3.318 \(\int \frac {e^{-4 i \tan ^{-1}(a x)}}{\sqrt {c+a^2 c x^2}} \, dx\)

Optimal. Leaf size=96 \[ \frac {2 i c (1-i a x)^3}{3 a \left (a^2 c x^2+c\right )^{3/2}}-\frac {2 i (1-i a x)}{a \sqrt {a^2 c x^2+c}}+\frac {\tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )}{a \sqrt {c}} \]

[Out]

2/3*I*c*(1-I*a*x)^3/a/(a^2*c*x^2+c)^(3/2)+arctanh(a*x*c^(1/2)/(a^2*c*x^2+c)^(1/2))/a/c^(1/2)-2*I*(1-I*a*x)/a/(
a^2*c*x^2+c)^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5074, 669, 653, 217, 206} \[ \frac {2 i c (1-i a x)^3}{3 a \left (a^2 c x^2+c\right )^{3/2}}-\frac {2 i (1-i a x)}{a \sqrt {a^2 c x^2+c}}+\frac {\tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )}{a \sqrt {c}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^((4*I)*ArcTan[a*x])*Sqrt[c + a^2*c*x^2]),x]

[Out]

(((2*I)/3)*c*(1 - I*a*x)^3)/(a*(c + a^2*c*x^2)^(3/2)) - ((2*I)*(1 - I*a*x))/(a*Sqrt[c + a^2*c*x^2]) + ArcTanh[
(a*Sqrt[c]*x)/Sqrt[c + a^2*c*x^2]]/(a*Sqrt[c])

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 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 653

Int[((d_) + (e_.)*(x_))^2*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)*(a + c*x^2)^(p + 1))/(c*(
p + 1)), x] - Dist[(e^2*(p + 2))/(c*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, p}, x] &&
EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && LtQ[p, -1]

Rule 669

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(p + 1)), x] - Dist[(e^2*(m + p))/(c*(p + 1)), Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]

Rule 5074

Int[E^(ArcTan[(a_.)*(x_)]*(n_))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[c^((I*n)/2), Int[(c + d*x^2)^(p
- (I*n)/2)*(1 - I*a*x)^(I*n), x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[d, a^2*c] &&  !(IntegerQ[p] || GtQ[c, 0]
) && IGtQ[(I*n)/2, 0]

Rubi steps

\begin {align*} \int \frac {e^{-4 i \tan ^{-1}(a x)}}{\sqrt {c+a^2 c x^2}} \, dx &=c^2 \int \frac {(1-i a x)^4}{\left (c+a^2 c x^2\right )^{5/2}} \, dx\\ &=\frac {2 i c (1-i a x)^3}{3 a \left (c+a^2 c x^2\right )^{3/2}}-c \int \frac {(1-i a x)^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx\\ &=\frac {2 i c (1-i a x)^3}{3 a \left (c+a^2 c x^2\right )^{3/2}}-\frac {2 i (1-i a x)}{a \sqrt {c+a^2 c x^2}}+\int \frac {1}{\sqrt {c+a^2 c x^2}} \, dx\\ &=\frac {2 i c (1-i a x)^3}{3 a \left (c+a^2 c x^2\right )^{3/2}}-\frac {2 i (1-i a x)}{a \sqrt {c+a^2 c x^2}}+\operatorname {Subst}\left (\int \frac {1}{1-a^2 c x^2} \, dx,x,\frac {x}{\sqrt {c+a^2 c x^2}}\right )\\ &=\frac {2 i c (1-i a x)^3}{3 a \left (c+a^2 c x^2\right )^{3/2}}-\frac {2 i (1-i a x)}{a \sqrt {c+a^2 c x^2}}+\frac {\tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c+a^2 c x^2}}\right )}{a \sqrt {c}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 132, normalized size = 1.38 \[ \frac {2 \sqrt {a^2 x^2+1} \left (2 i \sqrt {1+i a x} \left (2 a^2 x^2+i a x+1\right )+3 i \sqrt {1-i a x} (a x-i)^2 \sin ^{-1}\left (\frac {\sqrt {1-i a x}}{\sqrt {2}}\right )\right )}{3 a \sqrt {1-i a x} (a x-i)^2 \sqrt {a^2 c x^2+c}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^((4*I)*ArcTan[a*x])*Sqrt[c + a^2*c*x^2]),x]

[Out]

(2*Sqrt[1 + a^2*x^2]*((2*I)*Sqrt[1 + I*a*x]*(1 + I*a*x + 2*a^2*x^2) + (3*I)*Sqrt[1 - I*a*x]*(-I + a*x)^2*ArcSi
n[Sqrt[1 - I*a*x]/Sqrt[2]]))/(3*a*Sqrt[1 - I*a*x]*(-I + a*x)^2*Sqrt[c + a^2*c*x^2])

