2}\,\sqrt {3}}{2}-\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{\sqrt {2\,\mathrm {tan}\left (c+d\,x\right )-3}\,\left (\frac {2\,{\left (\frac {\sqrt {2}\,\sqrt {3}}{2}-\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}^2}{2\,\mathrm {tan}\left (c+d\,x\right )-3}+1\right )}\right )\,\sqrt {\frac {-\frac {1}{26}-\frac {3}{52}{}\mathrm {i}}{d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {8\,d\,\sqrt {\frac {-\frac {1}{26}+\frac {3}{52}{}\mathrm {i}}{d^2}}\,\left (\frac {\sqrt {2}\,\sqrt {3}}{2}-\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{\sqrt {2\,\mathrm {tan}\left (c+d\,x\right )-3}\,\left (\frac {2\,{\left (\frac {\sqrt {2}\,\sqrt {3}}{2}-\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}^2}{2\,\mathrm {tan}\left (c+d\,x\right )-3}+1\right )}\right )\,\sqrt {\frac {-\frac {1}{26}+\frac {3}{52}{}\mathrm {i}}{d^2}}\,2{}\mathrm {i} \]

[In]

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

[Out]

atan((8*d*((- 1/26 + 3i/52)/d^2)^(1/2)*((2^(1/2)*3^(1/2))/2 - tan(c + d*x)^(1/2)))/((2*tan(c + d*x) - 3)^(1/2)
*((2*((2^(1/2)*3^(1/2))/2 - tan(c + d*x)^(1/2))^2)/(2*tan(c + d*x) - 3) + 1)))*((- 1/26 + 3i/52)/d^2)^(1/2)*2i
- atan((8*d*((- 1/26 - 3i/52)/d^2)^(1/2)*((2^(1/2)*3^(1/2))/2 - tan(c + d*x)^(1/2)))/((2*tan(c + d*x) - 3)^(1
/2)*((2*((2^(1/2)*3^(1/2))/2 - tan(c + d*x)^(1/2))^2)/(2*tan(c + d*x) - 3) + 1)))*((- 1/26 - 3i/52)/d^2)^(1/2)
*2i