3.18.76 \(\int \frac {(a+\frac {b}{x})^{5/2}}{x^{5/2}} \, dx\) [1776]

Optimal. Leaf size=126 \[ -\frac {5 a^2 \sqrt {a+\frac {b}{x}}}{32 x^{3/2}}-\frac {5 a \left (a+\frac {b}{x}\right )^{3/2}}{24 x^{3/2}}-\frac {\left (a+\frac {b}{x}\right )^{5/2}}{4 x^{3/2}}-\frac {5 a^3 \sqrt {a+\frac {b}{x}}}{64 b \sqrt {x}}+\frac {5 a^4 \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x}} \sqrt {x}}\right )}{64 b^{3/2}} \]

[Out]

-5/24*a*(a+b/x)^(3/2)/x^(3/2)-1/4*(a+b/x)^(5/2)/x^(3/2)+5/64*a^4*arctanh(b^(1/2)/(a+b/x)^(1/2)/x^(1/2))/b^(3/2
)-5/32*a^2*(a+b/x)^(1/2)/x^(3/2)-5/64*a^3*(a+b/x)^(1/2)/b/x^(1/2)

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Rubi [A]
time = 0.09, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {344, 285, 327, 223, 212} \begin {gather*} \frac {5 a^4 \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {x} \sqrt {a+\frac {b}{x}}}\right )}{64 b^{3/2}}-\frac {5 a^3 \sqrt {a+\frac {b}{x}}}{64 b \sqrt {x}}-\frac {5 a^2 \sqrt {a+\frac {b}{x}}}{32 x^{3/2}}-\frac {5 a \left (a+\frac {b}{x}\right )^{3/2}}{24 x^{3/2}}-\frac {\left (a+\frac {b}{x}\right )^{5/2}}{4 x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^(5/2)/x^(5/2),x]

[Out]

(-5*a^2*Sqrt[a + b/x])/(32*x^(3/2)) - (5*a*(a + b/x)^(3/2))/(24*x^(3/2)) - (a + b/x)^(5/2)/(4*x^(3/2)) - (5*a^
3*Sqrt[a + b/x])/(64*b*Sqrt[x]) + (5*a^4*ArcTanh[Sqrt[b]/(Sqrt[a + b/x]*Sqrt[x])])/(64*b^(3/2))

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 285

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + n
*p + 1))), x] + Dist[a*n*(p/(m + n*p + 1)), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 344

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[-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]

Rubi steps

\begin {align*} \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{x^{5/2}} \, dx &=-\left (2 \text {Subst}\left (\int x^2 \left (a+b x^2\right )^{5/2} \, dx,x,\frac {1}{\sqrt {x}}\right )\right )\\ &=-\frac {\left (a+\frac {b}{x}\right )^{5/2}}{4 x^{3/2}}-\frac {1}{4} (5 a) \text {Subst}\left (\int x^2 \left (a+b x^2\right )^{3/2} \, dx,x,\frac {1}{\sqrt {x}}\right )\\ &=-\frac {5 a \left (a+\frac {b}{x}\right )^{3/2}}{24 x^{3/2}}-\frac {\left (a+\frac {b}{x}\right )^{5/2}}{4 x^{3/2}}-\frac {1}{8} \left (5 a^2\right ) \text {Subst}\left (\int x^2 \sqrt {a+b x^2} \, dx,x,\frac {1}{\sqrt {x}}\right )\\ &=-\frac {5 a^2 \sqrt {a+\frac {b}{x}}}{32 x^{3/2}}-\frac {5 a \left (a+\frac {b}{x}\right )^{3/2}}{24 x^{3/2}}-\frac {\left (a+\frac {b}{x}\right )^{5/2}}{4 x^{3/2}}-\frac {1}{32} \left (5 a^3\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{\sqrt {x}}\right )\\ &=-\frac {5 a^2 \sqrt {a+\frac {b}{x}}}{32 x^{3/2}}-\frac {5 a \left (a+\frac {b}{x}\right )^{3/2}}{24 x^{3/2}}-\frac {\left (a+\frac {b}{x}\right )^{5/2}}{4 x^{3/2}}-\frac {5 a^3 \sqrt {a+\frac {b}{x}}}{64 b \sqrt {x}}+\frac {\left (5 a^4\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{\sqrt {x}}\right )}{64 b}\\ &=-\frac {5 a^2 \sqrt {a+\frac {b}{x}}}{32 x^{3/2}}-\frac {5 a \left (a+\frac {b}{x}\right )^{3/2}}{24 x^{3/2}}-\frac {\left (a+\frac {b}{x}\right )^{5/2}}{4 x^{3/2}}-\frac {5 a^3 \sqrt {a+\frac {b}{x}}}{64 b \sqrt {x}}+\frac {\left (5 a^4\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {1}{\sqrt {a+\frac {b}{x}} \sqrt {x}}\right )}{64 b}\\ &=-\frac {5 a^2 \sqrt {a+\frac {b}{x}}}{32 x^{3/2}}-\frac {5 a \left (a+\frac {b}{x}\right )^{3/2}}{24 x^{3/2}}-\frac {\left (a+\frac {b}{x}\right )^{5/2}}{4 x^{3/2}}-\frac {5 a^3 \sqrt {a+\frac {b}{x}}}{64 b \sqrt {x}}+\frac {5 a^4 \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x}} \sqrt {x}}\right )}{64 b^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 5.82, size = 104, normalized size = 0.83 \begin {gather*} \frac {\sqrt {a+\frac {b}{x}} \sqrt {x} \left (\frac {\sqrt {b+a x} \left (-48 b^3-136 a b^2 x-118 a^2 b x^2-15 a^3 x^3\right )}{192 b x^4}+\frac {5 a^4 \tanh ^{-1}\left (\frac {\sqrt {b+a x}}{\sqrt {b}}\right )}{64 b^{3/2}}\right )}{\sqrt {b+a x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^(5/2)/x^(5/2),x]

