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

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

[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))

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Rubi [A]  time = 0.0516641, antiderivative size = 111, normalized size of antiderivative = 1., 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 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]]

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]

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*}

Mathematica [C]  time = 0.0183441, 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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Maple [A]  time = 0.006, size = 135, normalized size = 1.2 \begin{align*}{\frac{x}{384}\sqrt{{\frac{ax+b}{x}}} \left ( 96\,x \left ( a{x}^{2}+bx \right ) ^{3/2}{a}^{7/2}+176\,{a}^{5/2} \left ( a{x}^{2}+bx \right ) ^{3/2}b+60\,{a}^{5/2}\sqrt{a{x}^{2}+bx}x{b}^{2}+30\,{a}^{3/2}\sqrt{a{x}^{2}+bx}{b}^{3}-15\,\ln \left ( 1/2\,{\frac{2\,\sqrt{a{x}^{2}+bx}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ) a{b}^{4} \right ){a}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{ \left ( ax+b \right ) x}}}} \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.71405, size = 414, normalized size = 3.73 \begin{align*} \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 ] \end{align*}

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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Sympy [A]  time = 7.90546, size = 155, normalized size = 1.4 \begin{align*} \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}}} \end{align*}

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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Giac [A]  time = 1.15366, size = 144, normalized size = 1.3 \begin{align*} \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}\left (x\right )}{128 \, a^{\frac{3}{2}}} - \frac{5 \, b^{4} \log \left ({\left | b \right |}\right ) \mathrm{sgn}\left (x\right )}{128 \, a^{\frac{3}{2}}} + \frac{1}{192} \, \sqrt{a x^{2} + b x}{\left (\frac{15 \, b^{3} \mathrm{sgn}\left (x\right )}{a} + 2 \,{\left (59 \, b^{2} \mathrm{sgn}\left (x\right ) + 4 \,{\left (6 \, a^{2} x \mathrm{sgn}\left (x\right ) + 17 \, a b \mathrm{sgn}\left (x\right )\right )} x\right )} x\right )} \end{align*}

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)