3.25.43 \(\int \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x \, dx\)

Optimal. Leaf size=209 \[ -\frac {\left (4 a c-5 b^2 d\right ) \left (4 a c-b^2 d\right ) \tanh ^{-1}\left (\frac {2 a+b \sqrt {\frac {d}{x}}}{2 \sqrt {a} \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}\right )}{64 a^{7/2}}-\frac {x \left (4 a c-5 b^2 d\right ) \left (2 a+b \sqrt {\frac {d}{x}}\right ) \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}{32 a^3}-\frac {5 b d^2 \left (a+b \sqrt {\frac {d}{x}}+\frac {c}{x}\right )^{3/2}}{12 a^2 \left (\frac {d}{x}\right )^{3/2}}+\frac {x^2 \left (a+b \sqrt {\frac {d}{x}}+\frac {c}{x}\right )^{3/2}}{2 a} \]

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Rubi [A]  time = 0.28, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1970, 1357, 744, 806, 720, 724, 206} \begin {gather*} -\frac {x \left (4 a c-5 b^2 d\right ) \left (2 a+b \sqrt {\frac {d}{x}}\right ) \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}{32 a^3}-\frac {\left (4 a c-5 b^2 d\right ) \left (4 a c-b^2 d\right ) \tanh ^{-1}\left (\frac {2 a+b \sqrt {\frac {d}{x}}}{2 \sqrt {a} \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}\right )}{64 a^{7/2}}-\frac {5 b d^2 \left (a+b \sqrt {\frac {d}{x}}+\frac {c}{x}\right )^{3/2}}{12 a^2 \left (\frac {d}{x}\right )^{3/2}}+\frac {x^2 \left (a+b \sqrt {\frac {d}{x}}+\frac {c}{x}\right )^{3/2}}{2 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Sqrt[d/x] + c/x]*x,x]

[Out]

(-5*b*d^2*(a + b*Sqrt[d/x] + c/x)^(3/2))/(12*a^2*(d/x)^(3/2)) - ((4*a*c - 5*b^2*d)*(2*a + b*Sqrt[d/x])*Sqrt[a
+ b*Sqrt[d/x] + c/x]*x)/(32*a^3) + ((a + b*Sqrt[d/x] + c/x)^(3/2)*x^2)/(2*a) - ((4*a*c - 5*b^2*d)*(4*a*c - b^2
*d)*ArcTanh[(2*a + b*Sqrt[d/x])/(2*Sqrt[a]*Sqrt[a + b*Sqrt[d/x] + c/x])])/(64*a^(7/2))

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 720

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*
(d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^p)/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[(p*(b^2 -
4*a*c))/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[
{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && EqQ[m +
2*p + 2, 0] && GtQ[p, 0]

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 744

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1)
*(a + b*x + c*x^2)^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 - b*d*e + a*e^2)),
Int[(d + e*x)^(m + 1)*Simp[c*d*(m + 1) - b*e*(m + p + 2) - c*e*(m + 2*p + 3)*x, x]*(a + b*x + c*x^2)^p, x], x]
/; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e
, 0] && NeQ[m, -1] && ((LtQ[m, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]) || (SumSimplerQ[m, 1] && IntegerQ
[p]) || ILtQ[Simplify[m + 2*p + 3], 0])

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2)), x] - Dist[(b
*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x],
x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[Sim
plify[m + 2*p + 3], 0]

Rule 1357

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && NeQ[
b^2 - 4*a*c, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1970

Int[(x_)^(m_.)*((a_) + (b_.)*((d_.)/(x_))^(n_) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> -Dist[d^(m + 1), Subst
[Int[(a + b*x^n + (c*x^(2*n))/d^(2*n))^p/x^(m + 2), x], x, d/x], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[n2,
-2*n] && IntegerQ[2*n] && IntegerQ[m]

