3.152 \(\int \frac {(a+i a \tan (e+f x))^2}{\sqrt {d \tan (e+f x)}} \, dx\)

Optimal. Leaf size=66 \[ -\frac {4 \sqrt [4]{-1} a^2 \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {d} f}-\frac {2 a^2 \sqrt {d \tan (e+f x)}}{d f} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {3543, 3533, 205} \[ -\frac {4 \sqrt [4]{-1} a^2 \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {d} f}-\frac {2 a^2 \sqrt {d \tan (e+f x)}}{d f} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^2/Sqrt[d*Tan[e + f*x]],x]

[Out]

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

Rule 3543

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

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (e+f x))^2}{\sqrt {d \tan (e+f x)}} \, dx &=-\frac {2 a^2 \sqrt {d \tan (e+f x)}}{d f}+\int \frac {2 a^2+2 i a^2 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx\\ &=-\frac {2 a^2 \sqrt {d \tan (e+f x)}}{d f}+\frac {\left (8 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{2 a^2 d-2 i a^2 x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{f}\\ &=-\frac {4 \sqrt [4]{-1} a^2 \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {d} f}-\frac {2 a^2 \sqrt {d \tan (e+f x)}}{d f}\\ \end {align*}

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Mathematica [C]  time = 1.95, size = 157, normalized size = 2.38 \[ -\frac {2 a^2 e^{-2 i (e+f x)} \sqrt {\tan (e+f x)} (\cos (2 (e+f x))+i \sin (2 (e+f x))) \left (\tan (e+f x)+2 i \sqrt {i \tan (e+f x)} \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right )\right )}{f \sqrt {-\frac {i \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}} \sqrt {d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^2/Sqrt[d*Tan[e + f*x]],x]

[Out]

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

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fricas [C]  time = 0.50, size = 266, normalized size = 4.03 \[ \frac {d \sqrt {-\frac {16 i \, a^{4}}{d f^{2}}} f \log \left (\frac {{\left (-4 i \, a^{2} d e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (d f e^{\left (2 i \, f x + 2 i \, e\right )} + d f\right )} \sqrt {-\frac {16 i \, a^{4}}{d f^{2}}} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a^{2}}\right ) - d \sqrt {-\frac {16 i \, a^{4}}{d f^{2}}} f \log \left (\frac {{\left (-4 i \, a^{2} d e^{\left (2 i \, f x + 2 i \, e\right )} - {\left (d f e^{\left (2 i \, f x + 2 i \, e\right )} + d f\right )} \sqrt {-\frac {16 i \, a^{4}}{d f^{2}}} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a^{2}}\right ) - 8 \, a^{2} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{4 \, d f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(d*sqrt(-16*I*a^4/(d*f^2))*f*log(1/2*(-4*I*a^2*d*e^(2*I*f*x + 2*I*e) + (d*f*e^(2*I*f*x + 2*I*e) + d*f)*sqr
t(-16*I*a^4/(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^
2) - d*sqrt(-16*I*a^4/(d*f^2))*f*log(1/2*(-4*I*a^2*d*e^(2*I*f*x + 2*I*e) - (d*f*e^(2*I*f*x + 2*I*e) + d*f)*sqr
t(-16*I*a^4/(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^
2) - 8*a^2*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))/(d*f)

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giac [C]  time = 0.81, size = 92, normalized size = 1.39 \[ \frac {4 i \, \sqrt {2} a^{2} \arctan \left (\frac {16 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{8 i \, \sqrt {2} d^{\frac {3}{2}} + 8 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{\sqrt {d} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {2 \, \sqrt {d \tan \left (f x + e\right )} a^{2}}{d f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.20, size = 362, normalized size = 5.48 \[ -\frac {2 a^{2} \sqrt {d \tan \left (f x +e \right )}}{d f}+\frac {a^{2} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{2 f d}+\frac {a^{2} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{f d}-\frac {a^{2} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{f d}+\frac {i a^{2} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{2 f \left (d^{2}\right )^{\frac {1}{4}}}+\frac {i a^{2} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{f \left (d^{2}\right )^{\frac {1}{4}}}-\frac {i a^{2} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{f \left (d^{2}\right )^{\frac {1}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a^2*(d*tan(f*x+e))^(1/2)/d/f+1/2/f*a^2/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/f*a^2/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/f*a^2/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*I/f*a^2/(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)))+I/f*a^2/(d^
2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-I/f*a^2/(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 = 0.53, size = 177, normalized size = 2.68 \[ -\frac {a^{2} d {\left (-\frac {\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\left (i - 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\left (i - 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )} + 4 \, \sqrt {d \tan \left (f x + e\right )} a^{2}}{2 \, d f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.15, size = 59, normalized size = 0.89 \[ -\frac {2\,a^2\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{d\,f}+\frac {\sqrt {4{}\mathrm {i}}\,a^2\,\mathrm {atan}\left (\frac {\sqrt {4{}\mathrm {i}}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {-d}}\right )\,2{}\mathrm {i}}{\sqrt {-d}\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - a^{2} \left (\int \left (- \frac {1}{\sqrt {d \tan {\left (e + f x \right )}}}\right )\, dx + \int \frac {\tan ^{2}{\left (e + f x \right )}}{\sqrt {d \tan {\left (e + f x \right )}}}\, dx + \int \left (- \frac {2 i \tan {\left (e + f x \right )}}{\sqrt {d \tan {\left (e + f x \right )}}}\right )\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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