Integrand size = 24, antiderivative size = 209 \[ \int \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x \, dx=-\frac {5 b d^2 \left (a+b \sqrt {\frac {d}{x}}+\frac {c}{x}\right )^{3/2}}{12 a^2 \left (\frac {d}{x}\right )^{3/2}}-\frac {\left (4 a c-5 b^2 d\right ) \left (2 a+b \sqrt {\frac {d}{x}}\right ) \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x}{32 a^3}+\frac {\left (a+b \sqrt {\frac {d}{x}}+\frac {c}{x}\right )^{3/2} x^2}{2 a}-\frac {\left (4 a c-5 b^2 d\right ) \left (4 a c-b^2 d\right ) \text {arctanh}\left (\frac {2 a+b \sqrt {\frac {d}{x}}}{2 \sqrt {a} \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}\right )}{64 a^{7/2}} \] Output:
-5/12*b*d^2*(a+b*(d/x)^(1/2)+c/x)^(3/2)/a^2/(d/x)^(3/2)-1/32*(-5*b^2*d+4*a *c)*(2*a+b*(d/x)^(1/2))*(a+b*(d/x)^(1/2)+c/x)^(1/2)*x/a^3+1/2*(a+b*(d/x)^( 1/2)+c/x)^(3/2)*x^2/a-1/64*(-5*b^2*d+4*a*c)*(-b^2*d+4*a*c)*arctanh(1/2*(2* a+b*(d/x)^(1/2))/a^(1/2)/(a+b*(d/x)^(1/2)+c/x)^(1/2))/a^(7/2)
Time = 2.05 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.04 \[ \int \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x \, dx=\frac {\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} \left (\sqrt {a} x \left (-2 a b \left (5 b d+26 c \sqrt {\frac {d}{x}}\right )+15 b^3 d \sqrt {\frac {d}{x}}+48 a^3 x+8 a^2 \left (3 c+b \sqrt {\frac {d}{x}} x\right )\right )+\frac {3 \sqrt {d} \left (16 a^2 c^2-24 a b^2 c d+5 b^4 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {\frac {d}{x}}-\sqrt {\frac {d \left (c+a x+b \sqrt {\frac {d}{x}} x\right )}{x}}}{\sqrt {a} \sqrt {d}}\right )}{\sqrt {\frac {d \left (c+\left (a+b \sqrt {\frac {d}{x}}\right ) x\right )}{x}}}\right )}{96 a^{7/2}} \] Input:
Integrate[Sqrt[a + b*Sqrt[d/x] + c/x]*x,x]
Output:
(Sqrt[a + b*Sqrt[d/x] + c/x]*(Sqrt[a]*x*(-2*a*b*(5*b*d + 26*c*Sqrt[d/x]) + 15*b^3*d*Sqrt[d/x] + 48*a^3*x + 8*a^2*(3*c + b*Sqrt[d/x]*x)) + (3*Sqrt[d] *(16*a^2*c^2 - 24*a*b^2*c*d + 5*b^4*d^2)*ArcTanh[(Sqrt[c]*Sqrt[d/x] - Sqrt [(d*(c + a*x + b*Sqrt[d/x]*x))/x])/(Sqrt[a]*Sqrt[d])])/Sqrt[(d*(c + (a + b *Sqrt[d/x])*x))/x]))/(96*a^(7/2))
Time = 0.43 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.09, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2066, 1693, 1167, 27, 1228, 1152, 1154, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} \, dx\) |
\(\Big \downarrow \) 2066 |
\(\displaystyle -d^2 \int \frac {\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x^3}{d^3}d\frac {d}{x}\) |
\(\Big \downarrow \) 1693 |
\(\displaystyle -2 d^2 \int \frac {\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}} x^5}{d^5}d\sqrt {\frac {d}{x}}\) |
\(\Big \downarrow \) 1167 |
\(\displaystyle -2 d^2 \left (-\frac {\int \frac {\left (2 \sqrt {\frac {d}{x}} c+5 b d\right ) \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}} x^4}{2 d^5}d\sqrt {\frac {d}{x}}}{4 a}-\frac {x^4 \left (a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}\right )^{3/2}}{4 a d^4}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -2 d^2 \left (-\frac {\int \frac {\left (2 \sqrt {\frac {d}{x}} c+5 b d\right ) \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}} x^4}{d^4}d\sqrt {\frac {d}{x}}}{8 a d}-\frac {x^4 \left (a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}\right )^{3/2}}{4 a d^4}\right )\) |
\(\Big \downarrow \) 1228 |
\(\displaystyle -2 d^2 \left (-\frac {\frac {\left (4 a c-5 b^2 d\right ) \int \frac {\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}} x^3}{d^3}d\sqrt {\frac {d}{x}}}{2 a}-\frac {5 b x^3 \left (a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}\right )^{3/2}}{3 a d^2}}{8 a d}-\frac {x^4 \left (a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}\right )^{3/2}}{4 a d^4}\right )\) |
