\(\int \frac {\sqrt {a+c x^2}}{x^4 (d+e x)} \, dx\) [324]

Optimal result

Integrand size = 22, antiderivative size = 191 \[ \int \frac {\sqrt {a+c x^2}}{x^4 (d+e x)} \, dx=\frac {e \sqrt {a+c x^2}}{2 d^2 x^2}-\frac {e^2 \sqrt {a+c x^2}}{d^3 x}-\frac {\left (a+c x^2\right )^{3/2}}{3 a d x^3}-\frac {e^2 \sqrt {c d^2+a e^2} \text {arctanh}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{d^4}+\frac {c e \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 \sqrt {a} d^2}+\frac {\sqrt {a} e^3 \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{d^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.591, Rules used = {975, 270, 272, 43, 65, 214, 283, 223, 212, 52, 749, 858, 739} \[ \int \frac {\sqrt {a+c x^2}}{x^4 (d+e x)} \, dx=\frac {\sqrt {a} e^3 \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{d^4}+\frac {c e \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 \sqrt {a} d^2}-\frac {e^2 \sqrt {a e^2+c d^2} \text {arctanh}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{d^4}-\frac {e^2 \sqrt {a+c x^2}}{d^3 x}+\frac {e \sqrt {a+c x^2}}{2 d^2 x^2}-\frac {\left (a+c x^2\right )^{3/2}}{3 a d x^3} \]

[In]

Int[Sqrt[a + c*x^2]/(x^4*(d + e*x)),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.91 \[ \int \frac {\sqrt {a+c x^2}}{x^4 (d+e x)} \, dx=-\frac {\frac {d \sqrt {a+c x^2} \left (2 a d^2-3 a d e x+2 c d^2 x^2+6 a e^2 x^2\right )}{a x^3}-12 e^2 \sqrt {-c d^2-a e^2} \arctan \left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+c x^2}}{\sqrt {-c d^2-a e^2}}\right )+\frac {6 e \left (c d^2+2 a e^2\right ) \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a}}}{6 d^4} \]

[In]

Integrate[Sqrt[a + c*x^2]/(x^4*(d + e*x)),x]

[Out]

-1/6*((d*Sqrt[a + c*x^2]*(2*a*d^2 - 3*a*d*e*x + 2*c*d^2*x^2 + 6*a*e^2*x^2))/(a*x^3) - 12*e^2*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]] + (6*e*(c*d^2 + 2*a*e^2)*ArcTanh[
(Sqrt[c]*x - Sqrt[a + c*x^2])/Sqrt[a]])/Sqrt[a])/d^4

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.26

method	result	size
risch	\(-\frac {\sqrt {c \,x^{2}+a}\, \left (6 a \,e^{2} x^{2}+2 c \,d^{2} x^{2}-3 a d e x +2 a \,d^{2}\right )}{6 d^{3} x^{3} a}+\frac {e \left (-\frac {\left (-2 e^{2} a -c \,d^{2}\right ) \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )}{d \sqrt {a}}-\frac {2 \left (e^{2} a +c \,d^{2}\right ) \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{d \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}\right )}{2 d^{3}}\)	\(241\)
default	\(-\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}}}{3 a d \,x^{3}}+\frac {e^{2} \left (-\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}}}{a x}+\frac {2 c \left (\frac {x \sqrt {c \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}\right )}{a}\right )}{d^{3}}-\frac {e^{3} \left (\sqrt {c \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )\right )}{d^{4}}-\frac {e \left (-\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}}}{2 a \,x^{2}}+\frac {c \left (\sqrt {c \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )\right )}{2 a}\right )}{d^{2}}+\frac {e^{3} \left (\sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}-\frac {\sqrt {c}\, d \ln \left (\frac {-\frac {c d}{e}+c \left (x +\frac {d}{e}\right )}{\sqrt {c}}+\sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}\right )}{e}-\frac {\left (e^{2} a +c \,d^{2}\right ) \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{2} \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}\right )}{d^{4}}\)	\(465\)

