\(\int \frac {a-i a \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx\) [145]

Optimal result

Integrand size = 26, antiderivative size = 40 \[ \int \frac {a-i a \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx=-\frac {2 \sqrt [4]{-1} a \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {d} f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3614, 214} \[ \int \frac {a-i a \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx=-\frac {2 \sqrt [4]{-1} a \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {d} f} \]

[In]

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

[Out]

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

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 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]

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.25 \[ \int \frac {a-i a \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx=-\frac {2 \sqrt [4]{-1} a \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (e+f x)}\right ) \sqrt {\tan (e+f x)}}{f \sqrt {d \tan (e+f x)}} \]

[In]

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

[Out]

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.92 (sec) , antiderivative size = 274, normalized size of antiderivative = 6.85

method	result	size
derivativedivides	\(-\frac {a \left (-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d}+\frac {i \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f}\)	\(274\)
default	\(-\frac {a \left (-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d}+\frac {i \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f}\)	\(274\)
parts	\(\frac {a \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 f d}-\frac {i a \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 f \left (d^{2}\right )^{\frac {1}{4}}}\)	\(276\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.23 (sec) , antiderivative size = 169, normalized size of antiderivative = 4.22 \[ \int \frac {a-i a \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx=\frac {1}{2} \, \sqrt {\frac {i \, a^{2}}{d f^{2}}} \log \left (\frac {d f \sqrt {\frac {i \, a^{2}}{d f^{2}}} + {\left (a e^{\left (2 i \, f x + 2 i \, e\right )} + a\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{a}\right ) - \frac {1}{2} \, \sqrt {\frac {i \, a^{2}}{d f^{2}}} \log \left (-\frac {d f \sqrt {\frac {i \, a^{2}}{d f^{2}}} - {\left (a e^{\left (2 i \, f x + 2 i \, e\right )} + a\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{a}\right ) \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 154, normalized size of antiderivative = 3.85 \[ \int \frac {a-i a \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx=\frac {a {\left (-\frac {\left (2 i - 2\right ) \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (2 i - 2\right ) \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\left (i + 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\left (i + 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )}}{4 \, f} \]

[In]

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

[Out]

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

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.53 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.65 \[ \int \frac {a-i a \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx=-\frac {2 i \, \sqrt {2} a \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{-4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{\sqrt {d} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} \]

[In]

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

[Out]

-2*I*sqrt(2)*a*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)))/(
sqrt(d)*f*(-I*d/sqrt(d^2) + 1))

Mupad [B] (verification not implemented)

Time = 5.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.75 \[ \int \frac {a-i a \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx=-\frac {{\left (-1\right )}^{1/4}\,a\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )\,2{}\mathrm {i}}{\sqrt {d}\,f} \]

[In]

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

[Out]

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