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

Optimal. Leaf size=40 \[ -\frac {2 \sqrt [4]{-1} a \tanh ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {d} f} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3533, 208} \[ -\frac {2 \sqrt [4]{-1} a \tanh ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {d} f} \]

Antiderivative was successfully verified.

[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 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 3533

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*} \int \frac {a-i a \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx &=\frac {\left (2 a^2\right ) \operatorname {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 \tanh ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {d} f}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 50, normalized size = 1.25 \[ -\frac {2 \sqrt [4]{-1} a \sqrt {\tan (e+f x)} \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\tan (e+f x)}\right )}{f \sqrt {d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[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]])

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fricas [C]  time = 0.42, size = 169, normalized size = 4.22 \[ \frac {1}{2} \, \sqrt {\frac {i \, a^{2}}{d f^{2}}} \log \left (\frac {d f \sqrt {\frac {i \, a^{2}}{d f^{2}}} + {\left (a e^{\left (2 i \, f x + 2 i \, e\right )} + a\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{a}\right ) - \frac {1}{2} \, \sqrt {\frac {i \, a^{2}}{d f^{2}}} \log \left (-\frac {d f \sqrt {\frac {i \, a^{2}}{d f^{2}}} - {\left (a e^{\left (2 i \, f x + 2 i \, e\right )} + a\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{a}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [C]  time = 2.19, size = 67, normalized size = 1.68 \[ \frac {2 \, \sqrt {2} a \arctan \left (\frac {16 i \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{8 i \, \sqrt {2} d^{\frac {3}{2}} + 8 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{\sqrt {d} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(2)*a*arctan(16*I*sqrt(d^2)*sqrt(d*tan(f*x + e))/(8*I*sqrt(2)*d^(3/2) + 8*sqrt(2)*sqrt(d^2)*sqrt(d)))/(s
qrt(d)*f*(I*d/sqrt(d^2) + 1))

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maple [C]  time = 0.20, size = 330, normalized size = 8.25 \[ \frac {a \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{4 f d}+\frac {a \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f d}-\frac {a \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f d}-\frac {i a \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{4 f \left (d^{2}\right )^{\frac {1}{4}}}-\frac {i a \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f \left (d^{2}\right )^{\frac {1}{4}}}+\frac {i a \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f \left (d^{2}\right )^{\frac {1}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.60, size = 154, normalized size = 3.85 \[ \frac {a {\left (-\frac {\left (2 i - 2\right ) \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (2 i - 2\right ) \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\left (i + 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\left (i + 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )}}{4 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 4.22, size = 30, normalized size = 0.75 \[ -\frac {{\left (-1\right )}^{1/4}\,a\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )\,2{}\mathrm {i}}{\sqrt {d}\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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))

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