\(\int \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)} \, dx\) [395]

Optimal result

Integrand size = 30, antiderivative size = 323 \[ \int \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)} \, dx=\frac {i \sqrt {2} \sqrt {a} \sqrt {e} \arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{d}-\frac {i \sqrt {2} \sqrt {a} \sqrt {e} \arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{d}-\frac {i \sqrt {a} \sqrt {e} \log \left (a-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))\right )}{\sqrt {2} d}+\frac {i \sqrt {a} \sqrt {e} \log \left (a+\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))\right )}{\sqrt {2} d} \]

[Out]

-1/2*I*ln(a-2^(1/2)*a^(1/2)*e^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/(e*sec(d*x+c))^(1/2)+cos(d*x+c)*(a+I*a*tan(d*x+c)
))*a^(1/2)*e^(1/2)/d*2^(1/2)+1/2*I*ln(a+2^(1/2)*a^(1/2)*e^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/(e*sec(d*x+c))^(1/2)+
cos(d*x+c)*(a+I*a*tan(d*x+c)))*a^(1/2)*e^(1/2)/d*2^(1/2)+I*arctan(1-2^(1/2)*e^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/a
^(1/2)/(e*sec(d*x+c))^(1/2))*2^(1/2)*a^(1/2)*e^(1/2)/d-I*arctan(1+2^(1/2)*e^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/a^(
1/2)/(e*sec(d*x+c))^(1/2))*2^(1/2)*a^(1/2)*e^(1/2)/d

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {3576, 303, 1176, 631, 210, 1179, 642} \[ \int \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)} \, dx=\frac {i \sqrt {2} \sqrt {a} \sqrt {e} \arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{d}-\frac {i \sqrt {2} \sqrt {a} \sqrt {e} \arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{d}-\frac {i \sqrt {a} \sqrt {e} \log \left (-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{\sqrt {2} d}+\frac {i \sqrt {a} \sqrt {e} \log \left (\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{\sqrt {2} d} \]

[In]

Int[Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(I*Sqrt[2]*Sqrt[a]*Sqrt[e]*ArcTan[1 - (Sqrt[2]*Sqrt[e]*Sqrt[a + I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e*Sec[c + d*x
]])])/d - (I*Sqrt[2]*Sqrt[a]*Sqrt[e]*ArcTan[1 + (Sqrt[2]*Sqrt[e]*Sqrt[a + I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e*S
ec[c + d*x]])])/d - (I*Sqrt[a]*Sqrt[e]*Log[a - (Sqrt[2]*Sqrt[a]*Sqrt[e]*Sqrt[a + I*a*Tan[c + d*x]])/Sqrt[e*Sec
[c + d*x]] + Cos[c + d*x]*(a + I*a*Tan[c + d*x])])/(Sqrt[2]*d) + (I*Sqrt[a]*Sqrt[e]*Log[a + (Sqrt[2]*Sqrt[a]*S
qrt[e]*Sqrt[a + I*a*Tan[c + d*x]])/Sqrt[e*Sec[c + d*x]] + Cos[c + d*x]*(a + I*a*Tan[c + d*x])])/(Sqrt[2]*d)

