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

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

[Out]

ArcTan[(Sqrt[3 - 2*I]*Sqrt[Tan[c + d*x]])/Sqrt[-2 - 3*Tan[c + d*x]]]/(Sqrt[3 - 2*I]*d) + ArcTan[(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 = 0.10645, 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, 205} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{3-2 i} \sqrt{\tan (c+d x)}}{\sqrt{-3 \tan (c+d x)-2}}\right )}{\sqrt{3-2 i} d}+\frac{\tan ^{-1}\left (\frac{\sqrt{3+2 i} \sqrt{\tan (c+d x)}}{\sqrt{-3 \tan (c+d x)-2}}\right )}{\sqrt{3+2 i} d} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[-2 - 3*Tan[c + d*x]]*Sqrt[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) + ArcTan[(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 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{1}{\sqrt{-2-3 \tan (c+d x)} \sqrt{\tan (c+d x)}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{-2-3 x} \sqrt{x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{i}{2 \sqrt{-2-3 x} (i-x) \sqrt{x}}+\frac{i}{2 \sqrt{-2-3 x} \sqrt{x} (i+x)}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{i \operatorname{Subst}\left (\int \frac{1}{\sqrt{-2-3 x} (i-x) \sqrt{x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{i \operatorname{Subst}\left (\int \frac{1}{\sqrt{-2-3 x} \sqrt{x} (i+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{\tan ^{-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{\tan ^{-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 = 0.141398, size = 101, normalized size = 1.13 \[ \frac{\sqrt{-3+2 i} \tanh ^{-1}\left (\frac{\sqrt{-\frac{3}{13}+\frac{2 i}{13}} \sqrt{-3 \tan (c+d x)-2}}{\sqrt{\tan (c+d x)}}\right )-\sqrt{3+2 i} \tan ^{-1}\left (\frac{\sqrt{\frac{3}{13}+\frac{2 i}{13}} \sqrt{-3 \tan (c+d x)-2}}{\sqrt{\tan (c+d x)}}\right )}{\sqrt{13} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.078, size = 435, normalized size = 4.9 \begin{align*} -{\frac{\sqrt{13}-3+2\,\tan \left ( dx+c \right ) }{2\,d \left ( 2+3\,\tan \left ( dx+c \right ) \right ) \sqrt{2\,\sqrt{13}+6} \left ( 11\,\sqrt{13}-39 \right ) }\sqrt{-2-3\,\tan \left ( dx+c \right ) }\sqrt{-{\frac{\tan \left ( dx+c \right ) \left ( 2+3\,\tan \left ( dx+c \right ) \right ) }{ \left ( \sqrt{13}-3+2\,\tan \left ( dx+c \right ) \right ) ^{2}}}} \left ( 3\,\sqrt{13}\sqrt{2\,\sqrt{13}+6}\sqrt{-6+2\,\sqrt{13}}{\it Artanh} \left ({\frac{ \left ( \sqrt{13}+3 \right ) \left ( 11\,\sqrt{13}-39 \right ) \left ( \sqrt{13}+3-2\,\tan \left ( dx+c \right ) \right ) \sqrt{13}}{ \left ( 52\,\sqrt{13}-156+104\,\tan \left ( dx+c \right ) \right ) \sqrt{-6+2\,\sqrt{13}}}{\frac{1}{\sqrt{-{\frac{\tan \left ( dx+c \right ) \left ( 2+3\,\tan \left ( dx+c \right ) \right ) }{ \left ( \sqrt{13}-3+2\,\tan \left ( dx+c \right ) \right ) ^{2}}}}}}} \right ) -11\,\sqrt{2\,\sqrt{13}+6}\sqrt{-6+2\,\sqrt{13}}{\it Artanh} \left ({\frac{ \left ( \sqrt{13}+3 \right ) \left ( 11\,\sqrt{13}-39 \right ) \left ( \sqrt{13}+3-2\,\tan \left ( dx+c \right ) \right ) \sqrt{13}}{ \left ( 52\,\sqrt{13}-156+104\,\tan \left ( dx+c \right ) \right ) \sqrt{-6+2\,\sqrt{13}}}{\frac{1}{\sqrt{-{\frac{\tan \left ( dx+c \right ) \left ( 2+3\,\tan \left ( dx+c \right ) \right ) }{ \left ( \sqrt{13}-3+2\,\tan \left ( dx+c \right ) \right ) ^{2}}}}}}} \right ) -4\,\arctan \left ( 4\,{\frac{\sqrt{13}}{\sqrt{26\,\sqrt{13}+78}}\sqrt{-{\frac{\tan \left ( dx+c \right ) \left ( 2+3\,\tan \left ( dx+c \right ) \right ) }{ \left ( \sqrt{13}-3+2\,\tan \left ( dx+c \right ) \right ) ^{2}}}}} \right ) \sqrt{13}+12\,\arctan \left ( 4\,{\frac{\sqrt{13}}{\sqrt{26\,\sqrt{13}+78}}\sqrt{-{\frac{\tan \left ( dx+c \right ) \left ( 2+3\,\tan \left ( dx+c \right ) \right ) }{ \left ( \sqrt{13}-3+2\,\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(1/(-2-3*tan(d*x+c))^(1/2)/tan(d*x+c)^(1/2),x)

[Out]

-1/2/d*(-2-3*tan(d*x+c))^(1/2)*(-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)*arctanh(1/52*(13^(1/2)+3)*(11*13^(1/2)-39)*
(13^(1/2)+3-2*tan(d*x+c))/(13^(1/2)-3+2*tan(d*x+c))/(-6+2*13^(1/2))^(1/2)*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))-11*(2*13^(1/2)+6)^(1/2)*(-6+2*13^(1/2))^(1/2)*arctanh(1/52*(13^(1/2)+3)
*(11*13^(1/2)-39)*(13^(1/2)+3-2*tan(d*x+c))/(13^(1/2)-3+2*tan(d*x+c))/(-6+2*13^(1/2))^(1/2)*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))-4*arctan(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*arctan(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))/(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/(-2-3*tan(d*x+c))^(1/2)/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/(-2-3*tan(d*x+c))^(1/2)/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{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/(-2-3*tan(d*x+c))**(1/2)/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/(-2-3*tan(d*x+c))^(1/2)/tan(d*x+c)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError