3.1140 \(\int \frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {a+i a \tan (e+f x)}} \, dx\)

Optimal. Leaf size=121 \[ \frac {i \sqrt {c+d \tan (e+f x)}}{f \sqrt {a+i a \tan (e+f x)}}-\frac {i \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{\sqrt {2} \sqrt {a} f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.21, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {3546, 3544, 208} \[ \frac {i \sqrt {c+d \tan (e+f x)}}{f \sqrt {a+i a \tan (e+f x)}}-\frac {i \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{\sqrt {2} \sqrt {a} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*Sqrt[c - I*d]*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])/(Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e + f*x
]])])/(Sqrt[2]*Sqrt[a]*f) + (I*Sqrt[c + d*Tan[e + f*x]])/(f*Sqrt[a + I*a*Tan[e + f*x]])

Rule 208

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

Rule 3544

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*a*b)/f, Subst[Int[1/(a*c - b*d - 2*a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]

Rule 3546

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(a*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(2*b*f*m), x] - Dist[(a*c - b*d)/(2*b^2), Int[(a + b*Tan[e
+ f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && EqQ[m + n, 0] && LeQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {a+i a \tan (e+f x)}} \, dx &=\frac {i \sqrt {c+d \tan (e+f x)}}{f \sqrt {a+i a \tan (e+f x)}}+\frac {(c-i d) \int \frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}} \, dx}{2 a}\\ &=\frac {i \sqrt {c+d \tan (e+f x)}}{f \sqrt {a+i a \tan (e+f x)}}-\frac {(a (i c+d)) \operatorname {Subst}\left (\int \frac {1}{a c-i a d-2 a^2 x^2} \, dx,x,\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {a+i a \tan (e+f x)}}\right )}{f}\\ &=-\frac {i \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{\sqrt {2} \sqrt {a} f}+\frac {i \sqrt {c+d \tan (e+f x)}}{f \sqrt {a+i a \tan (e+f x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 3.49, size = 182, normalized size = 1.50 \[ \frac {i \left (\sqrt {1+e^{2 i (e+f x)}} \sqrt {c+d \tan (e+f x)}-\sqrt {c-i d} e^{i (e+f x)} \log \left (2 \left (\sqrt {c-i d} e^{i (e+f x)}+\sqrt {1+e^{2 i (e+f x)}} \sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )\right )\right )}{f \sqrt {1+e^{2 i (e+f x)}} \sqrt {a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I*(-(Sqrt[c - I*d]*E^(I*(e + f*x))*Log[2*(Sqrt[c - I*d]*E^(I*(e + f*x)) + Sqrt[1 + E^((2*I)*(e + f*x))]*Sqrt[
c - (I*d*(-1 + E^((2*I)*(e + f*x))))/(1 + E^((2*I)*(e + f*x)))])]) + Sqrt[1 + E^((2*I)*(e + f*x))]*Sqrt[c + d*
Tan[e + f*x]]))/(Sqrt[1 + E^((2*I)*(e + f*x))]*f*Sqrt[a + I*a*Tan[e + f*x]])

________________________________________________________________________________________

fricas [B]  time = 0.55, size = 351, normalized size = 2.90 \[ \frac {{\left (\sqrt {2} a f \sqrt {-\frac {c - i \, d}{a f^{2}}} e^{\left (i \, f x + i \, e\right )} \log \left (i \, \sqrt {2} a f \sqrt {-\frac {c - i \, d}{a f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}\right ) - \sqrt {2} a f \sqrt {-\frac {c - i \, d}{a f^{2}}} e^{\left (i \, f x + i \, e\right )} \log \left (-i \, \sqrt {2} a f \sqrt {-\frac {c - i \, d}{a f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}\right ) + \sqrt {2} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (2 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i\right )}\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Evaluation time: 0.57Error: Bad Argument Type

________________________________________________________________________________________

maple [B]  time = 0.44, size = 877, normalized size = 7.25 \[ \frac {\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (i \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \ln \left (\frac {3 c a +i a \tan \left (f x +e \right ) c -i d a +3 a \tan \left (f x +e \right ) d +2 \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}}{\tan \left (f x +e \right )+i}\right ) \left (\tan ^{2}\left (f x +e \right )\right ) d -2 i \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \ln \left (\frac {3 c a +i a \tan \left (f x +e \right ) c -i d a +3 a \tan \left (f x +e \right ) d +2 \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}}{\tan \left (f x +e \right )+i}\right ) \tan \left (f x +e \right ) c +\sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \ln \left (\frac {3 c a +i a \tan \left (f x +e \right ) c -i d a +3 a \tan \left (f x +e \right ) d +2 \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}}{\tan \left (f x +e \right )+i}\right ) \left (\tan ^{2}\left (f x +e \right )\right ) c -i \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \ln \left (\frac {3 c a +i a \tan \left (f x +e \right ) c -i d a +3 a \tan \left (f x +e \right ) d +2 \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}}{\tan \left (f x +e \right )+i}\right ) d +2 \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \ln \left (\frac {3 c a +i a \tan \left (f x +e \right ) c -i d a +3 a \tan \left (f x +e \right ) d +2 \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}}{\tan \left (f x +e \right )+i}\right ) \tan \left (f x +e \right ) d +4 i \tan \left (f x +e \right ) c \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}-\sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \ln \left (\frac {3 c a +i a \tan \left (f x +e \right ) c -i d a +3 a \tan \left (f x +e \right ) d +2 \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}}{\tan \left (f x +e \right )+i}\right ) c +4 i \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, d -4 \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, \tan \left (f x +e \right ) d +4 c \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\right )}{4 f a \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, \left (i c -d \right ) \left (-\tan \left (f x +e \right )+i\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/f*(c+d*tan(f*x+e))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)/a*(I*2^(1/2)*(-a*(I*d-c))^(1/2)*ln((3*c*a+I*a*tan(f*x+
e)*c-I*d*a+3*a*tan(f*x+e)*d+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x
+e)+I))*tan(f*x+e)^2*d-2*I*2^(1/2)*(-a*(I*d-c))^(1/2)*ln((3*c*a+I*a*tan(f*x+e)*c-I*d*a+3*a*tan(f*x+e)*d+2*2^(1
/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*tan(f*x+e)*c+2^(1/2)*(-a*(
I*d-c))^(1/2)*ln((3*c*a+I*a*tan(f*x+e)*c-I*d*a+3*a*tan(f*x+e)*d+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e
))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*tan(f*x+e)^2*c-I*2^(1/2)*(-a*(I*d-c))^(1/2)*ln((3*c*a+I*a*tan(f*x+
e)*c-I*d*a+3*a*tan(f*x+e)*d+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x
+e)+I))*d+2*2^(1/2)*(-a*(I*d-c))^(1/2)*ln((3*c*a+I*a*tan(f*x+e)*c-I*d*a+3*a*tan(f*x+e)*d+2*2^(1/2)*(-a*(I*d-c)
)^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*tan(f*x+e)*d+4*I*tan(f*x+e)*c*(a*(c+d*tan
(f*x+e))*(1+I*tan(f*x+e)))^(1/2)-2^(1/2)*(-a*(I*d-c))^(1/2)*ln((3*c*a+I*a*tan(f*x+e)*c-I*d*a+3*a*tan(f*x+e)*d+
2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*c+4*I*(a*(c+d*tan(f*
x+e))*(1+I*tan(f*x+e)))^(1/2)*d-4*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)*d+4*c*(a*(c+d*tan(f*x
+e))*(1+I*tan(f*x+e)))^(1/2))/(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)/(I*c-d)/(-tan(f*x+e)+I)^2

