3.4.34 \(\int \frac {(A+B x) (a+c x^2)^{3/2}}{x^4} \, dx\)

Optimal. Leaf size=109 \[ -\frac {c \sqrt {a+c x^2} (2 A-3 B x)}{2 x}-\frac {\left (a+c x^2\right )^{3/2} (2 A+3 B x)}{6 x^3}+A c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )-\frac {3}{2} \sqrt {a} B c \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {811, 813, 844, 217, 206, 266, 63, 208} \begin {gather*} -\frac {c \sqrt {a+c x^2} (2 A-3 B x)}{2 x}-\frac {\left (a+c x^2\right )^{3/2} (2 A+3 B x)}{6 x^3}+A c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )-\frac {3}{2} \sqrt {a} B c \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + c*x^2)^(3/2))/x^4,x]

[Out]

-(c*(2*A - 3*B*x)*Sqrt[a + c*x^2])/(2*x) - ((2*A + 3*B*x)*(a + c*x^2)^(3/2))/(6*x^3) + A*c^(3/2)*ArcTanh[(Sqrt
[c]*x)/Sqrt[a + c*x^2]] - (3*Sqrt[a]*B*c*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/2

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 811

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((d + e*x)^
(m + 1)*(a + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 + a*e^2) - 2*c*d^2*p*(e*f - d*g) - e*(g*(m + 1)*(c*d^2 + a*e
^2) + 2*c*d*p*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2 + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2
+ a*e^2)), Int[(d + e*x)^(m + 2)*(a + c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) - c*(2*c*d*(d*g*(2*p + 1
) - e*f*(m + 2*p + 2)) - 2*a*e^2*g*(m + 1))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2
, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

Rule 813

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + c*x^2)^p)/(e^2*(m + 1)*(m + 2*p + 2)), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {(A+B x) \left (a+c x^2\right )^{3/2}}{x^4} \, dx &=-\frac {(2 A+3 B x) \left (a+c x^2\right )^{3/2}}{6 x^3}-\frac {\int \frac {(-4 a A c-6 a B c x) \sqrt {a+c x^2}}{x^2} \, dx}{4 a}\\ &=-\frac {c (2 A-3 B x) \sqrt {a+c x^2}}{2 x}-\frac {(2 A+3 B x) \left (a+c x^2\right )^{3/2}}{6 x^3}+\frac {\int \frac {12 a^2 B c+8 a A c^2 x}{x \sqrt {a+c x^2}} \, dx}{8 a}\\ &=-\frac {c (2 A-3 B x) \sqrt {a+c x^2}}{2 x}-\frac {(2 A+3 B x) \left (a+c x^2\right )^{3/2}}{6 x^3}+\frac {1}{2} (3 a B c) \int \frac {1}{x \sqrt {a+c x^2}} \, dx+\left (A c^2\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx\\ &=-\frac {c (2 A-3 B x) \sqrt {a+c x^2}}{2 x}-\frac {(2 A+3 B x) \left (a+c x^2\right )^{3/2}}{6 x^3}+\frac {1}{4} (3 a B c) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,x^2\right )+\left (A c^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )\\ &=-\frac {c (2 A-3 B x) \sqrt {a+c x^2}}{2 x}-\frac {(2 A+3 B x) \left (a+c x^2\right )^{3/2}}{6 x^3}+A c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )+\frac {1}{2} (3 a B) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{c}+\frac {x^2}{c}} \, dx,x,\sqrt {a+c x^2}\right )\\ &=-\frac {c (2 A-3 B x) \sqrt {a+c x^2}}{2 x}-\frac {(2 A+3 B x) \left (a+c x^2\right )^{3/2}}{6 x^3}+A c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )-\frac {3}{2} \sqrt {a} B c \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 92, normalized size = 0.84 \begin {gather*} \frac {B c \left (a+c x^2\right )^{5/2} \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};\frac {c x^2}{a}+1\right )}{5 a^2}-\frac {a A \sqrt {a+c x^2} \, _2F_1\left (-\frac {3}{2},-\frac {3}{2};-\frac {1}{2};-\frac {c x^2}{a}\right )}{3 x^3 \sqrt {\frac {c x^2}{a}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + c*x^2)^(3/2))/x^4,x]

[Out]

-1/3*(a*A*Sqrt[a + c*x^2]*Hypergeometric2F1[-3/2, -3/2, -1/2, -((c*x^2)/a)])/(x^3*Sqrt[1 + (c*x^2)/a]) + (B*c*
(a + c*x^2)^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, 1 + (c*x^2)/a])/(5*a^2)

