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

Optimal. Leaf size=168 \[ \frac {c \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 a^{3/2} d}-\frac {e^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a} d^3}+\frac {e \sqrt {a+c x^2}}{a d^2 x}+\frac {e^3 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{d^3 \sqrt {a e^2+c d^2}}-\frac {\sqrt {a+c x^2}}{2 a d x^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.14, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {961, 266, 51, 63, 208, 264, 725, 206} \[ \frac {c \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 a^{3/2} d}+\frac {e^3 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{d^3 \sqrt {a e^2+c d^2}}-\frac {e^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a} d^3}+\frac {e \sqrt {a+c x^2}}{a d^2 x}-\frac {\sqrt {a+c x^2}}{2 a d x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[a + c*x^2]/(2*a*d*x^2) + (e*Sqrt[a + c*x^2])/(a*d^2*x) + (e^3*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]
*Sqrt[a + c*x^2])])/(d^3*Sqrt[c*d^2 + a*e^2]) + (c*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/(2*a^(3/2)*d) - (e^2*ArcT
anh[Sqrt[a + c*x^2]/Sqrt[a]])/(Sqrt[a]*d^3)

Rule 51

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 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 206

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

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 264

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 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 725

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 961

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

________________________________________________________________________________________

Mathematica [A]  time = 0.64, size = 163, normalized size = 0.97 \[ \frac {\frac {2 e^3 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{\sqrt {a e^2+c d^2}}+\frac {d \left (c d x^2 \sqrt {\frac {c x^2}{a}+1} \tanh ^{-1}\left (\sqrt {\frac {c x^2}{a}+1}\right )-\left (a+c x^2\right ) (d-2 e x)\right )}{a x^2 \sqrt {a+c x^2}}-\frac {2 e^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a}}}{2 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 1.04, size = 956, normalized size = 5.69 \[ \left [\frac {2 \, \sqrt {c d^{2} + a e^{2}} a^{2} e^{3} x^{2} \log \left (\frac {2 \, a c d e x - a c d^{2} - 2 \, a^{2} e^{2} - {\left (2 \, c^{2} d^{2} + a c e^{2}\right )} x^{2} + 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) - {\left (c^{2} d^{4} - a c d^{2} e^{2} - 2 \, a^{2} e^{4}\right )} \sqrt {a} x^{2} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (a c d^{4} + a^{2} d^{2} e^{2} - 2 \, {\left (a c d^{3} e + a^{2} d e^{3}\right )} x\right )} \sqrt {c x^{2} + a}}{4 \, {\left (a^{2} c d^{5} + a^{3} d^{3} e^{2}\right )} x^{2}}, \frac {4 \, \sqrt {-c d^{2} - a e^{2}} a^{2} e^{3} x^{2} \arctan \left (\frac {\sqrt {-c d^{2} - a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{a c d^{2} + a^{2} e^{2} + {\left (c^{2} d^{2} + a c e^{2}\right )} x^{2}}\right ) - {\left (c^{2} d^{4} - a c d^{2} e^{2} - 2 \, a^{2} e^{4}\right )} \sqrt {a} x^{2} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (a c d^{4} + a^{2} d^{2} e^{2} - 2 \, {\left (a c d^{3} e + a^{2} d e^{3}\right )} x\right )} \sqrt {c x^{2} + a}}{4 \, {\left (a^{2} c d^{5} + a^{3} d^{3} e^{2}\right )} x^{2}}, \frac {\sqrt {c d^{2} + a e^{2}} a^{2} e^{3} x^{2} \log \left (\frac {2 \, a c d e x - a c d^{2} - 2 \, a^{2} e^{2} - {\left (2 \, c^{2} d^{2} + a c e^{2}\right )} x^{2} + 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) - {\left (c^{2} d^{4} - a c d^{2} e^{2} - 2 \, a^{2} e^{4}\right )} \sqrt {-a} x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) - {\left (a c d^{4} + a^{2} d^{2} e^{2} - 2 \, {\left (a c d^{3} e + a^{2} d e^{3}\right )} x\right )} \sqrt {c x^{2} + a}}{2 \, {\left (a^{2} c d^{5} + a^{3} d^{3} e^{2}\right )} x^{2}}, \frac {2 \, \sqrt {-c d^{2} - a e^{2}} a^{2} e^{3} x^{2} \arctan \left (\frac {\sqrt {-c d^{2} - a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{a c d^{2} + a^{2} e^{2} + {\left (c^{2} d^{2} + a c e^{2}\right )} x^{2}}\right ) - {\left (c^{2} d^{4} - a c d^{2} e^{2} - 2 \, a^{2} e^{4}\right )} \sqrt {-a} x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) - {\left (a c d^{4} + a^{2} d^{2} e^{2} - 2 \, {\left (a c d^{3} e + a^{2} d e^{3}\right )} x\right )} \sqrt {c x^{2} + a}}{2 \, {\left (a^{2} c d^{5} + a^{3} d^{3} e^{2}\right )} x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.22, size = 239, normalized size = 1.42 \[ -c^{\frac {3}{2}} {\left (\frac {2 \, \arctan \left (-\frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} e + \sqrt {c} d}{\sqrt {-c d^{2} - a e^{2}}}\right ) e^{3}}{\sqrt {-c d^{2} - a e^{2}} c^{\frac {3}{2}} d^{3}} + \frac {{\left (c d^{2} - 2 \, a e^{2}\right )} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a c^{\frac {3}{2}} d^{3}} - \frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{3} \sqrt {c} d - 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} a e + {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} a \sqrt {c} d + 2 \, a^{2} e}{{\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} - a\right )}^{2} a c d^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c^(3/2)*(2*arctan(-((sqrt(c)*x - sqrt(c*x^2 + a))*e + sqrt(c)*d)/sqrt(-c*d^2 - a*e^2))*e^3/(sqrt(-c*d^2 - a*e
^2)*c^(3/2)*d^3) + (c*d^2 - 2*a*e^2)*arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))/sqrt(-a))/(sqrt(-a)*a*c^(3/2)*d^3)
- ((sqrt(c)*x - sqrt(c*x^2 + a))^3*sqrt(c)*d - 2*(sqrt(c)*x - sqrt(c*x^2 + a))^2*a*e + (sqrt(c)*x - sqrt(c*x^2
+ a))*a*sqrt(c)*d + 2*a^2*e)/(((sqrt(c)*x - sqrt(c*x^2 + a))^2 - a)^2*a*c*d^2))

________________________________________________________________________________________

maple [A]  time = 0.01, size = 236, normalized size = 1.40 \[ \frac {e^{2} \ln \left (\frac {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, d^{3}}-\frac {e^{2} \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )}{\sqrt {a}\, d^{3}}+\frac {c \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )}{2 a^{\frac {3}{2}} d}+\frac {\sqrt {c \,x^{2}+a}\, e}{a \,d^{2} x}-\frac {\sqrt {c \,x^{2}+a}}{2 a d \,x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________