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

mupad [B]  time = 19.83, size = 1724, normalized size = 14.25 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(c + d*1i)*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2)))/(d*f*((c + d*tan(e + f*x))^(1/2) - c^(1/2))*((a*1i)/d
+ ((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2))^2/((c + d*tan(e + f*x))^(1/2) - c^(1/2))^2 - (a^(1/2)*c^(1/2)*((a
+ a*tan(e + f*x)*1i)^(1/2) - a^(1/2))*2i)/(d*((c + d*tan(e + f*x))^(1/2) - c^(1/2))))) - (2^(1/2)*atan(((2^(1
/2)*(d*1i - c)^(1/2)*(4*d^7*(4*a^(3/2)*c^(3/2)*f - a^(3/2)*c^(1/2)*d*f*4i) + (16*d^7*((a + a*tan(e + f*x)*1i)^
(1/2) - a^(1/2))*(a*d^2*f - a*c^2*f + a*c*d*f*2i))/((c + d*tan(e + f*x))^(1/2) - c^(1/2)) - (4*d^8*(a^(1/2)*c^
(3/2)*f*4i + 4*a^(1/2)*c^(1/2)*d*f)*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2))^2)/((c + d*tan(e + f*x))^(1/2) -
c^(1/2))^2 - (2^(1/2)*(d*1i - c)^(1/2)*(4*d^7*(a^2*c*f^2*4i - 4*a^2*d*f^2) - (16*d^7*(a^(3/2)*c^(3/2)*f^2*2i
+ 6*a^(3/2)*c^(1/2)*d*f^2)*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2)))/((c + d*tan(e + f*x))^(1/2) - c^(1/2)) +
(4*d^8*(20*a*c*f^2 - a*d*f^2*12i)*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2))^2)/((c + d*tan(e + f*x))^(1/2) -
c^(1/2))^2))/(4*a^(1/2)*f))*1i)/(4*a^(1/2)*f) + (2^(1/2)*(d*1i - c)^(1/2)*(4*d^7*(4*a^(3/2)*c^(3/2)*f - a^(3/2
)*c^(1/2)*d*f*4i) + (16*d^7*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2))*(a*d^2*f - a*c^2*f + a*c*d*f*2i))/((c +
d*tan(e + f*x))^(1/2) - c^(1/2)) - (4*d^8*(a^(1/2)*c^(3/2)*f*4i + 4*a^(1/2)*c^(1/2)*d*f)*((a + a*tan(e + f*x)*
1i)^(1/2) - a^(1/2))^2)/((c + d*tan(e + f*x))^(1/2) - c^(1/2))^2 + (2^(1/2)*(d*1i - c)^(1/2)*(4*d^7*(a^2*c*f^2
*4i - 4*a^2*d*f^2) - (16*d^7*(a^(3/2)*c^(3/2)*f^2*2i + 6*a^(3/2)*c^(1/2)*d*f^2)*((a + a*tan(e + f*x)*1i)^(1/2)
- a^(1/2)))/((c + d*tan(e + f*x))^(1/2) - c^(1/2)) + (4*d^8*(20*a*c*f^2 - a*d*f^2*12i)*((a + a*tan(e + f*x)*1
i)^(1/2) - a^(1/2))^2)/((c + d*tan(e + f*x))^(1/2) - c^(1/2))^2))/(4*a^(1/2)*f))*1i)/(4*a^(1/2)*f))/(8*d^7*(a*
c^2*1i - a*d^2*1i + 2*a*c*d) + (8*d^8*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2))^2*(c*d*2i - c^2 + d^2))/((c +
d*tan(e + f*x))^(1/2) - c^(1/2))^2 + (2^(1/2)*(d*1i - c)^(1/2)*(4*d^7*(4*a^(3/2)*c^(3/2)*f - a^(3/2)*c^(1/2)*d
*f*4i) + (16*d^7*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2))*(a*d^2*f - a*c^2*f + a*c*d*f*2i))/((c + d*tan(e + f
*x))^(1/2) - c^(1/2)) - (4*d^8*(a^(1/2)*c^(3/2)*f*4i + 4*a^(1/2)*c^(1/2)*d*f)*((a + a*tan(e + f*x)*1i)^(1/2) -
a^(1/2))^2)/((c + d*tan(e + f*x))^(1/2) - c^(1/2))^2 - (2^(1/2)*(d*1i - c)^(1/2)*(4*d^7*(a^2*c*f^2*4i - 4*a^2
*d*f^2) - (16*d^7*(a^(3/2)*c^(3/2)*f^2*2i + 6*a^(3/2)*c^(1/2)*d*f^2)*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2))
)/((c + d*tan(e + f*x))^(1/2) - c^(1/2)) + (4*d^8*(20*a*c*f^2 - a*d*f^2*12i)*((a + a*tan(e + f*x)*1i)^(1/2) -
a^(1/2))^2)/((c + d*tan(e + f*x))^(1/2) - c^(1/2))^2))/(4*a^(1/2)*f)))/(4*a^(1/2)*f) - (2^(1/2)*(d*1i - c)^(1/
2)*(4*d^7*(4*a^(3/2)*c^(3/2)*f - a^(3/2)*c^(1/2)*d*f*4i) + (16*d^7*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2))*(
a*d^2*f - a*c^2*f + a*c*d*f*2i))/((c + d*tan(e + f*x))^(1/2) - c^(1/2)) - (4*d^8*(a^(1/2)*c^(3/2)*f*4i + 4*a^(
1/2)*c^(1/2)*d*f)*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2))^2)/((c + d*tan(e + f*x))^(1/2) - c^(1/2))^2 + (2^(
1/2)*(d*1i - c)^(1/2)*(4*d^7*(a^2*c*f^2*4i - 4*a^2*d*f^2) - (16*d^7*(a^(3/2)*c^(3/2)*f^2*2i + 6*a^(3/2)*c^(1/2
)*d*f^2)*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2)))/((c + d*tan(e + f*x))^(1/2) - c^(1/2)) + (4*d^8*(20*a*c*f^
2 - a*d*f^2*12i)*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2))^2)/((c + d*tan(e + f*x))^(1/2) - c^(1/2))^2))/(4*a^
(1/2)*f)))/(4*a^(1/2)*f)))*(d*1i - c)^(1/2)*1i)/(2*a^(1/2)*f)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________