3.1684 \(\int \frac {1}{(a+\frac {b}{x})^3 \sqrt {x}} \, dx\)

Optimal. Leaf size=82 \[ -\frac {15 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{4 a^{7/2}}+\frac {15 \sqrt {x}}{4 a^3}-\frac {5 x^{3/2}}{4 a^2 (a x+b)}-\frac {x^{5/2}}{2 a (a x+b)^2} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ -\frac {5 x^{3/2}}{4 a^2 (a x+b)}-\frac {15 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{4 a^{7/2}}+\frac {15 \sqrt {x}}{4 a^3}-\frac {x^{5/2}}{2 a (a x+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x)^3*Sqrt[x]),x]

[Out]

(15*Sqrt[x])/(4*a^3) - x^(5/2)/(2*a*(b + a*x)^2) - (5*x^(3/2))/(4*a^2*(b + a*x)) - (15*Sqrt[b]*ArcTan[(Sqrt[a]
*Sqrt[x])/Sqrt[b]])/(4*a^(7/2))

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 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]

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]

Rubi steps

\begin {align*} \int \frac {1}{\left (a+\frac {b}{x}\right )^3 \sqrt {x}} \, dx &=\int \frac {x^{5/2}}{(b+a x)^3} \, dx\\ &=-\frac {x^{5/2}}{2 a (b+a x)^2}+\frac {5 \int \frac {x^{3/2}}{(b+a x)^2} \, dx}{4 a}\\ &=-\frac {x^{5/2}}{2 a (b+a x)^2}-\frac {5 x^{3/2}}{4 a^2 (b+a x)}+\frac {15 \int \frac {\sqrt {x}}{b+a x} \, dx}{8 a^2}\\ &=\frac {15 \sqrt {x}}{4 a^3}-\frac {x^{5/2}}{2 a (b+a x)^2}-\frac {5 x^{3/2}}{4 a^2 (b+a x)}-\frac {(15 b) \int \frac {1}{\sqrt {x} (b+a x)} \, dx}{8 a^3}\\ &=\frac {15 \sqrt {x}}{4 a^3}-\frac {x^{5/2}}{2 a (b+a x)^2}-\frac {5 x^{3/2}}{4 a^2 (b+a x)}-\frac {(15 b) \operatorname {Subst}\left (\int \frac {1}{b+a x^2} \, dx,x,\sqrt {x}\right )}{4 a^3}\\ &=\frac {15 \sqrt {x}}{4 a^3}-\frac {x^{5/2}}{2 a (b+a x)^2}-\frac {5 x^{3/2}}{4 a^2 (b+a x)}-\frac {15 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{4 a^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 27, normalized size = 0.33 \[ \frac {2 x^{7/2} \, _2F_1\left (3,\frac {7}{2};\frac {9}{2};-\frac {a x}{b}\right )}{7 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x)^3*Sqrt[x]),x]

[Out]

(2*x^(7/2)*Hypergeometric2F1[3, 7/2, 9/2, -((a*x)/b)])/(7*b^3)

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fricas [A]  time = 1.03, size = 200, normalized size = 2.44 \[ \left [\frac {15 \, {\left (a^{2} x^{2} + 2 \, a b x + b^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {a x - 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - b}{a x + b}\right ) + 2 \, {\left (8 \, a^{2} x^{2} + 25 \, a b x + 15 \, b^{2}\right )} \sqrt {x}}{8 \, {\left (a^{5} x^{2} + 2 \, a^{4} b x + a^{3} b^{2}\right )}}, -\frac {15 \, {\left (a^{2} x^{2} + 2 \, a b x + b^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {x} \sqrt {\frac {b}{a}}}{b}\right ) - {\left (8 \, a^{2} x^{2} + 25 \, a b x + 15 \, b^{2}\right )} \sqrt {x}}{4 \, {\left (a^{5} x^{2} + 2 \, a^{4} b x + a^{3} b^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(15*(a^2*x^2 + 2*a*b*x + b^2)*sqrt(-b/a)*log((a*x - 2*a*sqrt(x)*sqrt(-b/a) - b)/(a*x + b)) + 2*(8*a^2*x^2
+ 25*a*b*x + 15*b^2)*sqrt(x))/(a^5*x^2 + 2*a^4*b*x + a^3*b^2), -1/4*(15*(a^2*x^2 + 2*a*b*x + b^2)*sqrt(b/a)*a
rctan(a*sqrt(x)*sqrt(b/a)/b) - (8*a^2*x^2 + 25*a*b*x + 15*b^2)*sqrt(x))/(a^5*x^2 + 2*a^4*b*x + a^3*b^2)]