Rubi steps

\begin {align*} \int \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x \, dx &=-\left (d^2 \operatorname {Subst}\left (\int \frac {\sqrt {a+b \sqrt {x}+\frac {c x}{d}}}{x^3} \, dx,x,\frac {d}{x}\right )\right )\\ &=-\left (\left (2 d^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x+\frac {c x^2}{d}}}{x^5} \, dx,x,\sqrt {\frac {d}{x}}\right )\right )\\ &=\frac {\left (a+b \sqrt {\frac {d}{x}}+\frac {c}{x}\right )^{3/2} x^2}{2 a}+\frac {d^2 \operatorname {Subst}\left (\int \frac {\left (\frac {5 b}{2}+\frac {c x}{d}\right ) \sqrt {a+b x+\frac {c x^2}{d}}}{x^4} \, dx,x,\sqrt {\frac {d}{x}}\right )}{2 a}\\ &=-\frac {5 b d^2 \left (a+b \sqrt {\frac {d}{x}}+\frac {c}{x}\right )^{3/2}}{12 a^2 \left (\frac {d}{x}\right )^{3/2}}+\frac {\left (a+b \sqrt {\frac {d}{x}}+\frac {c}{x}\right )^{3/2} x^2}{2 a}+\frac {\left (d \left (4 a c-5 b^2 d\right )\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x+\frac {c x^2}{d}}}{x^3} \, dx,x,\sqrt {\frac {d}{x}}\right )}{8 a^2}\\ &=-\frac {5 b d^2 \left (a+b \sqrt {\frac {d}{x}}+\frac {c}{x}\right )^{3/2}}{12 a^2 \left (\frac {d}{x}\right )^{3/2}}-\frac {\left (4 a c-5 b^2 d\right ) \left (2 a+b \sqrt {\frac {d}{x}}\right ) \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x}{32 a^3}+\frac {\left (a+b \sqrt {\frac {d}{x}}+\frac {c}{x}\right )^{3/2} x^2}{2 a}+\frac {\left (\left (4 a c-5 b^2 d\right ) \left (4 a c-b^2 d\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x+\frac {c x^2}{d}}} \, dx,x,\sqrt {\frac {d}{x}}\right )}{64 a^3}\\ &=-\frac {5 b d^2 \left (a+b \sqrt {\frac {d}{x}}+\frac {c}{x}\right )^{3/2}}{12 a^2 \left (\frac {d}{x}\right )^{3/2}}-\frac {\left (4 a c-5 b^2 d\right ) \left (2 a+b \sqrt {\frac {d}{x}}\right ) \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x}{32 a^3}+\frac {\left (a+b \sqrt {\frac {d}{x}}+\frac {c}{x}\right )^{3/2} x^2}{2 a}-\frac {\left (\left (4 a c-5 b^2 d\right ) \left (4 a c-b^2 d\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 a-x^2} \, dx,x,\frac {2 a+b \sqrt {\frac {d}{x}}}{\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}\right )}{32 a^3}\\ &=-\frac {5 b d^2 \left (a+b \sqrt {\frac {d}{x}}+\frac {c}{x}\right )^{3/2}}{12 a^2 \left (\frac {d}{x}\right )^{3/2}}-\frac {\left (4 a c-5 b^2 d\right ) \left (2 a+b \sqrt {\frac {d}{x}}\right ) \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x}{32 a^3}+\frac {\left (a+b \sqrt {\frac {d}{x}}+\frac {c}{x}\right )^{3/2} x^2}{2 a}-\frac {\left (4 a c-5 b^2 d\right ) \left (4 a c-b^2 d\right ) \tanh ^{-1}\left (\frac {2 a+b \sqrt {\frac {d}{x}}}{2 \sqrt {a} \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}\right )}{64 a^{7/2}}\\ \end {align*}

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Mathematica [F]  time = 0.33, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[Sqrt[a + b*Sqrt[d/x] + c/x]*x,x]

[Out]

Integrate[Sqrt[a + b*Sqrt[d/x] + c/x]*x, x]

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IntegrateAlgebraic [A]  time = 0.75, size = 204, normalized size = 0.98 \begin {gather*} \frac {\left (16 a^2 c^2-24 a b^2 c d+5 b^4 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {\frac {c}{d}} \sqrt {\frac {d}{x}}-\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}{\sqrt {a}}\right )}{32 a^{7/2}}+\frac {x^2 \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} \left (48 a^3 d^2+8 a^2 b d^2 \sqrt {\frac {d}{x}}+\frac {24 a^2 c d^2}{x}-\frac {10 a b^2 d^3}{x}-52 a b c d \left (\frac {d}{x}\right )^{3/2}+15 b^3 d^2 \left (\frac {d}{x}\right )^{3/2}\right )}{96 a^3 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[a + b*Sqrt[d/x] + c/x]*x,x]

[Out]

