\(\int \frac {1}{x^3 (d+e x)^2 \sqrt {a+c x^2}} \, dx\) [350]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 22, antiderivative size = 268 \[ \int \frac {1}{x^3 (d+e x)^2 \sqrt {a+c x^2}} \, dx=-\frac {\sqrt {a+c x^2}}{2 a d^2 x^2}+\frac {2 e \sqrt {a+c x^2}}{a d^3 x}+\frac {e^4 \sqrt {a+c x^2}}{d^3 \left (c d^2+a e^2\right ) (d+e x)}+\frac {c e^3 \text {arctanh}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{d^2 \left (c d^2+a e^2\right )^{3/2}}+\frac {3 e^3 \text {arctanh}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{d^4 \sqrt {c d^2+a e^2}}+\frac {c \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 a^{3/2} d^2}-\frac {3 e^2 \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a} d^4} \]

[Out]

c*e^3*arctanh((-c*d*x+a*e)/(a*e^2+c*d^2)^(1/2)/(c*x^2+a)^(1/2))/d^2/(a*e^2+c*d^2)^(3/2)+1/2*c*arctanh((c*x^2+a
)^(1/2)/a^(1/2))/a^(3/2)/d^2-3*e^2*arctanh((c*x^2+a)^(1/2)/a^(1/2))/d^4/a^(1/2)+3*e^3*arctanh((-c*d*x+a*e)/(a*
e^2+c*d^2)^(1/2)/(c*x^2+a)^(1/2))/d^4/(a*e^2+c*d^2)^(1/2)-1/2*(c*x^2+a)^(1/2)/a/d^2/x^2+2*e*(c*x^2+a)^(1/2)/a/
d^3/x+e^4*(c*x^2+a)^(1/2)/d^3/(a*e^2+c*d^2)/(e*x+d)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {975, 272, 44, 65, 214, 270, 745, 739, 212} \[ \int \frac {1}{x^3 (d+e x)^2 \sqrt {a+c x^2}} \, dx=\frac {c \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 a^{3/2} d^2}-\frac {3 e^2 \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a} d^4}+\frac {c e^3 \text {arctanh}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{d^2 \left (a e^2+c d^2\right )^{3/2}}+\frac {3 e^3 \text {arctanh}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{d^4 \sqrt {a e^2+c d^2}}+\frac {2 e \sqrt {a+c x^2}}{a d^3 x}-\frac {\sqrt {a+c x^2}}{2 a d^2 x^2}+\frac {e^4 \sqrt {a+c x^2}}{d^3 (d+e x) \left (a e^2+c d^2\right )} \]

[In]

Int[1/(x^3*(d + e*x)^2*Sqrt[a + c*x^2]),x]

[Out]

-1/2*Sqrt[a + c*x^2]/(a*d^2*x^2) + (2*e*Sqrt[a + c*x^2])/(a*d^3*x) + (e^4*Sqrt[a + c*x^2])/(d^3*(c*d^2 + a*e^2
)*(d + e*x)) + (c*e^3*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/(d^2*(c*d^2 + a*e^2)^(3/2)
) + (3*e^3*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/(d^4*Sqrt[c*d^2 + a*e^2]) + (c*ArcTan
h[Sqrt[a + c*x^2]/Sqrt[a]])/(2*a^(3/2)*d^2) - (3*e^2*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/(Sqrt[a]*d^4)

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
 /; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 214

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

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*
c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 272

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 739

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
 (a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 745