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giac [A]  time = 0.16, size = 59, normalized size = 0.72 \[ -\frac {15 \, b \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{3}} + \frac {2 \, \sqrt {x}}{a^{3}} + \frac {9 \, a b x^{\frac {3}{2}} + 7 \, b^{2} \sqrt {x}}{4 \, {\left (a x + b\right )}^{2} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/4*b*arctan(a*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3) + 2*sqrt(x)/a^3 + 1/4*(9*a*b*x^(3/2) + 7*b^2*sqrt(x))/((a*
x + b)^2*a^3)

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maple [A]  time = 0.02, size = 66, normalized size = 0.80 \[ \frac {9 b \,x^{\frac {3}{2}}}{4 \left (a x +b \right )^{2} a^{2}}+\frac {7 b^{2} \sqrt {x}}{4 \left (a x +b \right )^{2} a^{3}}-\frac {15 b \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}\, a^{3}}+\frac {2 \sqrt {x}}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x^(1/2)/a^3+9/4/a^2*b/(a*x+b)^2*x^(3/2)+7/4/a^3*b^2/(a*x+b)^2*x^(1/2)-15/4/a^3*b/(a*b)^(1/2)*arctan(1/(a*b)^
(1/2)*a*x^(1/2))

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maxima [A]  time = 2.41, size = 75, normalized size = 0.91 \[ \frac {8 \, a^{2} + \frac {25 \, a b}{x} + \frac {15 \, b^{2}}{x^{2}}}{4 \, {\left (\frac {a^{5}}{\sqrt {x}} + \frac {2 \, a^{4} b}{x^{\frac {3}{2}}} + \frac {a^{3} b^{2}}{x^{\frac {5}{2}}}\right )}} + \frac {15 \, b \arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{4 \, \sqrt {a b} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(8*a^2 + 25*a*b/x + 15*b^2/x^2)/(a^5/sqrt(x) + 2*a^4*b/x^(3/2) + a^3*b^2/x^(5/2)) + 15/4*b*arctan(b/(sqrt(
a*b)*sqrt(x)))/(sqrt(a*b)*a^3)

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mupad [B]  time = 0.08, size = 69, normalized size = 0.84 \[ \frac {\frac {7\,b^2\,\sqrt {x}}{4}+\frac {9\,a\,b\,x^{3/2}}{4}}{a^5\,x^2+2\,a^4\,b\,x+a^3\,b^2}+\frac {2\,\sqrt {x}}{a^3}-\frac {15\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{4\,a^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(1/2)*(a + b/x)^3),x)

[Out]

((7*b^2*x^(1/2))/4 + (9*a*b*x^(3/2))/4)/(a^3*b^2 + a^5*x^2 + 2*a^4*b*x) + (2*x^(1/2))/a^3 - (15*b^(1/2)*atan((
a^(1/2)*x^(1/2))/b^(1/2)))/(4*a^(7/2))

