3.84 \(\int \frac {x^4 (d+e x)^3}{(d^2-e^2 x^2)^{7/2}} \, dx\)

Optimal. Leaf size=142 \[ -\frac {6 d^2 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {24 d (d+e x)}{5 e^5 \sqrt {d^2-e^2 x^2}}+\frac {\sqrt {d^2-e^2 x^2}}{e^5}-\frac {3 d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5}+\frac {d^3 (d+e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}} \]

[Out]

1/5*d^3*(e*x+d)^3/e^5/(-e^2*x^2+d^2)^(5/2)-6/5*d^2*(e*x+d)^2/e^5/(-e^2*x^2+d^2)^(3/2)-3*d*arctan(e*x/(-e^2*x^2
+d^2)^(1/2))/e^5+24/5*d*(e*x+d)/e^5/(-e^2*x^2+d^2)^(1/2)+(-e^2*x^2+d^2)^(1/2)/e^5

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Rubi [A]  time = 0.32, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {1635, 641, 217, 203} \[ \frac {d^3 (d+e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {6 d^2 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {24 d (d+e x)}{5 e^5 \sqrt {d^2-e^2 x^2}}+\frac {\sqrt {d^2-e^2 x^2}}{e^5}-\frac {3 d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(d + e*x)^3)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(d^3*(d + e*x)^3)/(5*e^5*(d^2 - e^2*x^2)^(5/2)) - (6*d^2*(d + e*x)^2)/(5*e^5*(d^2 - e^2*x^2)^(3/2)) + (24*d*(d
+ e*x))/(5*e^5*Sqrt[d^2 - e^2*x^2]) + Sqrt[d^2 - e^2*x^2]/e^5 - (3*d*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^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 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[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^4 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac {d^3 (d+e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {(d+e x)^2 \left (\frac {3 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)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {6 d^2 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {(d+e x) \left (\frac {27 d^4}{e^4}+\frac {30 d^3 x}{e^3}+\frac {15 d^2 x^2}{e^2}\right )}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^2}\\ &=\frac {d^3 (d+e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {6 d^2 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {24 d (d+e x)}{5 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {\frac {45 d^4}{e^4}+\frac {15 d^3 x}{e^3}}{\sqrt {d^2-e^2 x^2}} \, dx}{15 d^3}\\ &=\frac {d^3 (d+e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {6 d^2 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {24 d (d+e x)}{5 e^5 \sqrt {d^2-e^2 x^2}}+\frac {\sqrt {d^2-e^2 x^2}}{e^5}-\frac {(3 d) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e^4}\\ &=\frac {d^3 (d+e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {6 d^2 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {24 d (d+e x)}{5 e^5 \sqrt {d^2-e^2 x^2}}+\frac {\sqrt {d^2-e^2 x^2}}{e^5}-\frac {(3 d) \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)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {6 d^2 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {24 d (d+e x)}{5 e^5 \sqrt {d^2-e^2 x^2}}+\frac {\sqrt {d^2-e^2 x^2}}{e^5}-\frac {3 d \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 = 119, normalized size = 0.84 \[ \frac {(d+e x) \left (\sqrt {1-\frac {e^2 x^2}{d^2}} \left (24 d^3-57 d^2 e x+39 d e^2 x^2-5 e^3 x^3\right )-15 (d-e x)^3 \sin ^{-1}\left (\frac {e x}{d}\right )\right )}{5 e^5 (d-e x)^2 \sqrt {d^2-e^2 x^2} \sqrt {1-\frac {e^2 x^2}{d^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(d + e*x)^3)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

((d + e*x)*(Sqrt[1 - (e^2*x^2)/d^2]*(24*d^3 - 57*d^2*e*x + 39*d*e^2*x^2 - 5*e^3*x^3) - 15*(d - e*x)^3*ArcSin[(
e*x)/d]))/(5*e^5*(d - e*x)^2*Sqrt[d^2 - e^2*x^2]*Sqrt[1 - (e^2*x^2)/d^2])

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fricas [A]  time = 0.89, size = 177, normalized size = 1.25 \[ \frac {24 \, d e^{3} x^{3} - 72 \, d^{2} e^{2} x^{2} + 72 \, d^{3} e x - 24 \, d^{4} + 30 \, {\left (d e^{3} x^{3} - 3 \, d^{2} e^{2} x^{2} + 3 \, d^{3} e x - d^{4}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (5 \, e^{3} x^{3} - 39 \, d e^{2} x^{2} + 57 \, d^{2} e x - 24 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{5 \, {\left (e^{8} x^{3} - 3 \, d e^{7} x^{2} + 3 \, d^{2} e^{6} x - d^{3} e^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