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

Rule 975

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

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{d^2 x^3 \sqrt {a+c x^2}}-\frac {2 e}{d^3 x^2 \sqrt {a+c x^2}}+\frac {3 e^2}{d^4 x \sqrt {a+c x^2}}-\frac {e^3}{d^3 (d+e x)^2 \sqrt {a+c x^2}}-\frac {3 e^3}{d^4 (d+e x) \sqrt {a+c x^2}}\right ) \, dx \\ & = \frac {\int \frac {1}{x^3 \sqrt {a+c x^2}} \, dx}{d^2}-\frac {(2 e) \int \frac {1}{x^2 \sqrt {a+c x^2}} \, dx}{d^3}+\frac {\left (3 e^2\right ) \int \frac {1}{x \sqrt {a+c x^2}} \, dx}{d^4}-\frac {\left (3 e^3\right ) \int \frac {1}{(d+e x) \sqrt {a+c x^2}} \, dx}{d^4}-\frac {e^3 \int \frac {1}{(d+e x)^2 \sqrt {a+c x^2}} \, dx}{d^3} \\ & = \frac {2 e \sqrt {a+c x^2}}{a d^3 x}+\frac {e^4 \sqrt {a+c x^2}}{d^3 \left (c d^2+a e^2\right ) (d+e x)}+\frac {\text {Subst}\left (\int \frac {1}{x^2 \sqrt {a+c x}} \, dx,x,x^2\right )}{2 d^2}+\frac {\left (3 e^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,x^2\right )}{2 d^4}+\frac {\left (3 e^3\right ) \text {Subst}\left (\int \frac {1}{c d^2+a e^2-x^2} \, dx,x,\frac {a e-c d x}{\sqrt {a+c x^2}}\right )}{d^4}-\frac {\left (c e^3\right ) \int \frac {1}{(d+e x) \sqrt {a+c x^2}} \, dx}{d^2 \left (c d^2+a e^2\right )} \\ & = -\frac {\sqrt {a+c x^2}}{2 a d^2 x^2}+\frac {2 e \sqrt {a+c x^2}}{a d^3 x}+\frac {e^4 \sqrt {a+c x^2}}{d^3 \left (c d^2+a e^2\right ) (d+e x)}+\frac {3 e^3 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{d^4 \sqrt {c d^2+a e^2}}-\frac {c \text {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,x^2\right )}{4 a d^2}+\frac {\left (3 e^2\right ) \text {Subst}\left (\int \frac {1}{-\frac {a}{c}+\frac {x^2}{c}} \, dx,x,\sqrt {a+c x^2}\right )}{c d^4}+\frac {\left (c e^3\right ) \text {Subst}\left (\int \frac {1}{c d^2+a e^2-x^2} \, dx,x,\frac {a e-c d x}{\sqrt {a+c x^2}}\right )}{d^2 \left (c d^2+a e^2\right )} \\ & = -\frac {\sqrt {a+c x^2}}{2 a d^2 x^2}+\frac {2 e \sqrt {a+c x^2}}{a d^3 x}+\frac {e^4 \sqrt {a+c x^2}}{d^3 \left (c d^2+a e^2\right ) (d+e x)}+\frac {c e^3 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{d^2 \left (c d^2+a e^2\right )^{3/2}}+\frac {3 e^3 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{d^4 \sqrt {c d^2+a e^2}}-\frac {3 e^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a} d^4}-\frac {\text {Subst}\left (\int \frac {1}{-\frac {a}{c}+\frac {x^2}{c}} \, dx,x,\sqrt {a+c x^2}\right )}{2 a d^2} \\ & = -\frac {\sqrt {a+c x^2}}{2 a d^2 x^2}+\frac {2 e \sqrt {a+c x^2}}{a d^3 x}+\frac {e^4 \sqrt {a+c x^2}}{d^3 \left (c d^2+a e^2\right ) (d+e x)}+\frac {c e^3 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{d^2 \left (c d^2+a e^2\right )^{3/2}}+\frac {3 e^3 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{d^4 \sqrt {c d^2+a e^2}}+\frac {c \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 a^{3/2} d^2}-\frac {3 e^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a} d^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.13 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x^3 (d+e x)^2 \sqrt {a+c x^2}} \, dx=\frac {\frac {d \sqrt {a+c x^2} \left (c d^2 \left (-d^2+3 d e x+4 e^2 x^2\right )+a e^2 \left (-d^2+3 d e x+6 e^2 x^2\right )\right )}{a \left (c d^2+a e^2\right ) x^2 (d+e x)}-\frac {4 e^3 \left (4 c d^2+3 a e^2\right ) \arctan \left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+c x^2}}{\sqrt {-c d^2-a e^2}}\right )}{\left (-c d^2-a e^2\right )^{3/2}}+\frac {2 \left (-c d^2+6 a e^2\right ) \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{3/2}}}{2 d^4} \]

