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

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

[Out]

arctan((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)+arctan((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)

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Rubi [A]  time = 0.10, 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} \[ \frac {\tan ^{-1}\left (\frac {\sqrt {2-3 i} \sqrt {\tan (c+d x)}}{\sqrt {3-2 \tan (c+d x)}}\right )}{\sqrt {2-3 i} d}+\frac {\tan ^{-1}\left (\frac {\sqrt {2+3 i} \sqrt {\tan (c+d x)}}{\sqrt {3-2 \tan (c+d x)}}\right )}{\sqrt {2+3 i} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[(Sqrt[2 - 3*I]*Sqrt[Tan[c + d*x]])/Sqrt[3 - 2*Tan[c + d*x]]]/(Sqrt[2 - 3*I]*d) + ArcTan[(Sqrt[2 + 3*I]*
Sqrt[Tan[c + d*x]])/Sqrt[3 - 2*Tan[c + d*x]]]/(Sqrt[2 + 3*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 {3-2 \tan (c+d x)} \sqrt {\tan (c+d x)}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {3-2 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 {3-2 x} (i-x) \sqrt {x}}+\frac {i}{2 \sqrt {3-2 x} \sqrt {x} (i+x)}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {i \operatorname {Subst}\left (\int \frac {1}{\sqrt {3-2 x} (i-x) \sqrt {x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {i \operatorname {Subst}\left (\int \frac {1}{\sqrt {3-2 x} \sqrt {x} (i+x)} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac {i \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 \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 {\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 {\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*}

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Mathematica [A]  time = 0.17, size = 101, normalized size = 1.13 \[ \frac {\sqrt {-2+3 i} \tanh ^{-1}\left (\frac {\sqrt {-\frac {2}{13}+\frac {3 i}{13}} \sqrt {3-2 \tan (c+d x)}}{\sqrt {\tan (c+d x)}}\right )-\sqrt {2+3 i} \tan ^{-1}\left (\frac {\sqrt {\frac {2}{13}+\frac {3 i}{13}} \sqrt {3-2 \tan (c+d x)}}{\sqrt {\tan (c+d x)}}\right )}{\sqrt {13} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(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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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/(3-2*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 \[ \int \frac {1}{\sqrt {-2 \, \tan \left (d x + c\right ) + 3} \sqrt {\tan \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.31, size = 435, normalized size = 4.89 \[ \frac {\sqrt {3-2 \tan \left (d x +c \right )}\, \sqrt {-\frac {\tan \left (d x +c \right ) \left (-3+2 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right )^{2}}}\, \left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right ) \left (4 \sqrt {-4+2 \sqrt {13}}\, \arctanh \left (\frac {\left (\sqrt {13}+2\right ) \left (\sqrt {13}+2+3 \tan \left (d x +c \right )\right ) \left (17 \sqrt {13}-52\right ) \sqrt {13}}{351 \sqrt {-4+2 \sqrt {13}}\, \left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right ) \sqrt {-\frac {\tan \left (d x +c \right ) \left (-3+2 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right )^{2}}}}\right ) \sqrt {13}\, \sqrt {2 \sqrt {13}+4}-17 \sqrt {-4+2 \sqrt {13}}\, \arctanh \left (\frac {\left (\sqrt {13}+2\right ) \left (\sqrt {13}+2+3 \tan \left (d x +c \right )\right ) \left (17 \sqrt {13}-52\right ) \sqrt {13}}{351 \sqrt {-4+2 \sqrt {13}}\, \left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right ) \sqrt {-\frac {\tan \left (d x +c \right ) \left (-3+2 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right )^{2}}}}\right ) \sqrt {2 \sqrt {13}+4}-18 \arctan \left (\frac {6 \sqrt {13}\, \sqrt {-\frac {\tan \left (d x +c \right ) \left (-3+2 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right )^{2}}}}{\sqrt {26 \sqrt {13}+52}}\right ) \sqrt {13}+36 \arctan \left (\frac {6 \sqrt {13}\, \sqrt {-\frac {\tan \left (d x +c \right ) \left (-3+2 \tan \left (d x +c \right )\right )}{\left (\sqrt {13}-2-3 \tan \left (d x +c \right )\right )^{2}}}}{\sqrt {26 \sqrt {13}+52}}\right )\right )}{2 d \sqrt {\tan \left (d x +c \right )}\, \sqrt {2 \sqrt {13}+4}\, \left (-3+2 \tan \left (d x +c \right )\right ) \left (17 \sqrt {13}-52\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/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*(-4+2*13^(1/2))^(1/2)*arctanh(1/351*(13^(1/2)+2)*(13^(1/2)+2+3*tan(d*x+c))*(17*13^(1/2)-52)/(-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)-17*(-4+2*13^(1/2))^(1/2)*arctanh(1/351*(13^(1/2)+2)*(13^(1/2)+2+3*tan(
d*x+c))*(17*13^(1/2)-52)/(-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)-18*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))*13^(1/2)+36*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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-2 \, \tan \left (d x + c\right ) + 3} \sqrt {\tan \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 6.36, size = 201, normalized size = 2.26 \[ -\mathrm {atan}\left (\frac {\sqrt {3}\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {-\frac {1}{26}+\frac {3}{52}{}\mathrm {i}}{d^2}}\,\left (6-4{}\mathrm {i}\right )+d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {-\frac {1}{26}+\frac {3}{52}{}\mathrm {i}}{d^2}}\,\sqrt {3-2\,\mathrm {tan}\left (c+d\,x\right )}\,\left (-6+4{}\mathrm {i}\right )}{2\,\mathrm {tan}\left (c+d\,x\right )+\sqrt {3}\,\sqrt {3-2\,\mathrm {tan}\left (c+d\,x\right )}-3}\right )\,\sqrt {\frac {-\frac {1}{26}+\frac {3}{52}{}\mathrm {i}}{d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {\sqrt {3}\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {-\frac {1}{26}-\frac {3}{52}{}\mathrm {i}}{d^2}}\,\left (6+4{}\mathrm {i}\right )+d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {-\frac {1}{26}-\frac {3}{52}{}\mathrm {i}}{d^2}}\,\sqrt {3-2\,\mathrm {tan}\left (c+d\,x\right )}\,\left (-6-4{}\mathrm {i}\right )}{2\,\mathrm {tan}\left (c+d\,x\right )+\sqrt {9-6\,\mathrm {tan}\left (c+d\,x\right )}-3}\right )\,\sqrt {\frac {-\frac {1}{26}-\frac {3}{52}{}\mathrm {i}}{d^2}}\,2{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atan((3^(1/2)*d*tan(c + d*x)^(1/2)*((- 1/26 - 3i/52)/d^2)^(1/2)*(6 + 4i) - d*tan(c + d*x)^(1/2)*((- 1/26 - 3i/
52)/d^2)^(1/2)*(3 - 2*tan(c + d*x))^(1/2)*(6 + 4i))/(2*tan(c + d*x) + (9 - 6*tan(c + d*x))^(1/2) - 3))*((- 1/2
6 - 3i/52)/d^2)^(1/2)*2i - atan((3^(1/2)*d*tan(c + d*x)^(1/2)*((- 1/26 + 3i/52)/d^2)^(1/2)*(6 - 4i) - d*tan(c
+ d*x)^(1/2)*((- 1/26 + 3i/52)/d^2)^(1/2)*(3 - 2*tan(c + d*x))^(1/2)*(6 - 4i))/(2*tan(c + d*x) + 3^(1/2)*(3 -
2*tan(c + d*x))^(1/2) - 3))*((- 1/26 + 3i/52)/d^2)^(1/2)*2i

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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