3.3.36 \(\int \frac {\sqrt {a+c x^2}}{x^5 (d+e x)} \, dx\)

Optimal. Leaf size=274 \[ \frac {c^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{8 a^{3/2} d}-\frac {\sqrt {a} e^4 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{d^5}+\frac {e^3 \sqrt {a+c x^2}}{d^4 x}-\frac {e^2 \sqrt {a+c x^2}}{2 d^3 x^2}-\frac {c e^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 \sqrt {a} d^3}+\frac {e \left (a+c x^2\right )^{3/2}}{3 a d^2 x^3}+\frac {e^3 \sqrt {a e^2+c d^2} \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{d^5}-\frac {c \sqrt {a+c x^2}}{8 a d x^2}-\frac {\sqrt {a+c x^2}}{4 d x^4} \]

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Rubi [A]  time = 0.30, antiderivative size = 274, normalized size of antiderivative = 1.00, number of steps used = 25, number of rules used = 14, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {961, 266, 47, 51, 63, 208, 264, 277, 217, 206, 50, 735, 844, 725} \begin {gather*} \frac {c^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{8 a^{3/2} d}+\frac {e^3 \sqrt {a+c x^2}}{d^4 x}-\frac {e^2 \sqrt {a+c x^2}}{2 d^3 x^2}-\frac {\sqrt {a} e^4 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{d^5}+\frac {e^3 \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^5}-\frac {c e^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 \sqrt {a} d^3}+\frac {e \left (a+c x^2\right )^{3/2}}{3 a d^2 x^3}-\frac {c \sqrt {a+c x^2}}{8 a d x^2}-\frac {\sqrt {a+c x^2}}{4 d x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + c*x^2]/(x^5*(d + e*x)),x]

[Out]

-Sqrt[a + c*x^2]/(4*d*x^4) - (c*Sqrt[a + c*x^2])/(8*a*d*x^2) - (e^2*Sqrt[a + c*x^2])/(2*d^3*x^2) + (e^3*Sqrt[a
+ c*x^2])/(d^4*x) + (e*(a + c*x^2)^(3/2))/(3*a*d^2*x^3) + (e^3*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^5 + (c^2*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/(8*a^(3/2)*d) - (c*e^2*ArcTan
h[Sqrt[a + c*x^2]/Sqrt[a]])/(2*Sqrt[a]*d^3) - (Sqrt[a]*e^4*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/d^5