[In]

Integrate[1/(x^3*(d + e*x)^2*Sqrt[a + c*x^2]),x]

[Out]

((d*Sqrt[a + c*x^2]*(c*d^2*(-d^2 + 3*d*e*x + 4*e^2*x^2) + a*e^2*(-d^2 + 3*d*e*x + 6*e^2*x^2)))/(a*(c*d^2 + a*e
^2)*x^2*(d + e*x)) - (4*e^3*(4*c*d^2 + 3*a*e^2)*ArcTan[(Sqrt[c]*(d + e*x) - e*Sqrt[a + c*x^2])/Sqrt[-(c*d^2) -
 a*e^2]])/(-(c*d^2) - a*e^2)^(3/2) + (2*(-(c*d^2) + 6*a*e^2)*ArcTanh[(Sqrt[c]*x - Sqrt[a + c*x^2])/Sqrt[a]])/a
^(3/2))/(2*d^4)

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 424, normalized size of antiderivative = 1.58

method result size
risch \(-\frac {\sqrt {c \,x^{2}+a}\, \left (-4 e x +d \right )}{2 a \,d^{3} x^{2}}-\frac {2 a e \left (-\frac {e^{2} \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{\left (e^{2} a +c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}-\frac {e c d \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (e^{2} a +c \,d^{2}\right ) \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}\right )-\frac {6 e^{2} a \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{d \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}-\frac {\left (-6 e^{2} a +c \,d^{2}\right ) \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )}{d \sqrt {a}}}{2 d^{3} a}\) \(424\)
default \(\frac {-\frac {\sqrt {c \,x^{2}+a}}{2 a \,x^{2}}+\frac {c \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )}{2 a^{\frac {3}{2}}}}{d^{2}}-\frac {3 e^{2} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )}{d^{4} \sqrt {a}}+\frac {2 e \sqrt {c \,x^{2}+a}}{a \,d^{3} x}-\frac {e \left (-\frac {e^{2} \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{\left (e^{2} a +c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}-\frac {e c d \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (e^{2} a +c \,d^{2}\right ) \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}\right )}{d^{3}}+\frac {3 e^{2} \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{d^{4} \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}\) \(453\)

[In]

int(1/x^3/(e*x+d)^2/(c*x^2+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/2*(c*x^2+a)^(1/2)*(-4*e*x+d)/a/d^3/x^2-1/2/d^3/a*(2*a*e*(-1/(a*e^2+c*d^2)*e^2/(x+d/e)*((x+d/e)^2*c-2/e*c*d*
(x+d/e)+(a*e^2+c*d^2)/e^2)^(1/2)-e*c*d/(a*e^2+c*d^2)/((a*e^2+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2+c*d^2)/e^2-2/e*c*d
*(x+d/e)+2*((a*e^2+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c-2/e*c*d*(x+d/e)+(a*e^2+c*d^2)/e^2)^(1/2))/(x+d/e)))-6*e^2*a/
d/((a*e^2+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2+c*d^2)/e^2-2/e*c*d*(x+d/e)+2*((a*e^2+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c-2
/e*c*d*(x+d/e)+(a*e^2+c*d^2)/e^2)^(1/2))/(x+d/e))-(-6*a*e^2+c*d^2)/d/a^(1/2)*ln((2*a+2*a^(1/2)*(c*x^2+a)^(1/2)
)/x))

