3.146 \(\int \frac {1}{x^3 (d+e x) (d^2-e^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=186 \[ \frac {1}{5 d^2 x^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {16 e \sqrt {d^2-e^2 x^2}}{5 d^8 x}-\frac {7 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^8}-\frac {7 \sqrt {d^2-e^2 x^2}}{2 d^7 x^2}+\frac {35 d-24 e x}{15 d^6 x^2 \sqrt {d^2-e^2 x^2}}+\frac {7 d-6 e x}{15 d^4 x^2 \left (d^2-e^2 x^2\right )^{3/2}} \]

[Out]

1/15*(-6*e*x+7*d)/d^4/x^2/(-e^2*x^2+d^2)^(3/2)+1/5/d^2/x^2/(e*x+d)/(-e^2*x^2+d^2)^(3/2)-7/2*e^2*arctanh((-e^2*
x^2+d^2)^(1/2)/d)/d^8+1/15*(-24*e*x+35*d)/d^6/x^2/(-e^2*x^2+d^2)^(1/2)-7/2*(-e^2*x^2+d^2)^(1/2)/d^7/x^2+16/5*e
*(-e^2*x^2+d^2)^(1/2)/d^8/x

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Rubi [A]  time = 0.17, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {857, 823, 835, 807, 266, 63, 208} \[ \frac {16 e \sqrt {d^2-e^2 x^2}}{5 d^8 x}-\frac {7 \sqrt {d^2-e^2 x^2}}{2 d^7 x^2}+\frac {35 d-24 e x}{15 d^6 x^2 \sqrt {d^2-e^2 x^2}}+\frac {7 d-6 e x}{15 d^4 x^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 x^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}-\frac {7 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^8} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(d + e*x)*(d^2 - e^2*x^2)^(5/2)),x]

[Out]

(7*d - 6*e*x)/(15*d^4*x^2*(d^2 - e^2*x^2)^(3/2)) + 1/(5*d^2*x^2*(d + e*x)*(d^2 - e^2*x^2)^(3/2)) + (35*d - 24*
e*x)/(15*d^6*x^2*Sqrt[d^2 - e^2*x^2]) - (7*Sqrt[d^2 - e^2*x^2])/(2*d^7*x^2) + (16*e*Sqrt[d^2 - e^2*x^2])/(5*d^
8*x) - (7*e^2*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(2*d^8)

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 807

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

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 835

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

Rule 857

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

Rubi steps

\begin {align*} \int \frac {1}{x^3 (d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx &=\frac {1}{5 d^2 x^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}-\frac {\int \frac {-7 d e^2+6 e^3 x}{x^3 \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2 e^2}\\ &=\frac {7 d-6 e x}{15 d^4 x^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 x^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}-\frac {\int \frac {-35 d^3 e^4+24 d^2 e^5 x}{x^3 \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^6 e^4}\\ &=\frac {7 d-6 e x}{15 d^4 x^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 x^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {35 d-24 e x}{15 d^6 x^2 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {-105 d^5 e^6+48 d^4 e^7 x}{x^3 \sqrt {d^2-e^2 x^2}} \, dx}{15 d^{10} e^6}\\ &=\frac {7 d-6 e x}{15 d^4 x^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 x^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {35 d-24 e x}{15 d^6 x^2 \sqrt {d^2-e^2 x^2}}-\frac {7 \sqrt {d^2-e^2 x^2}}{2 d^7 x^2}+\frac {\int \frac {-96 d^6 e^7+105 d^5 e^8 x}{x^2 \sqrt {d^2-e^2 x^2}} \, dx}{30 d^{12} e^6}\\ &=\frac {7 d-6 e x}{15 d^4 x^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 x^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {35 d-24 e x}{15 d^6 x^2 \sqrt {d^2-e^2 x^2}}-\frac {7 \sqrt {d^2-e^2 x^2}}{2 d^7 x^2}+\frac {16 e \sqrt {d^2-e^2 x^2}}{5 d^8 x}+\frac {\left (7 e^2\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{2 d^7}\\ &=\frac {7 d-6 e x}{15 d^4 x^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 x^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {35 d-24 e x}{15 d^6 x^2 \sqrt {d^2-e^2 x^2}}-\frac {7 \sqrt {d^2-e^2 x^2}}{2 d^7 x^2}+\frac {16 e \sqrt {d^2-e^2 x^2}}{5 d^8 x}+\frac {\left (7 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{4 d^7}\\ &=\frac {7 d-6 e x}{15 d^4 x^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 x^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {35 d-24 e x}{15 d^6 x^2 \sqrt {d^2-e^2 x^2}}-\frac {7 \sqrt {d^2-e^2 x^2}}{2 d^7 x^2}+\frac {16 e \sqrt {d^2-e^2 x^2}}{5 d^8 x}-\frac {7 \operatorname {Subst}\left (\int \frac {1}{\frac {d^2}{e^2}-\frac {x^2}{e^2}} \, dx,x,\sqrt {d^2-e^2 x^2}\right )}{2 d^7}\\ &=\frac {7 d-6 e x}{15 d^4 x^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 x^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {35 d-24 e x}{15 d^6 x^2 \sqrt {d^2-e^2 x^2}}-\frac {7 \sqrt {d^2-e^2 x^2}}{2 d^7 x^2}+\frac {16 e \sqrt {d^2-e^2 x^2}}{5 d^8 x}-\frac {7 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^8}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 137, normalized size = 0.74 \[ \frac {-105 e^2 \log \left (\sqrt {d^2-e^2 x^2}+d\right )+\frac {\sqrt {d^2-e^2 x^2} \left (-15 d^6+15 d^5 e x+176 d^4 e^2 x^2-4 d^3 e^3 x^3-249 d^2 e^4 x^4-9 d e^5 x^5+96 e^6 x^6\right )}{x^2 (d-e x)^2 (d+e x)^3}+105 e^2 \log (x)}{30 d^8} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(d + e*x)*(d^2 - e^2*x^2)^(5/2)),x]

