3.1679 \(\int \frac {1}{(a+\frac {b}{x})^2 x^{7/2}} \, dx\)

Optimal. Leaf size=56 \[ -\frac {3 \sqrt {a} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{b^{5/2}}+\frac {1}{b \sqrt {x} (a x+b)}-\frac {3}{b^2 \sqrt {x}} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {263, 51, 63, 205} \[ -\frac {3 \sqrt {a} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{b^{5/2}}+\frac {1}{b \sqrt {x} (a x+b)}-\frac {3}{b^2 \sqrt {x}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x)^2*x^(7/2)),x]

[Out]

-3/(b^2*Sqrt[x]) + 1/(b*Sqrt[x]*(b + a*x)) - (3*Sqrt[a]*ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]])/b^(5/2)

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

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Mathematica [C]  time = 0.00, size = 25, normalized size = 0.45 \[ -\frac {2 \, _2F_1\left (-\frac {1}{2},2;\frac {1}{2};-\frac {a x}{b}\right )}{b^2 \sqrt {x}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x)^2*x^(7/2)),x]

[Out]

(-2*Hypergeometric2F1[-1/2, 2, 1/2, -((a*x)/b)])/(b^2*Sqrt[x])

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fricas [A]  time = 0.72, size = 147, normalized size = 2.62 \[ \left [\frac {3 \, {\left (a x^{2} + b x\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {a x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - b}{a x + b}\right ) - 2 \, {\left (3 \, a x + 2 \, b\right )} \sqrt {x}}{2 \, {\left (a b^{2} x^{2} + b^{3} x\right )}}, \frac {3 \, {\left (a x^{2} + b x\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {\frac {a}{b}}}{a \sqrt {x}}\right ) - {\left (3 \, a x + 2 \, b\right )} \sqrt {x}}{a b^{2} x^{2} + b^{3} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^2/x^(7/2),x, algorithm="fricas")

[Out]

[1/2*(3*(a*x^2 + b*x)*sqrt(-a/b)*log((a*x - 2*b*sqrt(x)*sqrt(-a/b) - b)/(a*x + b)) - 2*(3*a*x + 2*b)*sqrt(x))/
(a*b^2*x^2 + b^3*x), (3*(a*x^2 + b*x)*sqrt(a/b)*arctan(b*sqrt(a/b)/(a*sqrt(x))) - (3*a*x + 2*b)*sqrt(x))/(a*b^
2*x^2 + b^3*x)]

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giac [A]  time = 0.16, size = 49, normalized size = 0.88 \[ -\frac {3 \, a \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} - \frac {3 \, a x + 2 \, b}{{\left (a x^{\frac {3}{2}} + b \sqrt {x}\right )} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^2/x^(7/2),x, algorithm="giac")

[Out]

-3*a*arctan(a*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^2) - (3*a*x + 2*b)/((a*x^(3/2) + b*sqrt(x))*b^2)

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maple [A]  time = 0.01, size = 48, normalized size = 0.86 \[ -\frac {3 a \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, b^{2}}-\frac {a \sqrt {x}}{\left (a x +b \right ) b^{2}}-\frac {2}{b^{2} \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b/x)^2/x^(7/2),x)

[Out]

-a/b^2*x^(1/2)/(a*x+b)-3*a/b^2/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*a*x^(1/2))-2/b^2/x^(1/2)

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maxima [A]  time = 2.35, size = 52, normalized size = 0.93 \[ -\frac {a}{{\left (a b^{2} + \frac {b^{3}}{x}\right )} \sqrt {x}} + \frac {3 \, a \arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{\sqrt {a b} b^{2}} - \frac {2}{b^{2} \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^2/x^(7/2),x, algorithm="maxima")

[Out]

-a/((a*b^2 + b^3/x)*sqrt(x)) + 3*a*arctan(b/(sqrt(a*b)*sqrt(x)))/(sqrt(a*b)*b^2) - 2/(b^2*sqrt(x))

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mupad [B]  time = 1.13, size = 48, normalized size = 0.86 \[ -\frac {\frac {2}{b}+\frac {3\,a\,x}{b^2}}{a\,x^{3/2}+b\,\sqrt {x}}-\frac {3\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{b^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(7/2)*(a + b/x)^2),x)

[Out]

- (2/b + (3*a*x)/b^2)/(a*x^(3/2) + b*x^(1/2)) - (3*a^(1/2)*atan((a^(1/2)*x^(1/2))/b^(1/2)))/b^(5/2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)**2/x**(7/2),x)

[Out]

Timed out

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