3.1.45 \(\int \frac {x^4 (d+e x)^2}{(d^2-e^2 x^2)^{7/2}} \, dx\)

Optimal. Leaf size=121 \[ -\frac {17 d^2 (d+e x)}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 (15 d+13 e x)}{15 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5}+\frac {d^3 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}} \]

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Rubi [A]  time = 0.21, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {1635, 1814, 12, 217, 203} \begin {gather*} \frac {d^3 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {17 d^2 (d+e x)}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 (15 d+13 e x)}{15 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*(d + e*x)^2)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(d^3*(d + e*x)^2)/(5*e^5*(d^2 - e^2*x^2)^(5/2)) - (17*d^2*(d + e*x))/(15*e^5*(d^2 - e^2*x^2)^(3/2)) + (2*(15*d
+ 13*e*x))/(15*e^5*Sqrt[d^2 - e^2*x^2]) - ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]]/e^5

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 1635

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, -Simp[(d*f*(d + e*x)^m*(a + c*x^2)^(p + 1))/(2*
a*e*(p + 1)), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)*Q
+ f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p +
1/2, 0] && GtQ[m, 0]

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {x^4 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac {d^3 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {(d+e x) \left (\frac {2 d^4}{e^4}+\frac {5 d^3 x}{e^3}+\frac {5 d^2 x^2}{e^2}+\frac {5 d x^3}{e}\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d}\\ &=\frac {d^3 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {17 d^2 (d+e x)}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {\frac {11 d^4}{e^4}+\frac {30 d^3 x}{e^3}+\frac {15 d^2 x^2}{e^2}}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^2}\\ &=\frac {d^3 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {17 d^2 (d+e x)}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 (15 d+13 e x)}{15 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {15 d^4}{e^4 \sqrt {d^2-e^2 x^2}} \, dx}{15 d^4}\\ &=\frac {d^3 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {17 d^2 (d+e x)}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 (15 d+13 e x)}{15 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e^4}\\ &=\frac {d^3 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {17 d^2 (d+e x)}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 (15 d+13 e x)}{15 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4}\\ &=\frac {d^3 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {17 d^2 (d+e x)}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 (15 d+13 e x)}{15 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 96, normalized size = 0.79 \begin {gather*} \frac {16 d^3-15 d (d-e x)^2 \sqrt {1-\frac {e^2 x^2}{d^2}} \sin ^{-1}\left (\frac {e x}{d}\right )-17 d^2 e x-22 d e^2 x^2+26 e^3 x^3}{15 e^5 (d-e x)^2 \sqrt {d^2-e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(d + e*x)^2)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(16*d^3 - 17*d^2*e*x - 22*d*e^2*x^2 + 26*e^3*x^3 - 15*d*(d - e*x)^2*Sqrt[1 - (e^2*x^2)/d^2]*ArcSin[(e*x)/d])/(
15*e^5*(d - e*x)^2*Sqrt[d^2 - e^2*x^2])

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IntegrateAlgebraic [A]  time = 0.62, size = 114, normalized size = 0.94 \begin {gather*} -\frac {\sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{e^6}-\frac {\sqrt {d^2-e^2 x^2} \left (16 d^3-17 d^2 e x-22 d e^2 x^2+26 e^3 x^3\right )}{15 e^5 (e x-d)^3 (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^4*(d + e*x)^2)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

-1/15*(Sqrt[d^2 - e^2*x^2]*(16*d^3 - 17*d^2*e*x - 22*d*e^2*x^2 + 26*e^3*x^3))/(e^5*(-d + e*x)^3*(d + e*x)) - (
Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/e^6

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fricas [A]  time = 0.42, size = 172, normalized size = 1.42 \begin {gather*} \frac {16 \, e^{4} x^{4} - 32 \, d e^{3} x^{3} + 32 \, d^{3} e x - 16 \, d^{4} + 30 \, {\left (e^{4} x^{4} - 2 \, d e^{3} x^{3} + 2 \, d^{3} e x - d^{4}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (26 \, e^{3} x^{3} - 22 \, d e^{2} x^{2} - 17 \, d^{2} e x + 16 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (e^{9} x^{4} - 2 \, d e^{8} x^{3} + 2 \, d^{3} e^{6} x - d^{4} e^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x+d)^2/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

1/15*(16*e^4*x^4 - 32*d*e^3*x^3 + 32*d^3*e*x - 16*d^4 + 30*(e^4*x^4 - 2*d*e^3*x^3 + 2*d^3*e*x - d^4)*arctan(-(
d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (26*e^3*x^3 - 22*d*e^2*x^2 - 17*d^2*e*x + 16*d^3)*sqrt(-e^2*x^2 + d^2))/(e^
9*x^4 - 2*d*e^8*x^3 + 2*d^3*e^6*x - d^4*e^5)

