3.1712 \(\int (a+\frac {b}{x})^{5/2} x^3 \, dx\)

Optimal. Leaf size=111 \[ -\frac {5 b^4 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{64 a^{3/2}}+\frac {5 b^3 x \sqrt {a+\frac {b}{x}}}{64 a}+\frac {5}{32} b^2 x^2 \sqrt {a+\frac {b}{x}}+\frac {1}{4} x^4 \left (a+\frac {b}{x}\right )^{5/2}+\frac {5}{24} b x^3 \left (a+\frac {b}{x}\right )^{3/2} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {266, 47, 51, 63, 208} \[ -\frac {5 b^4 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{64 a^{3/2}}+\frac {5}{32} b^2 x^2 \sqrt {a+\frac {b}{x}}+\frac {5 b^3 x \sqrt {a+\frac {b}{x}}}{64 a}+\frac {5}{24} b x^3 \left (a+\frac {b}{x}\right )^{3/2}+\frac {1}{4} x^4 \left (a+\frac {b}{x}\right )^{5/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^(5/2)*x^3,x]

[Out]

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

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 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 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 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]]

Rubi steps

\begin {align*} \int \left (a+\frac {b}{x}\right )^{5/2} x^3 \, dx &=-\operatorname {Subst}\left (\int \frac {(a+b x)^{5/2}}{x^5} \, dx,x,\frac {1}{x}\right )\\ &=\frac {1}{4} \left (a+\frac {b}{x}\right )^{5/2} x^4-\frac {1}{8} (5 b) \operatorname {Subst}\left (\int \frac {(a+b x)^{3/2}}{x^4} \, dx,x,\frac {1}{x}\right )\\ &=\frac {5}{24} b \left (a+\frac {b}{x}\right )^{3/2} x^3+\frac {1}{4} \left (a+\frac {b}{x}\right )^{5/2} x^4-\frac {1}{16} \left (5 b^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x^3} \, dx,x,\frac {1}{x}\right )\\ &=\frac {5}{32} b^2 \sqrt {a+\frac {b}{x}} x^2+\frac {5}{24} b \left (a+\frac {b}{x}\right )^{3/2} x^3+\frac {1}{4} \left (a+\frac {b}{x}\right )^{5/2} x^4-\frac {1}{64} \left (5 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {5 b^3 \sqrt {a+\frac {b}{x}} x}{64 a}+\frac {5}{32} b^2 \sqrt {a+\frac {b}{x}} x^2+\frac {5}{24} b \left (a+\frac {b}{x}\right )^{3/2} x^3+\frac {1}{4} \left (a+\frac {b}{x}\right )^{5/2} x^4+\frac {\left (5 b^4\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )}{128 a}\\ &=\frac {5 b^3 \sqrt {a+\frac {b}{x}} x}{64 a}+\frac {5}{32} b^2 \sqrt {a+\frac {b}{x}} x^2+\frac {5}{24} b \left (a+\frac {b}{x}\right )^{3/2} x^3+\frac {1}{4} \left (a+\frac {b}{x}\right )^{5/2} x^4+\frac {\left (5 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+\frac {b}{x}}\right )}{64 a}\\ &=\frac {5 b^3 \sqrt {a+\frac {b}{x}} x}{64 a}+\frac {5}{32} b^2 \sqrt {a+\frac {b}{x}} x^2+\frac {5}{24} b \left (a+\frac {b}{x}\right )^{3/2} x^3+\frac {1}{4} \left (a+\frac {b}{x}\right )^{5/2} x^4-\frac {5 b^4 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{64 a^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 39, normalized size = 0.35 \[ \frac {2 b^4 \left (a+\frac {b}{x}\right )^{7/2} \, _2F_1\left (\frac {7}{2},5;\frac {9}{2};\frac {b}{a x}+1\right )}{7 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^(5/2)*x^3,x]

[Out]

(2*b^4*(a + b/x)^(7/2)*Hypergeometric2F1[7/2, 5, 9/2, 1 + b/(a*x)])/(7*a^5)

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fricas [A]  time = 0.96, size = 173, normalized size = 1.56 \[ \left [\frac {15 \, \sqrt {a} b^{4} \log \left (2 \, a x - 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) + 2 \, {\left (48 \, a^{4} x^{4} + 136 \, a^{3} b x^{3} + 118 \, a^{2} b^{2} x^{2} + 15 \, a b^{3} x\right )} \sqrt {\frac {a x + b}{x}}}{384 \, a^{2}}, \frac {15 \, \sqrt {-a} b^{4} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) + {\left (48 \, a^{4} x^{4} + 136 \, a^{3} b x^{3} + 118 \, a^{2} b^{2} x^{2} + 15 \, a b^{3} x\right )} \sqrt {\frac {a x + b}{x}}}{192 \, a^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^(5/2)*x^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.23, size = 107, normalized size = 0.96 \[ \frac {5 \, b^{4} \log \left ({\left | -2 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )} \sqrt {a} - b \right |}\right ) \mathrm {sgn}\relax (x)}{128 \, a^{\frac {3}{2}}} - \frac {5 \, b^{4} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\relax (x)}{128 \, a^{\frac {3}{2}}} + \frac {1}{192} \, \sqrt {a x^{2} + b x} {\left (\frac {15 \, b^{3} \mathrm {sgn}\relax (x)}{a} + 2 \, {\left (59 \, b^{2} \mathrm {sgn}\relax (x) + 4 \, {\left (6 \, a^{2} x \mathrm {sgn}\relax (x) + 17 \, a b \mathrm {sgn}\relax (x)\right )} x\right )} x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^(5/2)*x^3,x, algorithm="giac")

