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

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

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Rubi [A]  time = 0.23, antiderivative size = 191, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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} \end {gather*}

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 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 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 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 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 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 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 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 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 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 {\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*}

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Mathematica [A]  time = 1.02, size = 301, normalized size = 1.58 \begin {gather*} -\frac {\frac {2 d^3 \left (a+c x^2\right )^{3/2}}{a x^3}+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 {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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(-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*Sqr
t[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)/Sqr
t[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*(Sqr
t[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]*Arc
Tanh[Sqrt[1 + (c*x^2)/a]]))/(x^2*Sqrt[a + c*x^2]))/d^4

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IntegrateAlgebraic [A]  time = 0.86, size = 214, normalized size = 1.12 \begin {gather*} \frac {\left (-2 a e^3-c d^2 e\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x-\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a} d^4}+\frac {2 e^2 \sqrt {-a e^2-c d^2} \tan ^{-1}\left (-\frac {e \sqrt {a+c x^2}}{\sqrt {-a e^2-c d^2}}+\frac {\sqrt {c} e x}{\sqrt {-a e^2-c d^2}}+\frac {\sqrt {c} d}{\sqrt {-a e^2-c d^2}}\right )}{d^4}+\frac {\sqrt {a+c x^2} \left (-2 a d^2+3 a d e x-6 a e^2 x^2-2 c d^2 x^2\right )}{6 a d^3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.51, size = 824, normalized size = 4.31 \begin {gather*} \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 ] \end {gather*}

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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giac [A]  time = 0.21, size = 309, normalized size = 1.62 \begin {gather*} \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 {gather*}

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)

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maple [B]  time = 0.02, size = 600, normalized size = 3.14 \begin {gather*} -\frac {a \,e^{3} \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^{4}}-\frac {c e \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^{2}}+\frac {\sqrt {a}\, e^{3} \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )}{d^{4}}+\frac {c e \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )}{2 \sqrt {a}\, d^{2}}+\frac {\sqrt {c}\, e^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{d^{3}}-\frac {\sqrt {c}\, e^{2} \ln \left (\frac {-\frac {c d}{e}+\left (x +\frac {d}{e}\right ) c}{\sqrt {c}}+\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}}}\right )}{d^{3}}+\frac {\sqrt {c \,x^{2}+a}\, c \,e^{2} x}{a \,d^{3}}-\frac {\sqrt {c \,x^{2}+a}\, c e}{2 a \,d^{2}}-\frac {\sqrt {c \,x^{2}+a}\, e^{3}}{d^{4}}+\frac {\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}}}\, e^{3}}{d^{4}}-\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} e^{2}}{a \,d^{3} x}+\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} e}{2 a \,d^{2} x^{2}}-\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}}}{3 a d \,x^{3}} \end {gather*}

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]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {c x^{2} + a}}{{\left (e x + d\right )} x^{4}}\,{d x} \end {gather*}

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {c\,x^2+a}}{x^4\,\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^4*(d + e*x)),x)

[Out]

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

[Out]

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

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