3.1.85 \(\int \frac {x^3 (d+e x)^3}{(d^2-e^2 x^2)^{7/2}} \, dx\) [85]

Optimal. Leaf size=118 \[ \frac {d^2 (d+e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {13 d (d+e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {32 (d+e x)}{15 e^4 \sqrt {d^2-e^2 x^2}}-\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4} \]

[Out]

1/5*d^2*(e*x+d)^3/e^4/(-e^2*x^2+d^2)^(5/2)-13/15*d*(e*x+d)^2/e^4/(-e^2*x^2+d^2)^(3/2)-arctan(e*x/(-e^2*x^2+d^2
)^(1/2))/e^4+32/15*(e*x+d)/e^4/(-e^2*x^2+d^2)^(1/2)

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Rubi [A]
time = 0.13, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {1649, 792, 223, 209} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4}+\frac {d^2 (d+e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {13 d (d+e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {32 (d+e x)}{15 e^4 \sqrt {d^2-e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*(d + e*x)^3)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(d^2*(d + e*x)^3)/(5*e^4*(d^2 - e^2*x^2)^(5/2)) - (13*d*(d + e*x)^2)/(15*e^4*(d^2 - e^2*x^2)^(3/2)) + (32*(d +
e*x))/(15*e^4*Sqrt[d^2 - e^2*x^2]) - ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]]/e^4

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 792

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

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]

Rubi steps

\begin {align*} \int \frac {x^3 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac {d^2 (d+e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {(d+e x)^2 \left (\frac {3 d^3}{e^3}+\frac {5 d^2 x}{e^2}+\frac {5 d x^2}{e}\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d}\\ &=\frac {d^2 (d+e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {13 d (d+e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {\left (\frac {17 d^3}{e^3}+\frac {15 d^2 x}{e^2}\right ) (d+e x)}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^2}\\ &=\frac {d^2 (d+e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {13 d (d+e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {32 (d+e x)}{15 e^4 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e^3}\\ &=\frac {d^2 (d+e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {13 d (d+e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {32 (d+e x)}{15 e^4 \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^3}\\ &=\frac {d^2 (d+e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {13 d (d+e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {32 (d+e x)}{15 e^4 \sqrt {d^2-e^2 x^2}}-\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4}\\ \end {align*}

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Mathematica [A]
time = 0.39, size = 95, normalized size = 0.81 \begin {gather*} \frac {\left (-22 d^2+51 d e x-32 e^2 x^2\right ) \sqrt {d^2-e^2 x^2}}{15 e^4 (-d+e x)^3}+\frac {\log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{e^3 \sqrt {-e^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(d + e*x)^3)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

((-22*d^2 + 51*d*e*x - 32*e^2*x^2)*Sqrt[d^2 - e^2*x^2])/(15*e^4*(-d + e*x)^3) + Log[-(Sqrt[-e^2]*x) + Sqrt[d^2
- e^2*x^2]]/(e^3*Sqrt[-e^2])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(376\) vs. \(2(104)=208\).
time = 0.08, size = 377, normalized size = 3.19

method	result	size
default	\(e^{3} \left (\frac {x^{5}}{5 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {\frac {x^{3}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {\frac {x}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}}{e^{2}}}{e^{2}}\right )+3 e^{2} d \left (\frac {x^{4}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {4 d^{2} \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )}{e^{2}}\right )+3 e \,d^{2} \left (\frac {x^{3}}{2 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {3 d^{2} \left (\frac {x}{4 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{4 e^{2}}\right )}{2 e^{2}}\right )+d^{3} \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )\)	\(377\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x,method=_RETURNVERBOSE)

[Out]

e^3*(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)))))+3*e^2*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)))+3*e*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)))))+d^3*(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))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 273 vs. \(2 (100) = 200\).
time = 0.50, size = 273, normalized size = 2.31 \begin {gather*} \frac {3 \, d^{2} x^{3} e^{\left (-1\right )}}{2 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} - \frac {11 \, d^{3} x^{2} e^{\left (-2\right )}}{3 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} - \frac {9 \, d^{4} x e^{\left (-3\right )}}{10 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {22 \, d^{5} e^{\left (-4\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^{3} - \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 e + \frac {3 \, d x^{4}}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {17 \, d^{2} x e^{\left (-3\right )}}{30 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}}} - \arcsin \left (\frac {x e}{d}\right ) e^{\left (-4\right )} + \frac {2 \, x e^{\left (-3\right )}}{15 \, \sqrt {-x^{2} e^{2} + d^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

