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

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Rubi [A]  time = 0.11, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.15, 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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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-3 \, \tan \left (d x + c\right ) - 2} \sqrt {\tan \left (d x + c\right )}}\,{d x} \]

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]

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

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maple [B]  time = 0.30, size = 435, normalized size = 4.89 \[ -\frac {\sqrt {-2-3 \tan \left (d x +c \right )}\, \sqrt {-\frac {\tan \left (d x +c \right ) \left (2+3 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-3+2 \tan \left (d x +c \right )\right )^{2}}}\, \left (\sqrt {13}-3+2 \tan \left (d x +c \right )\right ) \left (3 \sqrt {-6+2 \sqrt {13}}\, \sqrt {13}\, \sqrt {2 \sqrt {13}+6}\, \arctanh \left (\frac {\left (\sqrt {13}+3\right ) \left (\sqrt {13}+3-2 \tan \left (d x +c \right )\right ) \left (11 \sqrt {13}-39\right ) \sqrt {13}}{52 \left (\sqrt {13}-3+2 \tan \left (d x +c \right )\right ) \sqrt {-6+2 \sqrt {13}}\, \sqrt {-\frac {\tan \left (d x +c \right ) \left (2+3 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-3+2 \tan \left (d x +c \right )\right )^{2}}}}\right )-11 \sqrt {-6+2 \sqrt {13}}\, \sqrt {2 \sqrt {13}+6}\, \arctanh \left (\frac {\left (\sqrt {13}+3\right ) \left (\sqrt {13}+3-2 \tan \left (d x +c \right )\right ) \left (11 \sqrt {13}-39\right ) \sqrt {13}}{52 \left (\sqrt {13}-3+2 \tan \left (d x +c \right )\right ) \sqrt {-6+2 \sqrt {13}}\, \sqrt {-\frac {\tan \left (d x +c \right ) \left (2+3 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-3+2 \tan \left (d x +c \right )\right )^{2}}}}\right )-4 \arctan \left (\frac {4 \sqrt {13}\, \sqrt {-\frac {\tan \left (d x +c \right ) \left (2+3 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-3+2 \tan \left (d x +c \right )\right )^{2}}}}{\sqrt {26 \sqrt {13}+78}}\right ) \sqrt {13}+12 \arctan \left (\frac {4 \sqrt {13}\, \sqrt {-\frac {\tan \left (d x +c \right ) \left (2+3 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-3+2 \tan \left (d x +c \right )\right )^{2}}}}{\sqrt {26 \sqrt {13}+78}}\right )\right )}{2 d \sqrt {\tan \left (d x +c \right )}\, \left (2+3 \tan \left (d x +c \right )\right ) \sqrt {2 \sqrt {13}+6}\, \left (11 \sqrt {13}-39\right )} \]

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*(-6+2*13^(1/2))^(1/2)*13^(1/2)*(2*13^(1/2)+6)^(1/2)*arctanh(1/52*(13^(1/2)+3)*(13^(1/2)+3-2*tan
(d*x+c))*(11*13^(1/2)-39)/(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*(-6+2*13^(1/2))^(1/2)*(2*13^(1/2)+6)^(1/2)*arctanh(1/52*(13^(1/2)+3)
*(13^(1/2)+3-2*tan(d*x+c))*(11*13^(1/2)-39)/(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-3 \, \tan \left (d x + c\right ) - 2} \sqrt {\tan \left (d x + c\right )}}\,{d x} \]

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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mupad [B]  time = 6.08, size = 129, normalized size = 1.45 \[ -2\,\mathrm {atanh}\left (\frac {4\,d\,\sqrt {\frac {-\frac {3}{52}-\frac {1}{26}{}\mathrm {i}}{d^2}}+6\,d\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {\frac {-\frac {3}{52}-\frac {1}{26}{}\mathrm {i}}{d^2}}}{\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-3\,\mathrm {tan}\left (c+d\,x\right )-2}}\right )\,\sqrt {\frac {-\frac {3}{52}-\frac {1}{26}{}\mathrm {i}}{d^2}}-2\,\mathrm {atanh}\left (\frac {4\,d\,\sqrt {\frac {-\frac {3}{52}+\frac {1}{26}{}\mathrm {i}}{d^2}}+6\,d\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {\frac {-\frac {3}{52}+\frac {1}{26}{}\mathrm {i}}{d^2}}}{\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-3\,\mathrm {tan}\left (c+d\,x\right )-2}}\right )\,\sqrt {\frac {-\frac {3}{52}+\frac {1}{26}{}\mathrm {i}}{d^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- 2*atanh((4*d*((- 3/52 - 1i/26)/d^2)^(1/2) + 6*d*tan(c + d*x)*((- 3/52 - 1i/26)/d^2)^(1/2))/(tan(c + d*x)^(1/
2)*(- 3*tan(c + d*x) - 2)^(1/2)))*((- 3/52 - 1i/26)/d^2)^(1/2) - 2*atanh((4*d*((- 3/52 + 1i/26)/d^2)^(1/2) + 6
*d*tan(c + d*x)*((- 3/52 + 1i/26)/d^2)^(1/2))/(tan(c + d*x)^(1/2)*(- 3*tan(c + d*x) - 2)^(1/2)))*((- 3/52 + 1i
/26)/d^2)^(1/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {- 3 \tan {\left (c + d x \right )} - 2} \sqrt {\tan {\left (c + d x \right )}}}\, dx \]

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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