3.1.21 \(\int \frac {x^5 (d+e x)}{(d^2-e^2 x^2)^{7/2}} \, dx\)

Optimal. Leaf size=122 \[ \frac {x^4 (d+e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {8 d+15 e x}{15 e^6 \sqrt {d^2-e^2 x^2}}-\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^6}-\frac {x^2 (4 d+5 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {819, 778, 217, 203} \begin {gather*} \frac {x^4 (d+e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x^2 (4 d+5 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {8 d+15 e x}{15 e^6 \sqrt {d^2-e^2 x^2}}-\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^5*(d + e*x))/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(x^4*(d + e*x))/(5*e^2*(d^2 - e^2*x^2)^(5/2)) - (x^2*(4*d + 5*e*x))/(15*e^4*(d^2 - e^2*x^2)^(3/2)) + (8*d + 15
*e*x)/(15*e^6*Sqrt[d^2 - e^2*x^2]) - ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]]/e^6

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 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rubi steps

\begin {align*} \int \frac {x^5 (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac {x^4 (d+e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {x^3 \left (4 d^3+5 d^2 e x\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2 e^2}\\ &=\frac {x^4 (d+e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x^2 (4 d+5 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {x \left (8 d^5+15 d^4 e x\right )}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^4 e^4}\\ &=\frac {x^4 (d+e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x^2 (4 d+5 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {8 d+15 e x}{15 e^6 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e^5}\\ &=\frac {x^4 (d+e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x^2 (4 d+5 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {8 d+15 e x}{15 e^6 \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^5}\\ &=\frac {x^4 (d+e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x^2 (4 d+5 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {8 d+15 e x}{15 e^6 \sqrt {d^2-e^2 x^2}}-\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^6}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.08, size = 130, normalized size = 1.07 \begin {gather*} \frac {8 d^4+7 d^3 e x-27 d^2 e^2 x^2-15 (d-e x)^2 (d+e x) \sqrt {d^2-e^2 x^2} \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-8 d e^3 x^3+23 e^4 x^4}{15 e^6 (d-e x)^2 (d+e x) \sqrt {d^2-e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(d + e*x))/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(8*d^4 + 7*d^3*e*x - 27*d^2*e^2*x^2 - 8*d*e^3*x^3 + 23*e^4*x^4 - 15*(d - e*x)^2*(d + e*x)*Sqrt[d^2 - e^2*x^2]*
ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(15*e^6*(d - e*x)^2*(d + e*x)*Sqrt[d^2 - e^2*x^2])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.68, size = 125, normalized size = 1.02 \begin {gather*} -\frac {\sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{e^7}-\frac {\sqrt {d^2-e^2 x^2} \left (8 d^4+7 d^3 e x-27 d^2 e^2 x^2-8 d e^3 x^3+23 e^4 x^4\right )}{15 e^6 (e x-d)^3 (d+e x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^5*(d + e*x))/(d^2 - e^2*x^2)^(7/2),x]

[Out]

-1/15*(Sqrt[d^2 - e^2*x^2]*(8*d^4 + 7*d^3*e*x - 27*d^2*e^2*x^2 - 8*d*e^3*x^3 + 23*e^4*x^4))/(e^6*(-d + e*x)^3*
(d + e*x)^2) - (Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/e^7

________________________________________________________________________________________

fricas [B]  time = 0.41, size = 247, normalized size = 2.02 \begin {gather*} \frac {8 \, e^{5} x^{5} - 8 \, d e^{4} x^{4} - 16 \, d^{2} e^{3} x^{3} + 16 \, d^{3} e^{2} x^{2} + 8 \, d^{4} e x - 8 \, d^{5} + 30 \, {\left (e^{5} x^{5} - d e^{4} x^{4} - 2 \, d^{2} e^{3} x^{3} + 2 \, d^{3} e^{2} x^{2} + d^{4} e x - d^{5}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (23 \, e^{4} x^{4} - 8 \, d e^{3} x^{3} - 27 \, d^{2} e^{2} x^{2} + 7 \, d^{3} e x + 8 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (e^{11} x^{5} - d e^{10} x^{4} - 2 \, d^{2} e^{9} x^{3} + 2 \, d^{3} e^{8} x^{2} + d^{4} e^{7} x - d^{5} e^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(e*x+d)/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

1/15*(8*e^5*x^5 - 8*d*e^4*x^4 - 16*d^2*e^3*x^3 + 16*d^3*e^2*x^2 + 8*d^4*e*x - 8*d^5 + 30*(e^5*x^5 - d*e^4*x^4
- 2*d^2*e^3*x^3 + 2*d^3*e^2*x^2 + d^4*e*x - d^5)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (23*e^4*x^4 - 8*d
*e^3*x^3 - 27*d^2*e^2*x^2 + 7*d^3*e*x + 8*d^4)*sqrt(-e^2*x^2 + d^2))/(e^11*x^5 - d*e^10*x^4 - 2*d^2*e^9*x^3 +
2*d^3*e^8*x^2 + d^4*e^7*x - d^5*e^6)