\(\Big \downarrow \) 1152 |
\(\displaystyle -2 d^2 \left (-\frac {\frac {\left (4 a c-5 b^2 d\right ) \left (-\frac {\left (b^2-\frac {4 a c}{d}\right ) \int \frac {x}{d \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}}}d\sqrt {\frac {d}{x}}}{8 a}-\frac {x^2 \left (2 a+b \sqrt {\frac {d}{x}}\right ) \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}}}{4 a d^2}\right )}{2 a}-\frac {5 b x^3 \left (a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}\right )^{3/2}}{3 a d^2}}{8 a d}-\frac {x^4 \left (a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}\right )^{3/2}}{4 a d^4}\right )\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle -2 d^2 \left (-\frac {\frac {\left (4 a c-5 b^2 d\right ) \left (\frac {\left (b^2-\frac {4 a c}{d}\right ) \int \frac {1}{4 a-\frac {d^2}{x^2}}d\frac {2 a+b \sqrt {\frac {d}{x}}}{\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}}}}{4 a}-\frac {x^2 \left (2 a+b \sqrt {\frac {d}{x}}\right ) \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}}}{4 a d^2}\right )}{2 a}-\frac {5 b x^3 \left (a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}\right )^{3/2}}{3 a d^2}}{8 a d}-\frac {x^4 \left (a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}\right )^{3/2}}{4 a d^4}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -2 d^2 \left (-\frac {\frac {\left (4 a c-5 b^2 d\right ) \left (\frac {\left (b^2-\frac {4 a c}{d}\right ) \text {arctanh}\left (\frac {2 a+b \sqrt {\frac {d}{x}}}{2 \sqrt {a} \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}}}\right )}{8 a^{3/2}}-\frac {x^2 \left (2 a+b \sqrt {\frac {d}{x}}\right ) \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}}}{4 a d^2}\right )}{2 a}-\frac {5 b x^3 \left (a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}\right )^{3/2}}{3 a d^2}}{8 a d}-\frac {x^4 \left (a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}\right )^{3/2}}{4 a d^4}\right )\) |
Input:
Int[Sqrt[a + b*Sqrt[d/x] + c/x]*x,x]
Output:
-2*d^2*(-1/4*((a + b*Sqrt[d/x] + (c*d)/x^2)^(3/2)*x^4)/(a*d^4) - ((-5*b*(a + b*Sqrt[d/x] + (c*d)/x^2)^(3/2)*x^3)/(3*a*d^2) + ((4*a*c - 5*b^2*d)*(-1/ 4*((2*a + b*Sqrt[d/x])*Sqrt[a + b*Sqrt[d/x] + (c*d)/x^2]*x^2)/(a*d^2) + (( b^2 - (4*a*c)/d)*ArcTanh[(2*a + b*Sqrt[d/x])/(2*Sqrt[a]*Sqrt[a + b*Sqrt[d/ x] + (c*d)/x^2])])/(8*a^(3/2))))/(2*a))/(8*a*d))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> 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] + Simp[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] && EqQ[m + 2*p + 2, 0] && GtQ[p, 0]
Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym bol] :> Simp[-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]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> 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] + Simp[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[m , -1] && ((LtQ[m, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]) || (SumSimp lerQ[m, 1] && IntegerQ[p]) || ILtQ[Simplify[m + 2*p + 3], 0])
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(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] - Simp[(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 ] && EqQ[Simplify[m + 2*p + 3], 0]
Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol ] :> Simp[1/n Subst[Int[x^(Simplify[(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] && IntegerQ [Simplify[(m + 1)/n]]
Int[(x_)^(m_.)*((a_) + (b_.)*((d_.)/(x_))^(n_) + (c_.)*(x_)^(n2_.))^(p_), x _Symbol] :> Simp[-d^(m + 1) Subst[Int[(a + b*x^n + (c/d^(2*n))*x^(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]
Leaf count of result is larger than twice the leaf count of optimal. \(397\) vs. \(2(173)=346\).
