\(\int \frac {1}{x^3 (d+e x) (a+c x^2)^{3/2}} \, dx\) [341]

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

Optimal result

Integrand size = 22, antiderivative size = 276 \[ \int \frac {1}{x^3 (d+e x) \left (a+c x^2\right )^{3/2}} \, dx=-\frac {3 c}{2 a^2 d \sqrt {a+c x^2}}+\frac {e^2}{a d^3 \sqrt {a+c x^2}}-\frac {1}{2 a d x^2 \sqrt {a+c x^2}}+\frac {e}{a d^2 x \sqrt {a+c x^2}}+\frac {2 c e x}{a^2 d^2 \sqrt {a+c x^2}}-\frac {e^3 (a e+c d x)}{a d^3 \left (c d^2+a e^2\right ) \sqrt {a+c x^2}}+\frac {e^5 \text {arctanh}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{d^3 \left (c d^2+a e^2\right )^{3/2}}+\frac {3 c \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 a^{5/2} d}-\frac {e^2 \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{3/2} d^3} \]

[Out]

e^5*arctanh((-c*d*x+a*e)/(a*e^2+c*d^2)^(1/2)/(c*x^2+a)^(1/2))/d^3/(a*e^2+c*d^2)^(3/2)+3/2*c*arctanh((c*x^2+a)^
(1/2)/a^(1/2))/a^(5/2)/d-e^2*arctanh((c*x^2+a)^(1/2)/a^(1/2))/a^(3/2)/d^3-3/2*c/a^2/d/(c*x^2+a)^(1/2)+e^2/a/d^
3/(c*x^2+a)^(1/2)-1/2/a/d/x^2/(c*x^2+a)^(1/2)+e/a/d^2/x/(c*x^2+a)^(1/2)+2*c*e*x/a^2/d^2/(c*x^2+a)^(1/2)-e^3*(c
*d*x+a*e)/a/d^3/(a*e^2+c*d^2)/(c*x^2+a)^(1/2)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {975, 272, 44, 53, 65, 214, 277, 197, 755, 12, 739, 212} \[ \int \frac {1}{x^3 (d+e x) \left (a+c x^2\right )^{3/2}} \, dx=-\frac {e^2 \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{3/2} d^3}+\frac {3 c \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 a^{5/2} d}+\frac {2 c e x}{a^2 d^2 \sqrt {a+c x^2}}-\frac {3 c}{2 a^2 d \sqrt {a+c x^2}}+\frac {e^5 \text {arctanh}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{d^3 \left (a e^2+c d^2\right )^{3/2}}+\frac {e^2}{a d^3 \sqrt {a+c x^2}}+\frac {e}{a d^2 x \sqrt {a+c x^2}}-\frac {e^3 (a e+c d x)}{a d^3 \sqrt {a+c x^2} \left (a e^2+c d^2\right )}-\frac {1}{2 a d x^2 \sqrt {a+c x^2}} \]

[In]

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

[Out]

(-3*c)/(2*a^2*d*Sqrt[a + c*x^2]) + e^2/(a*d^3*Sqrt[a + c*x^2]) - 1/(2*a*d*x^2*Sqrt[a + c*x^2]) + e/(a*d^2*x*Sq
rt[a + c*x^2]) + (2*c*e*x)/(a^2*d^2*Sqrt[a + c*x^2]) - (e^3*(a*e + c*d*x))/(a*d^3*(c*d^2 + a*e^2)*Sqrt[a + c*x
^2]) + (e^5*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/(d^3*(c*d^2 + a*e^2)^(3/2)) + (3*c*A
rcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/(2*a^(5/2)*d) - (e^2*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/(a^(3/2)*d^3)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 53

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] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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 197

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

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 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 277

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

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 755

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^(m + 1))*(a*e + c*d*x)*
((a + c*x^2)^(p + 1)/(2*a*(p + 1)*(c*d^2 + a*e^2))), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^
m*Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[
{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

