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

Optimal. Leaf size=69 \[ \frac {5 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{7/2}}+\frac {5 b}{a^3 \sqrt {x}}-\frac {5}{3 a^2 x^{3/2}}+\frac {1}{a x^{3/2} (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {51, 63, 205} \[ \frac {5 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{7/2}}+\frac {5 b}{a^3 \sqrt {x}}-\frac {5}{3 a^2 x^{3/2}}+\frac {1}{a x^{3/2} (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(5/2)*(a + b*x)^2),x]

[Out]

-5/(3*a^2*x^(3/2)) + (5*b)/(a^3*Sqrt[x]) + 1/(a*x^(3/2)*(a + b*x)) + (5*b^(3/2)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[
a]])/a^(7/2)

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

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Mathematica [C]  time = 0.01, size = 27, normalized size = 0.39 \[ -\frac {2 \, _2F_1\left (-\frac {3}{2},2;-\frac {1}{2};-\frac {b x}{a}\right )}{3 a^2 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(5/2)*(a + b*x)^2),x]

[Out]

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

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fricas [A]  time = 0.46, size = 184, normalized size = 2.67 \[ \left [\frac {15 \, {\left (b^{2} x^{3} + a b x^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x + 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (15 \, b^{2} x^{2} + 10 \, a b x - 2 \, a^{2}\right )} \sqrt {x}}{6 \, {\left (a^{3} b x^{3} + a^{4} x^{2}\right )}}, -\frac {15 \, {\left (b^{2} x^{3} + a b x^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (15 \, b^{2} x^{2} + 10 \, a b x - 2 \, a^{2}\right )} \sqrt {x}}{3 \, {\left (a^{3} b x^{3} + a^{4} x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(15*(b^2*x^3 + a*b*x^2)*sqrt(-b/a)*log((b*x + 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) + 2*(15*b^2*x^2 + 10
*a*b*x - 2*a^2)*sqrt(x))/(a^3*b*x^3 + a^4*x^2), -1/3*(15*(b^2*x^3 + a*b*x^2)*sqrt(b/a)*arctan(a*sqrt(b/a)/(b*s
qrt(x))) - (15*b^2*x^2 + 10*a*b*x - 2*a^2)*sqrt(x))/(a^3*b*x^3 + a^4*x^2)]

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giac [A]  time = 0.96, size = 58, normalized size = 0.84 \[ \frac {5 \, b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} + \frac {b^{2} \sqrt {x}}{{\left (b x + a\right )} a^{3}} + \frac {2 \, {\left (6 \, b x - a\right )}}{3 \, a^{3} x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*b^2*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3) + b^2*sqrt(x)/((b*x + a)*a^3) + 2/3*(6*b*x - a)/(a^3*x^(3/2)
)

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maple [A]  time = 0.02, size = 60, normalized size = 0.87 \[ \frac {5 b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, a^{3}}+\frac {b^{2} \sqrt {x}}{\left (b x +a \right ) a^{3}}+\frac {4 b}{a^{3} \sqrt {x}}-\frac {2}{3 a^{2} x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.88, size = 64, normalized size = 0.93 \[ \frac {15 \, b^{2} x^{2} + 10 \, a b x - 2 \, a^{2}}{3 \, {\left (a^{3} b x^{\frac {5}{2}} + a^{4} x^{\frac {3}{2}}\right )}} + \frac {5 \, b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(15*b^2*x^2 + 10*a*b*x - 2*a^2)/(a^3*b*x^(5/2) + a^4*x^(3/2)) + 5*b^2*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*
b)*a^3)

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mupad [B]  time = 0.15, size = 58, normalized size = 0.84 \[ \frac {\frac {5\,b^2\,x^2}{a^3}-\frac {2}{3\,a}+\frac {10\,b\,x}{3\,a^2}}{a\,x^{3/2}+b\,x^{5/2}}+\frac {5\,b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{a^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(5/2)*(a + b*x)^2),x)

[Out]

((5*b^2*x^2)/a^3 - 2/(3*a) + (10*b*x)/(3*a^2))/(a*x^(3/2) + b*x^(5/2)) + (5*b^(3/2)*atan((b^(1/2)*x^(1/2))/a^(
1/2)))/a^(7/2)

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sympy [A]  time = 50.52, size = 507, normalized size = 7.35 \[ \begin {cases} \frac {\tilde {\infty }}{x^{\frac {7}{2}}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {2}{3 a^{2} x^{\frac {3}{2}}} & \text {for}\: b = 0 \\- \frac {2}{7 b^{2} x^{\frac {7}{2}}} & \text {for}\: a = 0 \\- \frac {4 i a^{\frac {5}{2}} \sqrt {\frac {1}{b}}}{6 i a^{\frac {9}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 6 i a^{\frac {7}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} + \frac {20 i a^{\frac {3}{2}} b x \sqrt {\frac {1}{b}}}{6 i a^{\frac {9}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 6 i a^{\frac {7}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} + \frac {30 i \sqrt {a} b^{2} x^{2} \sqrt {\frac {1}{b}}}{6 i a^{\frac {9}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 6 i a^{\frac {7}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} + \frac {15 a b x^{\frac {3}{2}} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{6 i a^{\frac {9}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 6 i a^{\frac {7}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} - \frac {15 a b x^{\frac {3}{2}} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{6 i a^{\frac {9}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 6 i a^{\frac {7}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} + \frac {15 b^{2} x^{\frac {5}{2}} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{6 i a^{\frac {9}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 6 i a^{\frac {7}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} - \frac {15 b^{2} x^{\frac {5}{2}} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{6 i a^{\frac {9}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 6 i a^{\frac {7}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo/x**(7/2), Eq(a, 0) & Eq(b, 0)), (-2/(3*a**2*x**(3/2)), Eq(b, 0)), (-2/(7*b**2*x**(7/2)), Eq(a,
0)), (-4*I*a**(5/2)*sqrt(1/b)/(6*I*a**(9/2)*x**(3/2)*sqrt(1/b) + 6*I*a**(7/2)*b*x**(5/2)*sqrt(1/b)) + 20*I*a**
(3/2)*b*x*sqrt(1/b)/(6*I*a**(9/2)*x**(3/2)*sqrt(1/b) + 6*I*a**(7/2)*b*x**(5/2)*sqrt(1/b)) + 30*I*sqrt(a)*b**2*
x**2*sqrt(1/b)/(6*I*a**(9/2)*x**(3/2)*sqrt(1/b) + 6*I*a**(7/2)*b*x**(5/2)*sqrt(1/b)) + 15*a*b*x**(3/2)*log(-I*
sqrt(a)*sqrt(1/b) + sqrt(x))/(6*I*a**(9/2)*x**(3/2)*sqrt(1/b) + 6*I*a**(7/2)*b*x**(5/2)*sqrt(1/b)) - 15*a*b*x*
*(3/2)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(6*I*a**(9/2)*x**(3/2)*sqrt(1/b) + 6*I*a**(7/2)*b*x**(5/2)*sqrt(1/b)
) + 15*b**2*x**(5/2)*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(6*I*a**(9/2)*x**(3/2)*sqrt(1/b) + 6*I*a**(7/2)*b*x**
(5/2)*sqrt(1/b)) - 15*b**2*x**(5/2)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(6*I*a**(9/2)*x**(3/2)*sqrt(1/b) + 6*I*
a**(7/2)*b*x**(5/2)*sqrt(1/b)), True))

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