3.6.67 \(\int \frac {1}{(d+e x)^3 \sqrt {a+c x^2}} \, dx\) [567]

Optimal. Leaf size=145 \[ -\frac {e \sqrt {a+c x^2}}{2 \left (c d^2+a e^2\right ) (d+e x)^2}-\frac {3 c d e \sqrt {a+c x^2}}{2 \left (c d^2+a e^2\right )^2 (d+e x)}-\frac {c \left (2 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{2 \left (c d^2+a e^2\right )^{5/2}} \]

[Out]

-1/2*c*(-a*e^2+2*c*d^2)*arctanh((-c*d*x+a*e)/(a*e^2+c*d^2)^(1/2)/(c*x^2+a)^(1/2))/(a*e^2+c*d^2)^(5/2)-1/2*e*(c
*x^2+a)^(1/2)/(a*e^2+c*d^2)/(e*x+d)^2-3/2*c*d*e*(c*x^2+a)^(1/2)/(a*e^2+c*d^2)^2/(e*x+d)

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Rubi [A]
time = 0.04, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {759, 821, 739, 212} \begin {gather*} -\frac {3 c d e \sqrt {a+c x^2}}{2 (d+e x) \left (a e^2+c d^2\right )^2}-\frac {e \sqrt {a+c x^2}}{2 (d+e x)^2 \left (a e^2+c d^2\right )}-\frac {c \left (2 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{2 \left (a e^2+c d^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)^3*Sqrt[a + c*x^2]),x]

[Out]

-1/2*(e*Sqrt[a + c*x^2])/((c*d^2 + a*e^2)*(d + e*x)^2) - (3*c*d*e*Sqrt[a + c*x^2])/(2*(c*d^2 + a*e^2)^2*(d + e
*x)) - (c*(2*c*d^2 - a*e^2)*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/(2*(c*d^2 + a*e^2)^(
5/2))

Rule 212

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

Rule 739

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 759

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m + 1)*((a + c*x^2)^(p
+ 1)/((m + 1)*(c*d^2 + a*e^2))), x] + Dist[c/((m + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^(m + 1)*Simp[d*(m + 1)
- e*(m + 2*p + 3)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[
m, -1] && ((LtQ[m, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]) || (SumSimplerQ[m, 1] && IntegerQ[p]) || ILtQ
[Simplify[m + 2*p + 3], 0])

Rule 821

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(e*f - d*g
))*(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 + a*e^2))), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e
^2), Int[(d + e*x)^(m + 1)*(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
] && EqQ[Simplify[m + 2*p + 3], 0]

Rubi steps

\begin {align*} \int \frac {1}{(d+e x)^3 \sqrt {a+c x^2}} \, dx &=-\frac {e \sqrt {a+c x^2}}{2 \left (c d^2+a e^2\right ) (d+e x)^2}-\frac {c \int \frac {-2 d+e x}{(d+e x)^2 \sqrt {a+c x^2}} \, dx}{2 \left (c d^2+a e^2\right )}\\ &=-\frac {e \sqrt {a+c x^2}}{2 \left (c d^2+a e^2\right ) (d+e x)^2}-\frac {3 c d e \sqrt {a+c x^2}}{2 \left (c d^2+a e^2\right )^2 (d+e x)}+\frac {\left (c \left (2 c d^2-a e^2\right )\right ) \int \frac {1}{(d+e x) \sqrt {a+c x^2}} \, dx}{2 \left (c d^2+a e^2\right )^2}\\ &=-\frac {e \sqrt {a+c x^2}}{2 \left (c d^2+a e^2\right ) (d+e x)^2}-\frac {3 c d e \sqrt {a+c x^2}}{2 \left (c d^2+a e^2\right )^2 (d+e x)}-\frac {\left (c \left (2 c d^2-a e^2\right )\right ) \text {Subst}\left (\int \frac {1}{c d^2+a e^2-x^2} \, dx,x,\frac {a e-c d x}{\sqrt {a+c x^2}}\right )}{2 \left (c d^2+a e^2\right )^2}\\ &=-\frac {e \sqrt {a+c x^2}}{2 \left (c d^2+a e^2\right ) (d+e x)^2}-\frac {3 c d e \sqrt {a+c x^2}}{2 \left (c d^2+a e^2\right )^2 (d+e x)}-\frac {c \left (2 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{2 \left (c d^2+a e^2\right )^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.77, size = 132, normalized size = 0.91 \begin {gather*} -\frac {e \sqrt {a+c x^2} \left (a e^2+c d (4 d+3 e x)\right )}{2 \left (c d^2+a e^2\right )^2 (d+e x)^2}-\frac {c \left (2 c d^2-a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+c x^2}}{\sqrt {-c d^2-a e^2}}\right )}{\left (-c d^2-a e^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)^3*Sqrt[a + c*x^2]),x]