[Out]

(Sqrt[a + b/x]*Sqrt[x]*((Sqrt[b + a*x]*(-48*b^3 - 136*a*b^2*x - 118*a^2*b*x^2 - 15*a^3*x^3))/(192*b*x^4) + (5*
a^4*ArcTanh[Sqrt[b + a*x]/Sqrt[b]])/(64*b^(3/2))))/Sqrt[b + a*x]

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Maple [A]
time = 0.04, size = 110, normalized size = 0.87

method	result	size
risch	\(-\frac {\left (15 a^{3} x^{3}+118 a^{2} b \,x^{2}+136 a \,b^{2} x +48 b^{3}\right ) \sqrt {\frac {a x +b}{x}}}{192 x^{\frac {7}{2}} b}+\frac {5 a^{4} \arctanh \left (\frac {\sqrt {a x +b}}{\sqrt {b}}\right ) \sqrt {\frac {a x +b}{x}}\, \sqrt {x}}{64 b^{\frac {3}{2}} \sqrt {a x +b}}\)	\(92\)
default	\(-\frac {\sqrt {\frac {a x +b}{x}}\, \left (-15 \arctanh \left (\frac {\sqrt {a x +b}}{\sqrt {b}}\right ) a^{4} x^{4}+48 \sqrt {a x +b}\, b^{\frac {7}{2}}+136 a \,b^{\frac {5}{2}} x \sqrt {a x +b}+118 a^{2} b^{\frac {3}{2}} x^{2} \sqrt {a x +b}+15 a^{3} x^{3} \sqrt {a x +b}\, \sqrt {b}\right )}{192 x^{\frac {7}{2}} b^{\frac {3}{2}} \sqrt {a x +b}}\)	\(110\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+1/x*b)^(5/2)/x^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/192*((a*x+b)/x)^(1/2)*(-15*arctanh((a*x+b)^(1/2)/b^(1/2))*a^4*x^4+48*(a*x+b)^(1/2)*b^(7/2)+136*a*b^(5/2)*x*
(a*x+b)^(1/2)+118*a^2*b^(3/2)*x^2*(a*x+b)^(1/2)+15*a^3*x^3*(a*x+b)^(1/2)*b^(1/2))/x^(7/2)/b^(3/2)/(a*x+b)^(1/2
)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 194 vs. \(2 (92) = 184\).
time = 0.50, size = 194, normalized size = 1.54 \begin {gather*} -\frac {5 \, a^{4} \log \left (\frac {\sqrt {a + \frac {b}{x}} \sqrt {x} - \sqrt {b}}{\sqrt {a + \frac {b}{x}} \sqrt {x} + \sqrt {b}}\right )}{128 \, b^{\frac {3}{2}}} - \frac {15 \, {\left (a + \frac {b}{x}\right )}^{\frac {7}{2}} a^{4} x^{\frac {7}{2}} + 73 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} a^{4} b x^{\frac {5}{2}} - 55 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a^{4} b^{2} x^{\frac {3}{2}} + 15 \, \sqrt {a + \frac {b}{x}} a^{4} b^{3} \sqrt {x}}{192 \, {\left ({\left (a + \frac {b}{x}\right )}^{4} b x^{4} - 4 \, {\left (a + \frac {b}{x}\right )}^{3} b^{2} x^{3} + 6 \, {\left (a + \frac {b}{x}\right )}^{2} b^{3} x^{2} - 4 \, {\left (a + \frac {b}{x}\right )} b^{4} x + b^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^(5/2)/x^(5/2),x, algorithm="maxima")

[Out]

-5/128*a^4*log((sqrt(a + b/x)*sqrt(x) - sqrt(b))/(sqrt(a + b/x)*sqrt(x) + sqrt(b)))/b^(3/2) - 1/192*(15*(a + b
/x)^(7/2)*a^4*x^(7/2) + 73*(a + b/x)^(5/2)*a^4*b*x^(5/2) - 55*(a + b/x)^(3/2)*a^4*b^2*x^(3/2) + 15*sqrt(a + b/
x)*a^4*b^3*sqrt(x))/((a + b/x)^4*b*x^4 - 4*(a + b/x)^3*b^2*x^3 + 6*(a + b/x)^2*b^3*x^2 - 4*(a + b/x)*b^4*x + b
^5)