[In]

int((c*x^2+a)^(1/2)/x^4/(e*x+d),x,method=_RETURNVERBOSE)

[Out]

-1/6*(c*x^2+a)^(1/2)*(6*a*e^2*x^2+2*c*d^2*x^2-3*a*d*e*x+2*a*d^2)/d^3/x^3/a+1/2*e/d^3*(-(-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 = 0.45 (sec) , antiderivative size = 824, normalized size of antiderivative = 4.31 \[ \int \frac {\sqrt {a+c x^2}}{x^4 (d+e x)} \, dx=\left [\frac {6 \, \sqrt {c d^{2} + a e^{2}} a e^{2} x^{3} \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 ) + 3 \, {\left (c d^{2} e + 2 \, a e^{3}\right )} \sqrt {a} x^{3} \log \left (-\frac {c x^{2} + 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (3 \, a d^{2} e x - 2 \, a d^{3} - 2 \, {\left (c d^{3} + 3 \, a d e^{2}\right )} x^{2}\right )} \sqrt {c x^{2} + a}}{12 \, a d^{4} x^{3}}, -\frac {12 \, \sqrt {-c d^{2} - a e^{2}} a e^{2} x^{3} \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 ) - 3 \, {\left (c d^{2} e + 2 \, a e^{3}\right )} \sqrt {a} x^{3} \log \left (-\frac {c x^{2} + 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (3 \, a d^{2} e x - 2 \, a d^{3} - 2 \, {\left (c d^{3} + 3 \, a d e^{2}\right )} x^{2}\right )} \sqrt {c x^{2} + a}}{12 \, a d^{4} x^{3}}, \frac {3 \, \sqrt {c d^{2} + a e^{2}} a e^{2} x^{3} \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 ) - 3 \, {\left (c d^{2} e + 2 \, a e^{3}\right )} \sqrt {-a} x^{3} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) + {\left (3 \, a d^{2} e x - 2 \, a d^{3} - 2 \, {\left (c d^{3} + 3 \, a d e^{2}\right )} x^{2}\right )} \sqrt {c x^{2} + a}}{6 \, a d^{4} x^{3}}, -\frac {6 \, \sqrt {-c d^{2} - a e^{2}} a e^{2} x^{3} \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 ) + 3 \, {\left (c d^{2} e + 2 \, a e^{3}\right )} \sqrt {-a} x^{3} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) - {\left (3 \, a d^{2} e x - 2 \, a d^{3} - 2 \, {\left (c d^{3} + 3 \, a d e^{2}\right )} x^{2}\right )} \sqrt {c x^{2} + a}}{6 \, a d^{4} x^{3}}\right ] \]

[In]

integrate((c*x^2+a)^(1/2)/x^4/(e*x+d),x, algorithm="fricas")

[Out]

[1/12*(6*sqrt(c*d^2 + a*e^2)*a*e^2*x^3*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*d^2*e + 2*a*e^3)*sqrt(a)*
x^3*log(-(c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(3*a*d^2*e*x - 2*a*d^3 - 2*(c*d^3 + 3*a*d*e^2)*x^2
)*sqrt(c*x^2 + a))/(a*d^4*x^3), -1/12*(12*sqrt(-c*d^2 - a*e^2)*a*e^2*x^3*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*d^2*e + 2*a*e^3)*sqrt(a)*x^3*log(-(
c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(3*a*d^2*e*x - 2*a*d^3 - 2*(c*d^3 + 3*a*d*e^2)*x^2)*sqrt(c*x
^2 + a))/(a*d^4*x^3), 1/6*(3*sqrt(c*d^2 + a*e^2)*a*e^2*x^3*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*d^2*e
+ 2*a*e^3)*sqrt(-a)*x^3*arctan(sqrt(-a)/sqrt(c*x^2 + a)) + (3*a*d^2*e*x - 2*a*d^3 - 2*(c*d^3 + 3*a*d*e^2)*x^2
)*sqrt(c*x^2 + a))/(a*d^4*x^3), -1/6*(6*sqrt(-c*d^2 - a*e^2)*a*e^2*x^3*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*d^2*e + 2*a*e^3)*sqrt(-a)*x^3*arctan(
sqrt(-a)/sqrt(c*x^2 + a)) - (3*a*d^2*e*x - 2*a*d^3 - 2*(c*d^3 + 3*a*d*e^2)*x^2)*sqrt(c*x^2 + a))/(a*d^4*x^3)]

