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)) + (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)

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Rubi [A]  time = 0.0218529, antiderivative size = 69, normalized size of antiderivative = 1., 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*}

Mathematica [C]  time = 0.0049991, 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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Maple [A]  time = 0.014, size = 60, normalized size = 0.9 \begin{align*} -{\frac{2}{3\,{a}^{2}}{x}^{-{\frac{3}{2}}}}+4\,{\frac{b}{{a}^{3}\sqrt{x}}}+{\frac{{b}^{2}}{{a}^{3} \left ( bx+a \right ) }\sqrt{x}}+5\,{\frac{{b}^{2}}{{a}^{3}\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)^2,x)

[Out]

-2/3/a^2/x^(3/2)+4*b/a^3/x^(1/2)+b^2/a^3*x^(1/2)/(b*x+a)+5*b^2/a^3/(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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.33587, size = 402, normalized size = 5.83 \begin{align*} \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 ] \end{align*}

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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Sympy [A]  time = 116.113, size = 507, normalized size = 7.35 \begin{align*} \begin{cases} \frac{\tilde{\infty }}{x^{\frac{7}{2}}} & \text{for}\: a = 0 \wedge b = 0 \\- \frac{2}{7 b^{2} x^{\frac{7}{2}}} & \text{for}\: a = 0 \\- \frac{2}{3 a^{2} x^{\frac{3}{2}}} & \text{for}\: b = 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} \end{align*}

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/(7*b**2*x**(7/2)), Eq(a, 0)), (-2/(3*a**2*x**(3/2)), Eq(b,
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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Giac [A]  time = 1.23766, size = 78, normalized size = 1.13 \begin{align*} \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}}} \end{align*}

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)
)