3.312 \(\int \frac {e^{3 i \tan ^{-1}(a x)}}{\sqrt {c+a^2 c x^2}} \, dx\)

Optimal. Leaf size=84 \[ \frac {2 \sqrt {a^2 x^2+1}}{a (a x+i) \sqrt {a^2 c x^2+c}}-\frac {i \sqrt {a^2 x^2+1} \log (a x+i)}{a \sqrt {a^2 c x^2+c}} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {5076, 5073, 43} \[ \frac {2 \sqrt {a^2 x^2+1}}{a (a x+i) \sqrt {a^2 c x^2+c}}-\frac {i \sqrt {a^2 x^2+1} \log (a x+i)}{a \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

Int[E^((3*I)*ArcTan[a*x])/Sqrt[c + a^2*c*x^2],x]

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 5073

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

Rule 5076

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 55, normalized size = 0.65 \[ \frac {\sqrt {a^2 x^2+1} \left (\frac {2}{a x+i}-i \log (a x+i)\right )}{a \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^((3*I)*ArcTan[a*x])/Sqrt[c + a^2*c*x^2],x]

[Out]

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

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fricas [B]  time = 0.57, size = 360, normalized size = 4.29 \[ \frac {{\left (-i \, a^{3} c x^{3} + a^{2} c x^{2} - i \, a c x + c\right )} \sqrt {\frac {1}{a^{2} c}} \log \left (\frac {{\left (i \, a^{6} x^{2} - 2 \, a^{5} x - 2 i \, a^{4}\right )} \sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1} + {\left (i \, a^{9} c x^{4} - 2 \, a^{8} c x^{3} + i \, a^{7} c x^{2} - 2 \, a^{6} c x\right )} \sqrt {\frac {1}{a^{2} c}}}{8 \, a^{3} x^{3} + 8 i \, a^{2} x^{2} + 8 \, a x + 8 i}\right ) + {\left (i \, a^{3} c x^{3} - a^{2} c x^{2} + i \, a c x - c\right )} \sqrt {\frac {1}{a^{2} c}} \log \left (\frac {{\left (i \, a^{6} x^{2} - 2 \, a^{5} x - 2 i \, a^{4}\right )} \sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1} + {\left (-i \, a^{9} c x^{4} + 2 \, a^{8} c x^{3} - i \, a^{7} c x^{2} + 2 \, a^{6} c x\right )} \sqrt {\frac {1}{a^{2} c}}}{8 \, a^{3} x^{3} + 8 i \, a^{2} x^{2} + 8 \, a x + 8 i}\right ) + 4 i \, \sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1} x}{2 \, a^{3} c x^{3} + 2 i \, a^{2} c x^{2} + 2 \, a c x + 2 i \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^3/(a^2*x^2+1)^(3/2)/(a^2*c*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

((-I*a^3*c*x^3 + a^2*c*x^2 - I*a*c*x + c)*sqrt(1/(a^2*c))*log(((I*a^6*x^2 - 2*a^5*x - 2*I*a^4)*sqrt(a^2*c*x^2
+ c)*sqrt(a^2*x^2 + 1) + (I*a^9*c*x^4 - 2*a^8*c*x^3 + I*a^7*c*x^2 - 2*a^6*c*x)*sqrt(1/(a^2*c)))/(8*a^3*x^3 + 8
*I*a^2*x^2 + 8*a*x + 8*I)) + (I*a^3*c*x^3 - a^2*c*x^2 + I*a*c*x - c)*sqrt(1/(a^2*c))*log(((I*a^6*x^2 - 2*a^5*x
- 2*I*a^4)*sqrt(a^2*c*x^2 + c)*sqrt(a^2*x^2 + 1) + (-I*a^9*c*x^4 + 2*a^8*c*x^3 - I*a^7*c*x^2 + 2*a^6*c*x)*sqr
t(1/(a^2*c)))/(8*a^3*x^3 + 8*I*a^2*x^2 + 8*a*x + 8*I)) + 4*I*sqrt(a^2*c*x^2 + c)*sqrt(a^2*x^2 + 1)*x)/(2*a^3*c
*x^3 + 2*I*a^2*c*x^2 + 2*a*c*x + 2*I*c)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a x + 1\right )}^{3}}{\sqrt {a^{2} c x^{2} + c} {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^3/(a^2*x^2+1)^(3/2)/(a^2*c*x^2+c)^(1/2),x, algorithm="giac")

[Out]

integrate((I*a*x + 1)^3/(sqrt(a^2*c*x^2 + c)*(a^2*x^2 + 1)^(3/2)), x)

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maple [A]  time = 0.16, size = 61, normalized size = 0.73 \[ \frac {\left (-i \ln \left (a x +i\right ) x a +\ln \left (a x +i\right )+2\right ) \sqrt {c \left (a^{2} x^{2}+1\right )}}{\sqrt {a^{2} x^{2}+1}\, c a \left (a x +i\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+I*a*x)^3/(a^2*x^2+1)^(3/2)/(a^2*c*x^2+c)^(1/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a x + 1\right )}^{3}}{\sqrt {a^{2} c x^{2} + c} {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^3/(a^2*x^2+1)^(3/2)/(a^2*c*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((I*a*x + 1)^3/(sqrt(a^2*c*x^2 + c)*(a^2*x^2 + 1)^(3/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (1+a\,x\,1{}\mathrm {i}\right )}^3}{\sqrt {c\,a^2\,x^2+c}\,{\left (a^2\,x^2+1\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x*1i + 1)^3/((c + a^2*c*x^2)^(1/2)*(a^2*x^2 + 1)^(3/2)),x)

[Out]

int((a*x*1i + 1)^3/((c + a^2*c*x^2)^(1/2)*(a^2*x^2 + 1)^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - i \left (\int \frac {i}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c}}\, dx + \int \left (- \frac {3 a x}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c}}\right )\, dx + \int \frac {a^{3} x^{3}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c}}\, dx + \int \left (- \frac {3 i a^{2} x^{2}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c}}\right )\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)**3/(a**2*x**2+1)**(3/2)/(a**2*c*x**2+c)**(1/2),x)

[Out]

-I*(Integral(I/(a**2*x**2*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c) + sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c
)), x) + Integral(-3*a*x/(a**2*x**2*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c) + sqrt(a**2*x**2 + 1)*sqrt(a**2*
c*x**2 + c)), x) + Integral(a**3*x**3/(a**2*x**2*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c) + sqrt(a**2*x**2 +
1)*sqrt(a**2*c*x**2 + c)), x) + Integral(-3*I*a**2*x**2/(a**2*x**2*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c) +
sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c)), x))

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