Fricas [A] (verification not implemented)

none

Time = 1.02 (sec) , antiderivative size = 1867, normalized size of antiderivative = 6.97 \[ \int \frac {1}{x^3 (d+e x)^2 \sqrt {a+c x^2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[1/4*(2*((4*a^2*c*d^2*e^4 + 3*a^3*e^6)*x^3 + (4*a^2*c*d^3*e^3 + 3*a^3*d*e^5)*x^2)*sqrt(c*d^2 + a*e^2)*log((2*a
*c*d*e*x - a*c*d^2 - 2*a^2*e^2 - (2*c^2*d^2 + a*c*e^2)*x^2 + 2*sqrt(c*d^2 + a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 +
a))/(e^2*x^2 + 2*d*e*x + d^2)) - ((c^3*d^6*e - 4*a*c^2*d^4*e^3 - 11*a^2*c*d^2*e^5 - 6*a^3*e^7)*x^3 + (c^3*d^7
- 4*a*c^2*d^5*e^2 - 11*a^2*c*d^3*e^4 - 6*a^3*d*e^6)*x^2)*sqrt(a)*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a
)/x^2) - 2*(a*c^2*d^7 + 2*a^2*c*d^5*e^2 + a^3*d^3*e^4 - 2*(2*a*c^2*d^5*e^2 + 5*a^2*c*d^3*e^4 + 3*a^3*d*e^6)*x^
2 - 3*(a*c^2*d^6*e + 2*a^2*c*d^4*e^3 + a^3*d^2*e^5)*x)*sqrt(c*x^2 + a))/((a^2*c^2*d^8*e + 2*a^3*c*d^6*e^3 + a^
4*d^4*e^5)*x^3 + (a^2*c^2*d^9 + 2*a^3*c*d^7*e^2 + a^4*d^5*e^4)*x^2), 1/4*(4*((4*a^2*c*d^2*e^4 + 3*a^3*e^6)*x^3
 + (4*a^2*c*d^3*e^3 + 3*a^3*d*e^5)*x^2)*sqrt(-c*d^2 - a*e^2)*arctan(sqrt(-c*d^2 - a*e^2)*(c*d*x - a*e)*sqrt(c*
x^2 + a)/(a*c*d^2 + a^2*e^2 + (c^2*d^2 + a*c*e^2)*x^2)) - ((c^3*d^6*e - 4*a*c^2*d^4*e^3 - 11*a^2*c*d^2*e^5 - 6
*a^3*e^7)*x^3 + (c^3*d^7 - 4*a*c^2*d^5*e^2 - 11*a^2*c*d^3*e^4 - 6*a^3*d*e^6)*x^2)*sqrt(a)*log(-(c*x^2 - 2*sqrt
(c*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(a*c^2*d^7 + 2*a^2*c*d^5*e^2 + a^3*d^3*e^4 - 2*(2*a*c^2*d^5*e^2 + 5*a^2*c*
d^3*e^4 + 3*a^3*d*e^6)*x^2 - 3*(a*c^2*d^6*e + 2*a^2*c*d^4*e^3 + a^3*d^2*e^5)*x)*sqrt(c*x^2 + a))/((a^2*c^2*d^8
*e + 2*a^3*c*d^6*e^3 + a^4*d^4*e^5)*x^3 + (a^2*c^2*d^9 + 2*a^3*c*d^7*e^2 + a^4*d^5*e^4)*x^2), -1/2*(((c^3*d^6*
e - 4*a*c^2*d^4*e^3 - 11*a^2*c*d^2*e^5 - 6*a^3*e^7)*x^3 + (c^3*d^7 - 4*a*c^2*d^5*e^2 - 11*a^2*c*d^3*e^4 - 6*a^
3*d*e^6)*x^2)*sqrt(-a)*arctan(sqrt(-a)/sqrt(c*x^2 + a)) - ((4*a^2*c*d^2*e^4 + 3*a^3*e^6)*x^3 + (4*a^2*c*d^3*e^
3 + 3*a^3*d*e^5)*x^2)*sqrt(c*d^2 + a*e^2)*log((2*a*c*d*e*x - a*c*d^2 - 2*a^2*e^2 - (2*c^2*d^2 + a*c*e^2)*x^2 +
 2*sqrt(c*d^2 + a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a))/(e^2*x^2 + 2*d*e*x + d^2)) + (a*c^2*d^7 + 2*a^2*c*d^5*e^
2 + a^3*d^3*e^4 - 2*(2*a*c^2*d^5*e^2 + 5*a^2*c*d^3*e^4 + 3*a^3*d*e^6)*x^2 - 3*(a*c^2*d^6*e + 2*a^2*c*d^4*e^3 +
 a^3*d^2*e^5)*x)*sqrt(c*x^2 + a))/((a^2*c^2*d^8*e + 2*a^3*c*d^6*e^3 + a^4*d^4*e^5)*x^3 + (a^2*c^2*d^9 + 2*a^3*
c*d^7*e^2 + a^4*d^5*e^4)*x^2), 1/2*(2*((4*a^2*c*d^2*e^4 + 3*a^3*e^6)*x^3 + (4*a^2*c*d^3*e^3 + 3*a^3*d*e^5)*x^2
)*sqrt(-c*d^2 - a*e^2)*arctan(sqrt(-c*d^2 - a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a)/(a*c*d^2 + a^2*e^2 + (c^2*d^2
 + a*c*e^2)*x^2)) - ((c^3*d^6*e - 4*a*c^2*d^4*e^3 - 11*a^2*c*d^2*e^5 - 6*a^3*e^7)*x^3 + (c^3*d^7 - 4*a*c^2*d^5
*e^2 - 11*a^2*c*d^3*e^4 - 6*a^3*d*e^6)*x^2)*sqrt(-a)*arctan(sqrt(-a)/sqrt(c*x^2 + a)) - (a*c^2*d^7 + 2*a^2*c*d
^5*e^2 + a^3*d^3*e^4 - 2*(2*a*c^2*d^5*e^2 + 5*a^2*c*d^3*e^4 + 3*a^3*d*e^6)*x^2 - 3*(a*c^2*d^6*e + 2*a^2*c*d^4*
e^3 + a^3*d^2*e^5)*x)*sqrt(c*x^2 + a))/((a^2*c^2*d^8*e + 2*a^3*c*d^6*e^3 + a^4*d^4*e^5)*x^3 + (a^2*c^2*d^9 + 2
*a^3*c*d^7*e^2 + a^4*d^5*e^4)*x^2)]

