3.10.77 \(\int \frac {1}{(a+i a \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \, dx\) [977]

Optimal. Leaf size=124 \[ \frac {3 i \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{4 \sqrt {2} a \sqrt {c} f}-\frac {3 i}{4 a f \sqrt {c-i c \tan (e+f x)}}+\frac {i}{2 a f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3603, 3568, 44, 53, 65, 212} \begin {gather*} -\frac {3 i}{4 a f \sqrt {c-i c \tan (e+f x)}}+\frac {i}{2 a f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}+\frac {3 i \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{4 \sqrt {2} a \sqrt {c} f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((3*I)/4)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(Sqrt[2]*a*Sqrt[c]*f) - ((3*I)/4)/(a*f*Sqrt[
c - I*c*Tan[e + f*x]]) + (I/2)/(a*f*(1 + I*Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 53

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 65

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 212

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

Rule 3568

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 3603

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 {1}{(a+i a \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \, dx &=\frac {\int \cos ^2(e+f x) \sqrt {c-i c \tan (e+f x)} \, dx}{a c}\\ &=\frac {\left (i c^2\right ) \text {Subst}\left (\int \frac {1}{(c-x)^2 (c+x)^{3/2}} \, dx,x,-i c \tan (e+f x)\right )}{a f}\\ &=\frac {i}{2 a f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}+\frac {(3 i c) \text {Subst}\left (\int \frac {1}{(c-x) (c+x)^{3/2}} \, dx,x,-i c \tan (e+f x)\right )}{4 a f}\\ &=-\frac {3 i}{4 a f \sqrt {c-i c \tan (e+f x)}}+\frac {i}{2 a f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}+\frac {(3 i) \text {Subst}\left (\int \frac {1}{(c-x) \sqrt {c+x}} \, dx,x,-i c \tan (e+f x)\right )}{8 a f}\\ &=-\frac {3 i}{4 a f \sqrt {c-i c \tan (e+f x)}}+\frac {i}{2 a f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}+\frac {(3 i) \text {Subst}\left (\int \frac {1}{2 c-x^2} \, dx,x,\sqrt {c-i c \tan (e+f x)}\right )}{4 a f}\\ &=\frac {3 i \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{4 \sqrt {2} a \sqrt {c} f}-\frac {3 i}{4 a f \sqrt {c-i c \tan (e+f x)}}+\frac {i}{2 a f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 1.26, size = 117, normalized size = 0.94 \begin {gather*} -\frac {i e^{-2 i (e+f x)} \left (-1+e^{2 i (e+f x)}+2 e^{4 i (e+f x)}-3 e^{2 i (e+f x)} \sqrt {1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt {1+e^{2 i (e+f x)}}\right )\right ) \sqrt {c-i c \tan (e+f x)}}{8 a c f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/8*I)*(-1 + E^((2*I)*(e + f*x)) + 2*E^((4*I)*(e + f*x)) - 3*E^((2*I)*(e + f*x))*Sqrt[1 + E^((2*I)*(e + f*x
))]*ArcTanh[Sqrt[1 + E^((2*I)*(e + f*x))]])*Sqrt[c - I*c*Tan[e + f*x]])/(a*c*E^((2*I)*(e + f*x))*f)

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Maple [A]
time = 0.27, size = 102, normalized size = 0.82

method	result	size
derivativedivides	\(\frac {2 i c^{2} \left (\frac {\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{2 c +2 i c \tan \left (f x +e \right )}+\frac {3 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 \sqrt {c}}}{4 c^{2}}-\frac {1}{4 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f a}\)	\(102\)
default	\(\frac {2 i c^{2} \left (\frac {\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{2 c +2 i c \tan \left (f x +e \right )}+\frac {3 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 \sqrt {c}}}{4 c^{2}}-\frac {1}{4 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f a}\)	\(102\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 131, normalized size = 1.06 \begin {gather*} -\frac {i \, {\left (\frac {3 \, \sqrt {2} \sqrt {c} \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 \, {\left (3 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} c - 4 \, c^{2}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a - 2 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} a c}\right )}}{16 \, c f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*I*(3*sqrt(2)*sqrt(c)*log(-(sqrt(2)*sqrt(c) - sqrt(-I*c*tan(f*x + e) + c))/(sqrt(2)*sqrt(c) + sqrt(-I*c*t
an(f*x + e) + c)))/a + 4*(3*(-I*c*tan(f*x + e) + c)*c - 4*c^2)/((-I*c*tan(f*x + e) + c)^(3/2)*a - 2*sqrt(-I*c*
tan(f*x + e) + c)*a*c))/(c*f)

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 284 vs. \(2 (95) = 190\).
time = 1.66, size = 284, normalized size = 2.29 \begin {gather*} \frac {{\left (-3 i \, \sqrt {\frac {1}{2}} a c f \sqrt {\frac {1}{a^{2} c f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {3 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{2} c f^{2}}} - i\right )} e^{\left (-i \, f x - i \, e\right )}}{2 \, a f}\right ) + 3 i \, \sqrt {\frac {1}{2}} a c f \sqrt {\frac {1}{a^{2} c f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {3 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{2} c f^{2}}} - i\right )} e^{\left (-i \, f x - i \, e\right )}}{2 \, a f}\right ) + \sqrt {2} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-2 i \, e^{\left (4 i \, f x + 4 i \, e\right )} - i \, e^{\left (2 i \, f x + 2 i \, e\right )} + i\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a c f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {i \int \frac {1}{\sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} - i \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 5.03, size = 113, normalized size = 0.91 \begin {gather*} -\frac {\frac {c\,1{}\mathrm {i}}{a\,f}-\frac {\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{4\,a\,f}}{2\,c\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,3{}\mathrm {i}}{8\,a\,\sqrt {-c}\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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