Sympy [F]

\[ \int \frac {\sqrt {a+c x^2}}{x^4 (d+e x)} \, dx=\int \frac {\sqrt {a + c x^{2}}}{x^{4} \left (d + e x\right )}\, dx \]

[In]

integrate((c*x**2+a)**(1/2)/x**4/(e*x+d),x)

[Out]

Integral(sqrt(a + c*x**2)/(x**4*(d + e*x)), x)

Maxima [F]

\[ \int \frac {\sqrt {a+c x^2}}{x^4 (d+e x)} \, dx=\int { \frac {\sqrt {c x^{2} + a}}{{\left (e x + d\right )} x^{4}} \,d x } \]

[In]

integrate((c*x^2+a)^(1/2)/x^4/(e*x+d),x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^2 + a)/((e*x + d)*x^4), x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.64 \[ \int \frac {\sqrt {a+c x^2}}{x^4 (d+e x)} \, dx=\frac {2 \, {\left (c d^{2} e^{2} + a e^{4}\right )} \arctan \left (-\frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} e + \sqrt {c} d}{\sqrt {-c d^{2} - a e^{2}}}\right )}{\sqrt {-c d^{2} - a e^{2}} d^{4}} - \frac {{\left (c d^{2} e + 2 \, a e^{3}\right )} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} d^{4}} - \frac {3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{5} c d e - 6 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{4} c^{\frac {3}{2}} d^{2} - 6 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{4} a \sqrt {c} e^{2} + 12 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} a^{2} \sqrt {c} e^{2} - 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} a^{2} c d e - 2 \, a^{2} c^{\frac {3}{2}} d^{2} - 6 \, a^{3} \sqrt {c} e^{2}}{3 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} - a\right )}^{3} d^{3}} \]

[In]

integrate((c*x^2+a)^(1/2)/x^4/(e*x+d),x, algorithm="giac")

[Out]

2*(c*d^2*e^2 + a*e^4)*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^4) - (c*d^2*e + 2*a*e^3)*arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))/sqrt(-a))/(sqrt(-a)*d^4) - 1/3*(3*(
sqrt(c)*x - sqrt(c*x^2 + a))^5*c*d*e - 6*(sqrt(c)*x - sqrt(c*x^2 + a))^4*c^(3/2)*d^2 - 6*(sqrt(c)*x - sqrt(c*x
^2 + a))^4*a*sqrt(c)*e^2 + 12*(sqrt(c)*x - sqrt(c*x^2 + a))^2*a^2*sqrt(c)*e^2 - 3*(sqrt(c)*x - sqrt(c*x^2 + a)
)*a^2*c*d*e - 2*a^2*c^(3/2)*d^2 - 6*a^3*sqrt(c)*e^2)/(((sqrt(c)*x - sqrt(c*x^2 + a))^2 - a)^3*d^3)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+c x^2}}{x^4 (d+e x)} \, dx=\int \frac {\sqrt {c\,x^2+a}}{x^4\,\left (d+e\,x\right )} \,d x \]

[In]

int((a + c*x^2)^(1/2)/(x^4*(d + e*x)),x)

[Out]

int((a + c*x^2)^(1/2)/(x^4*(d + e*x)), x)