________________________________________________________________________________________

giac [A]  time = 0.31, size = 97, normalized size = 0.80 \begin {gather*} -\arcsin \left (\frac {x e}{d}\right ) e^{\left (-6\right )} \mathrm {sgn}\relax (d) - \frac {{\left (8 \, d^{5} e^{\left (-6\right )} + {\left (15 \, d^{4} e^{\left (-5\right )} - {\left (20 \, d^{3} e^{\left (-4\right )} + {\left (35 \, d^{2} e^{\left (-3\right )} - {\left (23 \, x e^{\left (-1\right )} + 15 \, d e^{\left (-2\right )}\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^5*(e*x+d)/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

-arcsin(x*e/d)*e^(-6)*sgn(d) - 1/15*(8*d^5*e^(-6) + (15*d^4*e^(-5) - (20*d^3*e^(-4) + (35*d^2*e^(-3) - (23*x*e
^(-1) + 15*d*e^(-2))*x)*x)*x)*x)*sqrt(-x^2*e^2 + d^2)/(x^2*e^2 - d^2)^3

________________________________________________________________________________________

maple [A]  time = 0.02, size = 166, normalized size = 1.36 \begin {gather*} \frac {x^{5}}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e}+\frac {d \,x^{4}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{2}}-\frac {4 d^{3} x^{2}}{3 \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^{3}}+\frac {8 d^{5}}{15 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{6}}+\frac {x}{\sqrt {-e^{2} x^{2}+d^{2}}\, e^{5}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}\, e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(e*x+d)/(-e^2*x^2+d^2)^(7/2),x)

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 1.01, size = 250, normalized size = 2.05 \begin {gather*} \frac {1}{15} \, e 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 {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 )}}{3 \, e} + \frac {d x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {4 \, d^{3} x^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}} + \frac {8 \, d^{5}}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{6}} + \frac {4 \, d^{2} x}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{5}} - \frac {7 \, x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{5}} - \frac {\arcsin \left (\frac {e x}{d}\right )}{e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(e*x+d)/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

1/15*e*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))/e + d*x^4/((
-e^2*x^2 + d^2)^(5/2)*e^2) - 4/3*d^3*x^2/((-e^2*x^2 + d^2)^(5/2)*e^4) + 8/15*d^5/((-e^2*x^2 + d^2)^(5/2)*e^6)
+ 4/15*d^2*x/((-e^2*x^2 + d^2)^(3/2)*e^5) - 7/15*x/(sqrt(-e^2*x^2 + d^2)*e^5) - arcsin(e*x/d)/e^6

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^5\,\left (d+e\,x\right )}{{\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^5*(d + e*x))/(d^2 - e^2*x^2)^(7/2),x)

[Out]

int((x^5*(d + e*x))/(d^2 - e^2*x^2)^(7/2), x)

________________________________________________________________________________________

sympy [B]  time = 73.14, size = 1739, normalized size = 14.25

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(e*x+d)/(-e**2*x**2+d**2)**(7/2),x)

[Out]

