3.467 \(\int \frac{1}{x^{3/2} (a+b x)^3} \, dx\)

Optimal. Leaf size=82 \[ \frac{5}{4 a^2 \sqrt{x} (a+b x)}-\frac{15 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 a^{7/2}}-\frac{15}{4 a^3 \sqrt{x}}+\frac{1}{2 a \sqrt{x} (a+b x)^2} \]

[Out]

-15/(4*a^3*Sqrt[x]) + 1/(2*a*Sqrt[x]*(a + b*x)^2) + 5/(4*a^2*Sqrt[x]*(a + b*x)) - (15*Sqrt[b]*ArcTan[(Sqrt[b]*
Sqrt[x])/Sqrt[a]])/(4*a^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0251132, antiderivative size = 82, 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} \[ \frac{5}{4 a^2 \sqrt{x} (a+b x)}-\frac{15 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 a^{7/2}}-\frac{15}{4 a^3 \sqrt{x}}+\frac{1}{2 a \sqrt{x} (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(3/2)*(a + b*x)^3),x]

[Out]

-15/(4*a^3*Sqrt[x]) + 1/(2*a*Sqrt[x]*(a + b*x)^2) + 5/(4*a^2*Sqrt[x]*(a + b*x)) - (15*Sqrt[b]*ArcTan[(Sqrt[b]*
Sqrt[x])/Sqrt[a]])/(4*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^{3/2} (a+b x)^3} \, dx &=\frac{1}{2 a \sqrt{x} (a+b x)^2}+\frac{5 \int \frac{1}{x^{3/2} (a+b x)^2} \, dx}{4 a}\\ &=\frac{1}{2 a \sqrt{x} (a+b x)^2}+\frac{5}{4 a^2 \sqrt{x} (a+b x)}+\frac{15 \int \frac{1}{x^{3/2} (a+b x)} \, dx}{8 a^2}\\ &=-\frac{15}{4 a^3 \sqrt{x}}+\frac{1}{2 a \sqrt{x} (a+b x)^2}+\frac{5}{4 a^2 \sqrt{x} (a+b x)}-\frac{(15 b) \int \frac{1}{\sqrt{x} (a+b x)} \, dx}{8 a^3}\\ &=-\frac{15}{4 a^3 \sqrt{x}}+\frac{1}{2 a \sqrt{x} (a+b x)^2}+\frac{5}{4 a^2 \sqrt{x} (a+b x)}-\frac{(15 b) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\sqrt{x}\right )}{4 a^3}\\ &=-\frac{15}{4 a^3 \sqrt{x}}+\frac{1}{2 a \sqrt{x} (a+b x)^2}+\frac{5}{4 a^2 \sqrt{x} (a+b x)}-\frac{15 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 a^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0049071, size = 25, normalized size = 0.3 \[ -\frac{2 \, _2F_1\left (-\frac{1}{2},3;\frac{1}{2};-\frac{b x}{a}\right )}{a^3 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(3/2)*(a + b*x)^3),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.011, size = 66, normalized size = 0.8 \begin{align*} -2\,{\frac{1}{{a}^{3}\sqrt{x}}}-{\frac{7\,{b}^{2}}{4\,{a}^{3} \left ( bx+a \right ) ^{2}}{x}^{{\frac{3}{2}}}}-{\frac{9\,b}{4\,{a}^{2} \left ( bx+a \right ) ^{2}}\sqrt{x}}-{\frac{15\,b}{4\,{a}^{3}}\arctan \left ({b\sqrt{x}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(3/2)/(b*x+a)^3,x)

[Out]

-2/a^3/x^(1/2)-7/4/a^3*b^2/(b*x+a)^2*x^(3/2)-9/4/a^2*b/(b*x+a)^2*x^(1/2)-15/4/a^3*b/(a*b)^(1/2)*arctan(b*x^(1/
2)/(a*b)^(1/2))

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.44803, size = 466, normalized size = 5.68 \begin{align*} \left [\frac{15 \,{\left (b^{2} x^{3} + 2 \, a b x^{2} + a^{2} x\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} + 25 \, a b x + 8 \, a^{2}\right )} \sqrt{x}}{8 \,{\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}}, \frac{15 \,{\left (b^{2} x^{3} + 2 \, a b x^{2} + a^{2} x\right )} \sqrt{\frac{b}{a}} \arctan \left (\frac{a \sqrt{\frac{b}{a}}}{b \sqrt{x}}\right ) -{\left (15 \, b^{2} x^{2} + 25 \, a b x + 8 \, a^{2}\right )} \sqrt{x}}{4 \,{\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(3/2)/(b*x+a)^3,x, algorithm="fricas")

