3.1.22 \(\int \frac {(a+b \sin (c+d x^2))^2}{x^4} \, dx\) [22]

Optimal. Leaf size=239 \[ -\frac {2 a^2+b^2}{6 x^3}-\frac {4 a b d \cos \left (c+d x^2\right )}{3 x}+\frac {b^2 \cos \left (2 c+2 d x^2\right )}{6 x^3}+\frac {4}{3} b^2 d^{3/2} \sqrt {\pi } \cos (2 c) C\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right )-\frac {4}{3} a b d^{3/2} \sqrt {2 \pi } \cos (c) S\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )-\frac {4}{3} a b d^{3/2} \sqrt {2 \pi } C\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right ) \sin (c)-\frac {4}{3} b^2 d^{3/2} \sqrt {\pi } S\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right ) \sin (2 c)-\frac {2 a b \sin \left (c+d x^2\right )}{3 x^3}-\frac {2 b^2 d \sin \left (2 c+2 d x^2\right )}{3 x} \]

[Out]

1/6*(-2*a^2-b^2)/x^3-4/3*a*b*d*cos(d*x^2+c)/x+1/6*b^2*cos(2*d*x^2+2*c)/x^3-2/3*a*b*sin(d*x^2+c)/x^3-2/3*b^2*d*
sin(2*d*x^2+2*c)/x+4/3*b^2*d^(3/2)*cos(2*c)*FresnelC(2*x*d^(1/2)/Pi^(1/2))*Pi^(1/2)-4/3*b^2*d^(3/2)*FresnelS(2
*x*d^(1/2)/Pi^(1/2))*sin(2*c)*Pi^(1/2)-4/3*a*b*d^(3/2)*cos(c)*FresnelS(x*d^(1/2)*2^(1/2)/Pi^(1/2))*2^(1/2)*Pi^
(1/2)-4/3*a*b*d^(3/2)*FresnelC(x*d^(1/2)*2^(1/2)/Pi^(1/2))*sin(c)*2^(1/2)*Pi^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.13, antiderivative size = 239, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3484, 6, 3469, 3468, 3435, 3433, 3432, 3434} \begin {gather*} -\frac {2 a^2+b^2}{6 x^3}-\frac {4}{3} \sqrt {2 \pi } a b d^{3/2} \sin (c) \text {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {d} x\right )-\frac {4}{3} \sqrt {2 \pi } a b d^{3/2} \cos (c) S\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )-\frac {4 a b d \cos \left (c+d x^2\right )}{3 x}-\frac {2 a b \sin \left (c+d x^2\right )}{3 x^3}+\frac {4}{3} \sqrt {\pi } b^2 d^{3/2} \cos (2 c) \text {FresnelC}\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right )-\frac {4}{3} \sqrt {\pi } b^2 d^{3/2} \sin (2 c) S\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right )-\frac {2 b^2 d \sin \left (2 c+2 d x^2\right )}{3 x}+\frac {b^2 \cos \left (2 c+2 d x^2\right )}{6 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x^2])^2/x^4,x]

[Out]

-1/6*(2*a^2 + b^2)/x^3 - (4*a*b*d*Cos[c + d*x^2])/(3*x) + (b^2*Cos[2*c + 2*d*x^2])/(6*x^3) + (4*b^2*d^(3/2)*Sq
rt[Pi]*Cos[2*c]*FresnelC[(2*Sqrt[d]*x)/Sqrt[Pi]])/3 - (4*a*b*d^(3/2)*Sqrt[2*Pi]*Cos[c]*FresnelS[Sqrt[d]*Sqrt[2
/Pi]*x])/3 - (4*a*b*d^(3/2)*Sqrt[2*Pi]*FresnelC[Sqrt[d]*Sqrt[2/Pi]*x]*Sin[c])/3 - (4*b^2*d^(3/2)*Sqrt[Pi]*Fres
nelS[(2*Sqrt[d]*x)/Sqrt[Pi]]*Sin[2*c])/3 - (2*a*b*Sin[c + d*x^2])/(3*x^3) - (2*b^2*d*Sin[2*c + 2*d*x^2])/(3*x)