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 x^3 \left (a+c x^2\right )^{3/2}}-\frac {e}{d^2 x^2 \left (a+c x^2\right )^{3/2}}+\frac {e^2}{d^3 x \left (a+c x^2\right )^{3/2}}-\frac {e^3}{d^3 (d+e x) \left (a+c x^2\right )^{3/2}}\right ) \, dx \\ & = \frac {\int \frac {1}{x^3 \left (a+c x^2\right )^{3/2}} \, dx}{d}-\frac {e \int \frac {1}{x^2 \left (a+c x^2\right )^{3/2}} \, dx}{d^2}+\frac {e^2 \int \frac {1}{x \left (a+c x^2\right )^{3/2}} \, dx}{d^3}-\frac {e^3 \int \frac {1}{(d+e x) \left (a+c x^2\right )^{3/2}} \, dx}{d^3} \\ & = \frac {e}{a d^2 x \sqrt {a+c x^2}}-\frac {e^3 (a e+c d x)}{a d^3 \left (c d^2+a e^2\right ) \sqrt {a+c x^2}}+\frac {\text {Subst}\left (\int \frac {1}{x^2 (a+c x)^{3/2}} \, dx,x,x^2\right )}{2 d}+\frac {(2 c e) \int \frac {1}{\left (a+c x^2\right )^{3/2}} \, dx}{a d^2}+\frac {e^2 \text {Subst}\left (\int \frac {1}{x (a+c x)^{3/2}} \, dx,x,x^2\right )}{2 d^3}-\frac {e^3 \int \frac {a e^2}{(d+e x) \sqrt {a+c x^2}} \, dx}{a d^3 \left (c d^2+a e^2\right )} \\ & = \frac {e^2}{a d^3 \sqrt {a+c x^2}}-\frac {1}{2 a d x^2 \sqrt {a+c x^2}}+\frac {e}{a d^2 x \sqrt {a+c x^2}}+\frac {2 c e x}{a^2 d^2 \sqrt {a+c x^2}}-\frac {e^3 (a e+c d x)}{a d^3 \left (c d^2+a e^2\right ) \sqrt {a+c x^2}}-\frac {(3 c) \text {Subst}\left (\int \frac {1}{x (a+c x)^{3/2}} \, dx,x,x^2\right )}{4 a d}+\frac {e^2 \text {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,x^2\right )}{2 a d^3}-\frac {e^5 \int \frac {1}{(d+e x) \sqrt {a+c x^2}} \, dx}{d^3 \left (c d^2+a e^2\right )} \\ & = -\frac {3 c}{2 a^2 d \sqrt {a+c x^2}}+\frac {e^2}{a d^3 \sqrt {a+c x^2}}-\frac {1}{2 a d x^2 \sqrt {a+c x^2}}+\frac {e}{a d^2 x \sqrt {a+c x^2}}+\frac {2 c e x}{a^2 d^2 \sqrt {a+c x^2}}-\frac {e^3 (a e+c d x)}{a d^3 \left (c d^2+a e^2\right ) \sqrt {a+c x^2}}-\frac {(3 c) \text {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,x^2\right )}{4 a^2 d}+\frac {e^2 \text {Subst}\left (\int \frac {1}{-\frac {a}{c}+\frac {x^2}{c}} \, dx,x,\sqrt {a+c x^2}\right )}{a c d^3}+\frac {e^5 \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^3 \left (c d^2+a e^2\right )} \\ & = -\frac {3 c}{2 a^2 d \sqrt {a+c x^2}}+\frac {e^2}{a d^3 \sqrt {a+c x^2}}-\frac {1}{2 a d x^2 \sqrt {a+c x^2}}+\frac {e}{a d^2 x \sqrt {a+c x^2}}+\frac {2 c e x}{a^2 d^2 \sqrt {a+c x^2}}-\frac {e^3 (a e+c d x)}{a d^3 \left (c d^2+a e^2\right ) \sqrt {a+c x^2}}+\frac {e^5 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{d^3 \left (c d^2+a e^2\right )^{3/2}}-\frac {e^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{3/2} d^3}-\frac {3 \text {Subst}\left (\int \frac {1}{-\frac {a}{c}+\frac {x^2}{c}} \, dx,x,\sqrt {a+c x^2}\right )}{2 a^2 d} \\ & = -\frac {3 c}{2 a^2 d \sqrt {a+c x^2}}+\frac {e^2}{a d^3 \sqrt {a+c x^2}}-\frac {1}{2 a d x^2 \sqrt {a+c x^2}}+\frac {e}{a d^2 x \sqrt {a+c x^2}}+\frac {2 c e x}{a^2 d^2 \sqrt {a+c x^2}}-\frac {e^3 (a e+c d x)}{a d^3 \left (c d^2+a e^2\right ) \sqrt {a+c x^2}}+\frac {e^5 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{d^3 \left (c d^2+a e^2\right )^{3/2}}+\frac {3 c \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 a^{5/2} d}-\frac {e^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{3/2} d^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.76 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.79 \[ \int \frac {1}{x^3 (d+e x) \left (a+c x^2\right )^{3/2}} \, dx=-\frac {\frac {d \left (c^2 d^2 x^2 (3 d-4 e x)+a^2 e^2 (d-2 e x)+a c \left (d^3-2 d^2 e x+d e^2 x^2-2 e^3 x^3\right )\right )}{a^2 \left (c d^2+a e^2\right ) x^2 \sqrt {a+c x^2}}+\frac {4 e^5 \arctan \left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+c x^2}}{\sqrt {-c d^2-a e^2}}\right )}{\left (-c d^2-a e^2\right )^{3/2}}+\frac {2 \left (3 c d^2-2 a e^2\right ) \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{5/2}}}{2 d^3} \]

