3.7.55 \(\int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {2+3 \tan (c+d x)}} \, dx\) [655]

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

[Out]

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)+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 = 0.08, 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 = {3656, 926, 95, 214} \begin {gather*} \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} \end {gather*}

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 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 926

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

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Mathematica [A]
time = 0.19, size = 89, normalized size = 1.00 \begin {gather*} \frac {\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 {\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[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] Leaf count of result is larger than twice the leaf count of optimal. \(479\) vs. \(2(73)=146\).
time = 2.95, size = 480, normalized size = 5.39

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 (3 \sqrt {13}\, \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}-11 \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}+4 \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}-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 )\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 )}\)	\(480\)
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 (3 \sqrt {13}\, \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}-11 \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}+4 \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}-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 )\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 )}\)	\(480\)

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,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))*(3*13^(1/2)*(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)-11*(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)+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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

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]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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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(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {3 \tan {\left (c + d x \right )} + 2} \sqrt {\tan {\left (c + d x \right )}}}\, dx \end {gather*}

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 [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 489 vs. \(2 (65) = 130\).
time = 0.57, size = 489, normalized size = 5.49 \begin {gather*} -\frac {\sqrt {3} {\left (\left (3 i + 2\right ) \, \sqrt {6 \, \sqrt {13} - 18} {\left (-\frac {2 i}{\sqrt {13} - 3} + 1\right )} \log \left (\left (120 i + 40\right ) \, \sqrt {13} {\left (\sqrt {3} \sqrt {\tan \left (d x + c\right )} - \sqrt {3 \, \tan \left (d x + c\right ) + 2}\right )}^{2} + \left (432 i + 144\right ) \, {\left (\sqrt {3} \sqrt {\tan \left (d x + c\right )} - \sqrt {3 \, \tan \left (d x + c\right ) + 2}\right )}^{2} + 80 \, \sqrt {13} \sqrt {15 \, \sqrt {13} + 54} - 800 \, \sqrt {13} - \left (16 i - 288\right ) \, \sqrt {15 \, \sqrt {13} + 54} - 2880\right ) - \left (3 i + 2\right ) \, \sqrt {6 \, \sqrt {13} - 18} {\left (-\frac {2 i}{\sqrt {13} - 3} + 1\right )} \log \left (\left (120 i + 40\right ) \, \sqrt {13} {\left (\sqrt {3} \sqrt {\tan \left (d x + c\right )} - \sqrt {3 \, \tan \left (d x + c\right ) + 2}\right )}^{2} + \left (432 i + 144\right ) \, {\left (\sqrt {3} \sqrt {\tan \left (d x + c\right )} - \sqrt {3 \, \tan \left (d x + c\right ) + 2}\right )}^{2} - 80 \, \sqrt {13} \sqrt {15 \, \sqrt {13} + 54} - 800 \, \sqrt {13} + \left (16 i - 288\right ) \, \sqrt {15 \, \sqrt {13} + 54} - 2880\right ) + \left (2 i + 3\right ) \, \sqrt {6 \, \sqrt {13} + 18} {\left (-\frac {2 i}{\sqrt {13} + 3} + 1\right )} \log \left (8 \, \sqrt {13} {\left (\sqrt {3} \sqrt {\tan \left (d x + c\right )} - \sqrt {3 \, \tan \left (d x + c\right ) + 2}\right )}^{2} - 24 \, {\left (\sqrt {3} \sqrt {\tan \left (d x + c\right )} - \sqrt {3 \, \tan \left (d x + c\right ) + 2}\right )}^{2} + 8 \, \sqrt {13} \sqrt {6 \, \sqrt {13} - 18} - \left (48 i + 16\right ) \, \sqrt {13} + \left (16 i - 24\right ) \, \sqrt {6 \, \sqrt {13} - 18} + 144 i + 48\right ) - \left (2 i + 3\right ) \, \sqrt {6 \, \sqrt {13} + 18} {\left (-\frac {2 i}{\sqrt {13} + 3} + 1\right )} \log \left (8 \, \sqrt {13} {\left (\sqrt {3} \sqrt {\tan \left (d x + c\right )} - \sqrt {3 \, \tan \left (d x + c\right ) + 2}\right )}^{2} - 24 \, {\left (\sqrt {3} \sqrt {\tan \left (d x + c\right )} - \sqrt {3 \, \tan \left (d x + c\right ) + 2}\right )}^{2} - 8 \, \sqrt {13} \sqrt {6 \, \sqrt {13} - 18} - \left (48 i + 16\right ) \, \sqrt {13} - \left (16 i - 24\right ) \, \sqrt {6 \, \sqrt {13} - 18} + 144 i + 48\right )\right )}}{156 \, d} \end {gather*}