[Out]

5/128*b^4*log(abs(-2*(sqrt(a)*x - sqrt(a*x^2 + b*x))*sqrt(a) - b))*sgn(x)/a^(3/2) - 5/128*b^4*log(abs(b))*sgn(
x)/a^(3/2) + 1/192*sqrt(a*x^2 + b*x)*(15*b^3*sgn(x)/a + 2*(59*b^2*sgn(x) + 4*(6*a^2*x*sgn(x) + 17*a*b*sgn(x))*
x)*x)

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maple [A]  time = 0.01, size = 135, normalized size = 1.22 \[ \frac {\sqrt {\frac {a x +b}{x}}\, \left (-15 a \,b^{4} \ln \left (\frac {2 a x +b +2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}}{2 \sqrt {a}}\right )+60 \sqrt {a \,x^{2}+b x}\, a^{\frac {5}{2}} b^{2} x +96 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {7}{2}} x +30 \sqrt {a \,x^{2}+b x}\, a^{\frac {3}{2}} b^{3}+176 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {5}{2}} b \right ) x}{384 \sqrt {\left (a x +b \right ) x}\, a^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x)^(5/2)*x^3,x)

[Out]

1/384*((a*x+b)/x)^(1/2)*x/a^(5/2)*(96*(a*x^2+b*x)^(3/2)*a^(7/2)*x+176*(a*x^2+b*x)^(3/2)*a^(5/2)*b+60*(a*x^2+b*
x)^(1/2)*a^(5/2)*b^2*x+30*(a*x^2+b*x)^(1/2)*a^(3/2)*b^3-15*a*b^4*ln(1/2*(2*a*x+b+2*(a*x^2+b*x)^(1/2)*a^(1/2))/
a^(1/2)))/((a*x+b)*x)^(1/2)

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maxima [A]  time = 2.29, size = 164, normalized size = 1.48 \[ \frac {5 \, b^{4} \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right )}{128 \, a^{\frac {3}{2}}} + \frac {15 \, {\left (a + \frac {b}{x}\right )}^{\frac {7}{2}} b^{4} + 73 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} a b^{4} - 55 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a^{2} b^{4} + 15 \, \sqrt {a + \frac {b}{x}} a^{3} b^{4}}{192 \, {\left ({\left (a + \frac {b}{x}\right )}^{4} a - 4 \, {\left (a + \frac {b}{x}\right )}^{3} a^{2} + 6 \, {\left (a + \frac {b}{x}\right )}^{2} a^{3} - 4 \, {\left (a + \frac {b}{x}\right )} a^{4} + a^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^(5/2)*x^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.37, size = 89, normalized size = 0.80 \[ \frac {73\,x^4\,{\left (a+\frac {b}{x}\right )}^{5/2}}{192}-\frac {55\,a\,x^4\,{\left (a+\frac {b}{x}\right )}^{3/2}}{192}+\frac {5\,a^2\,x^4\,\sqrt {a+\frac {b}{x}}}{64}+\frac {5\,x^4\,{\left (a+\frac {b}{x}\right )}^{7/2}}{64\,a}+\frac {b^4\,\mathrm {atan}\left (\frac {\sqrt {a+\frac {b}{x}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{64\,a^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b/x)^(5/2),x)

[Out]

(73*x^4*(a + b/x)^(5/2))/192 + (b^4*atan(((a + b/x)^(1/2)*1i)/a^(1/2))*5i)/(64*a^(3/2)) - (55*a*x^4*(a + b/x)^
(3/2))/192 + (5*a^2*x^4*(a + b/x)^(1/2))/64 + (5*x^4*(a + b/x)^(7/2))/(64*a)

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sympy [A]  time = 8.59, size = 155, normalized size = 1.40 \[ \frac {a^{3} x^{\frac {9}{2}}}{4 \sqrt {b} \sqrt {\frac {a x}{b} + 1}} + \frac {23 a^{2} \sqrt {b} x^{\frac {7}{2}}}{24 \sqrt {\frac {a x}{b} + 1}} + \frac {127 a b^{\frac {3}{2}} x^{\frac {5}{2}}}{96 \sqrt {\frac {a x}{b} + 1}} + \frac {133 b^{\frac {5}{2}} x^{\frac {3}{2}}}{192 \sqrt {\frac {a x}{b} + 1}} + \frac {5 b^{\frac {7}{2}} \sqrt {x}}{64 a \sqrt {\frac {a x}{b} + 1}} - \frac {5 b^{4} \operatorname {asinh}{\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}} \right )}}{64 a^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)**(5/2)*x**3,x)

[Out]

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

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