3.2.19 \(\int \frac {x^4}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx\)

Optimal. Leaf size=118 \[ \frac {x^3 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}-\frac {d (16 d-9 e x) \sqrt {d^2-e^2 x^2}}{6 e^5}-\frac {4 x^2 \sqrt {d^2-e^2 x^2}}{3 e^3}-\frac {3 d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^5} \]

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Rubi [A]  time = 0.10, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {850, 819, 833, 780, 217, 203} \begin {gather*} -\frac {d (16 d-9 e x) \sqrt {d^2-e^2 x^2}}{6 e^5}-\frac {4 x^2 \sqrt {d^2-e^2 x^2}}{3 e^3}+\frac {x^3 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}-\frac {3 d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4/((d + e*x)*Sqrt[d^2 - e^2*x^2]),x]

[Out]

(x^3*(d - e*x))/(e^2*Sqrt[d^2 - e^2*x^2]) - (4*x^2*Sqrt[d^2 - e^2*x^2])/(3*e^3) - (d*(16*d - 9*e*x)*Sqrt[d^2 -
e^2*x^2])/(6*e^5) - (3*d^3*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(2*e^5)

Rule 203

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

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 780

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

Rule 819

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 + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 833

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

Rule 850

Int[((x_)^(n_.)*((a_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*(x_)), x_Symbol] :> Int[x^n*(a/d + (c*x)/e)*(a + c*x
^2)^(p - 1), x] /; FreeQ[{a, c, d, e, n, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ( !IntegerQ[n] ||
!IntegerQ[2*p] || IGtQ[n, 2] || (GtQ[p, 0] && NeQ[n, 2]))

Rubi steps

\begin {align*} \int \frac {x^4}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx &=\int \frac {x^4 (d-e x)}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx\\ &=\frac {x^3 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {x^2 \left (3 d^3-4 d^2 e x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{d^2 e^2}\\ &=\frac {x^3 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}-\frac {4 x^2 \sqrt {d^2-e^2 x^2}}{3 e^3}+\frac {\int \frac {x \left (8 d^4 e-9 d^3 e^2 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{3 d^2 e^4}\\ &=\frac {x^3 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}-\frac {4 x^2 \sqrt {d^2-e^2 x^2}}{3 e^3}-\frac {d (16 d-9 e x) \sqrt {d^2-e^2 x^2}}{6 e^5}-\frac {\left (3 d^3\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{2 e^4}\\ &=\frac {x^3 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}-\frac {4 x^2 \sqrt {d^2-e^2 x^2}}{3 e^3}-\frac {d (16 d-9 e x) \sqrt {d^2-e^2 x^2}}{6 e^5}-\frac {\left (3 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^4}\\ &=\frac {x^3 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}-\frac {4 x^2 \sqrt {d^2-e^2 x^2}}{3 e^3}-\frac {d (16 d-9 e x) \sqrt {d^2-e^2 x^2}}{6 e^5}-\frac {3 d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^5}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 91, normalized size = 0.77 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-16 d^3-7 d^2 e x+d e^2 x^2-2 e^3 x^3\right )-9 d^3 (d+e x) \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{6 e^5 (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4/((d + e*x)*Sqrt[d^2 - e^2*x^2]),x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-16*d^3 - 7*d^2*e*x + d*e^2*x^2 - 2*e^3*x^3) - 9*d^3*(d + e*x)*ArcTan[(e*x)/Sqrt[d^2 - e
^2*x^2]])/(6*e^5*(d + e*x))

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IntegrateAlgebraic [A]  time = 0.46, size = 109, normalized size = 0.92 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-16 d^3-7 d^2 e x+d e^2 x^2-2 e^3 x^3\right )}{6 e^5 (d+e x)}-\frac {3 d^3 \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{2 e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^4/((d + e*x)*Sqrt[d^2 - e^2*x^2]),x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-16*d^3 - 7*d^2*e*x + d*e^2*x^2 - 2*e^3*x^3))/(6*e^5*(d + e*x)) - (3*d^3*Sqrt[-e^2]*Log[
-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/(2*e^6)

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fricas [A]  time = 0.41, size = 112, normalized size = 0.95 \begin {gather*} -\frac {16 \, d^{3} e x + 16 \, d^{4} - 18 \, {\left (d^{3} e x + d^{4}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (2 \, e^{3} x^{3} - d e^{2} x^{2} + 7 \, d^{2} e x + 16 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{6 \, {\left (e^{6} x + d e^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(e*x+d)/(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

-1/6*(16*d^3*e*x + 16*d^4 - 18*(d^3*e*x + d^4)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (2*e^3*x^3 - d*e^2*
x^2 + 7*d^2*e*x + 16*d^3)*sqrt(-e^2*x^2 + d^2))/(e^6*x + d*e^5)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(e*x+d)/(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: -3/2*d^3*sign(d)*asin(x*exp(2)/d/exp(1))
/exp(1)^5-2*d^3*exp(2)*atan((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x+exp(2))/sqrt(-exp(1)^4+exp(2)^
2))/sqrt(-exp(1)^4+exp(2)^2)/exp(1)^4/exp(1)+2*((-16*exp(1)^13*1/96/exp(1)^16*x+24*exp(1)^12*d*1/96/exp(1)^16)
*x-80*exp(1)^11*d^2*1/96/exp(1)^16)*sqrt(-exp(2)*x^2+d^2)

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maple [A]  time = 0.01, size = 147, normalized size = 1.25 \begin {gather*} -\frac {3 d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}\, e^{4}}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, x^{2}}{3 e^{3}}+\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, d x}{2 e^{4}}-\frac {\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{3}}{\left (x +\frac {d}{e}\right ) e^{6}}-\frac {5 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{2}}{3 e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(e*x+d)/(-e^2*x^2+d^2)^(1/2),x)

[Out]

-1/3*x^2*(-e^2*x^2+d^2)^(1/2)/e^3-5/3/e^5*d^2*(-e^2*x^2+d^2)^(1/2)+1/2*d/e^4*x*(-e^2*x^2+d^2)^(1/2)-3/2*d^3/e^
4/(e^2)^(1/2)*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)-d^3/e^6/(x+d/e)*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)

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maxima [A]  time = 1.01, size = 113, normalized size = 0.96 \begin {gather*} -\frac {\sqrt {-e^{2} x^{2} + d^{2}} d^{3}}{e^{6} x + d e^{5}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} x^{2}}{3 \, e^{3}} - \frac {3 \, d^{3} \arcsin \left (\frac {e x}{d}\right )}{2 \, e^{5}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} d x}{2 \, e^{4}} - \frac {5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2}}{3 \, e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(e*x+d)/(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

-sqrt(-e^2*x^2 + d^2)*d^3/(e^6*x + d*e^5) - 1/3*sqrt(-e^2*x^2 + d^2)*x^2/e^3 - 3/2*d^3*arcsin(e*x/d)/e^5 + 1/2
*sqrt(-e^2*x^2 + d^2)*d*x/e^4 - 5/3*sqrt(-e^2*x^2 + d^2)*d^2/e^5

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/((d^2 - e^2*x^2)^(1/2)*(d + e*x)),x)

[Out]

int(x^4/((d^2 - e^2*x^2)^(1/2)*(d + e*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4}}{\sqrt {- \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(e*x+d)/(-e**2*x**2+d**2)**(1/2),x)

[Out]

Integral(x**4/(sqrt(-(-d + e*x)*(d + e*x))*(d + e*x)), x)

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