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

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

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Rubi [A]  time = 0.22, 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 = {1635, 778, 217, 203} \begin {gather*} \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 {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 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 778

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 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]

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 {\operatorname {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 = 0.15, size = 112, normalized size = 0.95 \begin {gather*} \frac {(d+e x) \left (d \left (22 d^2-51 d e x+32 e^2 x^2\right ) \sqrt {1-\frac {e^2 x^2}{d^2}}-15 (d-e x)^3 \sin ^{-1}\left (\frac {e x}{d}\right )\right )}{15 d e^4 (d-e x)^2 \sqrt {d^2-e^2 x^2} \sqrt {1-\frac {e^2 x^2}{d^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]

((d + e*x)*(d*(22*d^2 - 51*d*e*x + 32*e^2*x^2)*Sqrt[1 - (e^2*x^2)/d^2] - 15*(d - e*x)^3*ArcSin[(e*x)/d]))/(15*
d*e^4*(d - e*x)^2*Sqrt[d^2 - e^2*x^2]*Sqrt[1 - (e^2*x^2)/d^2])

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IntegrateAlgebraic [A]  time = 0.62, size = 96, normalized size = 0.81 \begin {gather*} -\frac {\sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{e^5}-\frac {\sqrt {d^2-e^2 x^2} \left (22 d^2-51 d e x+32 e^2 x^2\right )}{15 e^4 (e x-d)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15*(Sqrt[d^2 - e^2*x^2]*(22*d^2 - 51*d*e*x + 32*e^2*x^2))/(e^4*(-d + e*x)^3) - (Sqrt[-e^2]*Log[-(Sqrt[-e^2]
*x) + Sqrt[d^2 - e^2*x^2]])/e^5

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fricas [A]  time = 0.40, size = 161, normalized size = 1.36 \begin {gather*} \frac {22 \, e^{3} x^{3} - 66 \, d e^{2} x^{2} + 66 \, d^{2} e x - 22 \, d^{3} + 30 \, {\left (e^{3} x^{3} - 3 \, d e^{2} x^{2} + 3 \, d^{2} e x - d^{3}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (32 \, e^{2} x^{2} - 51 \, d e x + 22 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (e^{7} x^{3} - 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x - 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*e^3*x^3 - 66*d*e^2*x^2 + 66*d^2*e*x - 22*d^3 + 30*(e^3*x^3 - 3*d*e^2*x^2 + 3*d^2*e*x - d^3)*arctan(-(
d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (32*e^2*x^2 - 51*d*e*x + 22*d^2)*sqrt(-e^2*x^2 + d^2))/(e^7*x^3 - 3*d*e^6*x
^2 + 3*d^2*e^5*x - d^3*e^4)

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giac [A]  time = 0.29, size = 95, normalized size = 0.81 \begin {gather*} -\arcsin \left (\frac {x e}{d}\right ) e^{\left (-4\right )} \mathrm {sgn}\relax (d) - \frac {{\left (22 \, d^{5} e^{\left (-4\right )} + {\left (15 \, d^{4} e^{\left (-3\right )} - {\left (55 \, d^{3} e^{\left (-2\right )} + {\left (35 \, d^{2} e^{\left (-1\right )} - {\left (32 \, x e + 45 \, d\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^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) - 1/15*(22*d^5*e^(-4) + (15*d^4*e^(-3) - (55*d^3*e^(-2) + (35*d^2*e^(-1) - (32*x*
e + 45*d)*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 = 0.01, size = 234, normalized size = 1.98 \begin {gather*} \frac {e \,x^{5}}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {3 d \,x^{4}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {3 d^{2} x^{3}}{2 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e}-\frac {11 d^{3} x^{2}}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{2}}-\frac {9 d^{4} x}{10 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{3}}-\frac {x^{3}}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e}+\frac {22 d^{5}}{15 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{4}}+\frac {3 d^{2} x}{10 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{3}}+\frac {8 x}{5 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{3}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}\, e^{3}} \end {gather*}

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)

[Out]

1/5*e*x^5/(-e^2*x^2+d^2)^(5/2)-1/3/e*x^3/(-e^2*x^2+d^2)^(3/2)+8/5/e^3*x/(-e^2*x^2+d^2)^(1/2)-1/e^3/(e^2)^(1/2)
*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)+3*d*x^4/(-e^2*x^2+d^2)^(5/2)-11/3/e^2*d^3*x^2/(-e^2*x^2+d^2)^(5/2)
+22/15/e^4*d^5/(-e^2*x^2+d^2)^(5/2)+3/2/e*d^2*x^3/(-e^2*x^2+d^2)^(5/2)-9/10/e^3*d^4*x/(-e^2*x^2+d^2)^(5/2)+3/1
0/e^3*d^2*x/(-e^2*x^2+d^2)^(3/2)

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maxima [B]  time = 1.02, size = 296, normalized size = 2.51 \begin {gather*} \frac {1}{15} \, e^{3} 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} \, e 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 {3 \, d x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {3 \, d^{2} x^{3}}{2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e} - \frac {11 \, d^{3} x^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {9 \, d^{4} x}{10 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{3}} + \frac {22 \, d^{5}}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}} + \frac {17 \, d^{2} x}{30 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{3}} + \frac {2 \, x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{3}} - \frac {\arcsin \left (\frac {e x}{d}\right )}{e^{4}} \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]

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

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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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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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