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Fricas [A]
time = 0.37, size = 195, normalized size = 1.55 \begin {gather*} \left [\frac {15 \, a^{4} \sqrt {b} x^{4} \log \left (\frac {a x + 2 \, \sqrt {b} \sqrt {x} \sqrt {\frac {a x + b}{x}} + 2 \, b}{x}\right ) - 2 \, {\left (15 \, a^{3} b x^{3} + 118 \, a^{2} b^{2} x^{2} + 136 \, a b^{3} x + 48 \, b^{4}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{384 \, b^{2} x^{4}}, -\frac {15 \, a^{4} \sqrt {-b} x^{4} \arctan \left (\frac {\sqrt {-b} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{b}\right ) + {\left (15 \, a^{3} b x^{3} + 118 \, a^{2} b^{2} x^{2} + 136 \, a b^{3} x + 48 \, b^{4}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{192 \, b^{2} x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^(5/2)/x^(5/2),x, algorithm="fricas")

[Out]

[1/384*(15*a^4*sqrt(b)*x^4*log((a*x + 2*sqrt(b)*sqrt(x)*sqrt((a*x + b)/x) + 2*b)/x) - 2*(15*a^3*b*x^3 + 118*a^
2*b^2*x^2 + 136*a*b^3*x + 48*b^4)*sqrt(x)*sqrt((a*x + b)/x))/(b^2*x^4), -1/192*(15*a^4*sqrt(-b)*x^4*arctan(sqr
t(-b)*sqrt(x)*sqrt((a*x + b)/x)/b) + (15*a^3*b*x^3 + 118*a^2*b^2*x^2 + 136*a*b^3*x + 48*b^4)*sqrt(x)*sqrt((a*x
+ b)/x))/(b^2*x^4)]

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Sympy [A]
time = 12.02, size = 155, normalized size = 1.23 \begin {gather*} - \frac {5 a^{\frac {7}{2}}}{64 b \sqrt {x} \sqrt {1 + \frac {b}{a x}}} - \frac {133 a^{\frac {5}{2}}}{192 x^{\frac {3}{2}} \sqrt {1 + \frac {b}{a x}}} - \frac {127 a^{\frac {3}{2}} b}{96 x^{\frac {5}{2}} \sqrt {1 + \frac {b}{a x}}} - \frac {23 \sqrt {a} b^{2}}{24 x^{\frac {7}{2}} \sqrt {1 + \frac {b}{a x}}} + \frac {5 a^{4} \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} \sqrt {x}} \right )}}{64 b^{\frac {3}{2}}} - \frac {b^{3}}{4 \sqrt {a} x^{\frac {9}{2}} \sqrt {1 + \frac {b}{a x}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)**(5/2)/x**(5/2),x)

[Out]

-5*a**(7/2)/(64*b*sqrt(x)*sqrt(1 + b/(a*x))) - 133*a**(5/2)/(192*x**(3/2)*sqrt(1 + b/(a*x))) - 127*a**(3/2)*b/
(96*x**(5/2)*sqrt(1 + b/(a*x))) - 23*sqrt(a)*b**2/(24*x**(7/2)*sqrt(1 + b/(a*x))) + 5*a**4*asinh(sqrt(b)/(sqrt
(a)*sqrt(x)))/(64*b**(3/2)) - b**3/(4*sqrt(a)*x**(9/2)*sqrt(1 + b/(a*x)))

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Giac [A]
time = 0.57, size = 99, normalized size = 0.79 \begin {gather*} -\frac {\frac {15 \, a^{5} \arctan \left (\frac {\sqrt {a x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b} + \frac {15 \, {\left (a x + b\right )}^{\frac {7}{2}} a^{5} + 73 \, {\left (a x + b\right )}^{\frac {5}{2}} a^{5} b - 55 \, {\left (a x + b\right )}^{\frac {3}{2}} a^{5} b^{2} + 15 \, \sqrt {a x + b} a^{5} b^{3}}{a^{4} b x^{4}}}{192 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^(5/2)/x^(5/2),x, algorithm="giac")

[Out]

-1/192*(15*a^5*arctan(sqrt(a*x + b)/sqrt(-b))/(sqrt(-b)*b) + (15*(a*x + b)^(7/2)*a^5 + 73*(a*x + b)^(5/2)*a^5*
b - 55*(a*x + b)^(3/2)*a^5*b^2 + 15*sqrt(a*x + b)*a^5*b^3)/(a^4*b*x^4))/a

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+\frac {b}{x}\right )}^{5/2}}{x^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x)^(5/2)/x^(5/2),x)

[Out]

int((a + b/x)^(5/2)/x^(5/2), x)

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