[Out]

-1/2*(e*Sqrt[a + c*x^2]*(a*e^2 + c*d*(4*d + 3*e*x)))/((c*d^2 + a*e^2)^2*(d + e*x)^2) - (c*(2*c*d^2 - a*e^2)*Ar
cTan[(Sqrt[c]*(d + e*x) - e*Sqrt[a + c*x^2])/Sqrt[-(c*d^2) - a*e^2]])/(-(c*d^2) - a*e^2)^(5/2)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(443\) vs. \(2(129)=258\).
time = 0.44, size = 444, normalized size = 3.06

method	result	size
default	\(\frac {-\frac {e^{2} \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{2 \left (e^{2} a +c \,d^{2}\right ) \left (x +\frac {d}{e}\right )^{2}}+\frac {3 c d e \left (-\frac {e^{2} \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{\left (e^{2} a +c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}-\frac {c d e \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (e^{2} a +c \,d^{2}\right ) \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}\right )}{2 \left (e^{2} a +c \,d^{2}\right )}+\frac {c \,e^{2} \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{2 \left (e^{2} a +c \,d^{2}\right ) \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}}{e^{3}}\)	\(444\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^3/(c*x^2+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/e^3*(-1/2/(a*e^2+c*d^2)*e^2/(x+d/e)^2*(c*(x+d/e)^2-2*c*d/e*(x+d/e)+(a*e^2+c*d^2)/e^2)^(1/2)+3/2*c*d*e/(a*e^2
+c*d^2)*(-1/(a*e^2+c*d^2)*e^2/(x+d/e)*(c*(x+d/e)^2-2*c*d/e*(x+d/e)+(a*e^2+c*d^2)/e^2)^(1/2)-c*d*e/(a*e^2+c*d^2
)/((a*e^2+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2+c*d^2)/e^2-2*c*d/e*(x+d/e)+2*((a*e^2+c*d^2)/e^2)^(1/2)*(c*(x+d/e)^2-2
*c*d/e*(x+d/e)+(a*e^2+c*d^2)/e^2)^(1/2))/(x+d/e)))+1/2*c/(a*e^2+c*d^2)*e^2/((a*e^2+c*d^2)/e^2)^(1/2)*ln((2*(a*
e^2+c*d^2)/e^2-2*c*d/e*(x+d/e)+2*((a*e^2+c*d^2)/e^2)^(1/2)*(c*(x+d/e)^2-2*c*d/e*(x+d/e)+(a*e^2+c*d^2)/e^2)^(1/
2))/(x+d/e)))

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Maxima [A]
time = 0.35, size = 241, normalized size = 1.66 \begin {gather*} \frac {3 \, c^{2} d^{2} \operatorname {arsinh}\left (\frac {c d x}{\sqrt {a c} {\left | x e + d \right |}} - \frac {a e}{\sqrt {a c} {\left | x e + d \right |}}\right ) e^{\left (-5\right )}}{2 \, {\left (c d^{2} e^{\left (-2\right )} + a\right )}^{\frac {5}{2}}} - \frac {3 \, \sqrt {c x^{2} + a} c d}{2 \, {\left (c^{2} d^{5} e^{\left (-1\right )} + c^{2} d^{4} x + 2 \, a c d^{2} x e^{2} + 2 \, a c d^{3} e + a^{2} x e^{4} + a^{2} d e^{3}\right )}} - \frac {c \operatorname {arsinh}\left (\frac {c d x}{\sqrt {a c} {\left | x e + d \right |}} - \frac {a e}{\sqrt {a c} {\left | x e + d \right |}}\right ) e^{\left (-3\right )}}{2 \, {\left (c d^{2} e^{\left (-2\right )} + a\right )}^{\frac {3}{2}}} - \frac {\sqrt {c x^{2} + a}}{2 \, {\left (c d^{2} x^{2} e + c d^{4} e^{\left (-1\right )} + 2 \, c d^{3} x + a x^{2} e^{3} + 2 \, a d x e^{2} + a d^{2} e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^3/(c*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