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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^5 (d+e x)} \, dx &=\int \left (\frac {\sqrt {a+c x^2}}{d x^5}-\frac {e \sqrt {a+c x^2}}{d^2 x^4}+\frac {e^2 \sqrt {a+c x^2}}{d^3 x^3}-\frac {e^3 \sqrt {a+c x^2}}{d^4 x^2}+\frac {e^4 \sqrt {a+c x^2}}{d^5 x}-\frac {e^5 \sqrt {a+c x^2}}{d^5 (d+e x)}\right ) \, dx\\ &=\frac {\int \frac {\sqrt {a+c x^2}}{x^5} \, dx}{d}-\frac {e \int \frac {\sqrt {a+c x^2}}{x^4} \, dx}{d^2}+\frac {e^2 \int \frac {\sqrt {a+c x^2}}{x^3} \, dx}{d^3}-\frac {e^3 \int \frac {\sqrt {a+c x^2}}{x^2} \, dx}{d^4}+\frac {e^4 \int \frac {\sqrt {a+c x^2}}{x} \, dx}{d^5}-\frac {e^5 \int \frac {\sqrt {a+c x^2}}{d+e x} \, dx}{d^5}\\ &=-\frac {e^4 \sqrt {a+c x^2}}{d^5}+\frac {e^3 \sqrt {a+c x^2}}{d^4 x}+\frac {e \left (a+c x^2\right )^{3/2}}{3 a d^2 x^3}+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {a+c x}}{x^3} \, dx,x,x^2\right )}{2 d}+\frac {e^2 \operatorname {Subst}\left (\int \frac {\sqrt {a+c x}}{x^2} \, dx,x,x^2\right )}{2 d^3}-\frac {\left (c e^3\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{d^4}+\frac {e^4 \operatorname {Subst}\left (\int \frac {\sqrt {a+c x}}{x} \, dx,x,x^2\right )}{2 d^5}-\frac {e^4 \int \frac {a e-c d x}{(d+e x) \sqrt {a+c x^2}} \, dx}{d^5}\\ &=-\frac {\sqrt {a+c x^2}}{4 d x^4}-\frac {e^2 \sqrt {a+c x^2}}{2 d^3 x^2}+\frac {e^3 \sqrt {a+c x^2}}{d^4 x}+\frac {e \left (a+c x^2\right )^{3/2}}{3 a d^2 x^3}+\frac {c \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {a+c x}} \, dx,x,x^2\right )}{8 d}+\frac {\left (c e^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,x^2\right )}{4 d^3}+\frac {\left (c e^3\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{d^4}-\frac {\left (c e^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{d^4}+\frac {\left (a e^4\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,x^2\right )}{2 d^5}-\frac {\left (e^3 \left (c d^2+a e^2\right )\right ) \int \frac {1}{(d+e x) \sqrt {a+c x^2}} \, dx}{d^5}\\ &=-\frac {\sqrt {a+c x^2}}{4 d x^4}-\frac {c \sqrt {a+c x^2}}{8 a d x^2}-\frac {e^2 \sqrt {a+c x^2}}{2 d^3 x^2}+\frac {e^3 \sqrt {a+c x^2}}{d^4 x}+\frac {e \left (a+c x^2\right )^{3/2}}{3 a d^2 x^3}-\frac {\sqrt {c} e^3 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{d^4}-\frac {c^2 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,x^2\right )}{16 a d}+\frac {e^2 \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{c}+\frac {x^2}{c}} \, dx,x,\sqrt {a+c x^2}\right )}{2 d^3}+\frac {\left (c e^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{d^4}+\frac {\left (a e^4\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{c}+\frac {x^2}{c}} \, dx,x,\sqrt {a+c x^2}\right )}{c d^5}+\frac {\left (e^3 \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^5}\\ &=-\frac {\sqrt {a+c x^2}}{4 d x^4}-\frac {c \sqrt {a+c x^2}}{8 a d x^2}-\frac {e^2 \sqrt {a+c x^2}}{2 d^3 x^2}+\frac {e^3 \sqrt {a+c x^2}}{d^4 x}+\frac {e \left (a+c x^2\right )^{3/2}}{3 a d^2 x^3}+\frac {e^3 \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^5}-\frac {c e^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 \sqrt {a} d^3}-\frac {\sqrt {a} e^4 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{d^5}-\frac {c \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{c}+\frac {x^2}{c}} \, dx,x,\sqrt {a+c x^2}\right )}{8 a d}\\ &=-\frac {\sqrt {a+c x^2}}{4 d x^4}-\frac {c \sqrt {a+c x^2}}{8 a d x^2}-\frac {e^2 \sqrt {a+c x^2}}{2 d^3 x^2}+\frac {e^3 \sqrt {a+c x^2}}{d^4 x}+\frac {e \left (a+c x^2\right )^{3/2}}{3 a d^2 x^3}+\frac {e^3 \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^5}+\frac {c^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{8 a^{3/2} d}-\frac {c e^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 \sqrt {a} d^3}-\frac {\sqrt {a} e^4 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{d^5}\\ \end {align*}

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Mathematica [C]  time = 1.10, size = 344, normalized size = 1.26 \begin {gather*} \frac {-\frac {2 c^2 d^4 \left (a+c x^2\right )^{3/2} \, _2F_1\left (\frac {3}{2},3;\frac {5}{2};\frac {c x^2}{a}+1\right )}{a^3}+\frac {2 d^3 e \left (a+c x^2\right )^{3/2}}{a x^3}-\frac {3 d^2 e^2 \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}}+6 e^3 \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 {6 d e^3 \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^4 \sqrt {a+c x^2}+6 e^4 \left (\sqrt {a+c x^2}-\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )\right )}{6 d^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + c*x^2]/(x^5*(d + e*x)),x]

[Out]

(-6*e^4*Sqrt[a + c*x^2] + (2*d^3*e*(a + c*x^2)^(3/2))/(a*x^3) + (6*d*e^3*(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^3*(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^4*(Sqrt[a
+ c*x^2] - Sqrt[a]*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]]) - (3*d^2*e^2*(a + c*x^2 + c*x^2*Sqrt[1 + (c*x^2)/a]*ArcT
anh[Sqrt[1 + (c*x^2)/a]]))/(x^2*Sqrt[a + c*x^2]) - (2*c^2*d^4*(a + c*x^2)^(3/2)*Hypergeometric2F1[3/2, 3, 5/2,
1 + (c*x^2)/a])/a^3)/(6*d^5)