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 3432

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 3433

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 3434

Int[Sin[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Dist[Sin[c], Int[Cos[d*(e + f*x)^2], x], x] + Dist[
Cos[c], Int[Sin[d*(e + f*x)^2], x], x] /; FreeQ[{c, d, e, f}, x]

Rule 3435

Int[Cos[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Dist[Cos[c], Int[Cos[d*(e + f*x)^2], x], x] - Dist[
Sin[c], Int[Sin[d*(e + f*x)^2], x], x] /; FreeQ[{c, d, e, f}, x]

Rule 3468

Int[((e_.)*(x_))^(m_)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Simp[(e*x)^(m + 1)*(Sin[c + d*x^n]/(e*(m + 1)
)), x] - Dist[d*(n/(e^n*(m + 1))), Int[(e*x)^(m + n)*Cos[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x] && IGtQ[n,
0] && LtQ[m, -1]

Rule 3469

Int[Cos[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_), x_Symbol] :> Simp[(e*x)^(m + 1)*(Cos[c + d*x^n]/(e*(m + 1)
)), x] + Dist[d*(n/(e^n*(m + 1))), Int[(e*x)^(m + n)*Sin[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x] && IGtQ[n,
0] && LtQ[m, -1]

Rule 3484

Int[((e_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_), x_Symbol] :> Int[ExpandTrigReduce[(e
*x)^m, (a + b*Sin[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 1] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\left (a+b \sin \left (c+d x^2\right )\right )^2}{x^4} \, dx &=\int \left (\frac {a^2}{x^4}+\frac {b^2}{2 x^4}-\frac {b^2 \cos \left (2 c+2 d x^2\right )}{2 x^4}+\frac {2 a b \sin \left (c+d x^2\right )}{x^4}\right ) \, dx\\ &=\int \left (\frac {a^2+\frac {b^2}{2}}{x^4}-\frac {b^2 \cos \left (2 c+2 d x^2\right )}{2 x^4}+\frac {2 a b \sin \left (c+d x^2\right )}{x^4}\right ) \, dx\\ &=-\frac {2 a^2+b^2}{6 x^3}+(2 a b) \int \frac {\sin \left (c+d x^2\right )}{x^4} \, dx-\frac {1}{2} b^2 \int \frac {\cos \left (2 c+2 d x^2\right )}{x^4} \, dx\\ &=-\frac {2 a^2+b^2}{6 x^3}+\frac {b^2 \cos \left (2 c+2 d x^2\right )}{6 x^3}-\frac {2 a b \sin \left (c+d x^2\right )}{3 x^3}+\frac {1}{3} (4 a b d) \int \frac {\cos \left (c+d x^2\right )}{x^2} \, dx+\frac {1}{3} \left (2 b^2 d\right ) \int \frac {\sin \left (2 c+2 d x^2\right )}{x^2} \, dx\\ &=-\frac {2 a^2+b^2}{6 x^3}-\frac {4 a b d \cos \left (c+d x^2\right )}{3 x}+\frac {b^2 \cos \left (2 c+2 d x^2\right )}{6 