Integrand size = 17, antiderivative size = 102 \[ \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{\sqrt {x}} \, dx=-\frac {b^2 \sqrt {a+\frac {b}{x}}}{2 x^{3/2}}-\frac {9 a b \sqrt {a+\frac {b}{x}}}{4 \sqrt {x}}+2 a^2 \sqrt {a+\frac {b}{x}} \sqrt {x}-\frac {15}{4} a^2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x}} \sqrt {x}}\right ) \] Output:
-1/2*b^2*(a+b/x)^(1/2)/x^(3/2)-9/4*a*b*(a+b/x)^(1/2)/x^(1/2)+2*a^2*(a+b/x) ^(1/2)*x^(1/2)-15/4*a^2*b^(1/2)*arctanh(b^(1/2)/(a+b/x)^(1/2)/x^(1/2))
Time = 5.26 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.88 \[ \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{\sqrt {x}} \, dx=\frac {\sqrt {a+\frac {b}{x}} \sqrt {x} \left (\frac {\sqrt {b+a x} \left (-2 b^2-9 a b x+8 a^2 x^2\right )}{4 x^2}-\frac {15}{4} a^2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b+a x}}{\sqrt {b}}\right )\right )}{\sqrt {b+a x}} \] Input:
Integrate[(a + b/x)^(5/2)/Sqrt[x],x]
Output:
(Sqrt[a + b/x]*Sqrt[x]*((Sqrt[b + a*x]*(-2*b^2 - 9*a*b*x + 8*a^2*x^2))/(4* x^2) - (15*a^2*Sqrt[b]*ArcTanh[Sqrt[b + a*x]/Sqrt[b]])/4))/Sqrt[b + a*x]
Time = 0.33 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {860, 247, 211, 211, 224, 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.
\(\displaystyle \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{\sqrt {x}} \, dx\) |
\(\Big \downarrow \) 860 |
\(\displaystyle -2 \int \left (a+\frac {b}{x}\right )^{5/2} xd\frac {1}{\sqrt {x}}\) |
\(\Big \downarrow \) 247 |
\(\displaystyle -2 \left (5 b \int \left (a+\frac {b}{x}\right )^{3/2}d\frac {1}{\sqrt {x}}-\sqrt {x} \left (a+\frac {b}{x}\right )^{5/2}\right )\) |
\(\Big \downarrow \) 211 |
\(\displaystyle -2 \left (5 b \left (\frac {3}{4} a \int \sqrt {a+\frac {b}{x}}d\frac {1}{\sqrt {x}}+\frac {\left (a+\frac {b}{x}\right )^{3/2}}{4 \sqrt {x}}\right )-\sqrt {x} \left (a+\frac {b}{x}\right )^{5/2}\right )\) |
\(\Big \downarrow \) 211 |
\(\displaystyle -2 \left (5 b \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {a+\frac {b}{x}}}d\frac {1}{\sqrt {x}}+\frac {\sqrt {a+\frac {b}{x}}}{2 \sqrt {x}}\right )+\frac {\left (a+\frac {b}{x}\right )^{3/2}}{4 \sqrt {x}}\right )-\sqrt {x} \left (a+\frac {b}{x}\right )^{5/2}\right )\) |
\(\Big \downarrow \) 224 |
\(\displaystyle -2 \left (5 b \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{1-\frac {b}{x}}d\frac {1}{\sqrt {a+\frac {b}{x}} \sqrt {x}}+\frac {\sqrt {a+\frac {b}{x}}}{2 \sqrt {x}}\right )+\frac {\left (a+\frac {b}{x}\right )^{3/2}}{4 \sqrt {x}}\right )-\sqrt {x} \left (a+\frac {b}{x}\right )^{5/2}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -2 \left (5 b \left (\frac {3}{4} a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {x} \sqrt {a+\frac {b}{x}}}\right )}{2 \sqrt {b}}+\frac {\sqrt {a+\frac {b}{x}}}{2 \sqrt {x}}\right )+\frac {\left (a+\frac {b}{x}\right )^{3/2}}{4 \sqrt {x}}\right )-\sqrt {x} \left (a+\frac {b}{x}\right )^{5/2}\right )\) |
Input:
Int[(a + b/x)^(5/2)/Sqrt[x],x]
Output:
