3.7.64 \(\int \frac {\sqrt {\tan (c+d x)}}{\sqrt {-2+3 \tan (c+d x)}} \, dx\) [664]

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

[Out]

-I*arctanh((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)+I*arctanh((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 = 0.07, antiderivative size = 95, 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 = {3656, 924, 95, 214} \begin {gather*} \frac {i \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 {i \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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 95

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 214

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

Rule 924

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 3656

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

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Mathematica [A]
time = 0.07, size = 95, normalized size = 1.00 \begin {gather*} -\frac {i \text {ArcTan}\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 {i \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 {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*ArcTan[(Sqrt[-3 + 2*I]*Sqrt[Tan[c + d*x]])/Sqrt[-2 + 3*Tan[c + d*x]]])/(Sqrt[-3 + 2*I]*d) + (I*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] Leaf count of result is larger than twice the leaf count of optimal. \(478\) vs. \(2(77)=154\).
time = 0.85, size = 479, normalized size = 5.04

method	result	size
derivativedivides	\(-\frac {\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 (\sqrt {2 \sqrt {13}-6}\, \arctan \left (\frac {\sqrt {2 \sqrt {13}-6}\, \sqrt {\frac {\left (11 \sqrt {13}-39\right ) \tan \left (d x +c \right ) \left (39+11 \sqrt {13}\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 (3 \sqrt {13}+11\right ) \left (\sqrt {13}+3+2 \tan \left (d x +c \right )\right ) \left (11 \sqrt {13}-39\right ) \left (\sqrt {13}-3-2 \tan \left (d x +c \right )\right )}{416 \tan \left (d x +c \right ) \left (-2+3 \tan \left (d x +c \right )\right )}\right ) \sqrt {13}\, \sqrt {2 \sqrt {13}+6}-3 \sqrt {2 \sqrt {13}-6}\, \arctan \left (\frac {\sqrt {2 \sqrt {13}-6}\, \sqrt {\frac {\left (11 \sqrt {13}-39\right ) \tan \left (d x +c \right ) \left (39+11 \sqrt {13}\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 (3 \sqrt {13}+11\right ) \left (\sqrt {13}+3+2 \tan \left (d x +c \right )\right ) \left (11 \sqrt {13}-39\right ) \left (\sqrt {13}-3-2 \tan \left (d x +c \right )\right )}{416 \tan \left (d x +c \right ) \left (-2+3 \tan \left (d x +c \right )\right )}\right ) \sqrt {2 \sqrt {13}+6}-12 \arctanh \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}+44 \arctanh \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 )}\, \sqrt {-2+3 \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {13}+6}\, \left (11 \sqrt {13}-39\right )}\)	\(479\)
default	\(-\frac {\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 (\sqrt {2 \sqrt {13}-6}\, \arctan \left (\frac {\sqrt {2 \sqrt {13}-6}\, \sqrt {\frac {\left (11 \sqrt {13}-39\right ) \tan \left (d x +c \right ) \left (39+11 \sqrt {13}\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 (3 \sqrt {13}+11\right ) \left (\sqrt {13}+3+2 \tan \left (d x +c \right )\right ) \left (11 \sqrt {13}-39\right ) \left (\sqrt {13}-3-2 \tan \left (d x +c \right )\right )}{416 \tan \left (d x +c \right ) \left (-2+3 \tan \left (d x +c \right )\right )}\right ) \sqrt {13}\, \sqrt {2 \sqrt {13}+6}-3 \sqrt {2 \sqrt {13}-6}\, \arctan \left (\frac {\sqrt {2 \sqrt {13}-6}\, \sqrt {\frac {\left (11 \sqrt {13}-39\right ) \tan \left (d x +c \right ) \left (39+11 \sqrt {13}\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 (3 \sqrt {13}+11\right ) \left (\sqrt {13}+3+2 \tan \left (d x +c \right )\right ) \left (11 \sqrt {13}-39\right ) \left (\sqrt {13}-3-2 \tan \left (d x +c \right )\right )}{416 \tan \left (d x +c \right ) \left (-2+3 \tan \left (d x +c \right )\right )}\right ) \sqrt {2 \sqrt {13}+6}-12 \arctanh \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}+44 \arctanh \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 )}\, \sqrt {-2+3 \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {13}+6}\, \left (11 \sqrt {13}-39\right )}\)	\(479\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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))*((2*13^(1/2)
-6)^(1/2)*arctan(1/416*(2*13^(1/2)-6)^(1/2)*((11*13^(1/2)-39)*tan(d*x+c)*(39+11*13^(1/2))*(-2+3*tan(d*x+c))/(1
3^(1/2)-3-2*tan(d*x+c))^2)^(1/2)*(3*13^(1/2)+11)*(13^(1/2)+3+2*tan(d*x+c))*(11*13^(1/2)-39)*(13^(1/2)-3-2*tan(
d*x+c))/tan(d*x+c)/(-2+3*tan(d*x+c)))*13^(1/2)*(2*13^(1/2)+6)^(1/2)-3*(2*13^(1/2)-6)^(1/2)*arctan(1/416*(2*13^
(1/2)-6)^(1/2)*((11*13^(1/2)-39)*tan(d*x+c)*(39+11*13^(1/2))*(-2+3*tan(d*x+c))/(13^(1/2)-3-2*tan(d*x+c))^2)^(1
/2)*(3*13^(1/2)+11)*(13^(1/2)+3+2*tan(d*x+c))*(11*13^(1/2)-39)*(13^(1/2)-3-2*tan(d*x+c))/tan(d*x+c)/(-2+3*tan(
d*x+c)))*(2*13^(1/2)+6)^(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))*13^(1/2)+44*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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {\tan {\left (c + d x \right )}}}{\sqrt {3 \tan {\left (c + d x \right )} - 2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, choosing root of [1,0,-702,-2704,70473] at parameters values [0]Invalid _EXT in replace_ext Error:
Bad Argume

