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

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Rubi [A]  time = 0.02, antiderivative size = 53, normalized size of antiderivative = 1.00, 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*}

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Mathematica [C]  time = 0.01, 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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fricas [A]  time = 0.46, size = 118, normalized size = 2.23 \[ \left [\frac {3 \, b x^{2} \sqrt {-\frac {b}{a}} \log \left (\frac {b x + 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (3 \, b x - a\right )} \sqrt {x}}{3 \, a^{2} x^{2}}, -\frac {2 \, {\left (3 \, b x^{2} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (3 \, b x - a\right )} \sqrt {x}\right )}}{3 \, a^{2} x^{2}}\right ] \]

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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giac [A]  time = 1.16, size = 41, normalized size = 0.77 \[ \frac {2 \, b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2}} + \frac {2 \, {\left (3 \, b x - a\right )}}{3 \, a^{2} x^{\frac {3}{2}}} \]

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))

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maple [A]  time = 0.01, size = 43, normalized size = 0.81 \[ \frac {2 b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, a^{2}}+\frac {2 b}{a^{2} \sqrt {x}}-\frac {2}{3 a \,x^{\frac {3}{2}}} \]

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/a^2*b^2/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x^(1/2))

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maxima [A]  time = 3.02, size = 41, normalized size = 0.77 \[ \frac {2 \, b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2}} + \frac {2 \, {\left (3 \, b x - a\right )}}{3 \, a^{2} x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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))

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mupad [B]  time = 0.10, size = 38, normalized size = 0.72 \[ \frac {2\,b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{a^{5/2}}-\frac {\frac {2}{3\,a}-\frac {2\,b\,x}{a^2}}{x^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(5/2)*(a + b*x)),x)

[Out]

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

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sympy [A]  time = 7.83, size = 121, normalized size = 2.28 \[ \begin {cases} \frac {\tilde {\infty }}{x^{\frac {5}{2}}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {2}{3 a x^{\frac {3}{2}}} & \text {for}\: b = 0 \\- \frac {2}{5 b x^{\frac {5}{2}}} & \text {for}\: a = 0 \\- \frac {2}{3 a x^{\frac {3}{2}}} + \frac {2 b}{a^{2} \sqrt {x}} - \frac {i b \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{a^{\frac {5}{2}} \sqrt {\frac {1}{b}}} + \frac {i b \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{a^{\frac {5}{2}} \sqrt {\frac {1}{b}}} & \text {otherwise} \end {cases} \]

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/(3*a*x**(3/2)), Eq(b, 0)), (-2/(5*b*x**(5/2)), Eq(a, 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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