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IntegrateAlgebraic [A]  time = 0.49, size = 113, normalized size = 1.04 \begin {gather*} \frac {\sqrt {a+c x^2} \left (-2 a A-3 a B x-8 A c x^2+6 B c x^3\right )}{6 x^3}-A c^{3/2} \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )+3 \sqrt {a} B c \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}-\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(a + c*x^2)^(3/2))/x^4,x]

[Out]

(Sqrt[a + c*x^2]*(-2*a*A - 3*a*B*x - 8*A*c*x^2 + 6*B*c*x^3))/(6*x^3) + 3*Sqrt[a]*B*c*ArcTanh[(Sqrt[c]*x)/Sqrt[
a] - Sqrt[a + c*x^2]/Sqrt[a]] - A*c^(3/2)*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]]

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fricas [A]  time = 0.49, size = 426, normalized size = 3.91 \begin {gather*} \left [\frac {6 \, A c^{\frac {3}{2}} x^{3} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 9 \, B \sqrt {a} c x^{3} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (6 \, B c x^{3} - 8 \, A c x^{2} - 3 \, B a x - 2 \, A a\right )} \sqrt {c x^{2} + a}}{12 \, x^{3}}, -\frac {12 \, A \sqrt {-c} c x^{3} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - 9 \, B \sqrt {a} c x^{3} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (6 \, B c x^{3} - 8 \, A c x^{2} - 3 \, B a x - 2 \, A a\right )} \sqrt {c x^{2} + a}}{12 \, x^{3}}, \frac {9 \, B \sqrt {-a} c x^{3} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) + 3 \, A c^{\frac {3}{2}} x^{3} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + {\left (6 \, B c x^{3} - 8 \, A c x^{2} - 3 \, B a x - 2 \, A a\right )} \sqrt {c x^{2} + a}}{6 \, x^{3}}, -\frac {6 \, A \sqrt {-c} c x^{3} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - 9 \, B \sqrt {-a} c x^{3} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) - {\left (6 \, B c x^{3} - 8 \, A c x^{2} - 3 \, B a x - 2 \, A a\right )} \sqrt {c x^{2} + a}}{6 \, x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(3/2)/x^4,x, algorithm="fricas")

[Out]

[1/12*(6*A*c^(3/2)*x^3*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 9*B*sqrt(a)*c*x^3*log(-(c*x^2 - 2*sqr
t(c*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(6*B*c*x^3 - 8*A*c*x^2 - 3*B*a*x - 2*A*a)*sqrt(c*x^2 + a))/x^3, -1/12*(12
*A*sqrt(-c)*c*x^3*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - 9*B*sqrt(a)*c*x^3*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(
a) + 2*a)/x^2) - 2*(6*B*c*x^3 - 8*A*c*x^2 - 3*B*a*x - 2*A*a)*sqrt(c*x^2 + a))/x^3, 1/6*(9*B*sqrt(-a)*c*x^3*arc
tan(sqrt(-a)/sqrt(c*x^2 + a)) + 3*A*c^(3/2)*x^3*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + (6*B*c*x^3 -
8*A*c*x^2 - 3*B*a*x - 2*A*a)*sqrt(c*x^2 + a))/x^3, -1/6*(6*A*sqrt(-c)*c*x^3*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)
) - 9*B*sqrt(-a)*c*x^3*arctan(sqrt(-a)/sqrt(c*x^2 + a)) - (6*B*c*x^3 - 8*A*c*x^2 - 3*B*a*x - 2*A*a)*sqrt(c*x^2
+ a))/x^3]

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giac [B]  time = 0.22, size = 211, normalized size = 1.94 \begin {gather*} \frac {3 \, B a c \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - A c^{\frac {3}{2}} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right ) + \sqrt {c x^{2} + a} B c + \frac {3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{5} B a c + 12 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{4} A a c^{\frac {3}{2}} - 12 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} A a^{2} c^{\frac {3}{2}} - 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} B a^{3} c + 8 \, A a^{3} c^{\frac {3}{2}}}{3 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} - a\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(3/2)/x^4,x, algorithm="giac")

[Out]

3*B*a*c*arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))/sqrt(-a))/sqrt(-a) - A*c^(3/2)*log(abs(-sqrt(c)*x + sqrt(c*x^2 +
a))) + sqrt(c*x^2 + a)*B*c + 1/3*(3*(sqrt(c)*x - sqrt(c*x^2 + a))^5*B*a*c + 12*(sqrt(c)*x - sqrt(c*x^2 + a))^
4*A*a*c^(3/2) - 12*(sqrt(c)*x - sqrt(c*x^2 + a))^2*A*a^2*c^(3/2) - 3*(sqrt(c)*x - sqrt(c*x^2 + a))*B*a^3*c + 8
*A*a^3*c^(3/2))/((sqrt(c)*x - sqrt(c*x^2 + a))^2 - a)^3

