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

Optimal. Leaf size=89 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{3-2 i} \sqrt{\tan (c+d x)}}{\sqrt{3 \tan (c+d x)-2}}\right )}{\sqrt{3-2 i} d}+\frac{\tanh ^{-1}\left (\frac{\sqrt{3+2 i} \sqrt{\tan (c+d x)}}{\sqrt{3 \tan (c+d x)-2}}\right )}{\sqrt{3+2 i} d} \]

[Out]

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

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Rubi [A]  time = 0.106433, antiderivative size = 89, 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, 912, 93, 208} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{3-2 i} \sqrt{\tan (c+d x)}}{\sqrt{3 \tan (c+d x)-2}}\right )}{\sqrt{3-2 i} d}+\frac{\tanh ^{-1}\left (\frac{\sqrt{3+2 i} \sqrt{\tan (c+d x)}}{\sqrt{3 \tan (c+d x)-2}}\right )}{\sqrt{3+2 i} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[(Sqrt[3 - 2*I]*Sqrt[Tan[c + d*x]])/Sqrt[-2 + 3*Tan[c + d*x]]]/(Sqrt[3 - 2*I]*d) + ArcTanh[(Sqrt[3 + 2*
I]*Sqrt[Tan[c + d*x]])/Sqrt[-2 + 3*Tan[c + d*x]]]/(Sqrt[3 + 2*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 912

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

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 208

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

Rubi steps

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

Mathematica [A]  time = 0.132056, size = 89, normalized size = 1. \[ \frac{\tan ^{-1}\left (\frac{\sqrt{-3+2 i} \sqrt{\tan (c+d x)}}{\sqrt{3 \tan (c+d x)-2}}\right )}{\sqrt{-3+2 i} d}+\frac{\tanh ^{-1}\left (\frac{\sqrt{3+2 i} \sqrt{\tan (c+d x)}}{\sqrt{3 \tan (c+d x)-2}}\right )}{\sqrt{3+2 i} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.105, size = 480, normalized size = 5.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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))*(3*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)*(-2
+3*tan(d*x+c))*(39+11*13^(1/2))/(13^(1/2)-3-2*tan(d*x+c))^2)^(1/2)*(11+3*13^(1/2))*(11*13^(1/2)-39)*(13^(1/2)+
3+2*tan(d*x+c))*(13^(1/2)-3-2*tan(d*x+c))/tan(d*x+c)/(-2+3*tan(d*x+c)))-11*(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)*(-2+3*tan(d*x+c))*(39+11*13^(1/2))/(13
^(1/2)-3-2*tan(d*x+c))^2)^(1/2)*(11+3*13^(1/2))*(11*13^(1/2)-39)*(13^(1/2)+3+2*tan(d*x+c))*(13^(1/2)-3-2*tan(d
*x+c))/tan(d*x+c)/(-2+3*tan(d*x+c)))+4*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)-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)))/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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{3 \, \tan \left (d x + c\right ) - 2} \sqrt{\tan \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(3*tan(d*x + c) - 2)*sqrt(tan(d*x + c))), 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(1/tan(d*x+c)^(1/2)/(-2+3*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{1}{\sqrt{3 \tan{\left (c + d x \right )} - 2} \sqrt{\tan{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: TypeError