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 303

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 3576

Int[Sqrt[(d_.)*sec[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-4*b*(d^
2/f), Subst[Int[x^2/(a^2 + d^2*x^4), x], x, Sqrt[a + b*Tan[e + f*x]]/Sqrt[d*Sec[e + f*x]]], x] /; FreeQ[{a, b,
d, e, f}, x] && EqQ[a^2 + b^2, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (4 i a e^2\right ) \text {Subst}\left (\int \frac {x^2}{a^2+e^2 x^4} \, dx,x,\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{d} \\ & = \frac {(2 i a e) \text {Subst}\left (\int \frac {a-e x^2}{a^2+e^2 x^4} \, dx,x,\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{d}-\frac {(2 i a e) \text {Subst}\left (\int \frac {a+e x^2}{a^2+e^2 x^4} \, dx,x,\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{d} \\ & = -\frac {(i a) \text {Subst}\left (\int \frac {1}{\frac {a}{e}-\frac {\sqrt {2} \sqrt {a} x}{\sqrt {e}}+x^2} \, dx,x,\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{d}-\frac {(i a) \text {Subst}\left (\int \frac {1}{\frac {a}{e}+\frac {\sqrt {2} \sqrt {a} x}{\sqrt {e}}+x^2} \, dx,x,\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{d}-\frac {\left (i \sqrt {a} \sqrt {e}\right ) \text {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt {a}}{\sqrt {e}}+2 x}{-\frac {a}{e}-\frac {\sqrt {2} \sqrt {a} x}{\sqrt {e}}-x^2} \, dx,x,\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} d}-\frac {\left (i \sqrt {a} \sqrt {e}\right ) \text {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt {a}}{\sqrt {e}}-2 x}{-\frac {a}{e}+\frac {\sqrt {2} \sqrt {a} x}{\sqrt {e}}-x^2} \, dx,x,\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} d} \\ & = -\frac {i \sqrt {a} \sqrt {e} \log \left (a-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))\right )}{\sqrt {2} d}+\frac {i \sqrt {a} \sqrt {e} \log \left (a+\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))\right )}{\sqrt {2} d}-\frac {\left (i \sqrt {2} \sqrt {a} \sqrt {e}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{d}+\frac {\left (i \sqrt {2} \sqrt {a} \sqrt {e}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{d} \\ & = \frac {i \sqrt {2} \sqrt {a} \sqrt {e} \arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{d}-\frac {i \sqrt {2} \sqrt {a} \sqrt {e} \arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{d}-\frac {i \sqrt {a} \sqrt {e} \log \left (a-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))\right )}{\sqrt {2} d}+\frac {i \sqrt {a} \sqrt {e} \log \left (a+\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))\right )}{\sqrt {2} d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.90 (sec) , antiderivative size = 277, normalized size of antiderivative = 0.86 \[ \int \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {2 e \left (\text {arctanh}\left (\frac {\sqrt {1-i \cos (c)+\sin (c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}{\sqrt {-1-i \cos (c)-\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}}\right ) \sqrt {-1-i \cos (c)-\sin (c)} \sqrt {1+i \cos (c)-\sin (c)}-\text {arctanh}\left (\frac {\sqrt {1+i \cos (c)-\sin (c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}{\sqrt {-1+i \cos (c)+\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}}\right ) \sqrt {1-i \cos (c)+\sin (c)} \sqrt {-1+i \cos (c)+\sin (c)}\right ) \sqrt {i+\tan \left (\frac {d x}{2}\right )} \sqrt {a+i a \tan (c+d x)}}{d \sqrt {e \sec (c+d x)} \sqrt {1+\cos (2 c)+i \sin (2 c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}} \]

[In]

Integrate[Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(-2*e*(ArcTanh[(Sqrt[1 - I*Cos[c] + Sin[c]]*Sqrt[I - Tan[(d*x)/2]])/(Sqrt[-1 - I*Cos[c] - Sin[c]]*Sqrt[I + Tan
[(d*x)/2]])]*Sqrt[-1 - I*Cos[c] - Sin[c]]*Sqrt[1 + I*Cos[c] - Sin[c]] - ArcTanh[(Sqrt[1 + I*Cos[c] - Sin[c]]*S
qrt[I - Tan[(d*x)/2]])/(Sqrt[-1 + I*Cos[c] + Sin[c]]*Sqrt[I + Tan[(d*x)/2]])]*Sqrt[1 - I*Cos[c] + Sin[c]]*Sqrt
[-1 + I*Cos[c] + Sin[c]])*Sqrt[I + Tan[(d*x)/2]]*Sqrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[e*Sec[c + d*x]]*Sqrt[1 +
Cos[2*c] + I*Sin[2*c]]*Sqrt[I - Tan[(d*x)/2]])

Maple [A] (verified)

Time = 11.34 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.49

method	result	size
default	\(\frac {\left (-1+i\right ) \sqrt {e \sec \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (i \operatorname {arctanh}\left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )-\operatorname {arctanh}\left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )\right ) \cos \left (d x +c \right )}{d \left (-i \cos \left (d x +c \right )+\sin \left (d x +c \right )-i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\)	\(157\)

[In]

int((e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

(-1+I)/d*(e*sec(d*x+c))^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)*(I*arctanh(1/2*(-cos(d*x+c)+sin(d*x+c)-1)/(cos(d*x+c)
+1)/(1/(cos(d*x+c)+1))^(1/2))-arctanh(1/2*(cos(d*x+c)+sin(d*x+c)+1)/(cos(d*x+c)+1)/(1/(cos(d*x+c)+1))^(1/2)))*
cos(d*x+c)/(-I*cos(d*x+c)+sin(d*x+c)-I)/(1/(cos(d*x+c)+1))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.00 \[ \int \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)} \, dx=\frac {1}{2} \, \sqrt {\frac {4 i \, a e}{d^{2}}} \log \left (2 \, \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + d \sqrt {\frac {4 i \, a e}{d^{2}}}\right ) - \frac {1}{2} \, \sqrt {\frac {4 i \, a e}{d^{2}}} \log \left (2 \, \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} - d \sqrt {\frac {4 i \, a e}{d^{2}}}\right ) - \frac {1}{2} \, \sqrt {-\frac {4 i \, a e}{d^{2}}} \log \left (2 \, \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + d \sqrt {-\frac {4 i \, a e}{d^{2}}}\right ) + \frac {1}{2} \, \sqrt {-\frac {4 i \, a e}{d^{2}}} \log \left (2 \, \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} - d \sqrt {-\frac {4 i \, a e}{d^{2}}}\right ) \]

