3.4.11 \(\int \frac {e^{4 i \text {ArcTan}(a x)}}{\sqrt {c+a^2 c x^2}} \, dx\) [311]

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

[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 = 0.06, 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 = {5183, 683, 667, 223, 212} \begin {gather*} -\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[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 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 223

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 667

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 683

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 5183

Int[E^(ArcTan[(a_.)*(x_)]*(n_))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/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]) && ILtQ[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}}+\text {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 [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.02, size = 71, normalized size = 0.74 \begin {gather*} -\frac {4 i \sqrt {2+2 a^2 x^2} \, _2F_1\left (-\frac {3}{2},-\frac {3}{2};-\frac {1}{2};\frac {1}{2} (1-i a x)\right )}{3 a (1-i a x)^{3/2} \sqrt {c+a^2 c x^2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(((-4*I)/3)*Sqrt[2 + 2*a^2*x^2]*Hypergeometric2F1[-3/2, -3/2, -1/2, (1 - I*a*x)/2])/(a*(1 - I*a*x)^(3/2)*Sqrt[
c + a^2*c*x^2])

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 525 vs. \(2 (80 ) = 160\).
time = 0.14, size = 526, normalized size = 5.48

method	result	size
default	\(\frac {\ln \left (\frac {a^{2} c x}{\sqrt {a^{2} c}}+\sqrt {a^{2} c \,x^{2}+c}\right )}{\sqrt {a^{2} c}}+\frac {2 \left (i \sqrt {-a^{2}}-a \right ) \sqrt {\left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right )^{2} a^{2} c -2 c \sqrt {-a^{2}}\, \left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right )}}{a^{3} c \left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right )}-\frac {2 \left (i \sqrt {-a^{2}}+a \right ) \left (-\frac {\sqrt {\left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right )^{2} a^{2} c +2 c \sqrt {-a^{2}}\, \left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right )}}{3 c \sqrt {-a^{2}}\, \left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right )^{2}}-\frac {\sqrt {\left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right )^{2} a^{2} c +2 c \sqrt {-a^{2}}\, \left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right )}}{3 c \left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right )}\right )}{a^{3}}-\frac {2 \left (i \sqrt {-a^{2}}+a \right ) \sqrt {\left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right )^{2} a^{2} c +2 c \sqrt {-a^{2}}\, \left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right )}}{a^{3} c \left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right )}+\frac {2 \left (i \sqrt {-a^{2}}-a \right ) \left (\frac {\sqrt {\left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right )^{2} a^{2} c -2 c \sqrt {-a^{2}}\, \left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right )}}{3 c \sqrt {-a^{2}}\, \left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right )^{2}}-\frac {\sqrt {\left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right )^{2} a^{2} c -2 c \sqrt {-a^{2}}\, \left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right )}}{3 c \left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right )}\right )}{a^{3}}\)	\(526\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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. 186 vs. \(2 (75) = 150\).
time = 2.81, size = 186, normalized size = 1.94 \begin {gather*} \frac {3 \, {\left (a^{3} c x^{2} + 2 i \, a^{2} c x - 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 ) - 3 \, {\left (a^{3} c x^{2} + 2 i \, a^{2} c x - 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 ) - 8 \, \sqrt {a^{2} c x^{2} + c} {\left (2 \, a x + i\right )}}{6 \, {\left (a^{3} c x^{2} + 2 i \, a^{2} c x - a c\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.51, size = 132, normalized size = 1.38 \begin {gather*} -\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} + 3 i \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} + c}\right )} \sqrt {c} - 2 \, c\right )}}{3 \, {\left (i \, \sqrt {a^{2} c} x - i \, \sqrt {a^{2} c x^{2} + c} - \sqrt {c}\right )}^{3} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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 +
3*I*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 + c))*sqrt(c) - 2*c)/((I*sqrt(a^2*c)*x - I*sqrt(a^2*c*x^2 + c) - sqrt(c))
^3*a)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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