[Out]

((Sqrt[d^2 - e^2*x^2]*(-15*d^6 + 15*d^5*e*x + 176*d^4*e^2*x^2 - 4*d^3*e^3*x^3 - 249*d^2*e^4*x^4 - 9*d*e^5*x^5
+ 96*e^6*x^6))/(x^2*(d - e*x)^2*(d + e*x)^3) + 105*e^2*Log[x] - 105*e^2*Log[d + Sqrt[d^2 - e^2*x^2]])/(30*d^8)

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fricas [A]  time = 1.06, size = 286, normalized size = 1.54 \[ \frac {116 \, e^{7} x^{7} + 116 \, d e^{6} x^{6} - 232 \, d^{2} e^{5} x^{5} - 232 \, d^{3} e^{4} x^{4} + 116 \, d^{4} e^{3} x^{3} + 116 \, d^{5} e^{2} x^{2} + 105 \, {\left (e^{7} x^{7} + d e^{6} x^{6} - 2 \, d^{2} e^{5} x^{5} - 2 \, d^{3} e^{4} x^{4} + d^{4} e^{3} x^{3} + d^{5} e^{2} x^{2}\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + {\left (96 \, e^{6} x^{6} - 9 \, d e^{5} x^{5} - 249 \, d^{2} e^{4} x^{4} - 4 \, d^{3} e^{3} x^{3} + 176 \, d^{4} e^{2} x^{2} + 15 \, d^{5} e x - 15 \, d^{6}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{30 \, {\left (d^{8} e^{5} x^{7} + d^{9} e^{4} x^{6} - 2 \, d^{10} e^{3} x^{5} - 2 \, d^{11} e^{2} x^{4} + d^{12} e x^{3} + d^{13} x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x, algorithm="fricas")

[Out]

1/30*(116*e^7*x^7 + 116*d*e^6*x^6 - 232*d^2*e^5*x^5 - 232*d^3*e^4*x^4 + 116*d^4*e^3*x^3 + 116*d^5*e^2*x^2 + 10
5*(e^7*x^7 + d*e^6*x^6 - 2*d^2*e^5*x^5 - 2*d^3*e^4*x^4 + d^4*e^3*x^3 + d^5*e^2*x^2)*log(-(d - sqrt(-e^2*x^2 +
d^2))/x) + (96*e^6*x^6 - 9*d*e^5*x^5 - 249*d^2*e^4*x^4 - 4*d^3*e^3*x^3 + 176*d^4*e^2*x^2 + 15*d^5*e*x - 15*d^6
)*sqrt(-e^2*x^2 + d^2))/(d^8*e^5*x^7 + d^9*e^4*x^6 - 2*d^10*e^3*x^5 - 2*d^11*e^2*x^4 + d^12*e*x^3 + d^13*x^2)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to transpose Error: Bad Argument Value

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maple [A]  time = 0.02, size = 298, normalized size = 1.60 \[ -\frac {4 e^{3} x}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{6}}-\frac {4 e^{3} x}{15 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} d^{6}}+\frac {e}{5 \left (x +\frac {d}{e}\right ) \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} d^{4}}+\frac {7 e^{2}}{6 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{5}}+\frac {e}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{4} x}-\frac {7 e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 \sqrt {d^{2}}\, d^{7}}-\frac {8 e^{3} x}{3 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{8}}-\frac {8 e^{3} x}{15 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{8}}-\frac {1}{2 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{3} x^{2}}+\frac {7 e^{2}}{2 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} {\left (e x + d\right )} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x, algorithm="maxima")

[Out]

integrate(1/((-e^2*x^2 + d^2)^(5/2)*(e*x + d)*x^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^3\,{\left (d^2-e^2\,x^2\right )}^{5/2}\,\left (d+e\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(d^2 - e^2*x^2)^(5/2)*(d + e*x)),x)

[Out]

int(1/(x^3*(d^2 - e^2*x^2)^(5/2)*(d + e*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{3} \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}} \left (d + e x\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(e*x+d)/(-e**2*x**2+d**2)**(5/2),x)

[Out]

Integral(1/(x**3*(-(-d + e*x)*(d + e*x))**(5/2)*(d + e*x)), x)

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