[In]

integrate((e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/2*sqrt(4*I*a*e/d^2)*log(2*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x +
2*I*c) + 1)*e^(1/2*I*d*x + 1/2*I*c) + d*sqrt(4*I*a*e/d^2)) - 1/2*sqrt(4*I*a*e/d^2)*log(2*sqrt(a/(e^(2*I*d*x +
2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(1/2*I*d*x + 1/2*I*c) - d*sqrt(4*I*
a*e/d^2)) - 1/2*sqrt(-4*I*a*e/d^2)*log(2*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(
e^(2*I*d*x + 2*I*c) + 1)*e^(1/2*I*d*x + 1/2*I*c) + d*sqrt(-4*I*a*e/d^2)) + 1/2*sqrt(-4*I*a*e/d^2)*log(2*sqrt(a
/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(1/2*I*d*x + 1/2*I*c
) - d*sqrt(-4*I*a*e/d^2))

Sympy [F]

\[ \int \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)} \, dx=\int \sqrt {e \sec {\left (c + d x \right )}} \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}\, dx \]

[In]

integrate((e*sec(d*x+c))**(1/2)*(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(e*sec(c + d*x))*sqrt(I*a*(tan(c + d*x) - I)), x)

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1400 vs. \(2 (239) = 478\).

Time = 0.48 (sec) , antiderivative size = 1400, normalized size of antiderivative = 4.33 \[ \int \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)} \, dx=\text {Too large to display} \]

[In]

integrate((e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

1/4*(-2*I*sqrt(2)*arctan2(sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1, sqrt(2)*si
n(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) - 2*I*sqrt(2)*arctan2(sqrt(2)*cos(1/3*arctan2(
sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1, -sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x +
3/2*c))) + 1) - 2*I*sqrt(2)*arctan2(sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 1,
sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) - 2*I*sqrt(2)*arctan2(sqrt(2)*cos(1
/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 1, -sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), co
s(3/2*d*x + 3/2*c))) + 1) - 2*sqrt(2)*arctan2(sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*
c))) + sin(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))), sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2
*c), cos(3/2*d*x + 3/2*c))) + cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) + 2*sqrt(2)*ar
ctan2(-sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + sin(2/3*arctan2(sin(3/2*d*x + 3/
2*c), cos(3/2*d*x + 3/2*c))), -sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + cos(2/3*
arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) + I*sqrt(2)*log(2*sqrt(2)*sin(2/3*arctan2(sin(3/2*d*
x + 3/2*c), cos(3/2*d*x + 3/2*c)))*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2*(sqrt(2)*c
os(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1)*cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2
*d*x + 3/2*c))) + cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*cos(1/3*arctan2(sin(3/2*d
*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + sin(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin
(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c),
cos(3/2*d*x + 3/2*c))) + 1) - I*sqrt(2)*log(-2*sqrt(2)*sin(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2
*c)))*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 2*(sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x +
3/2*c), cos(3/2*d*x + 3/2*c))) - 1)*cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + cos(2/3*arc
tan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/
2*c)))^2 + sin(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/
2*c), cos(3/2*d*x + 3/2*c)))^2 - 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) +
sqrt(2)*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x
+ 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) +
2*sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2) - sqrt(2)*log(2*cos(1/3*arctan2(si
n(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^
2 + 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 2*sqrt(2)*sin(1/3*arctan2(sin(3/2
*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2) + sqrt(2)*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x +
3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 - 2*sqrt(2)*cos(1/3*arctan2(sin
(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2*sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*
c))) + 2) - sqrt(2)*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(s
in(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 - 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x +
3/2*c))) - 2*sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2))*sqrt(a)*sqrt(e)/d

Giac [F]

\[ \int \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)} \, dx=\int { \sqrt {e \sec \left (d x + c\right )} \sqrt {i \, a \tan \left (d x + c\right ) + a} \,d x } \]

[In]

integrate((e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(e*sec(d*x + c))*sqrt(I*a*tan(d*x + c) + a), x)

Mupad [F(-1)]

Timed out. \[ \int \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)} \, dx=\int \sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \]

[In]

int((e/cos(c + d*x))^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2),x)

[Out]

int((e/cos(c + d*x))^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2), x)