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

Optimal. Leaf size=61 \[ \frac {2 (-1)^{3/4} a \sqrt {d} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}+\frac {2 i a \sqrt {d \tan (e+f x)}}{f} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {3528, 3533, 205} \[ \frac {2 (-1)^{3/4} a \sqrt {d} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}+\frac {2 i a \sqrt {d \tan (e+f x)}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 205

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

Rule 3528

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

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

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Mathematica [A]  time = 0.81, size = 85, normalized size = 1.39 \[ \frac {2 i a \sqrt {d \tan (e+f x)} \left (\sqrt {i \tan (e+f x)}-\tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right )\right )}{f \sqrt {i \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.44, size = 244, normalized size = 4.00 \[ \frac {\sqrt {\frac {4 i \, a^{2} d}{f^{2}}} f \log \left (\frac {{\left (-2 i \, a d e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, f\right )} \sqrt {\frac {4 i \, a^{2} d}{f^{2}}} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a}\right ) - \sqrt {\frac {4 i \, a^{2} d}{f^{2}}} f \log \left (\frac {{\left (-2 i \, a d e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, f\right )} \sqrt {\frac {4 i \, a^{2} d}{f^{2}}} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a}\right ) + 8 i \, a \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{4 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.53, size = 89, normalized size = 1.46 \[ \frac {2 \, a {\left (\frac {\sqrt {2} d^{\frac {3}{2}} \arctan \left (\frac {16 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{8 i \, \sqrt {2} d^{\frac {3}{2}} + 8 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {i \, \sqrt {d \tan \left (f x + e\right )} d}{f}\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.17, size = 341, normalized size = 5.59 \[ \frac {2 i a \sqrt {d \tan \left (f x +e \right )}}{f}-\frac {i a \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{4 f}-\frac {i a \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f}+\frac {i a \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f}+\frac {a d \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{4 f \left (d^{2}\right )^{\frac {1}{4}}}+\frac {a d \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f \left (d^{2}\right )^{\frac {1}{4}}}-\frac {a d \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f \left (d^{2}\right )^{\frac {1}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.50, size = 176, normalized size = 2.89 \[ -\frac {a d^{2} {\left (\frac {\left (2 i - 2\right ) \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\left (2 i - 2\right ) \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\left (i + 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\left (i + 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )} - 8 i \, \sqrt {d \tan \left (f x + e\right )} a d}{4 \, d f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.26, size = 128, normalized size = 2.10 \[ \frac {{\left (-1\right )}^{1/4}\,a\,\sqrt {d}\,\left (\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )-\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )\right )}{f}-\frac {{\left (-1\right )}^{1/4}\,a\,\sqrt {d}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{f}-\frac {{\left (-1\right )}^{1/4}\,a\,\sqrt {d}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{f}+\frac {a\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,2{}\mathrm {i}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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