3/2*c^2*d^2*arcsinh(c*d*x/(sqrt(a*c)*abs(x*e + d)) - a*e/(sqrt(a*c)*abs(x*e + d)))*e^(-5)/(c*d^2*e^(-2) + a)^(
5/2) - 3/2*sqrt(c*x^2 + a)*c*d/(c^2*d^5*e^(-1) + c^2*d^4*x + 2*a*c*d^2*x*e^2 + 2*a*c*d^3*e + a^2*x*e^4 + a^2*d
*e^3) - 1/2*c*arcsinh(c*d*x/(sqrt(a*c)*abs(x*e + d)) - a*e/(sqrt(a*c)*abs(x*e + d)))*e^(-3)/(c*d^2*e^(-2) + a)
^(3/2) - 1/2*sqrt(c*x^2 + a)/(c*d^2*x^2*e + c*d^4*e^(-1) + 2*c*d^3*x + a*x^2*e^3 + 2*a*d*x*e^2 + a*d^2*e)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 334 vs. \(2 (130) = 260\).
time = 2.04, size = 694, normalized size = 4.79 \begin {gather*} \left [-\frac {{\left (4 \, c^{2} d^{3} x e + 2 \, c^{2} d^{4} - a c x^{2} e^{4} - 2 \, a c d x e^{3} + {\left (2 \, c^{2} d^{2} x^{2} - a c d^{2}\right )} e^{2}\right )} \sqrt {c d^{2} + a e^{2}} \log \left (-\frac {2 \, c^{2} d^{2} x^{2} - 2 \, a c d x e + a c d^{2} - 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a} + {\left (a c x^{2} + 2 \, a^{2}\right )} e^{2}}{x^{2} e^{2} + 2 \, d x e + d^{2}}\right ) + 2 \, {\left (3 \, c^{2} d^{3} x e^{2} + 4 \, c^{2} d^{4} e + 3 \, a c d x e^{4} + 5 \, a c d^{2} e^{3} + a^{2} e^{5}\right )} \sqrt {c x^{2} + a}}{4 \, {\left (2 \, c^{3} d^{7} x e + c^{3} d^{8} + 6 \, a c^{2} d^{5} x e^{3} + 6 \, a^{2} c d^{3} x e^{5} + a^{3} x^{2} e^{8} + 2 \, a^{3} d x e^{7} + {\left (3 \, a^{2} c d^{2} x^{2} + a^{3} d^{2}\right )} e^{6} + 3 \, {\left (a c^{2} d^{4} x^{2} + a^{2} c d^{4}\right )} e^{4} + {\left (c^{3} d^{6} x^{2} + 3 \, a c^{2} d^{6}\right )} e^{2}\right )}}, \frac {{\left (4 \, c^{2} d^{3} x e + 2 \, c^{2} d^{4} - a c x^{2} e^{4} - 2 \, a c d x e^{3} + {\left (2 \, c^{2} d^{2} x^{2} - a c d^{2}\right )} e^{2}\right )} \sqrt {-c d^{2} - a e^{2}} \arctan \left (-\frac {\sqrt {-c d^{2} - a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{c^{2} d^{2} x^{2} + a c d^{2} + {\left (a c x^{2} + a^{2}\right )} e^{2}}\right ) - {\left (3 \, c^{2} d^{3} x e^{2} + 4 \, c^{2} d^{4} e + 3 \, a c d x e^{4} + 5 \, a c d^{2} e^{3} + a^{2} e^{5}\right )} \sqrt {c x^{2} + a}}{2 \, {\left (2 \, c^{3} d^{7} x e + c^{3} d^{8} + 6 \, a c^{2} d^{5} x e^{3} + 6 \, a^{2} c d^{3} x e^{5} + a^{3} x^{2} e^{8} + 2 \, a^{3} d x e^{7} + {\left (3 \, a^{2} c d^{2} x^{2} + a^{3} d^{2}\right )} e^{6} + 3 \, {\left (a c^{2} d^{4} x^{2} + a^{2} c d^{4}\right )} e^{4} + {\left (c^{3} d^{6} x^{2} + 3 \, a c^{2} d^{6}\right )} e^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^3/(c*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/4*((4*c^2*d^3*x*e + 2*c^2*d^4 - a*c*x^2*e^4 - 2*a*c*d*x*e^3 + (2*c^2*d^2*x^2 - a*c*d^2)*e^2)*sqrt(c*d^2 +
a*e^2)*log(-(2*c^2*d^2*x^2 - 2*a*c*d*x*e + a*c*d^2 - 2*sqrt(c*d^2 + a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a) + (a*
c*x^2 + 2*a^2)*e^2)/(x^2*e^2 + 2*d*x*e + d^2)) + 2*(3*c^2*d^3*x*e^2 + 4*c^2*d^4*e + 3*a*c*d*x*e^4 + 5*a*c*d^2*
e^3 + a^2*e^5)*sqrt(c*x^2 + a))/(2*c^3*d^7*x*e + c^3*d^8 + 6*a*c^2*d^5*x*e^3 + 6*a^2*c*d^3*x*e^5 + a^3*x^2*e^8
+ 2*a^3*d*x*e^7 + (3*a^2*c*d^2*x^2 + a^3*d^2)*e^6 + 3*(a*c^2*d^4*x^2 + a^2*c*d^4)*e^4 + (c^3*d^6*x^2 + 3*a*c^
2*d^6)*e^2), 1/2*((4*c^2*d^3*x*e + 2*c^2*d^4 - a*c*x^2*e^4 - 2*a*c*d*x*e^3 + (2*c^2*d^2*x^2 - a*c*d^2)*e^2)*sq
rt(-c*d^2 - a*e^2)*arctan(-sqrt(-c*d^2 - a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a)/(c^2*d^2*x^2 + a*c*d^2 + (a*c*x^
2 + a^2)*e^2)) - (3*c^2*d^3*x*e^2 + 4*c^2*d^4*e + 3*a*c*d*x*e^4 + 5*a*c*d^2*e^3 + a^2*e^5)*sqrt(c*x^2 + a))/(2
*c^3*d^7*x*e + c^3*d^8 + 6*a*c^2*d^5*x*e^3 + 6*a^2*c*d^3*x*e^5 + a^3*x^2*e^8 + 2*a^3*d*x*e^7 + (3*a^2*c*d^2*x^
2 + a^3*d^2)*e^6 + 3*(a*c^2*d^4*x^2 + a^2*c*d^4)*e^4 + (c^3*d^6*x^2 + 3*a*c^2*d^6)*e^2)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {a + c x^{2}} \left (d + e x\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**3/(c*x**2+a)**(1/2),x)

