\(\int \frac {\sqrt {a+c x^2}}{x^3 (d+e x)} \, dx\) [323]

Optimal result

Integrand size = 22, antiderivative size = 160 \[ \int \frac {\sqrt {a+c x^2}}{x^3 (d+e x)} \, dx=-\frac {\sqrt {a+c x^2}}{2 d x^2}+\frac {e \sqrt {a+c x^2}}{d^2 x}+\frac {e \sqrt {c d^2+a e^2} \text {arctanh}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{d^3}-\frac {c \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 \sqrt {a} d}-\frac {\sqrt {a} e^2 \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{d^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {975, 272, 43, 65, 214, 283, 223, 212, 52, 749, 858, 739} \[ \int \frac {\sqrt {a+c x^2}}{x^3 (d+e x)} \, dx=-\frac {\sqrt {a} e^2 \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{d^3}+\frac {e \sqrt {a e^2+c d^2} \text {arctanh}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{d^3}-\frac {c \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 \sqrt {a} d}+\frac {e \sqrt {a+c x^2}}{d^2 x}-\frac {\sqrt {a+c x^2}}{2 d x^2} \]

[In]

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

[Out]

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

Rule 43

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

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 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 223

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 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 283

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

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 749

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((a + c*x^2)^p/(e
*(m + 2*p + 1))), x] + Dist[2*(p/(e*(m + 2*p + 1))), Int[(d + e*x)^m*Simp[a*e - c*d*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] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !Ration
alQ[m] || LtQ[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 858

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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] &&  !IGtQ[m, 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 {\sqrt {a+c x^2}}{d x^3}-\frac {e \sqrt {a+c x^2}}{d^2 x^2}+\frac {e^2 \sqrt {a+c x^2}}{d^3 x}-\frac {e^3 \sqrt {a+c x^2}}{d^3 (d+e x)}\right ) \, dx \\ & = \frac {\int \frac {\sqrt {a+c x^2}}{x^3} \, dx}{d}-\frac {e \int \frac {\sqrt {a+c x^2}}{x^2} \, dx}{d^2}+\frac {e^2 \int \frac {\sqrt {a+c x^2}}{x} \, dx}{d^3}-\frac {e^3 \int \frac {\sqrt {a+c x^2}}{d+e x} \, dx}{d^3} \\ & = -\frac {e^2 \sqrt {a+c x^2}}{d^3}+\frac {e \sqrt {a+c x^2}}{d^2 x}+\frac {\text {Subst}\left (\int \frac {\sqrt {a+c x}}{x^2} \, dx,x,x^2\right )}{2 d}-\frac {(c e) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{d^2}+\frac {e^2 \text {Subst}\left (\int \frac {\sqrt {a+c x}}{x} \, dx,x,x^2\right )}{2 d^3}-\frac {e^2 \int \frac {a e-c d x}{(d+e x) \sqrt {a+c x^2}} \, dx}{d^3} \\ & = -\frac {\sqrt {a+c x^2}}{2 d x^2}+\frac {e \sqrt {a+c x^2}}{d^2 x}+\frac {c \text {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,x^2\right )}{4 d}+\frac {(c e) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{d^2}-\frac {(c e) \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{d^2}+\frac {\left (a e^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,x^2\right )}{2 d^3}-\frac {\left (e \left (c d^2+a e^2\right )\right ) \int \frac {1}{(d+e x) \sqrt {a+c x^2}} \, dx}{d^3} \\ & = -\frac {\sqrt {a+c x^2}}{2 d x^2}+\frac {e \sqrt {a+c x^2}}{d^2 x}-\frac {\sqrt {c} e \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{d^2}+\frac {\text {Subst}\left (\int \frac {1}{-\frac {a}{c}+\frac {x^2}{c}} \, dx,x,\sqrt {a+c x^2}\right )}{2 d}+\frac {(c e) \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{d^2}+\frac {\left (a 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^3}+\frac {\left (e \left (c d^2+a e^2\right )\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^3} \\ & = -\frac {\sqrt {a+c x^2}}{2 d x^2}+\frac {e \sqrt {a+c x^2}}{d^2 x}+\frac {e \sqrt {c d^2+a e^2} \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{d^3}-\frac {c \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 \sqrt {a} d}-\frac {\sqrt {a} e^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{d^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.90 \[ \int \frac {\sqrt {a+c x^2}}{x^3 (d+e x)} \, dx=\frac {\frac {d (-d+2 e x) \sqrt {a+c x^2}}{x^2}-4 e \sqrt {-c d^2-a e^2} \arctan \left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+c x^2}}{\sqrt {-c d^2-a e^2}}\right )+\frac {2 \left (c d^2+2 a e^2\right ) \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a}}}{2 d^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.32

