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

Optimal. Leaf size=95 \[ \frac {i \sqrt {c-i c \tan (e+f x)}}{2 a f (1+i \tan (e+f x))}+\frac {i \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{2 \sqrt {2} a f} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3522, 3487, 51, 63, 206} \[ \frac {i \sqrt {c-i c \tan (e+f x)}}{2 a f (1+i \tan (e+f x))}+\frac {i \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{2 \sqrt {2} a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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 206

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

Rule 3487

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

Rule 3522

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[m, 0] || GtQ[m, n]))

Rubi steps

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

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Mathematica [A]  time = 1.17, size = 110, normalized size = 1.16 \[ \frac {(\sin (e+f x)+i \cos (e+f x)) \left (2 \cos (e+f x) \sqrt {c-i c \tan (e+f x)}+\sqrt {2} \sqrt {c} (\cos (e+f x)+i \sin (e+f x)) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )\right )}{4 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I*Cos[e + f*x] + Sin[e + f*x])*(Sqrt[2]*Sqrt[c]*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])]*(Cos[e
+ f*x] + I*Sin[e + f*x]) + 2*Cos[e + f*x]*Sqrt[c - I*c*Tan[e + f*x]]))/(4*a*f)

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fricas [B]  time = 0.47, size = 250, normalized size = 2.63 \[ \frac {{\left (\sqrt {\frac {1}{2}} a f \sqrt {-\frac {c}{a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {c}{a^{2} f^{2}}} + i \, c\right )} e^{\left (-i \, f x - i \, e\right )}}{a f}\right ) - \sqrt {\frac {1}{2}} a f \sqrt {-\frac {c}{a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {c}{a^{2} f^{2}}} - i \, c\right )} e^{\left (-i \, f x - i \, e\right )}}{a f}\right ) + \sqrt {2} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (i \, e^{\left (2 i \, f x + 2 i \, e\right )} + i\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a f} \]

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)),x, algorithm="fricas")

[Out]

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

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

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)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.34, size = 80, normalized size = 0.84 \[ \frac {2 i c^{2} \left (-\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{4 c \left (-c -i c \tan \left (f x +e \right )\right )}+\frac {\sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 c^{\frac {3}{2}}}\right )}{f a} \]

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)),x)

[Out]

2*I/f/a*c^2*(-1/4*(c-I*c*tan(f*x+e))^(1/2)/c/(-c-I*c*tan(f*x+e))+1/8/c^(3/2)*2^(1/2)*arctanh(1/2*(c-I*c*tan(f*
x+e))^(1/2)*2^(1/2)/c^(1/2)))

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maxima [A]  time = 1.10, size = 106, normalized size = 1.12 \[ -\frac {i \, {\left (\frac {\sqrt {2} c^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a} + \frac {4 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} c^{2}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )} a - 2 \, a c}\right )}}{8 \, c f} \]

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)),x, algorithm="maxima")

[Out]

-1/8*I*(sqrt(2)*c^(3/2)*log(-(sqrt(2)*sqrt(c) - sqrt(-I*c*tan(f*x + e) + c))/(sqrt(2)*sqrt(c) + sqrt(-I*c*tan(
f*x + e) + c)))/a + 4*sqrt(-I*c*tan(f*x + e) + c)*c^2/((-I*c*tan(f*x + e) + c)*a - 2*a*c))/(c*f)

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mupad [B]  time = 0.30, size = 81, normalized size = 0.85 \[ \frac {\sqrt {2}\,\sqrt {-c}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,1{}\mathrm {i}}{4\,a\,f}+\frac {c\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,a\,f\,\left (c+c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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)),x)

[Out]

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

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