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sympy [A]  time = 39.81, size = 816, normalized size = 9.95 \[ \begin {cases} \tilde {\infty } x^{\frac {7}{2}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 \sqrt {x}}{a^{3}} & \text {for}\: b = 0 \\\frac {2 x^{\frac {7}{2}}}{7 b^{3}} & \text {for}\: a = 0 \\\frac {16 i a^{3} \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {1}{a}}}{8 i a^{6} \sqrt {b} x^{2} \sqrt {\frac {1}{a}} + 16 i a^{5} b^{\frac {3}{2}} x \sqrt {\frac {1}{a}} + 8 i a^{4} b^{\frac {5}{2}} \sqrt {\frac {1}{a}}} + \frac {50 i a^{2} b^{\frac {3}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{a}}}{8 i a^{6} \sqrt {b} x^{2} \sqrt {\frac {1}{a}} + 16 i a^{5} b^{\frac {3}{2}} x \sqrt {\frac {1}{a}} + 8 i a^{4} b^{\frac {5}{2}} \sqrt {\frac {1}{a}}} - \frac {15 a^{2} b x^{2} \log {\left (- i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{8 i a^{6} \sqrt {b} x^{2} \sqrt {\frac {1}{a}} + 16 i a^{5} b^{\frac {3}{2}} x \sqrt {\frac {1}{a}} + 8 i a^{4} b^{\frac {5}{2}} \sqrt {\frac {1}{a}}} + \frac {15 a^{2} b x^{2} \log {\left (i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{8 i a^{6} \sqrt {b} x^{2} \sqrt {\frac {1}{a}} + 16 i a^{5} b^{\frac {3}{2}} x \sqrt {\frac {1}{a}} + 8 i a^{4} b^{\frac {5}{2}} \sqrt {\frac {1}{a}}} + \frac {30 i a b^{\frac {5}{2}} \sqrt {x} \sqrt {\frac {1}{a}}}{8 i a^{6} \sqrt {b} x^{2} \sqrt {\frac {1}{a}} + 16 i a^{5} b^{\frac {3}{2}} x \sqrt {\frac {1}{a}} + 8 i a^{4} b^{\frac {5}{2}} \sqrt {\frac {1}{a}}} - \frac {30 a b^{2} x \log {\left (- i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{8 i a^{6} \sqrt {b} x^{2} \sqrt {\frac {1}{a}} + 16 i a^{5} b^{\frac {3}{2}} x \sqrt {\frac {1}{a}} + 8 i a^{4} b^{\frac {5}{2}} \sqrt {\frac {1}{a}}} + \frac {30 a b^{2} x \log {\left (i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{8 i a^{6} \sqrt {b} x^{2} \sqrt {\frac {1}{a}} + 16 i a^{5} b^{\frac {3}{2}} x \sqrt {\frac {1}{a}} + 8 i a^{4} b^{\frac {5}{2}} \sqrt {\frac {1}{a}}} - \frac {15 b^{3} \log {\left (- i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{8 i a^{6} \sqrt {b} x^{2} \sqrt {\frac {1}{a}} + 16 i a^{5} b^{\frac {3}{2}} x \sqrt {\frac {1}{a}} + 8 i a^{4} b^{\frac {5}{2}} \sqrt {\frac {1}{a}}} + \frac {15 b^{3} \log {\left (i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{8 i a^{6} \sqrt {b} x^{2} \sqrt {\frac {1}{a}} + 16 i a^{5} b^{\frac {3}{2}} x \sqrt {\frac {1}{a}} + 8 i a^{4} b^{\frac {5}{2}} \sqrt {\frac {1}{a}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x**(7/2), Eq(a, 0) & Eq(b, 0)), (2*sqrt(x)/a**3, Eq(b, 0)), (2*x**(7/2)/(7*b**3), Eq(a, 0)), (1
6*I*a**3*sqrt(b)*x**(5/2)*sqrt(1/a)/(8*I*a**6*sqrt(b)*x**2*sqrt(1/a) + 16*I*a**5*b**(3/2)*x*sqrt(1/a) + 8*I*a*
*4*b**(5/2)*sqrt(1/a)) + 50*I*a**2*b**(3/2)*x**(3/2)*sqrt(1/a)/(8*I*a**6*sqrt(b)*x**2*sqrt(1/a) + 16*I*a**5*b*
*(3/2)*x*sqrt(1/a) + 8*I*a**4*b**(5/2)*sqrt(1/a)) - 15*a**2*b*x**2*log(-I*sqrt(b)*sqrt(1/a) + sqrt(x))/(8*I*a*
*6*sqrt(b)*x**2*sqrt(1/a) + 16*I*a**5*b**(3/2)*x*sqrt(1/a) + 8*I*a**4*b**(5/2)*sqrt(1/a)) + 15*a**2*b*x**2*log
(I*sqrt(b)*sqrt(1/a) + sqrt(x))/(8*I*a**6*sqrt(b)*x**2*sqrt(1/a) + 16*I*a**5*b**(3/2)*x*sqrt(1/a) + 8*I*a**4*b
**(5/2)*sqrt(1/a)) + 30*I*a*b**(5/2)*sqrt(x)*sqrt(1/a)/(8*I*a**6*sqrt(b)*x**2*sqrt(1/a) + 16*I*a**5*b**(3/2)*x
*sqrt(1/a) + 8*I*a**4*b**(5/2)*sqrt(1/a)) - 30*a*b**2*x*log(-I*sqrt(b)*sqrt(1/a) + sqrt(x))/(8*I*a**6*sqrt(b)*
x**2*sqrt(1/a) + 16*I*a**5*b**(3/2)*x*sqrt(1/a) + 8*I*a**4*b**(5/2)*sqrt(1/a)) + 30*a*b**2*x*log(I*sqrt(b)*sqr
t(1/a) + sqrt(x))/(8*I*a**6*sqrt(b)*x**2*sqrt(1/a) + 16*I*a**5*b**(3/2)*x*sqrt(1/a) + 8*I*a**4*b**(5/2)*sqrt(1
/a)) - 15*b**3*log(-I*sqrt(b)*sqrt(1/a) + sqrt(x))/(8*I*a**6*sqrt(b)*x**2*sqrt(1/a) + 16*I*a**5*b**(3/2)*x*sqr
t(1/a) + 8*I*a**4*b**(5/2)*sqrt(1/a)) + 15*b**3*log(I*sqrt(b)*sqrt(1/a) + sqrt(x))/(8*I*a**6*sqrt(b)*x**2*sqrt
(1/a) + 16*I*a**5*b**(3/2)*x*sqrt(1/a) + 8*I*a**4*b**(5/2)*sqrt(1/a)), True))

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