\(\int \frac {\sqrt {d \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx\) [167]

Optimal result

Integrand size = 28, antiderivative size = 81 \[ \int \frac {\sqrt {d \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx=\frac {(-1)^{3/4} \sqrt {d} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a f}+\frac {i \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {3630, 3614, 211} \[ \int \frac {\sqrt {d \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx=\frac {(-1)^{3/4} \sqrt {d} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a f}+\frac {i \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))} \]

[In]

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

[Out]

((-1)^(3/4)*Sqrt[d]*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(2*a*f) + ((I/2)*Sqrt[d*Tan[e + f*x]])/
(f*(a + I*a*Tan[e + f*x]))

Rule 211

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

Rule 3614

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

Rule 3630

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(
a*c + b*d))*((c + d*Tan[e + f*x])^n/(2*(b*c - a*d)*f*(a + b*Tan[e + f*x]))), x] + Dist[1/(2*a*(b*c - a*d)), In
t[(c + d*Tan[e + f*x])^(n - 1)*Simp[a*c*d*(n - 1) + b*c^2 + b*d^2*n - d*(b*c - a*d)*(n - 1)*Tan[e + f*x], x],
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] && LtQ[0
, n, 1]

Rubi steps \begin{align*} \text {integral}& = \frac {i \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}-\frac {\int \frac {\frac {1}{2} i a d^2-\frac {1}{2} a d^2 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{2 a^2 d} \\ & = \frac {i \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}+\frac {d^3 \text {Subst}\left (\int \frac {1}{\frac {1}{2} i a d^3+\frac {1}{2} a d^2 x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{4 f} \\ & = \frac {(-1)^{3/4} \sqrt {d} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a f}+\frac {i \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.09 \[ \int \frac {\sqrt {d \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx=\frac {2 \sqrt {d \tan (e+f x)}-(1+i) \sqrt {2} \sqrt {d} \text {arctanh}\left (\frac {(1+i) \sqrt {d \tan (e+f x)}}{\sqrt {2} \sqrt {d}}\right ) (-i+\tan (e+f x))}{4 a f (-i+\tan (e+f x))} \]

[In]

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

[Out]

(2*Sqrt[d*Tan[e + f*x]] - (1 + I)*Sqrt[2]*Sqrt[d]*ArcTanh[((1 + I)*Sqrt[d*Tan[e + f*x]])/(Sqrt[2]*Sqrt[d])]*(-
I + Tan[e + f*x]))/(4*a*f*(-I + Tan[e + f*x]))

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.85

method	result	size
derivativedivides	\(\frac {d \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{2 f a \sqrt {i d}}+\frac {d \sqrt {d \tan \left (f x +e \right )}}{2 f a \left (d \tan \left (f x +e \right )-i d \right )}\)	\(69\)
default	\(\frac {d \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{2 f a \sqrt {i d}}+\frac {d \sqrt {d \tan \left (f x +e \right )}}{2 f a \left (d \tan \left (f x +e \right )-i d \right )}\)	\(69\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 288 vs. \(2 (61) = 122\).

Time = 0.25 (sec) , antiderivative size = 288, normalized size of antiderivative = 3.56 \[ \int \frac {\sqrt {d \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx=-\frac {{\left (a f \sqrt {\frac {i \, d}{4 \, a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-2 \, {\left (2 \, {\left (i \, a f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {i \, d}{4 \, a^{2} f^{2}}} + i \, d e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) - a f \sqrt {\frac {i \, d}{4 \, a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-2 \, {\left (2 \, {\left (-i \, a f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {i \, d}{4 \, a^{2} f^{2}}} + i \, d e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) - \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (i \, e^{\left (2 i \, f x + 2 i \, e\right )} + i\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a f} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {d \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [A] (verification not implemented)

none

Time = 0.49 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.32 \[ \int \frac {\sqrt {d \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx=\frac {1}{2} \, d^{2} {\left (\frac {\sqrt {2} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a d^{\frac {3}{2}} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {\sqrt {d \tan \left (f x + e\right )}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )} a d f}\right )} \]

[In]

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

[Out]

1/2*d^2*(sqrt(2)*arctan(8*sqrt(d^2)*sqrt(d*tan(f*x + e))/(4*I*sqrt(2)*d^(3/2) + 4*sqrt(2)*sqrt(d^2)*sqrt(d)))/
(a*d^(3/2)*f*(I*d/sqrt(d^2) + 1)) + sqrt(d*tan(f*x + e))/((d*tan(f*x + e) - I*d)*a*d*f))

Mupad [B] (verification not implemented)

Time = 4.69 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.88 \[ \int \frac {\sqrt {d \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx=-\frac {2\,\sqrt {\frac {1}{16}{}\mathrm {i}}\,\sqrt {d}\,\mathrm {atanh}\left (\frac {4\,\sqrt {\frac {1}{16}{}\mathrm {i}}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{a\,f}-\frac {d\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,a\,f\,\left (-d\,\mathrm {tan}\left (e+f\,x\right )+d\,1{}\mathrm {i}\right )} \]

[In]

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

[Out]

- (2*(1i/16)^(1/2)*d^(1/2)*atanh((4*(1i/16)^(1/2)*(d*tan(e + f*x))^(1/2))/d^(1/2)))/(a*f) - (d*(d*tan(e + f*x)
)^(1/2))/(2*a*f*(d*1i - d*tan(e + f*x)))