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maple [A]  time = 0.06, size = 174, normalized size = 1.60 \begin {gather*} A \,c^{\frac {3}{2}} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )-\frac {3 B \sqrt {a}\, c \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )}{2}+\frac {\sqrt {c \,x^{2}+a}\, A \,c^{2} x}{a}+\frac {2 \left (c \,x^{2}+a \right )^{\frac {3}{2}} A \,c^{2} x}{3 a^{2}}+\frac {3 \sqrt {c \,x^{2}+a}\, B c}{2}+\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} B c}{2 a}-\frac {2 \left (c \,x^{2}+a \right )^{\frac {5}{2}} A c}{3 a^{2} x}-\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} B}{2 a \,x^{2}}-\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} A}{3 a \,x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+a)^(3/2)/x^4,x)

[Out]

-1/3*A/a/x^3*(c*x^2+a)^(5/2)-2/3*A*c/a^2/x*(c*x^2+a)^(5/2)+2/3*A*c^2/a^2*x*(c*x^2+a)^(3/2)+A*c^2/a*x*(c*x^2+a)
^(1/2)+A*c^(3/2)*ln(c^(1/2)*x+(c*x^2+a)^(1/2))-1/2*B/a/x^2*(c*x^2+a)^(5/2)+1/2*B*c/a*(c*x^2+a)^(3/2)-3/2*B*c*a
^(1/2)*ln((2*a+2*(c*x^2+a)^(1/2)*a^(1/2))/x)+3/2*B*c*(c*x^2+a)^(1/2)

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maxima [A]  time = 0.53, size = 136, normalized size = 1.25 \begin {gather*} \frac {\sqrt {c x^{2} + a} A c^{2} x}{a} + A c^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right ) - \frac {3}{2} \, B \sqrt {a} c \operatorname {arsinh}\left (\frac {a}{\sqrt {a c} {\left | x \right |}}\right ) + \frac {3}{2} \, \sqrt {c x^{2} + a} B c + \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} B c}{2 \, a} - \frac {2 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} A c}{3 \, a x} - \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} B}{2 \, a x^{2}} - \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} A}{3 \, a x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(3/2)/x^4,x, algorithm="maxima")

[Out]

sqrt(c*x^2 + a)*A*c^2*x/a + A*c^(3/2)*arcsinh(c*x/sqrt(a*c)) - 3/2*B*sqrt(a)*c*arcsinh(a/(sqrt(a*c)*abs(x))) +
3/2*sqrt(c*x^2 + a)*B*c + 1/2*(c*x^2 + a)^(3/2)*B*c/a - 2/3*(c*x^2 + a)^(3/2)*A*c/(a*x) - 1/2*(c*x^2 + a)^(5/
2)*B/(a*x^2) - 1/3*(c*x^2 + a)^(5/2)*A/(a*x^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,x^2+a\right )}^{3/2}\,\left (A+B\,x\right )}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + c*x^2)^(3/2)*(A + B*x))/x^4,x)

[Out]

int(((a + c*x^2)^(3/2)*(A + B*x))/x^4, x)

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sympy [B]  time = 7.33, size = 202, normalized size = 1.85 \begin {gather*} - \frac {A \sqrt {a} c}{x \sqrt {1 + \frac {c x^{2}}{a}}} - \frac {A a \sqrt {c} \sqrt {\frac {a}{c x^{2}} + 1}}{3 x^{2}} - \frac {A c^{\frac {3}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{3} + A c^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )} - \frac {A c^{2} x}{\sqrt {a} \sqrt {1 + \frac {c x^{2}}{a}}} - \frac {3 B \sqrt {a} c \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {c} x} \right )}}{2} - \frac {B a \sqrt {c} \sqrt {\frac {a}{c x^{2}} + 1}}{2 x} + \frac {B a \sqrt {c}}{x \sqrt {\frac {a}{c x^{2}} + 1}} + \frac {B c^{\frac {3}{2}} x}{\sqrt {\frac {a}{c x^{2}} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+a)**(3/2)/x**4,x)

[Out]

-A*sqrt(a)*c/(x*sqrt(1 + c*x**2/a)) - A*a*sqrt(c)*sqrt(a/(c*x**2) + 1)/(3*x**2) - A*c**(3/2)*sqrt(a/(c*x**2) +
1)/3 + A*c**(3/2)*asinh(sqrt(c)*x/sqrt(a)) - A*c**2*x/(sqrt(a)*sqrt(1 + c*x**2/a)) - 3*B*sqrt(a)*c*asinh(sqrt
(a)/(sqrt(c)*x))/2 - B*a*sqrt(c)*sqrt(a/(c*x**2) + 1)/(2*x) + B*a*sqrt(c)/(x*sqrt(a/(c*x**2) + 1)) + B*c**(3/2
)*x/sqrt(a/(c*x**2) + 1)

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