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]

-1/156*sqrt(3)*((3*I + 2)*sqrt(6*sqrt(13) - 18)*(-2*I/(sqrt(13) - 3) + 1)*log((120*I + 40)*sqrt(13)*(sqrt(3)*s
qrt(tan(d*x + c)) - sqrt(3*tan(d*x + c) + 2))^2 + (432*I + 144)*(sqrt(3)*sqrt(tan(d*x + c)) - sqrt(3*tan(d*x +
c) + 2))^2 + 80*sqrt(13)*sqrt(15*sqrt(13) + 54) - 800*sqrt(13) - (16*I - 288)*sqrt(15*sqrt(13) + 54) - 2880)
- (3*I + 2)*sqrt(6*sqrt(13) - 18)*(-2*I/(sqrt(13) - 3) + 1)*log((120*I + 40)*sqrt(13)*(sqrt(3)*sqrt(tan(d*x +
c)) - sqrt(3*tan(d*x + c) + 2))^2 + (432*I + 144)*(sqrt(3)*sqrt(tan(d*x + c)) - sqrt(3*tan(d*x + c) + 2))^2 -
80*sqrt(13)*sqrt(15*sqrt(13) + 54) - 800*sqrt(13) + (16*I - 288)*sqrt(15*sqrt(13) + 54) - 2880) + (2*I + 3)*sq
rt(6*sqrt(13) + 18)*(-2*I/(sqrt(13) + 3) + 1)*log(8*sqrt(13)*(sqrt(3)*sqrt(tan(d*x + c)) - sqrt(3*tan(d*x + c)
+ 2))^2 - 24*(sqrt(3)*sqrt(tan(d*x + c)) - sqrt(3*tan(d*x + c) + 2))^2 + 8*sqrt(13)*sqrt(6*sqrt(13) - 18) - (
48*I + 16)*sqrt(13) + (16*I - 24)*sqrt(6*sqrt(13) - 18) + 144*I + 48) - (2*I + 3)*sqrt(6*sqrt(13) + 18)*(-2*I/
(sqrt(13) + 3) + 1)*log(8*sqrt(13)*(sqrt(3)*sqrt(tan(d*x + c)) - sqrt(3*tan(d*x + c) + 2))^2 - 24*(sqrt(3)*sqr
t(tan(d*x + c)) - sqrt(3*tan(d*x + c) + 2))^2 - 8*sqrt(13)*sqrt(6*sqrt(13) - 18) - (48*I + 16)*sqrt(13) - (16*
I - 24)*sqrt(6*sqrt(13) - 18) + 144*I + 48))/d

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Mupad [B]
time = 6.12, size = 207, normalized size = 2.33 \begin {gather*} -\mathrm {atan}\left (\frac {\sqrt {2}\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {\frac {3}{52}-\frac {1}{26}{}\mathrm {i}}{d^2}}\,\left (4-6{}\mathrm {i}\right )+d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {\frac {3}{52}-\frac {1}{26}{}\mathrm {i}}{d^2}}\,\sqrt {3\,\mathrm {tan}\left (c+d\,x\right )+2}\,\left (-4+6{}\mathrm {i}\right )}{3\,\mathrm {tan}\left (c+d\,x\right )-\sqrt {2}\,\sqrt {3\,\mathrm {tan}\left (c+d\,x\right )+2}+2}\right )\,\sqrt {\frac {\frac {3}{52}-\frac {1}{26}{}\mathrm {i}}{d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {\sqrt {2}\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {\frac {3}{52}+\frac {1}{26}{}\mathrm {i}}{d^2}}\,\left (4+6{}\mathrm {i}\right )+d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {\frac {3}{52}+\frac {1}{26}{}\mathrm {i}}{d^2}}\,\sqrt {3\,\mathrm {tan}\left (c+d\,x\right )+2}\,\left (-4-6{}\mathrm {i}\right )}{3\,\mathrm {tan}\left (c+d\,x\right )-\sqrt {2}\,\sqrt {3\,\mathrm {tan}\left (c+d\,x\right )+2}+2}\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(1/(tan(c + d*x)^(1/2)*(3*tan(c + d*x) + 2)^(1/2)),x)

[Out]

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

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