[Out]

[1/8*(15*(b^2*x^3 + 2*a*b*x^2 + a^2*x)*sqrt(-b/a)*log((b*x - 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) - 2*(15*b^
2*x^2 + 25*a*b*x + 8*a^2)*sqrt(x))/(a^3*b^2*x^3 + 2*a^4*b*x^2 + a^5*x), 1/4*(15*(b^2*x^3 + 2*a*b*x^2 + a^2*x)*
sqrt(b/a)*arctan(a*sqrt(b/a)/(b*sqrt(x))) - (15*b^2*x^2 + 25*a*b*x + 8*a^2)*sqrt(x))/(a^3*b^2*x^3 + 2*a^4*b*x^
2 + a^5*x)]

________________________________________________________________________________________

Sympy [A]  time = 155.787, size = 865, normalized size = 10.55 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(3/2)/(b*x+a)**3,x)

[Out]

Piecewise((zoo/x**(7/2), Eq(a, 0) & Eq(b, 0)), (-2/(a**3*sqrt(x)), Eq(b, 0)), (-2/(7*b**3*x**(7/2)), Eq(a, 0))
, (-16*I*a**(5/2)*sqrt(1/b)/(8*I*a**(11/2)*sqrt(x)*sqrt(1/b) + 16*I*a**(9/2)*b*x**(3/2)*sqrt(1/b) + 8*I*a**(7/
2)*b**2*x**(5/2)*sqrt(1/b)) - 50*I*a**(3/2)*b*x*sqrt(1/b)/(8*I*a**(11/2)*sqrt(x)*sqrt(1/b) + 16*I*a**(9/2)*b*x
**(3/2)*sqrt(1/b) + 8*I*a**(7/2)*b**2*x**(5/2)*sqrt(1/b)) - 30*I*sqrt(a)*b**2*x**2*sqrt(1/b)/(8*I*a**(11/2)*sq
rt(x)*sqrt(1/b) + 16*I*a**(9/2)*b*x**(3/2)*sqrt(1/b) + 8*I*a**(7/2)*b**2*x**(5/2)*sqrt(1/b)) - 15*a**2*sqrt(x)
*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(11/2)*sqrt(x)*sqrt(1/b) + 16*I*a**(9/2)*b*x**(3/2)*sqrt(1/b) + 8
*I*a**(7/2)*b**2*x**(5/2)*sqrt(1/b)) + 15*a**2*sqrt(x)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(11/2)*sqrt(
x)*sqrt(1/b) + 16*I*a**(9/2)*b*x**(3/2)*sqrt(1/b) + 8*I*a**(7/2)*b**2*x**(5/2)*sqrt(1/b)) - 30*a*b*x**(3/2)*lo
g(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(11/2)*sqrt(x)*sqrt(1/b) + 16*I*a**(9/2)*b*x**(3/2)*sqrt(1/b) + 8*I*
a**(7/2)*b**2*x**(5/2)*sqrt(1/b)) + 30*a*b*x**(3/2)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(11/2)*sqrt(x)*
sqrt(1/b) + 16*I*a**(9/2)*b*x**(3/2)*sqrt(1/b) + 8*I*a**(7/2)*b**2*x**(5/2)*sqrt(1/b)) - 15*b**2*x**(5/2)*log(
-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(11/2)*sqrt(x)*sqrt(1/b) + 16*I*a**(9/2)*b*x**(3/2)*sqrt(1/b) + 8*I*a*
*(7/2)*b**2*x**(5/2)*sqrt(1/b)) + 15*b**2*x**(5/2)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(11/2)*sqrt(x)*s
qrt(1/b) + 16*I*a**(9/2)*b*x**(3/2)*sqrt(1/b) + 8*I*a**(7/2)*b**2*x**(5/2)*sqrt(1/b)), True))

________________________________________________________________________________________

Giac [A]  time = 1.22685, size = 80, normalized size = 0.98 \begin{align*} -\frac{15 \, b \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{4 \, \sqrt{a b} a^{3}} - \frac{2}{a^{3} \sqrt{x}} - \frac{7 \, b^{2} x^{\frac{3}{2}} + 9 \, a b \sqrt{x}}{4 \,{\left (b x + a\right )}^{2} a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(3/2)/(b*x+a)^3,x, algorithm="giac")

[Out]

-15/4*b*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3) - 2/(a^3*sqrt(x)) - 1/4*(7*b^2*x^(3/2) + 9*a*b*sqrt(x))/((
b*x + a)^2*a^3)