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

Optimal. Leaf size=322 \[ \frac {x^2 \sqrt {a^2 x^2+1} (1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)}}{3 a^2 \sqrt {a^2 c x^2+c}}+\frac {2^{-\frac {1}{2}-\frac {i n}{2}} n \left (5-n^2\right ) \sqrt {a^2 x^2+1} (1-i a x)^{\frac {1}{2} (3+i n)} \, _2F_1\left (\frac {1}{2} (i n+1),\frac {1}{2} (i n+3);\frac {1}{2} (i n+5);\frac {1}{2} (1-i a x)\right )}{3 a^4 \left (4 n-i \left (3-n^2\right )\right ) \sqrt {a^2 c x^2+c}}-\frac {\sqrt {a^2 x^2+1} (1-i a x)^{\frac {1}{2} (1+i n)} \left (a (1+i n) n x-n^2-i n+4\right ) (1+i a x)^{\frac {1}{2} (1-i n)}}{6 a^4 (1+i n) \sqrt {a^2 c x^2+c}} \]

[Out]

1/3*x^2*(1-I*a*x)^(1/2+1/2*I*n)*(1+I*a*x)^(1/2-1/2*I*n)*(a^2*x^2+1)^(1/2)/a^2/(a^2*c*x^2+c)^(1/2)-1/6*(1-I*a*x
)^(1/2+1/2*I*n)*(1+I*a*x)^(1/2-1/2*I*n)*(4-I*n-n^2+a*(1+I*n)*n*x)*(a^2*x^2+1)^(1/2)/a^4/(1+I*n)/(a^2*c*x^2+c)^
(1/2)+1/3*2^(-1/2-1/2*I*n)*n*(-n^2+5)*(1-I*a*x)^(3/2+1/2*I*n)*hypergeom([3/2+1/2*I*n, 1/2+1/2*I*n],[5/2+1/2*I*
n],1/2-1/2*I*a*x)*(a^2*x^2+1)^(1/2)/a^4/(4*n-I*(-n^2+3))/(a^2*c*x^2+c)^(1/2)

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Rubi [A]  time = 0.36, antiderivative size = 322, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {5085, 5082, 100, 146, 69} \[ \frac {2^{-\frac {1}{2}-\frac {i n}{2}} n \left (5-n^2\right ) \sqrt {a^2 x^2+1} (1-i a x)^{\frac {1}{2} (3+i n)} \, _2F_1\left (\frac {1}{2} (i n+1),\frac {1}{2} (i n+3);\frac {1}{2} (i n+5);\frac {1}{2} (1-i a x)\right )}{3 a^4 \left (4 n-i \left (3-n^2\right )\right ) \sqrt {a^2 c x^2+c}}-\frac {\sqrt {a^2 x^2+1} (1-i a x)^{\frac {1}{2} (1+i n)} \left (a (1+i n) n x-n^2-i n+4\right ) (1+i a x)^{\frac {1}{2} (1-i n)}}{6 a^4 (1+i n) \sqrt {a^2 c x^2+c}}+\frac {x^2 \sqrt {a^2 x^2+1} (1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)}}{3 a^2 \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*(1 - I*a*x)^((1 + I*n)/2)*(1 + I*a*x)^((1 - I*n)/2)*Sqrt[1 + a^2*x^2])/(3*a^2*Sqrt[c + a^2*c*x^2]) - ((1
- I*a*x)^((1 + I*n)/2)*(1 + I*a*x)^((1 - I*n)/2)*(4 - I*n - n^2 + a*(1 + I*n)*n*x)*Sqrt[1 + a^2*x^2])/(6*a^4*(
1 + I*n)*Sqrt[c + a^2*c*x^2]) + (2^(-1/2 - (I/2)*n)*n*(5 - n^2)*(1 - I*a*x)^((3 + I*n)/2)*Sqrt[1 + a^2*x^2]*Hy
pergeometric2F1[(1 + I*n)/2, (3 + I*n)/2, (5 + I*n)/2, (1 - I*a*x)/2])/(3*a^4*(4*n - I*(3 - n^2))*Sqrt[c + a^2
*c*x^2])

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rule 100

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m - 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 1)), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]

Rule 146

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((a^2*d*f*h*(n + 2) + b^2*d*e*g*(m + n + 3) + a*b*(c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b*f*h*(
b*c - a*d)*(m + 1)*x)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b^2*d*(b*c - a*d)*(m + 1)*(m + n + 3)), x] - Dist[
(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m +
1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d*(b*c - a*d)*(m +
1)*(m + n + 3)), Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && ((Ge
Q[m, -2] && LtQ[m, -1]) || SumSimplerQ[m, 1]) && NeQ[m, -1] && NeQ[m + n + 3, 0]

