3.454 \(\int \frac{1}{x^{5/2} (a+b x)} \, dx\)

Optimal. Leaf size=53 \[ \frac{2 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{a^{5/2}}+\frac{2 b}{a^2 \sqrt{x}}-\frac{2}{3 a x^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0176221, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {51, 63, 205} \[ \frac{2 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{a^{5/2}}+\frac{2 b}{a^2 \sqrt{x}}-\frac{2}{3 a x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(5/2)*(a + b*x)),x]

[Out]

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

Rubi steps

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

Mathematica [C]  time = 0.0046637, size = 27, normalized size = 0.51 \[ -\frac{2 \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};-\frac{b x}{a}\right )}{3 a x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(5/2)*(a + b*x)),x]

[Out]

(-2*Hypergeometric2F1[-3/2, 1, -1/2, -((b*x)/a)])/(3*a*x^(3/2))

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Maple [A]  time = 0.009, size = 43, normalized size = 0.8 \begin{align*} -{\frac{2}{3\,a}{x}^{-{\frac{3}{2}}}}+2\,{\frac{b}{{a}^{2}\sqrt{x}}}+2\,{\frac{{b}^{2}}{{a}^{2}\sqrt{ab}}\arctan \left ({\frac{b\sqrt{x}}{\sqrt{ab}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(5/2)/(b*x+a),x)

[Out]

-2/3/a/x^(3/2)+2*b/a^2/x^(1/2)+2*b^2/a^2/(a*b)^(1/2)*arctan(b*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(1/x^(5/2)/(b*x+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.61631, size = 275, normalized size = 5.19 \begin{align*} \left [\frac{3 \, b x^{2} \sqrt{-\frac{b}{a}} \log \left (\frac{b x + 2 \, a \sqrt{x} \sqrt{-\frac{b}{a}} - a}{b x + a}\right ) + 2 \,{\left (3 \, b x - a\right )} \sqrt{x}}{3 \, a^{2} x^{2}}, -\frac{2 \,{\left (3 \, b x^{2} \sqrt{\frac{b}{a}} \arctan \left (\frac{a \sqrt{\frac{b}{a}}}{b \sqrt{x}}\right ) -{\left (3 \, b x - a\right )} \sqrt{x}\right )}}{3 \, a^{2} x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(b*x+a),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 15.8911, size = 121, normalized size = 2.28 \begin{align*} \begin{cases} \frac{\tilde{\infty }}{x^{\frac{5}{2}}} & \text{for}\: a = 0 \wedge b = 0 \\- \frac{2}{5 b x^{\frac{5}{2}}} & \text{for}\: a = 0 \\- \frac{2}{3 a x^{\frac{3}{2}}} & \text{for}\: b = 0 \\- \frac{2}{3 a x^{\frac{3}{2}}} + \frac{2 b}{a^{2} \sqrt{x}} - \frac{i b \log{\left (- i \sqrt{a} \sqrt{\frac{1}{b}} + \sqrt{x} \right )}}{a^{\frac{5}{2}} \sqrt{\frac{1}{b}}} + \frac{i b \log{\left (i \sqrt{a} \sqrt{\frac{1}{b}} + \sqrt{x} \right )}}{a^{\frac{5}{2}} \sqrt{\frac{1}{b}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(5/2)/(b*x+a),x)

[Out]

Piecewise((zoo/x**(5/2), Eq(a, 0) & Eq(b, 0)), (-2/(5*b*x**(5/2)), Eq(a, 0)), (-2/(3*a*x**(3/2)), Eq(b, 0)), (
-2/(3*a*x**(3/2)) + 2*b/(a**2*sqrt(x)) - I*b*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(a**(5/2)*sqrt(1/b)) + I*b*lo
g(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(a**(5/2)*sqrt(1/b)), True))

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Giac [A]  time = 1.2596, size = 55, normalized size = 1.04 \begin{align*} \frac{2 \, b^{2} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{\sqrt{a b} a^{2}} + \frac{2 \,{\left (3 \, b x - a\right )}}{3 \, a^{2} x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(b*x+a),x, algorithm="giac")

[Out]

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