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IntegrateAlgebraic [A]  time = 1.13, size = 252, normalized size = 0.92 \begin {gather*} \frac {\left (8 a^2 e^4+4 a c d^2 e^2-c^2 d^4\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x-\sqrt {a+c x^2}}{\sqrt {a}}\right )}{4 a^{3/2} d^5}-\frac {2 e^3 \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^5}+\frac {\sqrt {a+c x^2} \left (-6 a d^3+8 a d^2 e x-12 a d e^2 x^2+24 a e^3 x^3-3 c d^3 x^2+8 c d^2 e x^3\right )}{24 a d^4 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[a + c*x^2]/(x^5*(d + e*x)),x]

[Out]

(Sqrt[a + c*x^2]*(-6*a*d^3 + 8*a*d^2*e*x - 3*c*d^3*x^2 - 12*a*d*e^2*x^2 + 8*c*d^2*e*x^3 + 24*a*e^3*x^3))/(24*a
*d^4*x^4) - (2*e^3*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])/Sqrt[-(c*d^2) - a*e^2]])/d^5 + ((-(c^2*d^4) + 4*a*c*d^2*e^2 + 8*a^2*e^4)*Ar
cTanh[(Sqrt[c]*x - Sqrt[a + c*x^2])/Sqrt[a]])/(4*a^(3/2)*d^5)

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fricas [A]  time = 0.58, size = 1007, normalized size = 3.68 \begin {gather*} \left [\frac {24 \, \sqrt {c d^{2} + a e^{2}} a^{2} e^{3} x^{4} \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^{2} d^{4} - 4 \, a c d^{2} e^{2} - 8 \, a^{2} e^{4}\right )} \sqrt {a} x^{4} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (8 \, a^{2} d^{3} e x - 6 \, a^{2} d^{4} + 8 \, {\left (a c d^{3} e + 3 \, a^{2} d e^{3}\right )} x^{3} - 3 \, {\left (a c d^{4} + 4 \, a^{2} d^{2} e^{2}\right )} x^{2}\right )} \sqrt {c x^{2} + a}}{48 \, a^{2} d^{5} x^{4}}, \frac {48 \, \sqrt {-c d^{2} - a e^{2}} a^{2} e^{3} x^{4} \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^{2} d^{4} - 4 \, a c d^{2} e^{2} - 8 \, a^{2} e^{4}\right )} \sqrt {a} x^{4} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (8 \, a^{2} d^{3} e x - 6 \, a^{2} d^{4} + 8 \, {\left (a c d^{3} e + 3 \, a^{2} d e^{3}\right )} x^{3} - 3 \, {\left (a c d^{4} + 4 \, a^{2} d^{2} e^{2}\right )} x^{2}\right )} \sqrt {c x^{2} + a}}{48 \, a^{2} d^{5} x^{4}}, \frac {12 \, \sqrt {c d^{2} + a e^{2}} a^{2} e^{3} x^{4} \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^{2} d^{4} - 4 \, a c d^{2} e^{2} - 8 \, a^{2} e^{4}\right )} \sqrt {-a} x^{4} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) + {\left (8 \, a^{2} d^{3} e x - 6 \, a^{2} d^{4} + 8 \, {\left (a c d^{3} e + 3 \, a^{2} d e^{3}\right )} x^{3} - 3 \, {\left (a c d^{4} + 4 \, a^{2} d^{2} e^{2}\right )} x^{2}\right )} \sqrt {c x^{2} + a}}{24 \, a^{2} d^{5} x^{4}}, \frac {24 \, \sqrt {-c d^{2} - a e^{2}} a^{2} e^{3} x^{4} \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^{2} d^{4} - 4 \, a c d^{2} e^{2} - 8 \, a^{2} e^{4}\right )} \sqrt {-a} x^{4} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) + {\left (8 \, a^{2} d^{3} e x - 6 \, a^{2} d^{4} + 8 \, {\left (a c d^{3} e + 3 \, a^{2} d e^{3}\right )} x^{3} - 3 \, {\left (a c d^{4} + 4 \, a^{2} d^{2} e^{2}\right )} x^{2}\right )} \sqrt {c x^{2} + a}}{24 \, a^{2} d^{5} x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+a)^(1/2)/x^5/(e*x+d),x, algorithm="fricas")

[Out]