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fricas [B]  time = 0.51, size = 187, normalized size = 1.95 \[ \frac {{\left (3 \, a^{3} c x^{2} - 6 i \, a^{2} c x - 3 \, a c\right )} \sqrt {\frac {1}{a^{2} c}} \log \left (\frac {2 \, {\left (a^{2} c x + \sqrt {a^{2} c x^{2} + c} a^{2} c \sqrt {\frac {1}{a^{2} c}}\right )}}{x}\right ) - {\left (3 \, a^{3} c x^{2} - 6 i \, a^{2} c x - 3 \, a c\right )} \sqrt {\frac {1}{a^{2} c}} \log \left (\frac {2 \, {\left (a^{2} c x - \sqrt {a^{2} c x^{2} + c} a^{2} c \sqrt {\frac {1}{a^{2} c}}\right )}}{x}\right ) - \sqrt {a^{2} c x^{2} + c} {\left (16 \, a x - 8 i\right )}}{6 \, a^{3} c x^{2} - 12 i \, a^{2} c x - 6 \, a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*a*x)^4*(a^2*x^2+1)^2/(a^2*c*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

((3*a^3*c*x^2 - 6*I*a^2*c*x - 3*a*c)*sqrt(1/(a^2*c))*log(2*(a^2*c*x + sqrt(a^2*c*x^2 + c)*a^2*c*sqrt(1/(a^2*c)
))/x) - (3*a^3*c*x^2 - 6*I*a^2*c*x - 3*a*c)*sqrt(1/(a^2*c))*log(2*(a^2*c*x - sqrt(a^2*c*x^2 + c)*a^2*c*sqrt(1/
(a^2*c)))/x) - sqrt(a^2*c*x^2 + c)*(16*a*x - 8*I))/(6*a^3*c*x^2 - 12*I*a^2*c*x - 6*a*c)

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giac [A]  time = 0.25, size = 137, normalized size = 1.43 \[ -\frac {\log \left ({\left | -\sqrt {a^{2} c} x + \sqrt {a^{2} c x^{2} + c} \right |}\right )}{a \sqrt {c}} + \frac {8 \, {\left (3 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} + c}\right )}^{2} i - 2 \, c i + 3 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} + c}\right )} \sqrt {c}\right )}}{3 \, {\left ({\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} + c}\right )} i + \sqrt {c}\right )}^{3} a i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*a*x)^4*(a^2*x^2+1)^2/(a^2*c*x^2+c)^(1/2),x, algorithm="giac")

[Out]

-log(abs(-sqrt(a^2*c)*x + sqrt(a^2*c*x^2 + c)))/(a*sqrt(c)) + 8/3*(3*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 + c))^2*i
- 2*c*i + 3*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 + c))*sqrt(c))/(((sqrt(a^2*c)*x - sqrt(a^2*c*x^2 + c))*i + sqrt(c
))^3*a*i)

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maple [A]  time = 0.19, size = 136, normalized size = 1.42 \[ \frac {\ln \left (\frac {x \,a^{2} c}{\sqrt {a^{2} c}}+\sqrt {a^{2} c \,x^{2}+c}\right )}{\sqrt {a^{2} c}}-\frac {8 \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2} c +2 i a c \left (x -\frac {i}{a}\right )}}{3 a^{2} c \left (x -\frac {i}{a}\right )}-\frac {4 i \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2} c +2 i a c \left (x -\frac {i}{a}\right )}}{3 a^{3} c \left (x -\frac {i}{a}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1+I*a*x)^4*(a^2*x^2+1)^2/(a^2*c*x^2+c)^(1/2),x)

[Out]

ln(x*a^2*c/(a^2*c)^(1/2)+(a^2*c*x^2+c)^(1/2))/(a^2*c)^(1/2)-8/3/a^2/c/(x-I/a)*((x-I/a)^2*a^2*c+2*I*a*c*(x-I/a)
)^(1/2)-4/3*I/a^3/c/(x-I/a)^2*((x-I/a)^2*a^2*c+2*I*a*c*(x-I/a))^(1/2)

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maxima [A]  time = 0.42, size = 76, normalized size = 0.79 \[ -\frac {4 i \, \sqrt {a^{2} c x^{2} + c}}{3 \, {\left (a^{3} c x^{2} - 2 i \, a^{2} c x - a c\right )}} - \frac {8 i \, \sqrt {a^{2} c x^{2} + c}}{3 i \, a^{2} c x + 3 \, a c} + \frac {\operatorname {arsinh}\left (a x\right )}{a \sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*a*x)^4*(a^2*x^2+1)^2/(a^2*c*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

-4/3*I*sqrt(a^2*c*x^2 + c)/(a^3*c*x^2 - 2*I*a^2*c*x - a*c) - 8*I*sqrt(a^2*c*x^2 + c)/(3*I*a^2*c*x + 3*a*c) + a
rcsinh(a*x)/(a*sqrt(c))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a^2\,x^2+1\right )}^2}{\sqrt {c\,a^2\,x^2+c}\,{\left (1+a\,x\,1{}\mathrm {i}\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2*x^2 + 1)^2/((c + a^2*c*x^2)^(1/2)*(a*x*1i + 1)^4),x)

[Out]

int((a^2*x^2 + 1)^2/((c + a^2*c*x^2)^(1/2)*(a*x*1i + 1)^4), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a^{2} x^{2} + 1\right )^{2}}{\sqrt {c \left (a^{2} x^{2} + 1\right )} \left (a x - i\right )^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*a*x)**4*(a**2*x**2+1)**2/(a**2*c*x**2+c)**(1/2),x)

[Out]

Integral((a**2*x**2 + 1)**2/(sqrt(c*(a**2*x**2 + 1))*(a*x - I)**4), x)

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