(Sqrt[a + b*Sqrt[d/x] + c/x]*(48*a^3*d^2 + 8*a^2*b*d^2*Sqrt[d/x] - 52*a*b*c*d*(d/x)^(3/2) + 15*b^3*d^2*(d/x)^(
3/2) + (24*a^2*c*d^2)/x - (10*a*b^2*d^3)/x)*x^2)/(96*a^3*d^2) + ((16*a^2*c^2 - 24*a*b^2*c*d + 5*b^4*d^2)*ArcTa
nh[(-Sqrt[a + b*Sqrt[d/x] + c/x] + Sqrt[c/d]*Sqrt[d/x])/Sqrt[a]])/(32*a^(7/2))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [B]  time = 0.16, size = 398, normalized size = 1.90 \begin {gather*} \frac {\sqrt {\frac {a x +\sqrt {\frac {d}{x}}\, b x +c}{x}}\, \left (-15 a \,b^{4} d^{2} \ln \left (\frac {2 a \sqrt {x}+\sqrt {\frac {d}{x}}\, b \sqrt {x}+2 \sqrt {a x +\sqrt {\frac {d}{x}}\, b x +c}\, \sqrt {a}}{2 \sqrt {a}}\right )+72 a^{2} b^{2} c d \ln \left (\frac {2 a \sqrt {x}+\sqrt {\frac {d}{x}}\, b \sqrt {x}+2 \sqrt {a x +\sqrt {\frac {d}{x}}\, b x +c}\, \sqrt {a}}{2 \sqrt {a}}\right )-48 a^{3} c^{2} \ln \left (\frac {2 a \sqrt {x}+\sqrt {\frac {d}{x}}\, b \sqrt {x}+2 \sqrt {a x +\sqrt {\frac {d}{x}}\, b x +c}\, \sqrt {a}}{2 \sqrt {a}}\right )+60 \sqrt {a x +\sqrt {\frac {d}{x}}\, b x +c}\, a^{\frac {5}{2}} b^{2} d \sqrt {x}+30 \sqrt {a x +\sqrt {\frac {d}{x}}\, b x +c}\, \left (\frac {d}{x}\right )^{\frac {3}{2}} a^{\frac {3}{2}} b^{3} x^{\frac {3}{2}}-48 \sqrt {a x +\sqrt {\frac {d}{x}}\, b x +c}\, a^{\frac {7}{2}} c \sqrt {x}-24 \sqrt {a x +\sqrt {\frac {d}{x}}\, b x +c}\, \sqrt {\frac {d}{x}}\, a^{\frac {5}{2}} b c \sqrt {x}+96 \left (a x +\sqrt {\frac {d}{x}}\, b x +c \right )^{\frac {3}{2}} a^{\frac {7}{2}} \sqrt {x}-80 \left (a x +\sqrt {\frac {d}{x}}\, b x +c \right )^{\frac {3}{2}} \sqrt {\frac {d}{x}}\, a^{\frac {5}{2}} b \sqrt {x}\right ) \sqrt {x}}{192 \sqrt {a x +\sqrt {\frac {d}{x}}\, b x +c}\, a^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*((a*x+(d/x)^(1/2)*b*x+c)/x)^(1/2)*x^(1/2)*(30*(a*x+(d/x)^(1/2)*b*x+c)^(1/2)*a^(3/2)*(d/x)^(3/2)*x^(3/2)*
b^3+60*(a*x+(d/x)^(1/2)*b*x+c)^(1/2)*a^(5/2)*d*x^(1/2)*b^2-15*ln(1/2*(2*a*x^(1/2)+(d/x)^(1/2)*b*x^(1/2)+2*(a*x
+(d/x)^(1/2)*b*x+c)^(1/2)*a^(1/2))/a^(1/2))*d^2*a*b^4+96*x^(1/2)*(a*x+(d/x)^(1/2)*b*x+c)^(3/2)*a^(7/2)-80*(a*x
+(d/x)^(1/2)*b*x+c)^(3/2)*a^(5/2)*(d/x)^(1/2)*x^(1/2)*b-48*(a*x+(d/x)^(1/2)*b*x+c)^(1/2)*a^(7/2)*x^(1/2)*c-24*
(a*x+(d/x)^(1/2)*b*x+c)^(1/2)*a^(5/2)*(d/x)^(1/2)*x^(1/2)*b*c+72*ln(1/2*(2*a*x^(1/2)+(d/x)^(1/2)*b*x^(1/2)+2*(
a*x+(d/x)^(1/2)*b*x+c)^(1/2)*a^(1/2))/a^(1/2))*d*a^2*b^2*c-48*ln(1/2*(2*a*x^(1/2)+(d/x)^(1/2)*b*x^(1/2)+2*(a*x
+(d/x)^(1/2)*b*x+c)^(1/2)*a^(1/2))/a^(1/2))*a^3*c^2)/(a*x+(d/x)^(1/2)*b*x+c)^(1/2)/a^(9/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {b \sqrt {\frac {d}{x}} + a + \frac {c}{x}} x\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*sqrt(d/x) + a + c/x)*x, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x\,\sqrt {a+\frac {c}{x}+b\,\sqrt {\frac {d}{x}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \sqrt {a + b \sqrt {\frac {d}{x}} + \frac {c}{x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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