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

Optimal. Leaf size=280 \[ \frac {x \sqrt {a^2 c x^2+c} (1-i a x)^{\frac {1}{2} (3+i n)} (1+i a x)^{\frac {1}{2} (3-i n)}}{4 a^2 \sqrt {a^2 x^2+1}}+\frac {2^{-\frac {1}{2}-\frac {i n}{2}} \left (3-n^2\right ) \sqrt {a^2 c x^2+c} (1-i a x)^{\frac {1}{2} (3+i n)} \, _2F_1\left (\frac {1}{2} (i n-1),\frac {1}{2} (i n+3);\frac {1}{2} (i n+5);\frac {1}{2} (1-i a x)\right )}{3 a^3 (-n+3 i) \sqrt {a^2 x^2+1}}-\frac {n \sqrt {a^2 c x^2+c} (1-i a x)^{\frac {1}{2} (3+i n)} (1+i a x)^{\frac {1}{2} (3-i n)}}{12 a^3 \sqrt {a^2 x^2+1}} \]

[Out]

-1/12*n*(1-I*a*x)^(3/2+1/2*I*n)*(1+I*a*x)^(3/2-1/2*I*n)*(a^2*c*x^2+c)^(1/2)/a^3/(a^2*x^2+1)^(1/2)+1/4*x*(1-I*a
*x)^(3/2+1/2*I*n)*(1+I*a*x)^(3/2-1/2*I*n)*(a^2*c*x^2+c)^(1/2)/a^2/(a^2*x^2+1)^(1/2)+1/3*2^(-1/2-1/2*I*n)*(-n^2
+3)*(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*c*x^2+c)^(
1/2)/a^3/(3*I-n)/(a^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.29, antiderivative size = 280, 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, 90, 80, 69} \[ \frac {2^{-\frac {1}{2}-\frac {i n}{2}} \left (3-n^2\right ) \sqrt {a^2 c x^2+c} (1-i a x)^{\frac {1}{2} (3+i n)} \, _2F_1\left (\frac {1}{2} (i n-1),\frac {1}{2} (i n+3);\frac {1}{2} (i n+5);\frac {1}{2} (1-i a x)\right )}{3 a^3 (-n+3 i) \sqrt {a^2 x^2+1}}-\frac {n \sqrt {a^2 c x^2+c} (1-i a x)^{\frac {1}{2} (3+i n)} (1+i a x)^{\frac {1}{2} (3-i n)}}{12 a^3 \sqrt {a^2 x^2+1}}+\frac {x \sqrt {a^2 c x^2+c} (1-i a x)^{\frac {1}{2} (3+i n)} (1+i a x)^{\frac {1}{2} (3-i n)}}{4 a^2 \sqrt {a^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(n*(1 - I*a*x)^((3 + I*n)/2)*(1 + I*a*x)^((3 - I*n)/2)*Sqrt[c + a^2*c*x^2])/(12*a^3*Sqrt[1 + a^2*x^2]) + (x*(
1 - I*a*x)^((3 + I*n)/2)*(1 + I*a*x)^((3 - I*n)/2)*Sqrt[c + a^2*c*x^2])/(4*a^2*Sqrt[1 + a^2*x^2]) + (2^(-1/2 -
(I/2)*n)*(3 - n^2)*(1 - I*a*x)^((3 + I*n)/2)*Sqrt[c + a^2*c*x^2]*Hypergeometric2F1[(-1 + I*n)/2, (3 + I*n)/2,
(5 + I*n)/2, (1 - I*a*x)/2])/(3*a^3*(3*I - n)*Sqrt[1 + a^2*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 80

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

Rule 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 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 e^{n \tan ^{-1}(a x)} x^2 \sqrt {c+a^2 c x^2} \, dx &=\frac {\sqrt {c+a^2 c x^2} \int e^{n \tan ^{-1}(a x)} x^2 \sqrt {1+a^2 x^2} \, dx}{\sqrt {1+a^2 x^2}}\\ &=\frac {\sqrt {c+a^2 c x^2} \int x^2 (1-i a x)^{\frac {1}{2}+\frac {i n}{2}} (1+i a x)^{\frac {1}{2}-\frac {i n}{2}} \, dx}{\sqrt {1+a^2 x^2}}\\ &=\frac {x (1-i a x)^{\frac {1}{2} (3+i n)} (1+i a x)^{\frac {1}{2} (3-i n)} \sqrt {c+a^2 c x^2}}{4 a^2 \sqrt {1+a^2 x^2}}+\frac {\sqrt {c+a^2 c x^2} \int (1-i a x)^{\frac {1}{2}+\frac {i n}{2}} (1+i a x)^{\frac {1}{2}-\frac {i n}{2}} (-1-a n x) \, dx}{4 a^2 \sqrt {1+a^2 x^2}}\\ &=-\frac {n (1-i a x)^{\frac {1}{2} (3+i n)} (1+i a x)^{\frac {1}{2} (3-i n)} \sqrt {c+a^2 c x^2}}{12 a^3 \sqrt {1+a^2 x^2}}+\frac {x (1-i a x)^{\frac {1}{2} (3+i n)} (1+i a x)^{\frac {1}{2} (3-i n)} \sqrt {c+a^2 c x^2}}{4 a^2 \sqrt {1+a^2 x^2}}+\frac {\left (\left (-3+n^2\right ) \sqrt {c+a^2 c 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}{12 a^2 \sqrt {1+a^2 x^2}}\\ &=-\frac {n (1-i a x)^{\frac {1}{2} (3+i n)} (1+i a x)^{\frac {1}{2} (3-i n)} \sqrt {c+a^2 c x^2}}{12 a^3 \sqrt {1+a^2 x^2}}+\frac {x (1-i a x)^{\frac {1}{2} (3+i n)} (1+i a x)^{\frac {1}{2} (3-i n)} \sqrt {c+a^2 c x^2}}{4 a^2 \sqrt {1+a^2 x^2}}+\frac {2^{-\frac {1}{2}-\frac {i n}{2}} \left (3-n^2\right ) (1-i a x)^{\frac {1}{2} (3+i n)} \sqrt {c+a^2 c 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^3 (3 i-n) \sqrt {1+a^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 214, normalized size = 0.76 \[ \frac {2^{-2-\frac {i n}{2}} (a x+i) \sqrt {a^2 c x^2+c} (1-i a x)^{\frac {1}{2}+\frac {i n}{2}} (1+i a x)^{-\frac {i n}{2}} \left (2^{\frac {i n}{2}} (n-3 i) \sqrt {1+i a x} (a x-i) (3 a x-n)-2 i \sqrt {2} \left (n^2-3\right ) (1+i a x)^{\frac {i n}{2}} \, _2F_1\left (\frac {1}{2} (i n+3),\frac {1}{2} i (n+i);\frac {1}{2} (i n+5);\frac {1}{2} (1-i a x)\right )\right )}{3 a^3 (n-3 i) \sqrt {a^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {a^{2} c x^{2} + c} x^{2} e^{\left (n \arctan \left (a x\right )\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^2*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 \[ \int x^{2} \sqrt {c \left (a^{2} x^{2} + 1\right )} e^{n \operatorname {atan}{\left (a x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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