Sympy [F]

\[ \int \frac {1}{x^3 (d+e x)^2 \sqrt {a+c x^2}} \, dx=\int \frac {1}{x^{3} \sqrt {a + c x^{2}} \left (d + e x\right )^{2}}\, dx \]

[In]

integrate(1/x**3/(e*x+d)**2/(c*x**2+a)**(1/2),x)

[Out]

Integral(1/(x**3*sqrt(a + c*x**2)*(d + e*x)**2), x)

Maxima [F]

\[ \int \frac {1}{x^3 (d+e x)^2 \sqrt {a+c x^2}} \, dx=\int { \frac {1}{\sqrt {c x^{2} + a} {\left (e x + d\right )}^{2} x^{3}} \,d x } \]

[In]

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

[Out]

integrate(1/(sqrt(c*x^2 + a)*(e*x + d)^2*x^3), x)

Giac [F]

\[ \int \frac {1}{x^3 (d+e x)^2 \sqrt {a+c x^2}} \, dx=\int { \frac {1}{\sqrt {c x^{2} + a} {\left (e x + d\right )}^{2} x^{3}} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^3 (d+e x)^2 \sqrt {a+c x^2}} \, dx=\int \frac {1}{x^3\,\sqrt {c\,x^2+a}\,{\left (d+e\,x\right )}^2} \,d x \]

[In]

int(1/(x^3*(a + c*x^2)^(1/2)*(d + e*x)^2),x)

[Out]

int(1/(x^3*(a + c*x^2)^(1/2)*(d + e*x)^2), x)