Rule 5082

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

Rule 5085

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

Rubi steps

\begin {align*} \int \frac {e^{n \tan ^{-1}(a x)} x^3}{\sqrt {c+a^2 c x^2}} \, dx &=\frac {\sqrt {1+a^2 x^2} \int \frac {e^{n \tan ^{-1}(a x)} x^3}{\sqrt {1+a^2 x^2}} \, dx}{\sqrt {c+a^2 c x^2}}\\ &=\frac {\sqrt {1+a^2 x^2} \int x^3 (1-i a x)^{-\frac {1}{2}+\frac {i n}{2}} (1+i a x)^{-\frac {1}{2}-\frac {i n}{2}} \, dx}{\sqrt {c+a^2 c x^2}}\\ &=\frac {x^2 (1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)} \sqrt {1+a^2 x^2}}{3 a^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {1+a^2 x^2} \int x (1-i a x)^{-\frac {1}{2}+\frac {i n}{2}} (1+i a x)^{-\frac {1}{2}-\frac {i n}{2}} (-2-a n x) \, dx}{3 a^2 \sqrt {c+a^2 c x^2}}\\ &=\frac {x^2 (1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)} \sqrt {1+a^2 x^2}}{3 a^2 \sqrt {c+a^2 c x^2}}-\frac {(1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)} \left (4-i n-n^2+a (1+i n) n x\right ) \sqrt {1+a^2 x^2}}{6 a^4 (1+i n) \sqrt {c+a^2 c x^2}}+\frac {\left (n \left (5-n^2\right ) \sqrt {1+a^2 x^2}\right ) \int (1-i a x)^{\frac {1}{2}+\frac {i n}{2}} (1+i a x)^{-\frac {1}{2}-\frac {i n}{2}} \, dx}{6 a^3 (1+i n) \sqrt {c+a^2 c x^2}}\\ &=\frac {x^2 (1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)} \sqrt {1+a^2 x^2}}{3 a^2 \sqrt {c+a^2 c x^2}}-\frac {(1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)} \left (4-i n-n^2+a (1+i n) n x\right ) \sqrt {1+a^2 x^2}}{6 a^4 (1+i n) \sqrt {c+a^2 c x^2}}+\frac {2^{-\frac {1}{2}-\frac {i n}{2}} n \left (5-n^2\right ) (1-i a x)^{\frac {1}{2} (3+i n)} \sqrt {1+a^2 x^2} \, _2F_1\left (\frac {1}{2} (1+i n),\frac {1}{2} (3+i n);\frac {1}{2} (5+i n);\frac {1}{2} (1-i a x)\right )}{3 a^4 \left (4 n-i \left (3-n^2\right )\right ) \sqrt {c+a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.32, size = 248, normalized size = 0.77 \[ \frac {2^{-\frac {3}{2}-\frac {i n}{2}} \sqrt {a^2 x^2+1} (1-i a x)^{\frac {1}{2}+\frac {i n}{2}} (1+i a x)^{-\frac {i n}{2}} \left (2^{\frac {1}{2}+\frac {i n}{2}} (n-3 i) \sqrt {1+i a x} \left (n \left (2 a^2 x^2+i a x+1\right )-2 i \left (a^2 x^2-2\right )-\left (n^2 (a x+i)\right )\right )+2 n \left (n^2-5\right ) (a x+i) (1+i a x)^{\frac {i n}{2}} \, _2F_1\left (\frac {i n}{2}+\frac {1}{2},\frac {i n}{2}+\frac {3}{2};\frac {i n}{2}+\frac {5}{2};\frac {1}{2}-\frac {i a x}{2}\right )\right )}{3 a^4 \left (n^2-4 i n-3\right ) \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2^(-3/2 - (I/2)*n)*(1 - I*a*x)^(1/2 + (I/2)*n)*Sqrt[1 + a^2*x^2]*(2^(1/2 + (I/2)*n)*(-3*I + n)*Sqrt[1 + I*a*x
]*(-(n^2*(I + a*x)) - (2*I)*(-2 + a^2*x^2) + n*(1 + I*a*x + 2*a^2*x^2)) + 2*n*(-5 + n^2)*(1 + I*a*x)^((I/2)*n)
*(I + a*x)*Hypergeometric2F1[1/2 + (I/2)*n, 3/2 + (I/2)*n, 5/2 + (I/2)*n, 1/2 - (I/2)*a*x]))/(3*a^4*(-3 - (4*I
)*n + n^2)*(1 + I*a*x)^((I/2)*n)*Sqrt[c + a^2*c*x^2])

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{3} e^{\left (n \arctan \left (a x\right )\right )}}{\sqrt {a^{2} c x^{2} + c}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^3*e^(n*arctan(a*x))/sqrt(a^2*c*x^2 + c), x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [F]  time = 0.30, size = 0, normalized size = 0.00 \[ \int \frac {{\mathrm e}^{n \arctan \left (a x \right )} x^{3}}{\sqrt {a^{2} c \,x^{2}+c}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arctan(a*x))*x^3/(a^2*c*x^2+c)^(1/2),x)

[Out]

int(exp(n*arctan(a*x))*x^3/(a^2*c*x^2+c)^(1/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3*e^(n*arctan(a*x))/sqrt(a^2*c*x^2 + c), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*exp(n*atan(a*x)))/(c + a^2*c*x^2)^(1/2),x)

[Out]

int((x^3*exp(n*atan(a*x)))/(c + a^2*c*x^2)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*atan(a*x))*x**3/(a**2*c*x**2+c)**(1/2),x)

[Out]

Integral(x**3*exp(n*atan(a*x))/sqrt(c*(a**2*x**2 + 1)), x)

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