x^3}-\frac {2 a b \sin \left (c+d x^2\right )}{3 x^3}-\frac {2 b^2 d \sin \left (2 c+2 d x^2\right )}{3 x}-\frac {1}{3} \left (8 a b d^2\right ) \int \sin \left (c+d x^2\right ) \, dx+\frac {1}{3} \left (8 b^2 d^2\right ) \int \cos \left (2 c+2 d x^2\right ) \, dx\\ &=-\frac {2 a^2+b^2}{6 x^3}-\frac {4 a b d \cos \left (c+d x^2\right )}{3 x}+\frac {b^2 \cos \left (2 c+2 d x^2\right )}{6 x^3}-\frac {2 a b \sin \left (c+d x^2\right )}{3 x^3}-\frac {2 b^2 d \sin \left (2 c+2 d x^2\right )}{3 x}-\frac {1}{3} \left (8 a b d^2 \cos (c)\right ) \int \sin \left (d x^2\right ) \, dx+\frac {1}{3} \left (8 b^2 d^2 \cos (2 c)\right ) \int \cos \left (2 d x^2\right ) \, dx-\frac {1}{3} \left (8 a b d^2 \sin (c)\right ) \int \cos \left (d x^2\right ) \, dx-\frac {1}{3} \left (8 b^2 d^2 \sin (2 c)\right ) \int \sin \left (2 d x^2\right ) \, dx\\ &=-\frac {2 a^2+b^2}{6 x^3}-\frac {4 a b d \cos \left (c+d x^2\right )}{3 x}+\frac {b^2 \cos \left (2 c+2 d x^2\right )}{6 x^3}+\frac {4}{3} b^2 d^{3/2} \sqrt {\pi } \cos (2 c) C\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right )-\frac {4}{3} a b d^{3/2} \sqrt {2 \pi } \cos (c) S\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )-\frac {4}{3} a b d^{3/2} \sqrt {2 \pi } C\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right ) \sin (c)-\frac {4}{3} b^2 d^{3/2} \sqrt {\pi } S\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right ) \sin (2 c)-\frac {2 a b \sin \left (c+d x^2\right )}{3 x^3}-\frac {2 b^2 d \sin \left (2 c+2 d x^2\right )}{3 x}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.42, size = 226, normalized size = 0.95 \begin {gather*} -\frac {2 a^2+b^2+8 a b d x^2 \cos \left (c+d x^2\right )-b^2 \cos \left (2 \left (c+d x^2\right )\right )-8 b^2 d^{3/2} \sqrt {\pi } x^3 \cos (2 c) C\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right )+8 a b d^{3/2} \sqrt {2 \pi } x^3 \cos (c) S\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )+8 a b d^{3/2} \sqrt {2 \pi } x^3 C\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right ) \sin (c)+8 b^2 d^{3/2} \sqrt {\pi } x^3 S\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right ) \sin (2 c)+4 a b \sin \left (c+d x^2\right )+4 b^2 d x^2 \sin \left (2 \left (c+d x^2\right )\right )}{6 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d*x^2])^2/x^4,x]