Time = 0.09 (sec) , antiderivative size = 398, normalized size of antiderivative = 1.90
method | result | size |
default | \(\frac {\sqrt {\frac {b \sqrt {\frac {d}{x}}\, x +a x +c}{x}}\, \sqrt {x}\, \left (30 a^{\frac {3}{2}} \sqrt {b \sqrt {\frac {d}{x}}\, x +a x +c}\, \left (\frac {d}{x}\right )^{\frac {3}{2}} x^{\frac {3}{2}} b^{3}-15 d^{2} \ln \left (\frac {b \sqrt {\frac {d}{x}}\, \sqrt {x}+2 \sqrt {b \sqrt {\frac {d}{x}}\, x +a x +c}\, \sqrt {a}+2 a \sqrt {x}}{2 \sqrt {a}}\right ) a \,b^{4}+60 d \,a^{\frac {5}{2}} \sqrt {b \sqrt {\frac {d}{x}}\, x +a x +c}\, \sqrt {x}\, b^{2}+72 d \ln \left (\frac {b \sqrt {\frac {d}{x}}\, \sqrt {x}+2 \sqrt {b \sqrt {\frac {d}{x}}\, x +a x +c}\, \sqrt {a}+2 a \sqrt {x}}{2 \sqrt {a}}\right ) a^{2} b^{2} c -80 a^{\frac {5}{2}} \left (b \sqrt {\frac {d}{x}}\, x +a x +c \right )^{\frac {3}{2}} \sqrt {\frac {d}{x}}\, \sqrt {x}\, b -24 a^{\frac {5}{2}} \sqrt {b \sqrt {\frac {d}{x}}\, x +a x +c}\, \sqrt {\frac {d}{x}}\, \sqrt {x}\, b c +96 \sqrt {x}\, \left (b \sqrt {\frac {d}{x}}\, x +a x +c \right )^{\frac {3}{2}} a^{\frac {7}{2}}-48 a^{\frac {7}{2}} \sqrt {b \sqrt {\frac {d}{x}}\, x +a x +c}\, c \sqrt {x}-48 \ln \left (\frac {b \sqrt {\frac {d}{x}}\, \sqrt {x}+2 \sqrt {b \sqrt {\frac {d}{x}}\, x +a x +c}\, \sqrt {a}+2 a \sqrt {x}}{2 \sqrt {a}}\right ) a^{3} c^{2}\right )}{192 \sqrt {b \sqrt {\frac {d}{x}}\, x +a x +c}\, a^{\frac {9}{2}}}\) | \(398\) |
Input:
int((a+b*(d/x)^(1/2)+c/x)^(1/2)*x,x,method=_RETURNVERBOSE)
Output:
1/192*((b*(d/x)^(1/2)*x+a*x+c)/x)^(1/2)*x^(1/2)*(30*a^(3/2)*(b*(d/x)^(1/2) *x+a*x+c)^(1/2)*(d/x)^(3/2)*x^(3/2)*b^3-15*d^2*ln(1/2*(b*(d/x)^(1/2)*x^(1/ 2)+2*(b*(d/x)^(1/2)*x+a*x+c)^(1/2)*a^(1/2)+2*a*x^(1/2))/a^(1/2))*a*b^4+60* d*a^(5/2)*(b*(d/x)^(1/2)*x+a*x+c)^(1/2)*x^(1/2)*b^2+72*d*ln(1/2*(b*(d/x)^( 1/2)*x^(1/2)+2*(b*(d/x)^(1/2)*x+a*x+c)^(1/2)*a^(1/2)+2*a*x^(1/2))/a^(1/2)) *a^2*b^2*c-80*a^(5/2)*(b*(d/x)^(1/2)*x+a*x+c)^(3/2)*(d/x)^(1/2)*x^(1/2)*b- 24*a^(5/2)*(b*(d/x)^(1/2)*x+a*x+c)^(1/2)*(d/x)^(1/2)*x^(1/2)*b*c+96*x^(1/2 )*(b*(d/x)^(1/2)*x+a*x+c)^(3/2)*a^(7/2)-48*a^(7/2)*(b*(d/x)^(1/2)*x+a*x+c) ^(1/2)*c*x^(1/2)-48*ln(1/2*(b*(d/x)^(1/2)*x^(1/2)+2*(b*(d/x)^(1/2)*x+a*x+c )^(1/2)*a^(1/2)+2*a*x^(1/2))/a^(1/2))*a^3*c^2)/(b*(d/x)^(1/2)*x+a*x+c)^(1/ 2)/a^(9/2)
Timed out. \[ \int \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x \, dx=\text {Timed out} \] Input:
integrate((a+b*(d/x)^(1/2)+c/x)^(1/2)*x,x, algorithm="fricas")
Output:
Timed out
\[ \int \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x \, dx=\int x \sqrt {a + b \sqrt {\frac {d}{x}} + \frac {c}{x}}\, dx \] Input:
integrate((a+b*(d/x)**(1/2)+c/x)**(1/2)*x,x)
Output:
Integral(x*sqrt(a + b*sqrt(d/x) + c/x), x)
\[ \int \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x \, dx=\int { \sqrt {b \sqrt {\frac {d}{x}} + a + \frac {c}{x}} x \,d x } \] Input:
integrate((a+b*(d/x)^(1/2)+c/x)^(1/2)*x,x, algorithm="maxima")
Output:
integrate(sqrt(b*sqrt(d/x) + a + c/x)*x, x)
Timed out. \[ \int \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x \, dx=\text {Timed out} \] Input:
integrate((a+b*(d/x)^(1/2)+c/x)^(1/2)*x,x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x \, dx=\int x\,\sqrt {a+\frac {c}{x}+b\,\sqrt {\frac {d}{x}}} \,d x \] Input:
int(x*(a + c/x + b*(d/x)^(1/2))^(1/2),x)
Output:
int(x*(a + c/x + b*(d/x)^(1/2))^(1/2), x)
\[ \int \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x \, dx=\int \sqrt {a +b \sqrt {\frac {d}{x}}+\frac {c}{x}}\, x d x \] Input:
int((a+b*(d/x)^(1/2)+c/x)^(1/2)*x,x)
Output:
int((a+b*(d/x)^(1/2)+c/x)^(1/2)*x,x)