3.994 \(\int \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)} \, dx\)

Optimal. Leaf size=63 \[ -\frac {2 i \sqrt {a} \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {3523, 63, 217, 203} \[ -\frac {2 i \sqrt {a} \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]],x]

[Out]

((-2*I)*Sqrt[a]*Sqrt[c]*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/f

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 3523

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

Rubi steps

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

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Mathematica [A]  time = 1.51, size = 74, normalized size = 1.17 \[ -\frac {i \sqrt {2} c e^{-i (e+f x)} \tan ^{-1}\left (e^{i (e+f x)}\right ) \sqrt {a+i a \tan (e+f x)}}{f \sqrt {\frac {c}{1+e^{2 i (e+f x)}}}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]],x]

[Out]

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

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fricas [B]  time = 0.59, size = 215, normalized size = 3.41 \[ -\frac {1}{2} \, \sqrt {\frac {a c}{f^{2}}} \log \left (\frac {2 \, {\left (4 \, \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (e^{\left (3 i \, f x + 3 i \, e\right )} + e^{\left (i \, f x + i \, e\right )}\right )} + {\left (2 i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - 2 i \, f\right )} \sqrt {\frac {a c}{f^{2}}}\right )}}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right ) + \frac {1}{2} \, \sqrt {\frac {a c}{f^{2}}} \log \left (\frac {2 \, {\left (4 \, \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (e^{\left (3 i \, f x + 3 i \, e\right )} + e^{\left (i \, f x + i \, e\right )}\right )} + {\left (-2 i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, f\right )} \sqrt {\frac {a c}{f^{2}}}\right )}}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(a*c/f^2)*log(2*(4*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*(e^(3*I*f*x +
3*I*e) + e^(I*f*x + I*e)) + (2*I*f*e^(2*I*f*x + 2*I*e) - 2*I*f)*sqrt(a*c/f^2))/(e^(2*I*f*x + 2*I*e) + 1)) + 1/
2*sqrt(a*c/f^2)*log(2*(4*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*(e^(3*I*f*x + 3*I
*e) + e^(I*f*x + I*e)) + (-2*I*f*e^(2*I*f*x + 2*I*e) + 2*I*f)*sqrt(a*c/f^2))/(e^(2*I*f*x + 2*I*e) + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(I*a*tan(f*x + e) + a)*sqrt(-I*c*tan(f*x + e) + c), x)

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maple [A]  time = 0.34, size = 96, normalized size = 1.52 \[ \frac {\sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, a c \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right )}{f \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.02, size = 104, normalized size = 1.65 \[ \frac {\sqrt {a} \sqrt {c} {\left (-2 i \, \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) - 2 i \, \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) + \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) - \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right )\right )}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(a)*sqrt(c)*(-2*I*arctan2(cos(f*x + e), sin(f*x + e) + 1) - 2*I*arctan2(cos(f*x + e), -sin(f*x + e) +
1) + log(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*sin(f*x + e) + 1) - log(cos(f*x + e)^2 + sin(f*x + e)^2 - 2*sin(f
*x + e) + 1))/f

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mupad [B]  time = 5.77, size = 62, normalized size = 0.98 \[ -\frac {\sqrt {a}\,\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {c}\,\left (\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}{\sqrt {a}\,\left (\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {c}\right )}\right )\,4{}\mathrm {i}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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