[Out]

-1/6*(2*a^2 + b^2 + 8*a*b*d*x^2*Cos[c + d*x^2] - b^2*Cos[2*(c + d*x^2)] - 8*b^2*d^(3/2)*Sqrt[Pi]*x^3*Cos[2*c]*
FresnelC[(2*Sqrt[d]*x)/Sqrt[Pi]] + 8*a*b*d^(3/2)*Sqrt[2*Pi]*x^3*Cos[c]*FresnelS[Sqrt[d]*Sqrt[2/Pi]*x] + 8*a*b*
d^(3/2)*Sqrt[2*Pi]*x^3*FresnelC[Sqrt[d]*Sqrt[2/Pi]*x]*Sin[c] + 8*b^2*d^(3/2)*Sqrt[Pi]*x^3*FresnelS[(2*Sqrt[d]*
x)/Sqrt[Pi]]*Sin[2*c] + 4*a*b*Sin[c + d*x^2] + 4*b^2*d*x^2*Sin[2*(c + d*x^2)])/x^3

________________________________________________________________________________________

Maple [A]
time = 0.09, size = 175, normalized size = 0.73

method	result	size
default	\(-\frac {a^{2}+\frac {b^{2}}{2}}{3 x^{3}}-\frac {b^{2} \left (-\frac {\cos \left (2 d \,x^{2}+2 c \right )}{3 x^{3}}-\frac {4 d \left (-\frac {\sin \left (2 d \,x^{2}+2 c \right )}{x}+2 \sqrt {d}\, \sqrt {\pi }\, \left (\cos \left (2 c \right ) \FresnelC \left (\frac {2 x \sqrt {d}}{\sqrt {\pi }}\right )-\sin \left (2 c \right ) \mathrm {S}\left (\frac {2 x \sqrt {d}}{\sqrt {\pi }}\right )\right )\right )}{3}\right )}{2}+2 a b \left (-\frac {\sin \left (d \,x^{2}+c \right )}{3 x^{3}}+\frac {2 d \left (-\frac {\cos \left (d \,x^{2}+c \right )}{x}-\sqrt {d}\, \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (c \right ) \mathrm {S}\left (\frac {x \sqrt {d}\, \sqrt {2}}{\sqrt {\pi }}\right )+\sin \left (c \right ) \FresnelC \left (\frac {x \sqrt {d}\, \sqrt {2}}{\sqrt {\pi }}\right )\right )\right )}{3}\right )\)	\(175\)
risch	\(-\frac {2 i a b \,d^{2} \sqrt {\pi }\, \erf \left (\sqrt {i d}\, x \right ) {\mathrm e}^{-i c}}{3 \sqrt {i d}}-\frac {a^{2}}{3 x^{3}}-\frac {b^{2}}{6 x^{3}}+\frac {b^{2} d^{2} \sqrt {\pi }\, \sqrt {2}\, \erf \left (\sqrt {2}\, \sqrt {i d}\, x \right ) {\mathrm e}^{-2 i c}}{3 \sqrt {i d}}+\frac {2 b^{2} d^{2} \sqrt {\pi }\, \erf \left (\sqrt {-2 i d}\, x \right ) {\mathrm e}^{2 i c}}{3 \sqrt {-2 i d}}+\frac {2 i a b \,d^{2} \sqrt {\pi }\, \erf \left (\sqrt {-i d}\, x \right ) {\mathrm e}^{i c}}{3 \sqrt {-i d}}-\frac {4 a b d \cos \left (d \,x^{2}+c \right )}{3 x}-\frac {2 a b \sin \left (d \,x^{2}+c \right )}{3 x^{3}}+\frac {b^{2} \cos \left (2 d \,x^{2}+2 c \right )}{6 x^{3}}-\frac {2 b^{2} d \sin \left (2 d \,x^{2}+2 c \right )}{3 x}\)	\(218\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(d*x^2+c))^2/x^4,x,method=_RETURNVERBOSE)

[Out]

-1/3*(a^2+1/2*b^2)/x^3-1/2*b^2*(-1/3/x^3*cos(2*d*x^2+2*c)-4/3*d*(-1/x*sin(2*d*x^2+2*c)+2*d^(1/2)*Pi^(1/2)*(cos
(2*c)*FresnelC(2*x*d^(1/2)/Pi^(1/2))-sin(2*c)*FresnelS(2*x*d^(1/2)/Pi^(1/2)))))+2*a*b*(-1/3*sin(d*x^2+c)/x^3+2
/3*d*(-1/x*cos(d*x^2+c)-d^(1/2)*2^(1/2)*Pi^(1/2)*(cos(c)*FresnelS(x*d^(1/2)*2^(1/2)/Pi^(1/2))+sin(c)*FresnelC(
x*d^(1/2)*2^(1/2)/Pi^(1/2)))))

________________________________________________________________________________________

Maxima [C] Result contains complex when optimal does not.
time = 0.66, size = 176, normalized size = 0.74 \begin {gather*} -\frac {\sqrt {d x^{2}} {\left ({\left (-\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, i \, d x^{2}\right ) + \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, -i \, d x^{2}\right )\right )} \cos \left (c\right ) + {\left (\left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, i \, d x^{2}\right ) - \left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, -i \, d x^{2}\right )\right )} \sin \left (c\right )\right )} a b d}{4 \, x} - \frac {{\left (3 \, \sqrt {2} \sqrt {d x^{2}} {\left ({\left (-\left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, 2 i \, d x^{2}\right ) + \left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, -2 i \, d x^{2}\right )\right )} \cos \left (2 \, c\right ) + {\left (-\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, 2 i \, d x^{2}\right ) + \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, -2 i \, d x^{2}\right )\right )} \sin \left (2 \, c\right )\right )} d x^{2} + 4\right )} b^{2}}{24 \, x^{3}} - \frac {a^{2}}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x^2+c))^2/x^4,x, algorithm="maxima")

