3.310 \(\int \frac {e^{5 i \tan ^{-1}(a x)}}{\sqrt {c+a^2 c x^2}} \, dx\)

Optimal. Leaf size=131 \[ \frac {4 i \sqrt {a^2 x^2+1}}{a (1-i a x) \sqrt {a^2 c x^2+c}}-\frac {2 i \sqrt {a^2 x^2+1}}{a (1-i a x)^2 \sqrt {a^2 c x^2+c}}+\frac {i \sqrt {a^2 x^2+1} \log (a x+i)}{a \sqrt {a^2 c x^2+c}} \]

[Out]

-2*I*(a^2*x^2+1)^(1/2)/a/(1-I*a*x)^2/(a^2*c*x^2+c)^(1/2)+4*I*(a^2*x^2+1)^(1/2)/a/(1-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 = 131, 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} \[ \frac {4 i \sqrt {a^2 x^2+1}}{a (1-i a x) \sqrt {a^2 c x^2+c}}-\frac {2 i \sqrt {a^2 x^2+1}}{a (1-i a x)^2 \sqrt {a^2 c x^2+c}}+\frac {i \sqrt {a^2 x^2+1} \log (a x+i)}{a \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

Int[E^((5*I)*ArcTan[a*x])/Sqrt[c + a^2*c*x^2],x]

[Out]

((-2*I)*Sqrt[1 + a^2*x^2])/(a*(1 - I*a*x)^2*Sqrt[c + a^2*c*x^2]) + ((4*I)*Sqrt[1 + a^2*x^2])/(a*(1 - I*a*x)*Sq
rt[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^{5 i \tan ^{-1}(a x)}}{\sqrt {c+a^2 c x^2}} \, dx &=\frac {\sqrt {1+a^2 x^2} \int \frac {e^{5 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)^2}{(1-i a x)^3} \, dx}{\sqrt {c+a^2 c x^2}}\\ &=\frac {\sqrt {1+a^2 x^2} \int \left (\frac {4}{(1-i a x)^3}-\frac {4}{(1-i a x)^2}+\frac {1}{1-i a x}\right ) \, dx}{\sqrt {c+a^2 c x^2}}\\ &=-\frac {2 i \sqrt {1+a^2 x^2}}{a (1-i a x)^2 \sqrt {c+a^2 c x^2}}+\frac {4 i \sqrt {1+a^2 x^2}}{a (1-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 = 0.04, size = 69, normalized size = 0.53 \[ \frac {i \sqrt {a^2 x^2+1} \left (4 i a x+(a x+i)^2 \log (a x+i)-2\right )}{a (a x+i)^2 \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^((5*I)*ArcTan[a*x])/Sqrt[c + a^2*c*x^2],x]

[Out]

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

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fricas [B]  time = 0.60, size = 366, normalized size = 2.79 \[ \frac {-4 i \, \sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1} a x^{2} + {\left (i \, a^{4} c x^{4} - 2 \, a^{3} c x^{3} - 2 \, a c x - i \, c\right )} \sqrt {\frac {1}{a^{2} c}} \log \left (\frac {{\left (i \, a^{6} x^{2} - 2 \, a^{5} x - 2 i \, a^{4}\right )} \sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1} + {\left (i \, a^{9} c x^{4} - 2 \, a^{8} c x^{3} + i \, a^{7} c x^{2} - 2 \, a^{6} c x\right )} \sqrt {\frac {1}{a^{2} c}}}{8 \, a^{3} x^{3} + 8 i \, a^{2} x^{2} + 8 \, a x + 8 i}\right ) + {\left (-i \, a^{4} c x^{4} + 2 \, a^{3} c x^{3} + 2 \, a c x + i \, c\right )} \sqrt {\frac {1}{a^{2} c}} \log \left (\frac {{\left (i \, a^{6} x^{2} - 2 \, a^{5} x - 2 i \, a^{4}\right )} \sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1} + {\left (-i \, a^{9} c x^{4} + 2 \, a^{8} c x^{3} - i \, a^{7} c x^{2} + 2 \, a^{6} c x\right )} \sqrt {\frac {1}{a^{2} c}}}{8 \, a^{3} x^{3} + 8 i \, a^{2} x^{2} + 8 \, a x + 8 i}\right )}{2 \, a^{4} c x^{4} + 4 i \, a^{3} c x^{3} + 4 i \, a c x - 2 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^5/(a^2*x^2+1)^(5/2)/(a^2*c*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a x + 1\right )}^{5}}{\sqrt {a^{2} c x^{2} + c} {\left (a^{2} x^{2} + 1\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^5/(a^2*x^2+1)^(5/2)/(a^2*c*x^2+c)^(1/2),x, algorithm="giac")

[Out]

integrate((I*a*x + 1)^5/(sqrt(a^2*c*x^2 + c)*(a^2*x^2 + 1)^(5/2)), x)

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maple [A]  time = 0.20, size = 84, normalized size = 0.64 \[ \frac {\sqrt {c \left (a^{2} x^{2}+1\right )}\, \left (i \ln \left (a x +i\right ) x^{2} a^{2}-2 \ln \left (a x +i\right ) x a -i \ln \left (a x +i\right )-4 a x -2 i\right )}{\sqrt {a^{2} x^{2}+1}\, c a \left (a x +i\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+I*a*x)^5/(a^2*x^2+1)^(5/2)/(a^2*c*x^2+c)^(1/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a x + 1\right )}^{5}}{\sqrt {a^{2} c x^{2} + c} {\left (a^{2} x^{2} + 1\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^5/(a^2*x^2+1)^(5/2)/(a^2*c*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((I*a*x + 1)^5/(sqrt(a^2*c*x^2 + c)*(a^2*x^2 + 1)^(5/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (1+a\,x\,1{}\mathrm {i}\right )}^5}{\sqrt {c\,a^2\,x^2+c}\,{\left (a^2\,x^2+1\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x*1i + 1)^5/((c + a^2*c*x^2)^(1/2)*(a^2*x^2 + 1)^(5/2)),x)

[Out]

int((a*x*1i + 1)^5/((c + a^2*c*x^2)^(1/2)*(a^2*x^2 + 1)^(5/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ i \left (\int \left (- \frac {i}{a^{4} x^{4} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + 2 a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c}}\right )\, dx + \int \frac {5 a x}{a^{4} x^{4} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + 2 a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c}}\, dx + \int \left (- \frac {10 a^{3} x^{3}}{a^{4} x^{4} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + 2 a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c}}\right )\, dx + \int \frac {a^{5} x^{5}}{a^{4} x^{4} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + 2 a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c}}\, dx + \int \frac {10 i a^{2} x^{2}}{a^{4} x^{4} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + 2 a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c}}\, dx + \int \left (- \frac {5 i a^{4} x^{4}}{a^{4} x^{4} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + 2 a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c}}\right )\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)**5/(a**2*x**2+1)**(5/2)/(a**2*c*x**2+c)**(1/2),x)

[Out]

I*(Integral(-I/(a**4*x**4*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c) + 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) + Integral(5*a*x/(a**4*x**4*sqrt(a**2*x**2 + 1)
*sqrt(a**2*c*x**2 + c) + 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) + Integral(-10*a**3*x**3/(a**4*x**4*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c) + 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) + Integral(a**5*x**
5/(a**4*x**4*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c) + 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) + Integral(10*I*a**2*x**2/(a**4*x**4*sqrt(a**2*x**2 + 1)*sqr
t(a**2*c*x**2 + c) + 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) + Integral(-5*I*a**4*x**4/(a**4*x**4*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c) + 2*a**2*x**2*sqr
t(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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