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

Optimal. Leaf size=84 \[ \frac{5 x^{3/2}}{4 b^2 (a-b x)}-\frac{15 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 b^{7/2}}-\frac{x^{5/2}}{2 b (a-b x)^2}+\frac{15 \sqrt{x}}{4 b^3} \]

[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]*ArcTanh[(Sqrt[b
]*Sqrt[x])/Sqrt[a]])/(4*b^(7/2))

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Rubi [A]  time = 0.0260807, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {47, 50, 63, 208} \[ \frac{5 x^{3/2}}{4 b^2 (a-b x)}-\frac{15 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 b^{7/2}}-\frac{x^{5/2}}{2 b (a-b x)^2}+\frac{15 \sqrt{x}}{4 b^3} \]

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]*ArcTanh[(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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[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} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 b^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0050722, size = 26, normalized size = 0.31 \[ -\frac{2 x^{7/2} \, _2F_1\left (3,\frac{7}{2};\frac{9}{2};\frac{b x}{a}\right )}{7 a^3} \]

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 = 0.012, size = 58, normalized size = 0.7 \begin{align*} 2\,{\frac{\sqrt{x}}{{b}^{3}}}+2\,{\frac{a}{{b}^{3}} \left ({\frac{1}{ \left ( bx-a \right ) ^{2}} \left ( -{\frac{9\,b{x}^{3/2}}{8}}+{\frac{7\,a\sqrt{x}}{8}} \right ) }-{\frac{15}{8\,\sqrt{ab}}{\it Artanh} \left ({\frac{b\sqrt{x}}{\sqrt{ab}}} \right ) } \right ) } \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+2/b^3*a*((-9/8*b*x^(3/2)+7/8*a*x^(1/2))/(b*x-a)^2-15/8/(a*b)^(1/2)*arctanh(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 = 1.60673, size = 441, normalized size = 5.25 \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)*arc
tan(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 [A]  time = 174.976, size = 756, normalized size = 9. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*sqrt(x), Eq(a, 0) & Eq(b, 0)), (2*sqrt(x)/b**3, Eq(a, 0)), (-2*x**(7/2)/(7*a**3), Eq(b, 0)), (3
0*a**(5/2)*b*sqrt(x)*sqrt(1/b)/(8*a**(5/2)*b**4*sqrt(1/b) - 16*a**(3/2)*b**5*x*sqrt(1/b) + 8*sqrt(a)*b**6*x**2
*sqrt(1/b)) - 50*a**(3/2)*b**2*x**(3/2)*sqrt(1/b)/(8*a**(5/2)*b**4*sqrt(1/b) - 16*a**(3/2)*b**5*x*sqrt(1/b) +
8*sqrt(a)*b**6*x**2*sqrt(1/b)) + 16*sqrt(a)*b**3*x**(5/2)*sqrt(1/b)/(8*a**(5/2)*b**4*sqrt(1/b) - 16*a**(3/2)*b
**5*x*sqrt(1/b) + 8*sqrt(a)*b**6*x**2*sqrt(1/b)) + 15*a**3*log(-sqrt(a)*sqrt(1/b) + sqrt(x))/(8*a**(5/2)*b**4*
sqrt(1/b) - 16*a**(3/2)*b**5*x*sqrt(1/b) + 8*sqrt(a)*b**6*x**2*sqrt(1/b)) - 15*a**3*log(sqrt(a)*sqrt(1/b) + sq
rt(x))/(8*a**(5/2)*b**4*sqrt(1/b) - 16*a**(3/2)*b**5*x*sqrt(1/b) + 8*sqrt(a)*b**6*x**2*sqrt(1/b)) - 30*a**2*b*
x*log(-sqrt(a)*sqrt(1/b) + sqrt(x))/(8*a**(5/2)*b**4*sqrt(1/b) - 16*a**(3/2)*b**5*x*sqrt(1/b) + 8*sqrt(a)*b**6
*x**2*sqrt(1/b)) + 30*a**2*b*x*log(sqrt(a)*sqrt(1/b) + sqrt(x))/(8*a**(5/2)*b**4*sqrt(1/b) - 16*a**(3/2)*b**5*
x*sqrt(1/b) + 8*sqrt(a)*b**6*x**2*sqrt(1/b)) + 15*a*b**2*x**2*log(-sqrt(a)*sqrt(1/b) + sqrt(x))/(8*a**(5/2)*b*
*4*sqrt(1/b) - 16*a**(3/2)*b**5*x*sqrt(1/b) + 8*sqrt(a)*b**6*x**2*sqrt(1/b)) - 15*a*b**2*x**2*log(sqrt(a)*sqrt
(1/b) + sqrt(x))/(8*a**(5/2)*b**4*sqrt(1/b) - 16*a**(3/2)*b**5*x*sqrt(1/b) + 8*sqrt(a)*b**6*x**2*sqrt(1/b)), T
rue))

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Giac [A]  time = 1.20904, size = 85, normalized size = 1.01 \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)