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

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*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])

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Rubi [A]  time = 0.0845581, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {5075, 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[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 5075

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]

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 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 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 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])

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*}

Mathematica [C]  time = 0.0216242, size = 71, normalized size = 0.74 \[ -\frac{4 i \sqrt{2 a^2 x^2+2} \, _2F_1\left (-\frac{3}{2},-\frac{3}{2};-\frac{1}{2};\frac{1}{2} (1-i a x)\right )}{3 a (1-i a x)^{3/2} \sqrt{a^2 c x^2+c}} \]

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]  time = 0.229, size = 800, normalized size = 8.3 \begin{align*} \text{result too large to display} \end{align*}

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)

[Out]

ln(x*a^2*c/(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)/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)+2/3*I/a^3/c/(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)+2/3/a^2/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)+2/3*I/a^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)*(-a^2)^(1/2)+2/
3/a^2/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)+2/3*I/
a^3/c/(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)-2/3/a^
2/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)-2/3*I/a^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)*(-a^2)^(1/2)+2/3/a^2/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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i \, a x + 1\right )}^{4}}{\sqrt{a^{2} c x^{2} + c}{\left (a^{2} x^{2} + 1\right )}^{2}}\,{d x} \end{align*}

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]  time = 2.22531, size = 417, normalized size = 4.34 \begin{align*} \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} \end{align*}

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]

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

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

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]

Exception raised: TypeError