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)*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])

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Rubi [A]  time = 0.0831411, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, 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 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])

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 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])

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*}

Mathematica [A]  time = 0.0333269, 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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Maple [A]  time = 0.151, size = 84, normalized size = 0.6 \begin{align*}{\frac{i\ln \left ( ax+i \right ){x}^{2}{a}^{2}-2\,\ln \left ( ax+i \right ) xa-i\ln \left ( ax+i \right ) -4\,ax-2\,i}{ac \left ( ax+i \right ) ^{2}}\sqrt{c \left ({a}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \end{align*}

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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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Fricas [B]  time = 2.2175, size = 815, normalized size = 6.22 \begin{align*} \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} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (i a x + 1\right )^{5}}{\sqrt{c \left (a^{2} x^{2} + 1\right )} \left (a^{2} x^{2} + 1\right )^{\frac{5}{2}}}\, dx \end{align*}

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]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)