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Mupad [B]
time = 5.64, size = 191, normalized size = 2.01 \begin {gather*} -\mathrm {atan}\left (\frac {12\,d\,\sqrt {\frac {-\frac {3}{52}-\frac {1}{26}{}\mathrm {i}}{d^2}}\,\left (\frac {\sqrt {2}\,\sqrt {3}}{3}-\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{\sqrt {3\,\mathrm {tan}\left (c+d\,x\right )-2}\,\left (\frac {3\,{\left (\frac {\sqrt {2}\,\sqrt {3}}{3}-\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}^2}{3\,\mathrm {tan}\left (c+d\,x\right )-2}+1\right )}\right )\,\sqrt {\frac {-\frac {3}{52}-\frac {1}{26}{}\mathrm {i}}{d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {12\,d\,\sqrt {\frac {-\frac {3}{52}+\frac {1}{26}{}\mathrm {i}}{d^2}}\,\left (\frac {\sqrt {2}\,\sqrt {3}}{3}-\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{\sqrt {3\,\mathrm {tan}\left (c+d\,x\right )-2}\,\left (\frac {3\,{\left (\frac {\sqrt {2}\,\sqrt {3}}{3}-\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}^2}{3\,\mathrm {tan}\left (c+d\,x\right )-2}+1\right )}\right )\,\sqrt {\frac {-\frac {3}{52}+\frac {1}{26}{}\mathrm {i}}{d^2}}\,2{}\mathrm {i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atan((12*d*((- 3/52 + 1i/26)/d^2)^(1/2)*((2^(1/2)*3^(1/2))/3 - tan(c + d*x)^(1/2)))/((3*tan(c + d*x) - 2)^(1/2
)*((3*((2^(1/2)*3^(1/2))/3 - tan(c + d*x)^(1/2))^2)/(3*tan(c + d*x) - 2) + 1)))*((- 3/52 + 1i/26)/d^2)^(1/2)*2
i - atan((12*d*((- 3/52 - 1i/26)/d^2)^(1/2)*((2^(1/2)*3^(1/2))/3 - tan(c + d*x)^(1/2)))/((3*tan(c + d*x) - 2)^
(1/2)*((3*((2^(1/2)*3^(1/2))/3 - tan(c + d*x)^(1/2))^2)/(3*tan(c + d*x) - 2) + 1)))*((- 3/52 - 1i/26)/d^2)^(1/
2)*2i

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