[In]

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

[Out]

-1/2*((d*(c^2*d^2*x^2*(3*d - 4*e*x) + a^2*e^2*(d - 2*e*x) + a*c*(d^3 - 2*d^2*e*x + d*e^2*x^2 - 2*e^3*x^3)))/(a
^2*(c*d^2 + a*e^2)*x^2*Sqrt[a + c*x^2]) + (4*e^5*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*(3*c*d^2 - 2*a*e^2)*ArcTanh[(Sqrt[c]*x - Sqrt[a + c*x^2])/Sqrt[a]])/a
^(5/2))/d^3

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 407, normalized size of antiderivative = 1.47

method result size
risch \(-\frac {\sqrt {c \,x^{2}+a}\, \left (-2 e x +d \right )}{2 a^{2} d^{2} x^{2}}-\frac {-\frac {\left (-2 e^{2} a +3 c \,d^{2}\right ) \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )}{d \sqrt {a}}+\frac {c^{2} d^{2} \sqrt {\left (x +\frac {\sqrt {-a c}}{c}\right )^{2} c -2 \sqrt {-a c}\, \left (x +\frac {\sqrt {-a c}}{c}\right )}}{\left (e \sqrt {-a c}-c d \right ) \sqrt {-a c}\, \left (x +\frac {\sqrt {-a c}}{c}\right )}+\frac {c^{2} d^{2} \sqrt {\left (x -\frac {\sqrt {-a c}}{c}\right )^{2} c +2 \sqrt {-a c}\, \left (x -\frac {\sqrt {-a c}}{c}\right )}}{\left (e \sqrt {-a c}+c d \right ) \sqrt {-a c}\, \left (x -\frac {\sqrt {-a c}}{c}\right )}+\frac {2 c \,a^{2} e^{4} \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (e \sqrt {-a c}+c d \right ) \left (e \sqrt {-a c}-c d \right ) d \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}}{2 a^{2} d^{2}}\) \(407\)
default \(\frac {-\frac {1}{2 a \,x^{2} \sqrt {c \,x^{2}+a}}-\frac {3 c \left (\frac {1}{a \sqrt {c \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )}{a^{\frac {3}{2}}}\right )}{2 a}}{d}+\frac {e^{2} \left (\frac {1}{a \sqrt {c \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )}{a^{\frac {3}{2}}}\right )}{d^{3}}-\frac {e \left (-\frac {1}{a x \sqrt {c \,x^{2}+a}}-\frac {2 c x}{a^{2} \sqrt {c \,x^{2}+a}}\right )}{d^{2}}-\frac {e^{2} \left (\frac {e^{2}}{\left (e^{2} a +c \,d^{2}\right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}+\frac {2 e c d \left (2 c \left (x +\frac {d}{e}\right )-\frac {2 c d}{e}\right )}{\left (e^{2} a +c \,d^{2}\right ) \left (\frac {4 c \left (e^{2} a +c \,d^{2}\right )}{e^{2}}-\frac {4 c^{2} d^{2}}{e^{2}}\right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}-\frac {e^{2} \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (e^{2} a +c \,d^{2}\right ) \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}\right )}{d^{3}}\) \(479\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.70 (sec) , antiderivative size = 1943, normalized size of antiderivative = 7.04 \[ \int \frac {1}{x^3 (d+e x) \left (a+c x^2\right )^{3/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 365, normalized size of antiderivative = 1.32 \[ \int \frac {1}{x^3 (d+e x) \left (a+c x^2\right )^{3/2}} \, dx=-\frac {2 \, e^{5} \arctan \left (-\frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} e + \sqrt {c} d}{\sqrt {-c d^{2} - a e^{2}}}\right )}{{\left (c d^{5} + a d^{3} e^{2}\right )} \sqrt {-c d^{2} - a e^{2}}} + \frac {\frac {{\left (a^{2} c^{3} d^{2} e + a^{3} c^{2} e^{3}\right )} x}{a^{4} c^{2} d^{4} + 2 \, a^{5} c d^{2} e^{2} + a^{6} e^{4}} - \frac {a^{2} c^{3} d^{3} + a^{3} c^{2} d e^{2}}{a^{4} c^{2} d^{4} + 2 \, a^{5} c d^{2} e^{2} + a^{6} e^{4}}}{\sqrt {c x^{2} + a}} - \frac {{\left (3 \, c d^{2} - 2 \, a e^{2}\right )} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2} d^{3}} + \frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{3} c d - 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} a \sqrt {c} e + {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} a c d + 2 \, a^{2} \sqrt {c} e}{{\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} - a\right )}^{2} a^{2} d^{2}} \]

[In]

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

[Out]

-2*e^5*arctan(-((sqrt(c)*x - sqrt(c*x^2 + a))*e + sqrt(c)*d)/sqrt(-c*d^2 - a*e^2))/((c*d^5 + a*d^3*e^2)*sqrt(-
c*d^2 - a*e^2)) + ((a^2*c^3*d^2*e + a^3*c^2*e^3)*x/(a^4*c^2*d^4 + 2*a^5*c*d^2*e^2 + a^6*e^4) - (a^2*c^3*d^3 +
a^3*c^2*d*e^2)/(a^4*c^2*d^4 + 2*a^5*c*d^2*e^2 + a^6*e^4))/sqrt(c*x^2 + a) - (3*c*d^2 - 2*a*e^2)*arctan(-(sqrt(
c)*x - sqrt(c*x^2 + a))/sqrt(-a))/(sqrt(-a)*a^2*d^3) + ((sqrt(c)*x - sqrt(c*x^2 + a))^3*c*d - 2*(sqrt(c)*x - s
qrt(c*x^2 + a))^2*a*sqrt(c)*e + (sqrt(c)*x - sqrt(c*x^2 + a))*a*c*d + 2*a^2*sqrt(c)*e)/(((sqrt(c)*x - sqrt(c*x
^2 + a))^2 - a)^2*a^2*d^2)

Mupad [F(-1)]

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

[In]

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

[Out]

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