[1/48*(24*sqrt(c*d^2 + a*e^2)*a^2*e^3*x^4*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^2*d^4 - 4*a*c*d^2*e^2
- 8*a^2*e^4)*sqrt(a)*x^4*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(8*a^2*d^3*e*x - 6*a^2*d^4 +
8*(a*c*d^3*e + 3*a^2*d*e^3)*x^3 - 3*(a*c*d^4 + 4*a^2*d^2*e^2)*x^2)*sqrt(c*x^2 + a))/(a^2*d^5*x^4), 1/48*(48*sq
rt(-c*d^2 - a*e^2)*a^2*e^3*x^4*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^2*d^4 - 4*a*c*d^2*e^2 - 8*a^2*e^4)*sqrt(a)*x^4*log(-(c*x^2 - 2*sqrt(c*x^2 + a
)*sqrt(a) + 2*a)/x^2) + 2*(8*a^2*d^3*e*x - 6*a^2*d^4 + 8*(a*c*d^3*e + 3*a^2*d*e^3)*x^3 - 3*(a*c*d^4 + 4*a^2*d^
2*e^2)*x^2)*sqrt(c*x^2 + a))/(a^2*d^5*x^4), 1/24*(12*sqrt(c*d^2 + a*e^2)*a^2*e^3*x^4*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^2*d^4 - 4*a*c*d^2*e^2 - 8*a^2*e^4)*sqrt(-a)*x^4*arctan(sqrt(-a)/sqrt(c*x^2 + a)) + (8*a^2
*d^3*e*x - 6*a^2*d^4 + 8*(a*c*d^3*e + 3*a^2*d*e^3)*x^3 - 3*(a*c*d^4 + 4*a^2*d^2*e^2)*x^2)*sqrt(c*x^2 + a))/(a^
2*d^5*x^4), 1/24*(24*sqrt(-c*d^2 - a*e^2)*a^2*e^3*x^4*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^2*d^4 - 4*a*c*d^2*e^2 - 8*a^2*e^4)*sqrt(-a)*x^4*arctan
(sqrt(-a)/sqrt(c*x^2 + a)) + (8*a^2*d^3*e*x - 6*a^2*d^4 + 8*(a*c*d^3*e + 3*a^2*d*e^3)*x^3 - 3*(a*c*d^4 + 4*a^2
*d^2*e^2)*x^2)*sqrt(c*x^2 + a))/(a^2*d^5*x^4)]

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giac [B]  time = 0.26, size = 596, normalized size = 2.18 \begin {gather*} -\frac {2 \, {\left (c d^{2} e^{3} + a e^{5}\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^{5}} - \frac {{\left (c^{2} d^{4} - 4 \, a c d^{2} e^{2} - 8 \, a^{2} e^{4}\right )} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a d^{5}} + \frac {3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{7} c^{2} d^{3} - 24 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{6} a c^{\frac {3}{2}} d^{2} e + 21 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{5} a c^{2} d^{3} + 12 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{7} a c d e^{2} + 24 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{4} a^{2} c^{\frac {3}{2}} d^{2} e + 21 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{3} a^{2} c^{2} d^{3} - 12 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{5} a^{2} c d e^{2} - 24 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{6} a^{2} \sqrt {c} e^{3} - 8 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} a^{3} c^{\frac {3}{2}} d^{2} e + 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} a^{3} c^{2} d^{3} - 12 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{3} a^{3} c d e^{2} + 72 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{4} a^{3} \sqrt {c} e^{3} + 8 \, a^{4} c^{\frac {3}{2}} d^{2} e + 12 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} a^{4} c d e^{2} - 72 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} a^{4} \sqrt {c} e^{3} + 24 \, a^{5} \sqrt {c} e^{3}}{12 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} - a\right )}^{4} a d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+a)^(1/2)/x^5/(e*x+d),x, algorithm="giac")

[Out]