[Out]

Integral(1/(sqrt(a + c*x**2)*(d + e*x)**3), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 345 vs. \(2 (130) = 260\).
time = 0.83, size = 345, normalized size = 2.38 \begin {gather*} -c {\left (\frac {{\left (2 \, c d^{2} - a e^{2}\right )} \arctan \left (\frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} e + \sqrt {c} d}{\sqrt {-c d^{2} - a e^{2}}}\right )}{{\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt {-c d^{2} - a e^{2}}} + \frac {2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{3} c d^{2} e + 6 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} c^{\frac {3}{2}} d^{3} - 10 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} a c d^{2} e - 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} a \sqrt {c} d e^{2} - {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{3} a e^{3} + 3 \, a^{2} \sqrt {c} d e^{2} - {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} a^{2} e^{3}}{{\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} e + 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} \sqrt {c} d - a e\right )}^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^3/(c*x^2+a)^(1/2),x, algorithm="giac")

[Out]

-c*((2*c*d^2 - a*e^2)*arctan(((sqrt(c)*x - sqrt(c*x^2 + a))*e + sqrt(c)*d)/sqrt(-c*d^2 - a*e^2))/((c^2*d^4 + 2
*a*c*d^2*e^2 + a^2*e^4)*sqrt(-c*d^2 - a*e^2)) + (2*(sqrt(c)*x - sqrt(c*x^2 + a))^3*c*d^2*e + 6*(sqrt(c)*x - sq
rt(c*x^2 + a))^2*c^(3/2)*d^3 - 10*(sqrt(c)*x - sqrt(c*x^2 + a))*a*c*d^2*e - 3*(sqrt(c)*x - sqrt(c*x^2 + a))^2*
a*sqrt(c)*d*e^2 - (sqrt(c)*x - sqrt(c*x^2 + a))^3*a*e^3 + 3*a^2*sqrt(c)*d*e^2 - (sqrt(c)*x - sqrt(c*x^2 + a))*
a^2*e^3)/((c^2*d^4 + 2*a*c*d^2*e^2 + a^2*e^4)*((sqrt(c)*x - sqrt(c*x^2 + a))^2*e + 2*(sqrt(c)*x - sqrt(c*x^2 +
a))*sqrt(c)*d - a*e)^2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + c*x^2)^(1/2)*(d + e*x)^3),x)

[Out]

int(1/((a + c*x^2)^(1/2)*(d + e*x)^3), x)

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