3.1701 \(\int (a+\frac {b}{x})^{3/2} x^3 \, dx\)

Optimal. Leaf size=114 \[ \frac {3 b^4 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{64 a^{5/2}}-\frac {3 b^3 x \sqrt {a+\frac {b}{x}}}{64 a^2}+\frac {b^2 x^2 \sqrt {a+\frac {b}{x}}}{32 a}+\frac {1}{4} x^4 \left (a+\frac {b}{x}\right )^{3/2}+\frac {1}{8} b x^3 \sqrt {a+\frac {b}{x}} \]

[Out]

1/4*(a+b/x)^(3/2)*x^4+3/64*b^4*arctanh((a+b/x)^(1/2)/a^(1/2))/a^(5/2)-3/64*b^3*x*(a+b/x)^(1/2)/a^2+1/32*b^2*x^
2*(a+b/x)^(1/2)/a+1/8*b*x^3*(a+b/x)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 114, 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} \[ -\frac {3 b^3 x \sqrt {a+\frac {b}{x}}}{64 a^2}+\frac {3 b^4 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{64 a^{5/2}}+\frac {b^2 x^2 \sqrt {a+\frac {b}{x}}}{32 a}+\frac {1}{8} b x^3 \sqrt {a+\frac {b}{x}}+\frac {1}{4} x^4 \left (a+\frac {b}{x}\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^(3/2)*x^3,x]

[Out]

(-3*b^3*Sqrt[a + b/x]*x)/(64*a^2) + (b^2*Sqrt[a + b/x]*x^2)/(32*a) + (b*Sqrt[a + b/x]*x^3)/8 + ((a + b/x)^(3/2
)*x^4)/4 + (3*b^4*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/(64*a^(5/2))

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

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Mathematica [C]  time = 0.02, size = 39, normalized size = 0.34 \[ \frac {2 b^4 \left (a+\frac {b}{x}\right )^{5/2} \, _2F_1\left (\frac {5}{2},5;\frac {7}{2};\frac {b}{a x}+1\right )}{5 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^(3/2)*x^3,x]

[Out]

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

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fricas [A]  time = 0.89, size = 174, normalized size = 1.53 \[ \left [\frac {3 \, \sqrt {a} b^{4} \log \left (2 \, a x + 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) + 2 \, {\left (16 \, a^{4} x^{4} + 24 \, a^{3} b x^{3} + 2 \, a^{2} b^{2} x^{2} - 3 \, a b^{3} x\right )} \sqrt {\frac {a x + b}{x}}}{128 \, a^{3}}, -\frac {3 \, \sqrt {-a} b^{4} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) - {\left (16 \, a^{4} x^{4} + 24 \, a^{3} b x^{3} + 2 \, a^{2} b^{2} x^{2} - 3 \, a b^{3} x\right )} \sqrt {\frac {a x + b}{x}}}{64 \, a^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/128*(3*sqrt(a)*b^4*log(2*a*x + 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 2*(16*a^4*x^4 + 24*a^3*b*x^3 + 2*a^2*b^
2*x^2 - 3*a*b^3*x)*sqrt((a*x + b)/x))/a^3, -1/64*(3*sqrt(-a)*b^4*arctan(sqrt(-a)*sqrt((a*x + b)/x)/a) - (16*a^
4*x^4 + 24*a^3*b*x^3 + 2*a^2*b^2*x^2 - 3*a*b^3*x)*sqrt((a*x + b)/x))/a^3]

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giac [A]  time = 0.26, size = 106, normalized size = 0.93 \[ -\frac {3 \, b^{4} \log \left ({\left | -2 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )} \sqrt {a} - b \right |}\right ) \mathrm {sgn}\relax (x)}{128 \, a^{\frac {5}{2}}} + \frac {3 \, b^{4} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\relax (x)}{128 \, a^{\frac {5}{2}}} + \frac {1}{64} \, \sqrt {a x^{2} + b x} {\left (2 \, {\left (4 \, {\left (2 \, a x \mathrm {sgn}\relax (x) + 3 \, b \mathrm {sgn}\relax (x)\right )} x + \frac {b^{2} \mathrm {sgn}\relax (x)}{a}\right )} x - \frac {3 \, b^{3} \mathrm {sgn}\relax (x)}{a^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/128*b^4*log(abs(-2*(sqrt(a)*x - sqrt(a*x^2 + b*x))*sqrt(a) - b))*sgn(x)/a^(5/2) + 3/128*b^4*log(abs(b))*sgn
(x)/a^(5/2) + 1/64*sqrt(a*x^2 + b*x)*(2*(4*(2*a*x*sgn(x) + 3*b*sgn(x))*x + b^2*sgn(x)/a)*x - 3*b^3*sgn(x)/a^2)

