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

Optimal. Leaf size=191 \[ -\frac{e^2 \sqrt{a+c x^2}}{d^3 x}+\frac{\sqrt{a} e^3 \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{d^4}-\frac{e^2 \sqrt{a e^2+c d^2} \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{d^4}+\frac{e \sqrt{a+c x^2}}{2 d^2 x^2}+\frac{c e \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{2 \sqrt{a} d^2}-\frac{\left (a+c x^2\right )^{3/2}}{3 a d x^3} \]

[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

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Rubi [A]  time = 0.231933, antiderivative size = 191, normalized size of antiderivative = 1., number of steps used = 20, number of rules used = 13, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.591, Rules used = {961, 264, 266, 47, 63, 208, 277, 217, 206, 50, 735, 844, 725} \[ -\frac{e^2 \sqrt{a+c x^2}}{d^3 x}+\frac{\sqrt{a} e^3 \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{d^4}-\frac{e^2 \sqrt{a e^2+c d^2} \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{d^4}+\frac{e \sqrt{a+c x^2}}{2 d^2 x^2}+\frac{c e \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{2 \sqrt{a} d^2}-\frac{\left (a+c x^2\right )^{3/2}}{3 a d x^3} \]

Antiderivative was successfully verified.

[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 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])

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 47

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},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && 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 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 277

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 217

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 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 50

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 735

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 844

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 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]

Rubi steps

\begin{align*} \int \frac{\sqrt{a+c x^2}}{x^4 (d+e x)} \, dx &=\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 \operatorname{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 \operatorname{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) \operatorname{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 ) \operatorname{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 ) \operatorname{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 \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \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^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]  time = 1.17276, size = 301, normalized size = 1.58 \[ -\frac{6 e^2 \left (\sqrt{a e^2+c d^2} \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )+\sqrt{c} d \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )\right )-\frac{3 d^2 e \left (c x^2 \sqrt{\frac{c x^2}{a}+1} \tanh ^{-1}\left (\sqrt{\frac{c x^2}{a}+1}\right )+a+c x^2\right )}{x^2 \sqrt{a+c x^2}}+\frac{2 d^3 \left (a+c x^2\right )^{3/2}}{a x^3}+\frac{6 d e^2 \left (-\sqrt{a} \sqrt{c} x \sqrt{\frac{c x^2}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )+a+c x^2\right )}{x \sqrt{a+c x^2}}-6 e^3 \sqrt{a+c x^2}+6 e^3 \left (\sqrt{a+c x^2}-\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )\right )}{6 d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.272, size = 600, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c x^{2} + a}}{{\left (e x + d\right )} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 2.2252, size = 1791, normalized size = 9.38 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a + c x^{2}}}{x^{4} \left (d + e x\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]  time = 1.21167, size = 417, normalized size = 2.18 \begin{align*} \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} - 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} + 12 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{2} a^{2} \sqrt{c} e^{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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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 - 3*(sqrt(c)*x - sqrt(c*x^2 + a))*a^2*c*d*e - 2*a^2*c^(3/2)*d^2 + 12*(sqrt(c)*x - sqr
t(c*x^2 + a))^2*a^2*sqrt(c)*e^2 - 6*a^3*sqrt(c)*e^2)/(((sqrt(c)*x - sqrt(c*x^2 + a))^2 - a)^3*d^3)