3.1683 \(\int \frac{\sqrt{x}}{(a+\frac{b}{x})^3} \, dx\)

Optimal. Leaf size=95 \[ \frac{35 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{4 a^{9/2}}-\frac{7 x^{5/2}}{4 a^2 (a x+b)}-\frac{35 b \sqrt{x}}{4 a^4}+\frac{35 x^{3/2}}{12 a^3}-\frac{x^{7/2}}{2 a (a x+b)^2} \]

[Out]

(-35*b*Sqrt[x])/(4*a^4) + (35*x^(3/2))/(12*a^3) - x^(7/2)/(2*a*(b + a*x)^2) - (7*x^(5/2))/(4*a^2*(b + a*x)) +
(35*b^(3/2)*ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]])/(4*a^(9/2))

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Rubi [A]  time = 0.0336625, antiderivative size = 95, 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 = {263, 47, 50, 63, 205} \[ \frac{35 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{4 a^{9/2}}-\frac{7 x^{5/2}}{4 a^2 (a x+b)}-\frac{35 b \sqrt{x}}{4 a^4}+\frac{35 x^{3/2}}{12 a^3}-\frac{x^{7/2}}{2 a (a x+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/(a + b/x)^3,x]

[Out]

(-35*b*Sqrt[x])/(4*a^4) + (35*x^(3/2))/(12*a^3) - x^(7/2)/(2*a*(b + a*x)^2) - (7*x^(5/2))/(4*a^2*(b + a*x)) +
(35*b^(3/2)*ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]])/(4*a^(9/2))

Rule 263

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\sqrt{x}}{\left (a+\frac{b}{x}\right )^3} \, dx &=\int \frac{x^{7/2}}{(b+a x)^3} \, dx\\ &=-\frac{x^{7/2}}{2 a (b+a x)^2}+\frac{7 \int \frac{x^{5/2}}{(b+a x)^2} \, dx}{4 a}\\ &=-\frac{x^{7/2}}{2 a (b+a x)^2}-\frac{7 x^{5/2}}{4 a^2 (b+a x)}+\frac{35 \int \frac{x^{3/2}}{b+a x} \, dx}{8 a^2}\\ &=\frac{35 x^{3/2}}{12 a^3}-\frac{x^{7/2}}{2 a (b+a x)^2}-\frac{7 x^{5/2}}{4 a^2 (b+a x)}-\frac{(35 b) \int \frac{\sqrt{x}}{b+a x} \, dx}{8 a^3}\\ &=-\frac{35 b \sqrt{x}}{4 a^4}+\frac{35 x^{3/2}}{12 a^3}-\frac{x^{7/2}}{2 a (b+a x)^2}-\frac{7 x^{5/2}}{4 a^2 (b+a x)}+\frac{\left (35 b^2\right ) \int \frac{1}{\sqrt{x} (b+a x)} \, dx}{8 a^4}\\ &=-\frac{35 b \sqrt{x}}{4 a^4}+\frac{35 x^{3/2}}{12 a^3}-\frac{x^{7/2}}{2 a (b+a x)^2}-\frac{7 x^{5/2}}{4 a^2 (b+a x)}+\frac{\left (35 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{b+a x^2} \, dx,x,\sqrt{x}\right )}{4 a^4}\\ &=-\frac{35 b \sqrt{x}}{4 a^4}+\frac{35 x^{3/2}}{12 a^3}-\frac{x^{7/2}}{2 a (b+a x)^2}-\frac{7 x^{5/2}}{4 a^2 (b+a x)}+\frac{35 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{4 a^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.0046109, size = 27, normalized size = 0.28 \[ \frac{2 x^{9/2} \, _2F_1\left (3,\frac{9}{2};\frac{11}{2};-\frac{a x}{b}\right )}{9 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]/(a + b/x)^3,x]

[Out]

(2*x^(9/2)*Hypergeometric2F1[3, 9/2, 11/2, -((a*x)/b)])/(9*b^3)