1/5*(24*d*e^3*x^3 - 72*d^2*e^2*x^2 + 72*d^3*e*x - 24*d^4 + 30*(d*e^3*x^3 - 3*d^2*e^2*x^2 + 3*d^3*e*x - d^4)*ar
ctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (5*e^3*x^3 - 39*d*e^2*x^2 + 57*d^2*e*x - 24*d^3)*sqrt(-e^2*x^2 + d^2
))/(e^8*x^3 - 3*d*e^7*x^2 + 3*d^2*e^6*x - d^3*e^5)

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giac [A]  time = 0.29, size = 107, normalized size = 0.75 \[ -3 \, d \arcsin \left (\frac {x e}{d}\right ) e^{\left (-5\right )} \mathrm {sgn}\relax (d) - \frac {{\left (24 \, d^{6} e^{\left (-5\right )} + {\left (15 \, d^{5} e^{\left (-4\right )} - {\left (60 \, d^{4} e^{\left (-3\right )} + {\left (35 \, d^{3} e^{\left (-2\right )} - {\left (45 \, d^{2} e^{\left (-1\right )} - {\left (5 \, x e - 24 \, d\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{5 \, {\left (x^{2} e^{2} - d^{2}\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

-3*d*arcsin(x*e/d)*e^(-5)*sgn(d) - 1/5*(24*d^6*e^(-5) + (15*d^5*e^(-4) - (60*d^4*e^(-3) + (35*d^3*e^(-2) - (45
*d^2*e^(-1) - (5*x*e - 24*d)*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 = 0.01, size = 262, normalized size = 1.85 \[ -\frac {e \,x^{6}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {3 d \,x^{5}}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {9 d^{2} x^{4}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e}+\frac {d^{3} x^{3}}{2 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{2}}-\frac {12 d^{4} x^{2}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{3}}-\frac {3 d^{5} x}{10 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{4}}-\frac {d \,x^{3}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{2}}+\frac {24 d^{6}}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{5}}+\frac {d^{3} x}{10 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{4}}+\frac {16 d x}{5 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{4}}-\frac {3 d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}\, e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x)

[Out]

-e*x^6/(-e^2*x^2+d^2)^(5/2)+9/e*d^2*x^4/(-e^2*x^2+d^2)^(5/2)-12/e^3*d^4*x^2/(-e^2*x^2+d^2)^(5/2)+24/5/e^5*d^6/
(-e^2*x^2+d^2)^(5/2)+3/5*d*x^5/(-e^2*x^2+d^2)^(5/2)-d/e^2*x^3/(-e^2*x^2+d^2)^(3/2)+16/5*d/e^4*x/(-e^2*x^2+d^2)
^(1/2)-3*d/e^4/(e^2)^(1/2)*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)+1/2*d^3*x^3/e^2/(-e^2*x^2+d^2)^(5/2)-3/1
0*d^5/e^4*x/(-e^2*x^2+d^2)^(5/2)+1/10*d^3/e^4*x/(-e^2*x^2+d^2)^(3/2)

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maxima [B]  time = 1.02, size = 324, normalized size = 2.28 \[ \frac {1}{5} \, d e^{2} x {\left (\frac {15 \, x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {20 \, d^{2} x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}} + \frac {8 \, d^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{6}}\right )} - \frac {e x^{6}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} - d x {\left (\frac {3 \, x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}} - \frac {2 \, d^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}}\right )} + \frac {9 \, d^{2} x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e} + \frac {d^{3} x^{3}}{2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {12 \, d^{4} x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{3}} - \frac {3 \, d^{5} x}{10 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}} + \frac {24 \, d^{6}}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{5}} + \frac {9 \, d^{3} x}{10 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}} - \frac {6 \, d x}{5 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{4}} - \frac {3 \, d \arcsin \left (\frac {e x}{d}\right )}{e^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

1/5*d*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)) - e*x^6/(-e^2*x^2 + d^2)^(5/2) - d*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)) + 9*d^2*x^4/((-e^2*x^2 + d^2)^(5/2)*e) + 1/2*d^3*x^3/((-e^2*x^2 + d^2)^(5/2)*e^2) - 12*d^4
*x^2/((-e^2*x^2 + d^2)^(5/2)*e^3) - 3/10*d^5*x/((-e^2*x^2 + d^2)^(5/2)*e^4) + 24/5*d^6/((-e^2*x^2 + d^2)^(5/2)
*e^5) + 9/10*d^3*x/((-e^2*x^2 + d^2)^(3/2)*e^4) - 6/5*d*x/(sqrt(-e^2*x^2 + d^2)*e^4) - 3*d*arcsin(e*x/d)/e^5

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^4\,{\left (d+e\,x\right )}^3}{{\left (d^2-e^2\,x^2\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(d + e*x)^3)/(d^2 - e^2*x^2)^(7/2),x)

[Out]

int((x^4*(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 \[ \int \frac {x^{4} \left (d + e x\right )^{3}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(e*x+d)**3/(-e**2*x**2+d**2)**(7/2),x)

[Out]

Integral(x**4*(d + e*x)**3/(-(-d + e*x)*(d + e*x))**(7/2), x)

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