[Out]

-1/4*sqrt(d*x^2)*((-(I + 1)*sqrt(2)*gamma(-3/2, I*d*x^2) + (I - 1)*sqrt(2)*gamma(-3/2, -I*d*x^2))*cos(c) + ((I
- 1)*sqrt(2)*gamma(-3/2, I*d*x^2) - (I + 1)*sqrt(2)*gamma(-3/2, -I*d*x^2))*sin(c))*a*b*d/x - 1/24*(3*sqrt(2)*
sqrt(d*x^2)*((-(I - 1)*sqrt(2)*gamma(-3/2, 2*I*d*x^2) + (I + 1)*sqrt(2)*gamma(-3/2, -2*I*d*x^2))*cos(2*c) + (-
(I + 1)*sqrt(2)*gamma(-3/2, 2*I*d*x^2) + (I - 1)*sqrt(2)*gamma(-3/2, -2*I*d*x^2))*sin(2*c))*d*x^2 + 4)*b^2/x^3
- 1/3*a^2/x^3

________________________________________________________________________________________

Fricas [A]
time = 0.40, size = 206, normalized size = 0.86 \begin {gather*} -\frac {4 \, \sqrt {2} \pi a b d x^{3} \sqrt {\frac {d}{\pi }} \cos \left (c\right ) \operatorname {S}\left (\sqrt {2} x \sqrt {\frac {d}{\pi }}\right ) + 4 \, \sqrt {2} \pi a b d x^{3} \sqrt {\frac {d}{\pi }} \operatorname {C}\left (\sqrt {2} x \sqrt {\frac {d}{\pi }}\right ) \sin \left (c\right ) - 4 \, \pi b^{2} d x^{3} \sqrt {\frac {d}{\pi }} \cos \left (2 \, c\right ) \operatorname {C}\left (2 \, x \sqrt {\frac {d}{\pi }}\right ) + 4 \, \pi b^{2} d x^{3} \sqrt {\frac {d}{\pi }} \operatorname {S}\left (2 \, x \sqrt {\frac {d}{\pi }}\right ) \sin \left (2 \, c\right ) + 4 \, a b d x^{2} \cos \left (d x^{2} + c\right ) - b^{2} \cos \left (d x^{2} + c\right )^{2} + a^{2} + b^{2} + 2 \, {\left (2 \, b^{2} d x^{2} \cos \left (d x^{2} + c\right ) + a b\right )} \sin \left (d x^{2} + c\right )}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x^2+c))^2/x^4,x, algorithm="fricas")

[Out]

-1/3*(4*sqrt(2)*pi*a*b*d*x^3*sqrt(d/pi)*cos(c)*fresnel_sin(sqrt(2)*x*sqrt(d/pi)) + 4*sqrt(2)*pi*a*b*d*x^3*sqrt
(d/pi)*fresnel_cos(sqrt(2)*x*sqrt(d/pi))*sin(c) - 4*pi*b^2*d*x^3*sqrt(d/pi)*cos(2*c)*fresnel_cos(2*x*sqrt(d/pi
)) + 4*pi*b^2*d*x^3*sqrt(d/pi)*fresnel_sin(2*x*sqrt(d/pi))*sin(2*c) + 4*a*b*d*x^2*cos(d*x^2 + c) - b^2*cos(d*x
^2 + c)^2 + a^2 + b^2 + 2*(2*b^2*d*x^2*cos(d*x^2 + c) + a*b)*sin(d*x^2 + c))/x^3

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \sin {\left (c + d x^{2} \right )}\right )^{2}}{x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x**2+c))**2/x**4,x)

[Out]

Integral((a + b*sin(c + d*x**2))**2/x**4, x)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x^2+c))^2/x^4,x, algorithm="giac")

[Out]

integrate((b*sin(d*x^2 + c) + a)^2/x^4, x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\sin \left (d\,x^2+c\right )\right )}^2}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(c + d*x^2))^2/x^4,x)

[Out]

int((a + b*sin(c + d*x^2))^2/x^4, x)

________________________________________________________________________________________