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Maple [A]  time = 0.013, size = 79, normalized size = 0.8 \begin{align*}{\frac{2}{3\,{a}^{3}}{x}^{{\frac{3}{2}}}}-6\,{\frac{b\sqrt{x}}{{a}^{4}}}-{\frac{13\,{b}^{2}}{4\,{a}^{3} \left ( ax+b \right ) ^{2}}{x}^{{\frac{3}{2}}}}-{\frac{11\,{b}^{3}}{4\,{a}^{4} \left ( ax+b \right ) ^{2}}\sqrt{x}}+{\frac{35\,{b}^{2}}{4\,{a}^{4}}\arctan \left ({a\sqrt{x}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)/(a+b/x)^3,x)

[Out]

2/3*x^(3/2)/a^3-6*b*x^(1/2)/a^4-13/4/a^3*b^2/(a*x+b)^2*x^(3/2)-11/4/a^4*b^3/(a*x+b)^2*x^(1/2)+35/4/a^4*b^2/(a*
b)^(1/2)*arctan(a*x^(1/2)/(a*b)^(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(x^(1/2)/(a+b/x)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.82909, size = 509, normalized size = 5.36 \begin{align*} \left [\frac{105 \,{\left (a^{2} b x^{2} + 2 \, a b^{2} x + b^{3}\right )} \sqrt{-\frac{b}{a}} \log \left (\frac{a x + 2 \, a \sqrt{x} \sqrt{-\frac{b}{a}} - b}{a x + b}\right ) + 2 \,{\left (8 \, a^{3} x^{3} - 56 \, a^{2} b x^{2} - 175 \, a b^{2} x - 105 \, b^{3}\right )} \sqrt{x}}{24 \,{\left (a^{6} x^{2} + 2 \, a^{5} b x + a^{4} b^{2}\right )}}, \frac{105 \,{\left (a^{2} b x^{2} + 2 \, a b^{2} x + b^{3}\right )} \sqrt{\frac{b}{a}} \arctan \left (\frac{a \sqrt{x} \sqrt{\frac{b}{a}}}{b}\right ) +{\left (8 \, a^{3} x^{3} - 56 \, a^{2} b x^{2} - 175 \, a b^{2} x - 105 \, b^{3}\right )} \sqrt{x}}{12 \,{\left (a^{6} x^{2} + 2 \, a^{5} b x + a^{4} b^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)/(a+b/x)^3,x, algorithm="fricas")

[Out]

[1/24*(105*(a^2*b*x^2 + 2*a*b^2*x + b^3)*sqrt(-b/a)*log((a*x + 2*a*sqrt(x)*sqrt(-b/a) - b)/(a*x + b)) + 2*(8*a
^3*x^3 - 56*a^2*b*x^2 - 175*a*b^2*x - 105*b^3)*sqrt(x))/(a^6*x^2 + 2*a^5*b*x + a^4*b^2), 1/12*(105*(a^2*b*x^2
+ 2*a*b^2*x + b^3)*sqrt(b/a)*arctan(a*sqrt(x)*sqrt(b/a)/b) + (8*a^3*x^3 - 56*a^2*b*x^2 - 175*a*b^2*x - 105*b^3
)*sqrt(x))/(a^6*x^2 + 2*a^5*b*x + a^4*b^2)]

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Sympy [A]  time = 41.951, size = 906, normalized size = 9.54 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(1/2)/(a+b/x)**3,x)

[Out]

