3.463 \(\int \frac{x^{5/2}}{(a+b x)^3} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0234385, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {47, 50, 63, 205} \[ -\frac{5 x^{3/2}}{4 b^2 (a+b x)}-\frac{15 \sqrt{a} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 b^{7/2}}-\frac{x^{5/2}}{2 b (a+b x)^2}+\frac{15 \sqrt{x}}{4 b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/(a + b*x)^3,x]

[Out]

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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{x^{5/2}}{(a+b x)^3} \, dx &=-\frac{x^{5/2}}{2 b (a+b x)^2}+\frac{5 \int \frac{x^{3/2}}{(a+b x)^2} \, dx}{4 b}\\ &=-\frac{x^{5/2}}{2 b (a+b x)^2}-\frac{5 x^{3/2}}{4 b^2 (a+b x)}+\frac{15 \int \frac{\sqrt{x}}{a+b x} \, dx}{8 b^2}\\ &=\frac{15 \sqrt{x}}{4 b^3}-\frac{x^{5/2}}{2 b (a+b x)^2}-\frac{5 x^{3/2}}{4 b^2 (a+b x)}-\frac{(15 a) \int \frac{1}{\sqrt{x} (a+b x)} \, dx}{8 b^3}\\ &=\frac{15 \sqrt{x}}{4 b^3}-\frac{x^{5/2}}{2 b (a+b x)^2}-\frac{5 x^{3/2}}{4 b^2 (a+b x)}-\frac{(15 a) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\sqrt{x}\right )}{4 b^3}\\ &=\frac{15 \sqrt{x}}{4 b^3}-\frac{x^{5/2}}{2 b (a+b x)^2}-\frac{5 x^{3/2}}{4 b^2 (a+b x)}-\frac{15 \sqrt{a} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 b^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0047717, size = 27, normalized size = 0.33 \[ \frac{2 x^{7/2} \, _2F_1\left (3,\frac{7}{2};\frac{9}{2};-\frac{b x}{a}\right )}{7 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/(a + b*x)^3,x]

[Out]

(2*x^(7/2)*Hypergeometric2F1[3, 7/2, 9/2, -((b*x)/a)])/(7*a^3)

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Maple [A]  time = 0.011, size = 66, normalized size = 0.8 \begin{align*} 2\,{\frac{\sqrt{x}}{{b}^{3}}}+{\frac{9\,a}{4\,{b}^{2} \left ( bx+a \right ) ^{2}}{x}^{{\frac{3}{2}}}}+{\frac{7\,{a}^{2}}{4\,{b}^{3} \left ( bx+a \right ) ^{2}}\sqrt{x}}-{\frac{15\,a}{4\,{b}^{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(x^(5/2)/(b*x+a)^3,x)

[Out]

2*x^(1/2)/b^3+9/4/b^2*a/(b*x+a)^2*x^(3/2)+7/4/b^3*a^2/(b*x+a)^2*x^(1/2)-15/4/b^3*a/(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(x^(5/2)/(b*x+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.32697, size = 443, normalized size = 5.4 \begin{align*} \left [\frac{15 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt{-\frac{a}{b}} \log \left (\frac{b x - 2 \, b \sqrt{x} \sqrt{-\frac{a}{b}} - a}{b x + a}\right ) + 2 \,{\left (8 \, b^{2} x^{2} + 25 \, a b x + 15 \, a^{2}\right )} \sqrt{x}}{8 \,{\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}, -\frac{15 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt{\frac{a}{b}} \arctan \left (\frac{b \sqrt{x} \sqrt{\frac{a}{b}}}{a}\right ) -{\left (8 \, b^{2} x^{2} + 25 \, a b x + 15 \, a^{2}\right )} \sqrt{x}}{4 \,{\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(b*x+a)^3,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)/(b*x+a)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.25916, size = 80, normalized size = 0.98 \begin{align*} -\frac{15 \, a \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{4 \, \sqrt{a b} b^{3}} + \frac{2 \, \sqrt{x}}{b^{3}} + \frac{9 \, a b x^{\frac{3}{2}} + 7 \, a^{2} \sqrt{x}}{4 \,{\left (b x + a\right )}^{2} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(b*x+a)^3,x, algorithm="giac")

[Out]

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