3.1.45 \(\int \frac {x^4 (d+e x)^2}{(d^2-e^2 x^2)^{7/2}} \, dx\) [45]

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

[Out]

1/5*d^3*(e*x+d)^2/e^5/(-e^2*x^2+d^2)^(5/2)-17/15*d^2*(e*x+d)/e^5/(-e^2*x^2+d^2)^(3/2)-arctan(e*x/(-e^2*x^2+d^2
)^(1/2))/e^5+2/15*(13*e*x+15*d)/e^5/(-e^2*x^2+d^2)^(1/2)

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Rubi [A]
time = 0.13, 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 = {1649, 1828, 12, 223, 209} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5}-\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 {d^3 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}} \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 209

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

Rule 223

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 1649

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 1828

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 {\text {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 = 0.40, size = 113, normalized size = 0.93 \begin {gather*} \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 (-d+e x)^3 (d+e x)}+\frac {\log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{e^4 \sqrt {-e^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]

(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))/(15*e^5*(-d + e*x)^3*(d + e*x)) + Log
[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]]/(e^4*Sqrt[-e^2])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(323\) vs. \(2(107)=214\).
time = 0.06, size = 324, normalized size = 2.68

method	result	size
default	\(e^{2} \left (\frac {x^{5}}{5 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {\frac {x^{3}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {\frac {x}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}}{e^{2}}}{e^{2}}\right )+2 e d \left (\frac {x^{4}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {4 d^{2} \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )}{e^{2}}\right )+d^{2} \left (\frac {x^{3}}{2 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {3 d^{2} \left (\frac {x}{4 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{4 e^{2}}\right )}{2 e^{2}}\right )\)	\(324\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 273 vs. \(2 (103) = 206\).
time = 0.49, size = 273, normalized size = 2.26 \begin {gather*} \frac {2 \, d x^{4} e^{\left (-1\right )}}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {d^{2} x^{3} e^{\left (-2\right )}}{2 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} - \frac {8 \, d^{3} x^{2} e^{\left (-3\right )}}{3 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} - \frac {3 \, d^{4} x e^{\left (-4\right )}}{10 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {16 \, d^{5} e^{\left (-5\right )}}{15 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {1}{15} \, {\left (\frac {15 \, x^{4} e^{\left (-2\right )}}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} - \frac {20 \, d^{2} x^{2} e^{\left (-4\right )}}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {8 \, d^{4} e^{\left (-6\right )}}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}}\right )} x e^{2} + \frac {11 \, d^{2} x e^{\left (-4\right )}}{30 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}}} - \frac {1}{3} \, {\left (\frac {3 \, x^{2} e^{\left (-2\right )}}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, d^{2} e^{\left (-4\right )}}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}}}\right )} x - \arcsin \left (\frac {x e}{d}\right ) e^{\left (-5\right )} - \frac {4 \, x e^{\left (-4\right )}}{15 \, \sqrt {-x^{2} e^{2} + d^{2}}} \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]

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

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Fricas [A]
time = 2.11, size = 162, normalized size = 1.34 \begin {gather*} \frac {16 \, x^{4} e^{4} - 32 \, d x^{3} e^{3} + 32 \, d^{3} x e - 16 \, d^{4} + 30 \, {\left (x^{4} e^{4} - 2 \, d x^{3} e^{3} + 2 \, d^{3} x e - d^{4}\right )} \arctan \left (-\frac {{\left (d - \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-1\right )}}{x}\right ) - {\left (26 \, x^{3} e^{3} - 22 \, d x^{2} e^{2} - 17 \, d^{2} x e + 16 \, d^{3}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{15 \, {\left (x^{4} e^{9} - 2 \, d x^{3} e^{8} + 2 \, d^{3} x e^{6} - 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*x^4*e^4 - 32*d*x^3*e^3 + 32*d^3*x*e - 16*d^4 + 30*(x^4*e^4 - 2*d*x^3*e^3 + 2*d^3*x*e - d^4)*arctan(-(
d - sqrt(-x^2*e^2 + d^2))*e^(-1)/x) - (26*x^3*e^3 - 22*d*x^2*e^2 - 17*d^2*x*e + 16*d^3)*sqrt(-x^2*e^2 + d^2))/
(x^4*e^9 - 2*d*x^3*e^8 + 2*d^3*x*e^6 - d^4*e^5)

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

[Out]

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

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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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