-2*(-((a + b/x)^(5/2)*Sqrt[x]) + 5*b*((a + b/x)^(3/2)/(4*Sqrt[x]) + (3*a*( Sqrt[a + b/x]/(2*Sqrt[x]) + (a*ArcTanh[Sqrt[b]/(Sqrt[a + b/x]*Sqrt[x])])/( 2*Sqrt[b])))/4))
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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[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]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1))) Int[ (c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Simp[-k/c Subst[Int[(a + b/(c^n*x^(k*n)))^p/x^(k*(m + 1 ) + 1), x], x, 1/(c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && ILtQ[n, 0] && FractionQ[m]
Time = 0.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.78
method | result | size |
risch | \(-\frac {b \left (9 a x +2 b \right ) \sqrt {\frac {a x +b}{x}}}{4 x^{\frac {3}{2}}}+\frac {a^{2} \left (16 \sqrt {a x +b}-30 \sqrt {b}\, \operatorname {arctanh}\left (\frac {\sqrt {a x +b}}{\sqrt {b}}\right )\right ) \sqrt {\frac {a x +b}{x}}\, \sqrt {x}}{8 \sqrt {a x +b}}\) | \(80\) |
default | \(-\frac {\sqrt {\frac {a x +b}{x}}\, \left (15 \,\operatorname {arctanh}\left (\frac {\sqrt {a x +b}}{\sqrt {b}}\right ) a^{2} b \,x^{2}-8 a^{2} x^{2} \sqrt {a x +b}\, \sqrt {b}+9 a \,b^{\frac {3}{2}} x \sqrt {a x +b}+2 \sqrt {a x +b}\, b^{\frac {5}{2}}\right )}{4 x^{\frac {3}{2}} \sqrt {a x +b}\, \sqrt {b}}\) | \(93\) |
Input:
int((a+b/x)^(5/2)/x^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/4*b*(9*a*x+2*b)/x^(3/2)*((a*x+b)/x)^(1/2)+1/8*a^2*(16*(a*x+b)^(1/2)-30* b^(1/2)*arctanh((a*x+b)^(1/2)/b^(1/2)))*((a*x+b)/x)^(1/2)/(a*x+b)^(1/2)*x^ (1/2)
Time = 0.08 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.62 \[ \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{\sqrt {x}} \, dx=\left [\frac {15 \, a^{2} \sqrt {b} x^{2} \log \left (\frac {a x - 2 \, \sqrt {b} \sqrt {x} \sqrt {\frac {a x + b}{x}} + 2 \, b}{x}\right ) + 2 \, {\left (8 \, a^{2} x^{2} - 9 \, a b x - 2 \, b^{2}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{8 \, x^{2}}, \frac {15 \, a^{2} \sqrt {-b} x^{2} \arctan \left (\frac {\sqrt {-b} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{a x + b}\right ) + {\left (8 \, a^{2} x^{2} - 9 \, a b x - 2 \, b^{2}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{4 \, x^{2}}\right ] \] Input:
integrate((a+b/x)^(5/2)/x^(1/2),x, algorithm="fricas")
Output:
[1/8*(15*a^2*sqrt(b)*x^2*log((a*x - 2*sqrt(b)*sqrt(x)*sqrt((a*x + b)/x) + 2*b)/x) + 2*(8*a^2*x^2 - 9*a*b*x - 2*b^2)*sqrt(x)*sqrt((a*x + b)/x))/x^2, 1/4*(15*a^2*sqrt(-b)*x^2*arctan(sqrt(-b)*sqrt(x)*sqrt((a*x + b)/x)/(a*x + b)) + (8*a^2*x^2 - 9*a*b*x - 2*b^2)*sqrt(x)*sqrt((a*x + b)/x))/x^2]
Time = 7.08 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.24 \[ \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{\sqrt {x}} \, dx=\frac {2 a^{\frac {5}{2}} \sqrt {x}}{\sqrt {1 + \frac {b}{a x}}} - \frac {a^{\frac {3}{2}} b}{4 \sqrt {x} \sqrt {1 + \frac {b}{a x}}} - \frac {11 \sqrt {a} b^{2}}{4 x^{\frac {3}{2}} \sqrt {1 + \frac {b}{a x}}} - \frac {15 a^{2} \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} \sqrt {x}} \right )}}{4} - \frac {b^{3}}{2 \sqrt {a} x^{\frac {5}{2}} \sqrt {1 + \frac {b}{a x}}} \] Input:
integrate((a+b/x)**(5/2)/x**(1/2),x)
Output:
2*a**(5/2)*sqrt(x)/sqrt(1 + b/(a*x)) - a**(3/2)*b/(4*sqrt(x)*sqrt(1 + b/(a *x))) - 11*sqrt(a)*b**2/(4*x**(3/2)*sqrt(1 + b/(a*x))) - 15*a**2*sqrt(b)*a sinh(sqrt(b)/(sqrt(a)*sqrt(x)))/4 - b**3/(2*sqrt(a)*x**(5/2)*sqrt(1 + b/(a *x)))
Time = 0.11 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.34 \[ \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{\sqrt {x}} \, dx=\frac {15}{8} \, a^{2} \sqrt {b} \log \left (\frac {\sqrt {a + \frac {b}{x}} \sqrt {x} - \sqrt {b}}{\sqrt {a + \frac {b}{x}} \sqrt {x} + \sqrt {b}}\right ) + 2 \, \sqrt {a + \frac {b}{x}} a^{2} \sqrt {x} - \frac {9 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a^{2} b x^{\frac {3}{2}} - 7 \, \sqrt {a + \frac {b}{x}} a^{2} b^{2} \sqrt {x}}{4 \, {\left ({\left (a + \frac {b}{x}\right )}^{2} x^{2} - 2 \, {\left (a + \frac {b}{x}\right )} b x + b^{2}\right )}} \] Input:
integrate((a+b/x)^(5/2)/x^(1/2),x, algorithm="maxima")
Output:
15/8*a^2*sqrt(b)*log((sqrt(a + b/x)*sqrt(x) - sqrt(b))/(sqrt(a + b/x)*sqrt (x) + sqrt(b))) + 2*sqrt(a + b/x)*a^2*sqrt(x) - 1/4*(9*(a + b/x)^(3/2)*a^2 *b*x^(3/2) - 7*sqrt(a + b/x)*a^2*b^2*sqrt(x))/((a + b/x)^2*x^2 - 2*(a + b/ x)*b*x + b^2)
Time = 0.14 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{\sqrt {x}} \, dx=\frac {\frac {15 \, a^{3} b \arctan \left (\frac {\sqrt {a x + b}}{\sqrt {-b}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-b}} + 8 \, \sqrt {a x + b} a^{3} \mathrm {sgn}\left (x\right ) - \frac {9 \, {\left (a x + b\right )}^{\frac {3}{2}} a^{3} b \mathrm {sgn}\left (x\right ) - 7 \, \sqrt {a x + b} a^{3} b^{2} \mathrm {sgn}\left (x\right )}{a^{2} x^{2}}}{4 \, a} \] Input:
integrate((a+b/x)^(5/2)/x^(1/2),x, algorithm="giac")
Output:
1/4*(15*a^3*b*arctan(sqrt(a*x + b)/sqrt(-b))*sgn(x)/sqrt(-b) + 8*sqrt(a*x + b)*a^3*sgn(x) - (9*(a*x + b)^(3/2)*a^3*b*sgn(x) - 7*sqrt(a*x + b)*a^3*b^ 2*sgn(x))/(a^2*x^2))/a
Timed out. \[ \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{\sqrt {x}} \, dx=\int \frac {{\left (a+\frac {b}{x}\right )}^{5/2}}{\sqrt {x}} \,d x \] Input:
int((a + b/x)^(5/2)/x^(1/2),x)
Output:
int((a + b/x)^(5/2)/x^(1/2), x)
Time = 0.21 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.82 \[ \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{\sqrt {x}} \, dx=\frac {16 \sqrt {a x +b}\, a^{2} x^{2}-18 \sqrt {a x +b}\, a b x -4 \sqrt {a x +b}\, b^{2}+15 \sqrt {b}\, \mathrm {log}\left (\sqrt {a x +b}-\sqrt {b}\right ) a^{2} x^{2}-15 \sqrt {b}\, \mathrm {log}\left (\sqrt {a x +b}+\sqrt {b}\right ) a^{2} x^{2}}{8 x^{2}} \] Input:
int((a+b/x)^(5/2)/x^(1/2),x)
Output:
(16*sqrt(a*x + b)*a**2*x**2 - 18*sqrt(a*x + b)*a*b*x - 4*sqrt(a*x + b)*b** 2 + 15*sqrt(b)*log(sqrt(a*x + b) - sqrt(b))*a**2*x**2 - 15*sqrt(b)*log(sqr t(a*x + b) + sqrt(b))*a**2*x**2)/(8*x**2)