Piecewise((zoo*x**(9/2), Eq(a, 0) & Eq(b, 0)), (2*x**(9/2)/(9*b**3), Eq(a, 0)), (2*x**(3/2)/(3*a**3), Eq(b, 0)
), (16*I*a**4*sqrt(b)*x**(7/2)*sqrt(1/a)/(24*I*a**7*sqrt(b)*x**2*sqrt(1/a) + 48*I*a**6*b**(3/2)*x*sqrt(1/a) +
24*I*a**5*b**(5/2)*sqrt(1/a)) - 112*I*a**3*b**(3/2)*x**(5/2)*sqrt(1/a)/(24*I*a**7*sqrt(b)*x**2*sqrt(1/a) + 48*
I*a**6*b**(3/2)*x*sqrt(1/a) + 24*I*a**5*b**(5/2)*sqrt(1/a)) - 350*I*a**2*b**(5/2)*x**(3/2)*sqrt(1/a)/(24*I*a**
7*sqrt(b)*x**2*sqrt(1/a) + 48*I*a**6*b**(3/2)*x*sqrt(1/a) + 24*I*a**5*b**(5/2)*sqrt(1/a)) + 105*a**2*b**2*x**2
*log(-I*sqrt(b)*sqrt(1/a) + sqrt(x))/(24*I*a**7*sqrt(b)*x**2*sqrt(1/a) + 48*I*a**6*b**(3/2)*x*sqrt(1/a) + 24*I
*a**5*b**(5/2)*sqrt(1/a)) - 105*a**2*b**2*x**2*log(I*sqrt(b)*sqrt(1/a) + sqrt(x))/(24*I*a**7*sqrt(b)*x**2*sqrt
(1/a) + 48*I*a**6*b**(3/2)*x*sqrt(1/a) + 24*I*a**5*b**(5/2)*sqrt(1/a)) - 210*I*a*b**(7/2)*sqrt(x)*sqrt(1/a)/(2
4*I*a**7*sqrt(b)*x**2*sqrt(1/a) + 48*I*a**6*b**(3/2)*x*sqrt(1/a) + 24*I*a**5*b**(5/2)*sqrt(1/a)) + 210*a*b**3*
x*log(-I*sqrt(b)*sqrt(1/a) + sqrt(x))/(24*I*a**7*sqrt(b)*x**2*sqrt(1/a) + 48*I*a**6*b**(3/2)*x*sqrt(1/a) + 24*
I*a**5*b**(5/2)*sqrt(1/a)) - 210*a*b**3*x*log(I*sqrt(b)*sqrt(1/a) + sqrt(x))/(24*I*a**7*sqrt(b)*x**2*sqrt(1/a)
+ 48*I*a**6*b**(3/2)*x*sqrt(1/a) + 24*I*a**5*b**(5/2)*sqrt(1/a)) + 105*b**4*log(-I*sqrt(b)*sqrt(1/a) + sqrt(x
))/(24*I*a**7*sqrt(b)*x**2*sqrt(1/a) + 48*I*a**6*b**(3/2)*x*sqrt(1/a) + 24*I*a**5*b**(5/2)*sqrt(1/a)) - 105*b*
*4*log(I*sqrt(b)*sqrt(1/a) + sqrt(x))/(24*I*a**7*sqrt(b)*x**2*sqrt(1/a) + 48*I*a**6*b**(3/2)*x*sqrt(1/a) + 24*
I*a**5*b**(5/2)*sqrt(1/a)), True))

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Giac [A]  time = 1.09522, size = 104, normalized size = 1.09 \begin{align*} \frac{35 \, b^{2} \arctan \left (\frac{a \sqrt{x}}{\sqrt{a b}}\right )}{4 \, \sqrt{a b} a^{4}} - \frac{13 \, a b^{2} x^{\frac{3}{2}} + 11 \, b^{3} \sqrt{x}}{4 \,{\left (a x + b\right )}^{2} a^{4}} + \frac{2 \,{\left (a^{6} x^{\frac{3}{2}} - 9 \, a^{5} b \sqrt{x}\right )}}{3 \, a^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)/(a+b/x)^3,x, algorithm="giac")

[Out]

35/4*b^2*arctan(a*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^4) - 1/4*(13*a*b^2*x^(3/2) + 11*b^3*sqrt(x))/((a*x + b)^2*a^
4) + 2/3*(a^6*x^(3/2) - 9*a^5*b*sqrt(x))/a^9