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giac [A]  time = 0.28, size = 95, normalized size = 0.79 \begin {gather*} -\arcsin \left (\frac {x e}{d}\right ) e^{\left (-5\right )} \mathrm {sgn}\relax (d) - \frac {{\left (16 \, d^{5} e^{\left (-5\right )} + {\left (15 \, d^{4} e^{\left (-4\right )} - {\left (40 \, d^{3} e^{\left (-3\right )} + {\left (35 \, d^{2} e^{\left (-2\right )} - 2 \, {\left (15 \, d e^{\left (-1\right )} + 13 \, x\right )} x\right )} x\right )} x\right )} x\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{15 \, {\left (x^{2} e^{2} - d^{2}\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x+d)^2/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

-arcsin(x*e/d)*e^(-5)*sgn(d) - 1/15*(16*d^5*e^(-5) + (15*d^4*e^(-4) - (40*d^3*e^(-3) + (35*d^2*e^(-2) - 2*(15*
d*e^(-1) + 13*x)*x)*x)*x)*x)*sqrt(-x^2*e^2 + d^2)/(x^2*e^2 - d^2)^3

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maple [B]  time = 0.02, size = 236, normalized size = 1.95 \begin {gather*} \frac {x^{5}}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {2 d \,x^{4}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e}+\frac {d^{2} x^{3}}{2 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{2}}-\frac {8 d^{3} x^{2}}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{3}}-\frac {3 d^{4} x}{10 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{4}}-\frac {x^{3}}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{2}}+\frac {16 d^{5}}{15 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{5}}+\frac {d^{2} x}{10 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{4}}+\frac {6 x}{5 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{4}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}\, e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(e*x+d)^2/(-e^2*x^2+d^2)^(7/2),x)

[Out]

1/5*x^5/(-e^2*x^2+d^2)^(5/2)-1/3*x^3/e^2/(-e^2*x^2+d^2)^(3/2)+6/5/e^4*x/(-e^2*x^2+d^2)^(1/2)-1/e^4/(e^2)^(1/2)
*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)+2*d/e*x^4/(-e^2*x^2+d^2)^(5/2)-8/3*d^3/e^3*x^2/(-e^2*x^2+d^2)^(5/2
)+16/15*d^5/e^5/(-e^2*x^2+d^2)^(5/2)+1/2*d^2*x^3/e^2/(-e^2*x^2+d^2)^(5/2)-3/10*d^4/e^4*x/(-e^2*x^2+d^2)^(5/2)+
1/10*d^2/e^4*x/(-e^2*x^2+d^2)^(3/2)

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maxima [B]  time = 1.00, size = 298, normalized size = 2.46 \begin {gather*} \frac {1}{15} \, e^{2} x {\left (\frac {15 \, x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {20 \, d^{2} x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}} + \frac {8 \, d^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{6}}\right )} - \frac {1}{3} \, x {\left (\frac {3 \, x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}} - \frac {2 \, d^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}}\right )} + \frac {2 \, d x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e} + \frac {d^{2} x^{3}}{2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {8 \, d^{3} x^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{3}} - \frac {3 \, d^{4} x}{10 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}} + \frac {16 \, d^{5}}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{5}} + \frac {11 \, d^{2} x}{30 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}} - \frac {4 \, x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{4}} - \frac {\arcsin \left (\frac {e x}{d}\right )}{e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x+d)^2/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

1/15*e^2*x*(15*x^4/((-e^2*x^2 + d^2)^(5/2)*e^2) - 20*d^2*x^2/((-e^2*x^2 + d^2)^(5/2)*e^4) + 8*d^4/((-e^2*x^2 +
d^2)^(5/2)*e^6)) - 1/3*x*(3*x^2/((-e^2*x^2 + d^2)^(3/2)*e^2) - 2*d^2/((-e^2*x^2 + d^2)^(3/2)*e^4)) + 2*d*x^4/
((-e^2*x^2 + d^2)^(5/2)*e) + 1/2*d^2*x^3/((-e^2*x^2 + d^2)^(5/2)*e^2) - 8/3*d^3*x^2/((-e^2*x^2 + d^2)^(5/2)*e^
3) - 3/10*d^4*x/((-e^2*x^2 + d^2)^(5/2)*e^4) + 16/15*d^5/((-e^2*x^2 + d^2)^(5/2)*e^5) + 11/30*d^2*x/((-e^2*x^2
+ d^2)^(3/2)*e^4) - 4/15*x/(sqrt(-e^2*x^2 + d^2)*e^4) - arcsin(e*x/d)/e^5

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^4\,{\left (d+e\,x\right )}^2}{{\left (d^2-e^2\,x^2\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(d + e*x)^2)/(d^2 - e^2*x^2)^(7/2),x)

[Out]

int((x^4*(d + e*x)^2)/(d^2 - e^2*x^2)^(7/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4} \left (d + e x\right )^{2}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(e*x+d)**2/(-e**2*x**2+d**2)**(7/2),x)

[Out]

Integral(x**4*(d + e*x)**2/(-(-d + e*x)*(d + e*x))**(7/2), x)

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