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

Optimal. Leaf size=95 \[ \frac{i \tan ^{-1}\left (\frac{\sqrt{2+3 i} \sqrt{\tan (c+d x)}}{\sqrt{3-2 \tan (c+d x)}}\right )}{\sqrt{2+3 i} d}-\frac{i \tan ^{-1}\left (\frac{\sqrt{2-3 i} \sqrt{\tan (c+d x)}}{\sqrt{3-2 \tan (c+d x)}}\right )}{\sqrt{2-3 i} d} \]

[Out]

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

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Rubi [A]  time = 0.096192, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {3575, 910, 93, 205} \[ \frac{i \tan ^{-1}\left (\frac{\sqrt{2+3 i} \sqrt{\tan (c+d x)}}{\sqrt{3-2 \tan (c+d x)}}\right )}{\sqrt{2+3 i} d}-\frac{i \tan ^{-1}\left (\frac{\sqrt{2-3 i} \sqrt{\tan (c+d x)}}{\sqrt{3-2 \tan (c+d x)}}\right )}{\sqrt{2-3 i} d} \]

Antiderivative was successfully verified.

[In]

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

Rule 3575

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]

Rule 910

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 93

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 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\sqrt{\tan (c+d x)}}{\sqrt{3-2 \tan (c+d x)}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sqrt{x}}{\sqrt{3-2 x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{1}{2 \sqrt{3-2 x} (i-x) \sqrt{x}}+\frac{1}{2 \sqrt{3-2 x} \sqrt{x} (i+x)}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{3-2 x} (i-x) \sqrt{x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{3-2 x} \sqrt{x} (i+x)} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=-\frac{\operatorname{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{\operatorname{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 \tan ^{-1}\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 \tan ^{-1}\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]  time = 0.0918224, size = 103, normalized size = 1.08 \[ \frac{i \left (\sqrt{2+3 i} \tan ^{-1}\left (\frac{\sqrt{\frac{2}{13}+\frac{3 i}{13}} \sqrt{3-2 \tan (c+d x)}}{\sqrt{\tan (c+d x)}}\right )+\sqrt{-2+3 i} \tanh ^{-1}\left (\frac{\sqrt{-\frac{2}{13}+\frac{3 i}{13}} \sqrt{3-2 \tan (c+d x)}}{\sqrt{\tan (c+d x)}}\right )\right )}{\sqrt{13} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.054, size = 434, normalized size = 4.6 \begin{align*}{\frac{3\,\sqrt{13}-6-9\,\tan \left ( dx+c \right ) }{2\,d\sqrt{2\,\sqrt{13}+4} \left ( -3+2\,\tan \left ( dx+c \right ) \right ) \left ( 17\,\sqrt{13}-52 \right ) }\sqrt{3-2\,\tan \left ( dx+c \right ) }\sqrt{-{\frac{\tan \left ( dx+c \right ) \left ( -3+2\,\tan \left ( dx+c \right ) \right ) }{ \left ( \sqrt{13}-2-3\,\tan \left ( dx+c \right ) \right ) ^{2}}}} \left ( \sqrt{-4+2\,\sqrt{13}}{\it Artanh} \left ({\frac{ \left ( \sqrt{13}+2 \right ) \left ( 17\,\sqrt{13}-52 \right ) \left ( \sqrt{13}+2+3\,\tan \left ( dx+c \right ) \right ) \sqrt{13}}{351\,\sqrt{-4+2\,\sqrt{13}} \left ( \sqrt{13}-2-3\,\tan \left ( dx+c \right ) \right ) }{\frac{1}{\sqrt{-{\frac{\tan \left ( dx+c \right ) \left ( -3+2\,\tan \left ( dx+c \right ) \right ) }{ \left ( \sqrt{13}-2-3\,\tan \left ( dx+c \right ) \right ) ^{2}}}}}}} \right ) \sqrt{13}\sqrt{2\,\sqrt{13}+4}-2\,\sqrt{-4+2\,\sqrt{13}}{\it Artanh} \left ({\frac{ \left ( \sqrt{13}+2 \right ) \left ( 17\,\sqrt{13}-52 \right ) \left ( \sqrt{13}+2+3\,\tan \left ( dx+c \right ) \right ) \sqrt{13}}{351\,\sqrt{-4+2\,\sqrt{13}} \left ( \sqrt{13}-2-3\,\tan \left ( dx+c \right ) \right ) }{\frac{1}{\sqrt{-{\frac{\tan \left ( dx+c \right ) \left ( -3+2\,\tan \left ( dx+c \right ) \right ) }{ \left ( \sqrt{13}-2-3\,\tan \left ( dx+c \right ) \right ) ^{2}}}}}}} \right ) \sqrt{2\,\sqrt{13}+4}+8\,\arctan \left ( 6\,{\frac{\sqrt{13}}{\sqrt{26\,\sqrt{13}+52}}\sqrt{-{\frac{\tan \left ( dx+c \right ) \left ( -3+2\,\tan \left ( dx+c \right ) \right ) }{ \left ( \sqrt{13}-2-3\,\tan \left ( dx+c \right ) \right ) ^{2}}}}} \right ) \sqrt{13}-34\,\arctan \left ( 6\,{\frac{\sqrt{13}}{\sqrt{26\,\sqrt{13}+52}}\sqrt{-{\frac{\tan \left ( dx+c \right ) \left ( -3+2\,\tan \left ( dx+c \right ) \right ) }{ \left ( \sqrt{13}-2-3\,\tan \left ( dx+c \right ) \right ) ^{2}}}}} \right ) \right ){\frac{1}{\sqrt{\tan \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2/d*(3-2*tan(d*x+c))^(1/2)*(-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*t
an(d*x+c))*((-4+2*13^(1/2))^(1/2)*arctanh(1/351*(13^(1/2)+2)*(17*13^(1/2)-52)*(13^(1/2)+2+3*tan(d*x+c))/(-4+2*
13^(1/2))^(1/2)/(13^(1/2)-2-3*tan(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))*13^(1/2)*(2*13^(1/2)+4)^(1/2)-2*(-4+2*13^(1/2))^(1/2)*arctanh(1/351*(13^(1/2)+2)*(17*13^(1/2)-52)*(13^
(1/2)+2+3*tan(d*x+c))/(-4+2*13^(1/2))^(1/2)/(13^(1/2)-2-3*tan(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))*(2*13^(1/2)+4)^(1/2)+8*arctan(6*13^(1/2)*(-tan(d*x+c)*(-3+2*tan(d*x+c))/(1
3^(1/2)-2-3*tan(d*x+c))^2)^(1/2)/(26*13^(1/2)+52)^(1/2))*13^(1/2)-34*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)))/tan(d*x+c)^(1/2)/(2*13^(1/2)+4)^(1/2)/(-3+
2*tan(d*x+c))/(17*13^(1/2)-52)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\tan \left (d x + c\right )}}{\sqrt{-2 \, \tan \left (d x + c\right ) + 3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\tan{\left (c + d x \right )}}}{\sqrt{3 - 2 \tan{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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(3 - 2*tan(c + d*x)), x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError