\(\int \frac {e^{5 i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx\) [310]

Optimal result

Integrand size = 25, antiderivative size = 131 \[ \int \frac {e^{5 i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=-\frac {2 i \sqrt {1+a^2 x^2}}{a (1-i a x)^2 \sqrt {c+a^2 c x^2}}+\frac {4 i \sqrt {1+a^2 x^2}}{a (1-i a x) \sqrt {c+a^2 c x^2}}+\frac {i \sqrt {1+a^2 x^2} \log (i+a x)}{a \sqrt {c+a^2 c x^2}} \]

[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)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {5184, 5181, 45} \[ \int \frac {e^{5 i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=\frac {4 i \sqrt {a^2 x^2+1}}{a (1-i a x) \sqrt {a^2 c x^2+c}}-\frac {2 i \sqrt {a^2 x^2+1}}{a (1-i a x)^2 \sqrt {a^2 c x^2+c}}+\frac {i \sqrt {a^2 x^2+1} \log (a x+i)}{a \sqrt {a^2 c x^2+c}} \]

[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 45

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 5181

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 5184

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*} \text {integral}& = \frac {\sqrt {1+a^2 x^2} \int \frac {e^{5 i \arctan (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] (verified)

Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.53 \[ \int \frac {e^{5 i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=\frac {i \sqrt {1+a^2 x^2} \left (-2+4 i a x+(i+a x)^2 \log (i+a x)\right )}{a (i+a x)^2 \sqrt {c+a^2 c x^2}} \]

[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])

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.63

method	result	size
risch	\(\frac {\sqrt {a^{2} x^{2}+1}\, \left (-4 x -\frac {2 i}{a}\right )}{\sqrt {c \left (a^{2} x^{2}+1\right )}\, \left (a x +i\right )^{2}}+\frac {i \sqrt {a^{2} x^{2}+1}\, \ln \left (a x +i\right )}{\sqrt {c \left (a^{2} x^{2}+1\right )}\, a}\)	\(82\)
default	\(\frac {\sqrt {c \left (a^{2} x^{2}+1\right )}\, \left (i \ln \left (a x +i\right ) x^{2} a^{2}-2 \ln \left (a x +i\right ) a x -i \ln \left (a x +i\right )-4 a x -2 i\right )}{\sqrt {a^{2} x^{2}+1}\, c a \left (a x +i\right )^{2}}\)	\(84\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 364 vs. \(2 (107) = 214\).

Time = 0.29 (sec) , antiderivative size = 364, normalized size of antiderivative = 2.78 \[ \int \frac {e^{5 i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=\frac {-4 i \, \sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1} a x^{2} + {\left (i \, a^{4} c x^{4} - 2 \, a^{3} c x^{3} - 2 \, a c x - i \, c\right )} \sqrt {\frac {1}{a^{2} c}} \log \left (\frac {{\left (i \, a^{6} x^{2} - 2 \, a^{5} x - 2 i \, a^{4}\right )} \sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1} + {\left (i \, a^{9} c x^{4} - 2 \, a^{8} c x^{3} + i \, a^{7} c x^{2} - 2 \, a^{6} c x\right )} \sqrt {\frac {1}{a^{2} c}}}{8 \, {\left (a^{3} x^{3} + i \, a^{2} x^{2} + a x + i\right )}}\right ) + {\left (-i \, a^{4} c x^{4} + 2 \, a^{3} c x^{3} + 2 \, a c x + i \, c\right )} \sqrt {\frac {1}{a^{2} c}} \log \left (\frac {{\left (i \, a^{6} x^{2} - 2 \, a^{5} x - 2 i \, a^{4}\right )} \sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1} + {\left (-i \, a^{9} c x^{4} + 2 \, a^{8} c x^{3} - i \, a^{7} c x^{2} + 2 \, a^{6} c x\right )} \sqrt {\frac {1}{a^{2} c}}}{8 \, {\left (a^{3} x^{3} + i \, a^{2} x^{2} + a x + i\right )}}\right )}{2 \, {\left (a^{4} c x^{4} + 2 i \, a^{3} c x^{3} + 2 i \, a c x - c\right )}} \]

[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]

1/2*(-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(1/8*((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)))/(a^3*x^3 + I*a^2*x^2 + a*x + 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(1/8*((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)))/(a^3*x^3 + I*a^2*x^2 + a*x + I)))
/(a^4*c*x^4 + 2*I*a^3*c*x^3 + 2*I*a*c*x - c)

Sympy [F]

\[ \int \frac {e^{5 i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=i \left (\int \left (- \frac {i}{a^{4} x^{4} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + 2 a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c}}\right )\, dx + \int \frac {5 a x}{a^{4} x^{4} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + 2 a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c}}\, dx + \int \left (- \frac {10 a^{3} x^{3}}{a^{4} x^{4} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + 2 a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c}}\right )\, dx + \int \frac {a^{5} x^{5}}{a^{4} x^{4} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + 2 a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c}}\, dx + \int \frac {10 i a^{2} x^{2}}{a^{4} x^{4} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + 2 a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c}}\, dx + \int \left (- \frac {5 i a^{4} x^{4}}{a^{4} x^{4} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + 2 a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c}}\right )\, dx\right ) \]

[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))

Maxima [F]

\[ \int \frac {e^{5 i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=\int { \frac {{\left (i \, a x + 1\right )}^{5}}{\sqrt {a^{2} c x^{2} + c} {\left (a^{2} x^{2} + 1\right )}^{\frac {5}{2}}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {e^{5 i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=\int { \frac {{\left (i \, a x + 1\right )}^{5}}{\sqrt {a^{2} c x^{2} + c} {\left (a^{2} x^{2} + 1\right )}^{\frac {5}{2}}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{5 i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=\int \frac {{\left (1+a\,x\,1{}\mathrm {i}\right )}^5}{\sqrt {c\,a^2\,x^2+c}\,{\left (a^2\,x^2+1\right )}^{5/2}} \,d x \]

[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)