d*Piecewise((8*d**4/(15*d**4*e**6*sqrt(d**2 - e**2*x**2) - 30*d**2*e**8*x**2*sqrt(d**2 - e**2*x**2) + 15*e**10
*x**4*sqrt(d**2 - e**2*x**2)) - 20*d**2*e**2*x**2/(15*d**4*e**6*sqrt(d**2 - e**2*x**2) - 30*d**2*e**8*x**2*sqr
t(d**2 - e**2*x**2) + 15*e**10*x**4*sqrt(d**2 - e**2*x**2)) + 15*e**4*x**4/(15*d**4*e**6*sqrt(d**2 - e**2*x**2
) - 30*d**2*e**8*x**2*sqrt(d**2 - e**2*x**2) + 15*e**10*x**4*sqrt(d**2 - e**2*x**2)), Ne(e, 0)), (x**6/(6*(d**
2)**(7/2)), True)) + e*Piecewise((-30*I*d**5*sqrt(-1 + e**2*x**2/d**2)*acosh(e*x/d)/(-30*d**5*e**7*sqrt(-1 + e
**2*x**2/d**2) + 60*d**3*e**9*x**2*sqrt(-1 + e**2*x**2/d**2) - 30*d*e**11*x**4*sqrt(-1 + e**2*x**2/d**2)) + 15
*pi*d**5*sqrt(-1 + e**2*x**2/d**2)/(-30*d**5*e**7*sqrt(-1 + e**2*x**2/d**2) + 60*d**3*e**9*x**2*sqrt(-1 + e**2
*x**2/d**2) - 30*d*e**11*x**4*sqrt(-1 + e**2*x**2/d**2)) + 30*I*d**4*e*x/(-30*d**5*e**7*sqrt(-1 + e**2*x**2/d*
*2) + 60*d**3*e**9*x**2*sqrt(-1 + e**2*x**2/d**2) - 30*d*e**11*x**4*sqrt(-1 + e**2*x**2/d**2)) + 60*I*d**3*e**
2*x**2*sqrt(-1 + e**2*x**2/d**2)*acosh(e*x/d)/(-30*d**5*e**7*sqrt(-1 + e**2*x**2/d**2) + 60*d**3*e**9*x**2*sqr
t(-1 + e**2*x**2/d**2) - 30*d*e**11*x**4*sqrt(-1 + e**2*x**2/d**2)) - 30*pi*d**3*e**2*x**2*sqrt(-1 + e**2*x**2
/d**2)/(-30*d**5*e**7*sqrt(-1 + e**2*x**2/d**2) + 60*d**3*e**9*x**2*sqrt(-1 + e**2*x**2/d**2) - 30*d*e**11*x**
4*sqrt(-1 + e**2*x**2/d**2)) - 70*I*d**2*e**3*x**3/(-30*d**5*e**7*sqrt(-1 + e**2*x**2/d**2) + 60*d**3*e**9*x**
2*sqrt(-1 + e**2*x**2/d**2) - 30*d*e**11*x**4*sqrt(-1 + e**2*x**2/d**2)) - 30*I*d*e**4*x**4*sqrt(-1 + e**2*x**
2/d**2)*acosh(e*x/d)/(-30*d**5*e**7*sqrt(-1 + e**2*x**2/d**2) + 60*d**3*e**9*x**2*sqrt(-1 + e**2*x**2/d**2) -
30*d*e**11*x**4*sqrt(-1 + e**2*x**2/d**2)) + 15*pi*d*e**4*x**4*sqrt(-1 + e**2*x**2/d**2)/(-30*d**5*e**7*sqrt(-
1 + e**2*x**2/d**2) + 60*d**3*e**9*x**2*sqrt(-1 + e**2*x**2/d**2) - 30*d*e**11*x**4*sqrt(-1 + e**2*x**2/d**2))
+ 46*I*e**5*x**5/(-30*d**5*e**7*sqrt(-1 + e**2*x**2/d**2) + 60*d**3*e**9*x**2*sqrt(-1 + e**2*x**2/d**2) - 30*
d*e**11*x**4*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2/d**2) > 1), (-15*d**5*sqrt(1 - e**2*x**2/d**2)*asin(e*x
/d)/(15*d**5*e**7*sqrt(1 - e**2*x**2/d**2) - 30*d**3*e**9*x**2*sqrt(1 - e**2*x**2/d**2) + 15*d*e**11*x**4*sqrt
(1 - e**2*x**2/d**2)) + 15*d**4*e*x/(15*d**5*e**7*sqrt(1 - e**2*x**2/d**2) - 30*d**3*e**9*x**2*sqrt(1 - e**2*x
**2/d**2) + 15*d*e**11*x**4*sqrt(1 - e**2*x**2/d**2)) + 30*d**3*e**2*x**2*sqrt(1 - e**2*x**2/d**2)*asin(e*x/d)
/(15*d**5*e**7*sqrt(1 - e**2*x**2/d**2) - 30*d**3*e**9*x**2*sqrt(1 - e**2*x**2/d**2) + 15*d*e**11*x**4*sqrt(1
- e**2*x**2/d**2)) - 35*d**2*e**3*x**3/(15*d**5*e**7*sqrt(1 - e**2*x**2/d**2) - 30*d**3*e**9*x**2*sqrt(1 - e**
2*x**2/d**2) + 15*d*e**11*x**4*sqrt(1 - e**2*x**2/d**2)) - 15*d*e**4*x**4*sqrt(1 - e**2*x**2/d**2)*asin(e*x/d)
/(15*d**5*e**7*sqrt(1 - e**2*x**2/d**2) - 30*d**3*e**9*x**2*sqrt(1 - e**2*x**2/d**2) + 15*d*e**11*x**4*sqrt(1
- e**2*x**2/d**2)) + 23*e**5*x**5/(15*d**5*e**7*sqrt(1 - e**2*x**2/d**2) - 30*d**3*e**9*x**2*sqrt(1 - e**2*x**
2/d**2) + 15*d*e**11*x**4*sqrt(1 - e**2*x**2/d**2)), True))

________________________________________________________________________________________