method	result	size
risch	\(-\frac {\sqrt {c \,x^{2}+a}\, \left (-2 e x +d \right )}{2 d^{2} x^{2}}-\frac {-\frac {\left (-2 e^{2} a -c \,d^{2}\right ) \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )}{d \sqrt {a}}-\frac {2 \left (e^{2} a +c \,d^{2}\right ) \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{d \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}}{2 d^{2}}\)	\(212\)
default	\(\frac {-\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}}}{2 a \,x^{2}}+\frac {c \left (\sqrt {c \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )\right )}{2 a}}{d}+\frac {e^{2} \left (\sqrt {c \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )\right )}{d^{3}}-\frac {e \left (-\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}}}{a x}+\frac {2 c \left (\frac {x \sqrt {c \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}\right )}{a}\right )}{d^{2}}-\frac {e^{2} \left (\sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}-\frac {\sqrt {c}\, d \ln \left (\frac {-\frac {c d}{e}+c \left (x +\frac {d}{e}\right )}{\sqrt {c}}+\sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}\right )}{e}-\frac {\left (e^{2} a +c \,d^{2}\right ) \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{2} \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}\right )}{d^{3}}\)	\(442\)

[In]

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

[Out]

-1/2*(c*x^2+a)^(1/2)*(-2*e*x+d)/d^2/x^2-1/2/d^2*(-(-2*a*e^2-c*d^2)/d/a^(1/2)*ln((2*a+2*a^(1/2)*(c*x^2+a)^(1/2)
)/x)-2*(a*e^2+c*d^2)/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 = 0.46 (sec) , antiderivative size = 726, normalized size of antiderivative = 4.54 \[ \int \frac {\sqrt {a+c x^2}}{x^3 (d+e x)} \, dx=\left [\frac {2 \, \sqrt {c d^{2} + a e^{2}} a e x^{2} \log \left (\frac {2 \, a c d e x - a c d^{2} - 2 \, a^{2} e^{2} - {\left (2 \, c^{2} d^{2} + a c e^{2}\right )} x^{2} + 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) + {\left (c d^{2} + 2 \, a e^{2}\right )} \sqrt {a} x^{2} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (2 \, a d e x - a d^{2}\right )} \sqrt {c x^{2} + a}}{4 \, a d^{3} x^{2}}, \frac {4 \, \sqrt {-c d^{2} - a e^{2}} a e x^{2} \arctan \left (\frac {\sqrt {-c d^{2} - a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{a c d^{2} + a^{2} e^{2} + {\left (c^{2} d^{2} + a c e^{2}\right )} x^{2}}\right ) + {\left (c d^{2} + 2 \, a e^{2}\right )} \sqrt {a} x^{2} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (2 \, a d e x - a d^{2}\right )} \sqrt {c x^{2} + a}}{4 \, a d^{3} x^{2}}, \frac {\sqrt {c d^{2} + a e^{2}} a e x^{2} \log \left (\frac {2 \, a c d e x - a c d^{2} - 2 \, a^{2} e^{2} - {\left (2 \, c^{2} d^{2} + a c e^{2}\right )} x^{2} + 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) + {\left (c d^{2} + 2 \, a e^{2}\right )} \sqrt {-a} x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) + {\left (2 \, a d e x - a d^{2}\right )} \sqrt {c x^{2} + a}}{2 \, a d^{3} x^{2}}, \frac {2 \, \sqrt {-c d^{2} - a e^{2}} a e x^{2} \arctan \left (\frac {\sqrt {-c d^{2} - a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{a c d^{2} + a^{2} e^{2} + {\left (c^{2} d^{2} + a c e^{2}\right )} x^{2}}\right ) + {\left (c d^{2} + 2 \, a e^{2}\right )} \sqrt {-a} x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) + {\left (2 \, a d e x - a d^{2}\right )} \sqrt {c x^{2} + a}}{2 \, a d^{3} x^{2}}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.44 \[ \int \frac {\sqrt {a+c x^2}}{x^3 (d+e x)} \, dx=-\frac {2 \, {\left (c d^{2} e + a e^{3}\right )} \arctan \left (-\frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} e + \sqrt {c} d}{\sqrt {-c d^{2} - a e^{2}}}\right )}{\sqrt {-c d^{2} - a e^{2}} d^{3}} + \frac {{\left (c d^{2} + 2 \, a e^{2}\right )} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} d^{3}} + \frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{3} c d - 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} a \sqrt {c} e + {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} a c d + 2 \, a^{2} \sqrt {c} e}{{\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} - a\right )}^{2} d^{2}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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