3.4.22 \(\int \frac {\sqrt {a+c x^2}}{x^2 (d+e x)} \, dx\) [322]

Optimal. Leaf size=105 \[ -\frac {\sqrt {a+c x^2}}{d x}-\frac {\sqrt {c d^2+a e^2} \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{d^2}+\frac {\sqrt {a} e \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{d^2} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 11, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {975, 283, 223, 212, 272, 52, 65, 214, 749, 858, 739} \begin {gather*} -\frac {\sqrt {a e^2+c d^2} \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{d^2}+\frac {\sqrt {a} e \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{d^2}-\frac {\sqrt {a+c x^2}}{d x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]
time = 0.26, size = 124, normalized size = 1.18 \begin {gather*} -\frac {d \sqrt {a+c x^2}-2 \sqrt {-c d^2-a e^2} x \tan ^{-1}\left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+c x^2}}{\sqrt {-c d^2-a e^2}}\right )+2 \sqrt {a} e x \tanh ^{-1}\left (\frac {\sqrt {c} x-\sqrt {a+c x^2}}{\sqrt {a}}\right )}{d^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(369\) vs. \(2(91)=182\).
time = 0.10, size = 370, normalized size = 3.52

method	result	size
risch	\(-\frac {\sqrt {c \,x^{2}+a}}{d x}-\frac {e \ln \left (\frac {\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right ) a}{d^{2} \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}-\frac {\ln \left (\frac {\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right ) c}{e \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}+\frac {\sqrt {a}\, e \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )}{d^{2}}\)	\(305\)
default	\(\frac {e \left (\sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}-\frac {\sqrt {c}\, d \ln \left (\frac {-\frac {c d}{e}+c \left (x +\frac {d}{e}\right )}{\sqrt {c}}+\sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\right )}{e}-\frac {\left (a \,e^{2}+c \,d^{2}\right ) \ln \left (\frac {\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{2} \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}\right )}{d^{2}}+\frac {-\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}}}{a x}+\frac {2 c \left (\frac {x \sqrt {c \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}\right )}{a}}{d}-\frac {e \left (\sqrt {c \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )\right )}{d^{2}}\)	\(370\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.62, size = 596, normalized size = 5.68 \begin {gather*} \left [\frac {\sqrt {a} x e \log \left (-\frac {c x^{2} + 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + \sqrt {c d^{2} + a e^{2}} x \log \left (-\frac {2 \, c^{2} d^{2} x^{2} - 2 \, a c d x e + a c d^{2} + 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a} + {\left (a c x^{2} + 2 \, a^{2}\right )} e^{2}}{x^{2} e^{2} + 2 \, d x e + d^{2}}\right ) - 2 \, \sqrt {c x^{2} + a} d}{2 \, d^{2} x}, \frac {\sqrt {a} x e \log \left (-\frac {c x^{2} + 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, \sqrt {-c d^{2} - a e^{2}} x \arctan \left (-\frac {\sqrt {-c d^{2} - a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{c^{2} d^{2} x^{2} + a c d^{2} + {\left (a c x^{2} + a^{2}\right )} e^{2}}\right ) - 2 \, \sqrt {c x^{2} + a} d}{2 \, d^{2} x}, -\frac {2 \, \sqrt {-a} x \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) e - \sqrt {c d^{2} + a e^{2}} x \log \left (-\frac {2 \, c^{2} d^{2} x^{2} - 2 \, a c d x e + a c d^{2} + 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a} + {\left (a c x^{2} + 2 \, a^{2}\right )} e^{2}}{x^{2} e^{2} + 2 \, d x e + d^{2}}\right ) + 2 \, \sqrt {c x^{2} + a} d}{2 \, d^{2} x}, -\frac {\sqrt {-a} x \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) e - \sqrt {-c d^{2} - a e^{2}} x \arctan \left (-\frac {\sqrt {-c d^{2} - a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{c^{2} d^{2} x^{2} + a c d^{2} + {\left (a c x^{2} + a^{2}\right )} e^{2}}\right ) + \sqrt {c x^{2} + a} d}{d^{2} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a + c x^{2}}}{x^{2} \left (d + e x\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.31, size = 145, normalized size = 1.38 \begin {gather*} -\frac {2 \, a \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right ) e}{\sqrt {-a} d^{2}} + \frac {2 \, a \sqrt {c}}{{\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} - a\right )} d} + \frac {2 \, {\left (c d^{2} + a e^{2}\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^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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