3/2*d^2*x^3*e^(-1)/(-x^2*e^2 + d^2)^(5/2) - 11/3*d^3*x^2*e^(-2)/(-x^2*e^2 + d^2)^(5/2) - 9/10*d^4*x*e^(-3)/(-x
^2*e^2 + d^2)^(5/2) + 22/15*d^5*e^(-4)/(-x^2*e^2 + d^2)^(5/2) + 1/15*(15*x^4*e^(-2)/(-x^2*e^2 + d^2)^(5/2) - 2
0*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^3 - 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*e + 3*d*x^4/(-x^2*e^2 + d^2)^(5/2) + 17/30*d^2*x*e^
(-3)/(-x^2*e^2 + d^2)^(3/2) - arcsin(x*e/d)*e^(-4) + 2/15*x*e^(-3)/sqrt(-x^2*e^2 + d^2)

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Fricas [A]
time = 2.98, size = 152, normalized size = 1.29 \begin {gather*} \frac {22 \, x^{3} e^{3} - 66 \, d x^{2} e^{2} + 66 \, d^{2} x e - 22 \, d^{3} + 30 \, {\left (x^{3} e^{3} - 3 \, d x^{2} e^{2} + 3 \, d^{2} x e - d^{3}\right )} \arctan \left (-\frac {{\left (d - \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-1\right )}}{x}\right ) - {\left (32 \, x^{2} e^{2} - 51 \, d x e + 22 \, d^{2}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{15 \, {\left (x^{3} e^{7} - 3 \, d x^{2} e^{6} + 3 \, d^{2} x e^{5} - d^{3} e^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

1/15*(22*x^3*e^3 - 66*d*x^2*e^2 + 66*d^2*x*e - 22*d^3 + 30*(x^3*e^3 - 3*d*x^2*e^2 + 3*d^2*x*e - d^3)*arctan(-(
d - sqrt(-x^2*e^2 + d^2))*e^(-1)/x) - (32*x^2*e^2 - 51*d*x*e + 22*d^2)*sqrt(-x^2*e^2 + d^2))/(x^3*e^7 - 3*d*x^
2*e^6 + 3*d^2*x*e^5 - d^3*e^4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3} \left (d + e x\right )^{3}}{\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**3*(e*x+d)**3/(-e**2*x**2+d**2)**(7/2),x)

[Out]

Integral(x**3*(d + e*x)**3/(-(-d + e*x)*(d + e*x))**(7/2), x)

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Giac [A]
time = 0.71, size = 170, normalized size = 1.44 \begin {gather*} -\arcsin \left (\frac {x e}{d}\right ) e^{\left (-4\right )} \mathrm {sgn}\left (d\right ) - \frac {2 \, {\left (\frac {95 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-2\right )}}{x} - \frac {145 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} e^{\left (-4\right )}}{x^{2}} + \frac {75 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} e^{\left (-6\right )}}{x^{3}} - \frac {15 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} e^{\left (-8\right )}}{x^{4}} - 22\right )} e^{\left (-4\right )}}{15 \, {\left (\frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-2\right )}}{x} - 1\right )}^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

-arcsin(x*e/d)*e^(-4)*sgn(d) - 2/15*(95*(d*e + sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/x - 145*(d*e + sqrt(-x^2*e^2 + d
^2)*e)^2*e^(-4)/x^2 + 75*(d*e + sqrt(-x^2*e^2 + d^2)*e)^3*e^(-6)/x^3 - 15*(d*e + sqrt(-x^2*e^2 + d^2)*e)^4*e^(
-8)/x^4 - 22)*e^(-4)/((d*e + sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/x - 1)^5

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3\,{\left (d+e\,x\right )}^3}{{\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^3*(d + e*x)^3)/(d^2 - e^2*x^2)^(7/2),x)

[Out]

int((x^3*(d + e*x)^3)/(d^2 - e^2*x^2)^(7/2), x)

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