-2*(c*d^2*e^3 + a*e^5)*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^5) - 1/4*(c^2*d^4 - 4*a*c*d^2*e^2 - 8*a^2*e^4)*arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))/sqrt(-a))/(s
qrt(-a)*a*d^5) + 1/12*(3*(sqrt(c)*x - sqrt(c*x^2 + a))^7*c^2*d^3 - 24*(sqrt(c)*x - sqrt(c*x^2 + a))^6*a*c^(3/2
)*d^2*e + 21*(sqrt(c)*x - sqrt(c*x^2 + a))^5*a*c^2*d^3 + 12*(sqrt(c)*x - sqrt(c*x^2 + a))^7*a*c*d*e^2 + 24*(sq
rt(c)*x - sqrt(c*x^2 + a))^4*a^2*c^(3/2)*d^2*e + 21*(sqrt(c)*x - sqrt(c*x^2 + a))^3*a^2*c^2*d^3 - 12*(sqrt(c)*
x - sqrt(c*x^2 + a))^5*a^2*c*d*e^2 - 24*(sqrt(c)*x - sqrt(c*x^2 + a))^6*a^2*sqrt(c)*e^3 - 8*(sqrt(c)*x - sqrt(
c*x^2 + a))^2*a^3*c^(3/2)*d^2*e + 3*(sqrt(c)*x - sqrt(c*x^2 + a))*a^3*c^2*d^3 - 12*(sqrt(c)*x - sqrt(c*x^2 + a
))^3*a^3*c*d*e^2 + 72*(sqrt(c)*x - sqrt(c*x^2 + a))^4*a^3*sqrt(c)*e^3 + 8*a^4*c^(3/2)*d^2*e + 12*(sqrt(c)*x -
sqrt(c*x^2 + a))*a^4*c*d*e^2 - 72*(sqrt(c)*x - sqrt(c*x^2 + a))^2*a^4*sqrt(c)*e^3 + 24*a^5*sqrt(c)*e^3)/(((sqr
t(c)*x - sqrt(c*x^2 + a))^2 - a)^4*a*d^4)

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maple [B]  time = 0.02, size = 703, normalized size = 2.57 \begin {gather*} \frac {a \,e^{4} \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^{5}}+\frac {c \,e^{2} \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^{3}}-\frac {\sqrt {a}\, e^{4} \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )}{d^{5}}-\frac {c \,e^{2} \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )}{2 \sqrt {a}\, d^{3}}+\frac {c^{2} \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )}{8 a^{\frac {3}{2}} d}-\frac {\sqrt {c}\, e^{3} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{d^{4}}+\frac {\sqrt {c}\, e^{3} \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^{4}}-\frac {\sqrt {c \,x^{2}+a}\, c \,e^{3} x}{a \,d^{4}}+\frac {\sqrt {c \,x^{2}+a}\, c \,e^{2}}{2 a \,d^{3}}-\frac {\sqrt {c \,x^{2}+a}\, c^{2}}{8 a^{2} d}+\frac {\sqrt {c \,x^{2}+a}\, e^{4}}{d^{5}}-\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^{4}}{d^{5}}+\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} e^{3}}{a \,d^{4} x}-\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} e^{2}}{2 a \,d^{3} x^{2}}+\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} c}{8 a^{2} d \,x^{2}}+\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} e}{3 a \,d^{2} x^{3}}-\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}}}{4 a d \,x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+a)^(1/2)/x^5/(e*x+d),x)

[Out]

1/d^4*e^3/a/x*(c*x^2+a)^(3/2)-1/d^4*e^3*c/a*x*(c*x^2+a)^(1/2)-1/d^4*e^3*c^(1/2)*ln(c^(1/2)*x+(c*x^2+a)^(1/2))-
1/2/d^3*e^2/a/x^2*(c*x^2+a)^(3/2)-1/2/d^3*e^2*c/a^(1/2)*ln((2*a+2*(c*x^2+a)^(1/2)*a^(1/2))/x)+1/2/d^3*e^2*c/a*
(c*x^2+a)^(1/2)+1/3*e*(c*x^2+a)^(3/2)/a/d^2/x^3-1/d^5*e^4*a^(1/2)*ln((2*a+2*(c*x^2+a)^(1/2)*a^(1/2))/x)+1/d^5*
e^4*(c*x^2+a)^(1/2)-1/d^5*e^4*(-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*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^5*e^4/((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^3*e^2/((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-1/4/d/a/x^4*(c*x^
2+a)^(3/2)+1/8/d*c/a^2/x^2*(c*x^2+a)^(3/2)+1/8/d*c^2/a^(3/2)*ln((2*a+2*(c*x^2+a)^(1/2)*a^(1/2))/x)-1/8/d*c^2/a
^2*(c*x^2+a)^(1/2)

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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^{5}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+a)^(1/2)/x^5/(e*x+d),x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^2 + a)/((e*x + d)*x^5), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sqrt {c\,x^2+a}}{x^5\,\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^5*(d + e*x)),x)

[Out]

int((a + c*x^2)^(1/2)/(x^5*(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^{5} \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**5/(e*x+d),x)

[Out]

Integral(sqrt(a + c*x**2)/(x**5*(d + e*x)), x)

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