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maple [A]  time = 0.01, size = 135, normalized size = 1.18 \[ \frac {\sqrt {\frac {a x +b}{x}}\, \left (3 a \,b^{4} \ln \left (\frac {2 a x +b +2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}}{2 \sqrt {a}}\right )-12 \sqrt {a \,x^{2}+b x}\, a^{\frac {5}{2}} b^{2} x +32 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {7}{2}} x -6 \sqrt {a \,x^{2}+b x}\, a^{\frac {3}{2}} b^{3}+16 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {5}{2}} b \right ) x}{128 \sqrt {\left (a x +b \right ) x}\, a^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128*((a*x+b)/x)^(1/2)*x/a^(7/2)*(32*(a*x^2+b*x)^(3/2)*a^(7/2)*x+16*(a*x^2+b*x)^(3/2)*a^(5/2)*b-12*(a*x^2+b*x
)^(1/2)*a^(5/2)*b^2*x-6*(a*x^2+b*x)^(1/2)*a^(3/2)*b^3+3*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 = 2.28, size = 166, normalized size = 1.46 \[ -\frac {3 \, b^{4} \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right )}{128 \, a^{\frac {5}{2}}} - \frac {3 \, {\left (a + \frac {b}{x}\right )}^{\frac {7}{2}} b^{4} - 11 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} a b^{4} - 11 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a^{2} b^{4} + 3 \, \sqrt {a + \frac {b}{x}} a^{3} b^{4}}{64 \, {\left ({\left (a + \frac {b}{x}\right )}^{4} a^{2} - 4 \, {\left (a + \frac {b}{x}\right )}^{3} a^{3} + 6 \, {\left (a + \frac {b}{x}\right )}^{2} a^{4} - 4 \, {\left (a + \frac {b}{x}\right )} a^{5} + a^{6}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/128*b^4*log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a)))/a^(5/2) - 1/64*(3*(a + b/x)^(7/2)*b^4 - 11
*(a + b/x)^(5/2)*a*b^4 - 11*(a + b/x)^(3/2)*a^2*b^4 + 3*sqrt(a + b/x)*a^3*b^4)/((a + b/x)^4*a^2 - 4*(a + b/x)^
3*a^3 + 6*(a + b/x)^2*a^4 - 4*(a + b/x)*a^5 + a^6)

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mupad [B]  time = 1.36, size = 89, normalized size = 0.78 \[ \frac {11\,x^4\,{\left (a+\frac {b}{x}\right )}^{3/2}}{64}-\frac {3\,a\,x^4\,\sqrt {a+\frac {b}{x}}}{64}+\frac {11\,x^4\,{\left (a+\frac {b}{x}\right )}^{5/2}}{64\,a}-\frac {3\,x^4\,{\left (a+\frac {b}{x}\right )}^{7/2}}{64\,a^2}-\frac {b^4\,\mathrm {atan}\left (\frac {\sqrt {a+\frac {b}{x}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,3{}\mathrm {i}}{64\,a^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b/x)^(3/2),x)

[Out]

(11*x^4*(a + b/x)^(3/2))/64 - (b^4*atan(((a + b/x)^(1/2)*1i)/a^(1/2))*3i)/(64*a^(5/2)) - (3*a*x^4*(a + b/x)^(1
/2))/64 + (11*x^4*(a + b/x)^(5/2))/(64*a) - (3*x^4*(a + b/x)^(7/2))/(64*a^2)

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sympy [A]  time = 8.94, size = 153, normalized size = 1.34 \[ \frac {a^{2} x^{\frac {9}{2}}}{4 \sqrt {b} \sqrt {\frac {a x}{b} + 1}} + \frac {5 a \sqrt {b} x^{\frac {7}{2}}}{8 \sqrt {\frac {a x}{b} + 1}} + \frac {13 b^{\frac {3}{2}} x^{\frac {5}{2}}}{32 \sqrt {\frac {a x}{b} + 1}} - \frac {b^{\frac {5}{2}} x^{\frac {3}{2}}}{64 a \sqrt {\frac {a x}{b} + 1}} - \frac {3 b^{\frac {7}{2}} \sqrt {x}}{64 a^{2} \sqrt {\frac {a x}{b} + 1}} + \frac {3 b^{4} \operatorname {asinh}{\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}} \right )}}{64 a^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*x**(9/2)/(4*sqrt(b)*sqrt(a*x/b + 1)) + 5*a*sqrt(b)*x**(7/2)/(8*sqrt(a*x/b + 1)) + 13*b**(3/2)*x**(5/2)/(3
2*sqrt(a*x/b + 1)) - b**(5/2)*x**(3/2)/(64*a*sqrt(a*x/b + 1)) - 3*b**(7/2)*sqrt(x)/(64*a**2*sqrt(a*x/b + 1)) +
3*b**4*asinh(sqrt(a)*sqrt(x)/sqrt(b))/(64*a**(5/2))

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