3.590 \(\int \frac{x^{5/2}}{\sqrt{a-b x}} \, dx\)

Optimal. Leaf size=105 \[ -\frac{5 a^2 \sqrt{x} \sqrt{a-b x}}{8 b^3}+\frac{5 a^3 \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a-b x}}\right )}{8 b^{7/2}}-\frac{5 a x^{3/2} \sqrt{a-b x}}{12 b^2}-\frac{x^{5/2} \sqrt{a-b x}}{3 b} \]

[Out]

(-5*a^2*Sqrt[x]*Sqrt[a - b*x])/(8*b^3) - (5*a*x^(3/2)*Sqrt[a - b*x])/(12*b^2) - (x^(5/2)*Sqrt[a - b*x])/(3*b)
+ (5*a^3*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a - b*x]])/(8*b^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0303057, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {50, 63, 217, 203} \[ -\frac{5 a^2 \sqrt{x} \sqrt{a-b x}}{8 b^3}+\frac{5 a^3 \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a-b x}}\right )}{8 b^{7/2}}-\frac{5 a x^{3/2} \sqrt{a-b x}}{12 b^2}-\frac{x^{5/2} \sqrt{a-b x}}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/Sqrt[a - b*x],x]

[Out]

(-5*a^2*Sqrt[x]*Sqrt[a - b*x])/(8*b^3) - (5*a*x^(3/2)*Sqrt[a - b*x])/(12*b^2) - (x^(5/2)*Sqrt[a - b*x])/(3*b)
+ (5*a^3*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a - b*x]])/(8*b^(7/2))

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{x^{5/2}}{\sqrt{a-b x}} \, dx &=-\frac{x^{5/2} \sqrt{a-b x}}{3 b}+\frac{(5 a) \int \frac{x^{3/2}}{\sqrt{a-b x}} \, dx}{6 b}\\ &=-\frac{5 a x^{3/2} \sqrt{a-b x}}{12 b^2}-\frac{x^{5/2} \sqrt{a-b x}}{3 b}+\frac{\left (5 a^2\right ) \int \frac{\sqrt{x}}{\sqrt{a-b x}} \, dx}{8 b^2}\\ &=-\frac{5 a^2 \sqrt{x} \sqrt{a-b x}}{8 b^3}-\frac{5 a x^{3/2} \sqrt{a-b x}}{12 b^2}-\frac{x^{5/2} \sqrt{a-b x}}{3 b}+\frac{\left (5 a^3\right ) \int \frac{1}{\sqrt{x} \sqrt{a-b x}} \, dx}{16 b^3}\\ &=-\frac{5 a^2 \sqrt{x} \sqrt{a-b x}}{8 b^3}-\frac{5 a x^{3/2} \sqrt{a-b x}}{12 b^2}-\frac{x^{5/2} \sqrt{a-b x}}{3 b}+\frac{\left (5 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a-b x^2}} \, dx,x,\sqrt{x}\right )}{8 b^3}\\ &=-\frac{5 a^2 \sqrt{x} \sqrt{a-b x}}{8 b^3}-\frac{5 a x^{3/2} \sqrt{a-b x}}{12 b^2}-\frac{x^{5/2} \sqrt{a-b x}}{3 b}+\frac{\left (5 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{1+b x^2} \, dx,x,\frac{\sqrt{x}}{\sqrt{a-b x}}\right )}{8 b^3}\\ &=-\frac{5 a^2 \sqrt{x} \sqrt{a-b x}}{8 b^3}-\frac{5 a x^{3/2} \sqrt{a-b x}}{12 b^2}-\frac{x^{5/2} \sqrt{a-b x}}{3 b}+\frac{5 a^3 \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a-b x}}\right )}{8 b^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.173854, size = 88, normalized size = 0.84 \[ \frac{\sqrt{a-b x} \left (\frac{15 a^{5/2} \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{\sqrt{1-\frac{b x}{a}}}-\sqrt{b} \sqrt{x} \left (15 a^2+10 a b x+8 b^2 x^2\right )\right )}{24 b^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/Sqrt[a - b*x],x]

[Out]

(Sqrt[a - b*x]*(-(Sqrt[b]*Sqrt[x]*(15*a^2 + 10*a*b*x + 8*b^2*x^2)) + (15*a^(5/2)*ArcSin[(Sqrt[b]*Sqrt[x])/Sqrt
[a]])/Sqrt[1 - (b*x)/a]))/(24*b^(7/2))

________________________________________________________________________________________

Maple [A]  time = 0.003, size = 108, normalized size = 1. \begin{align*} -{\frac{1}{3\,b}{x}^{{\frac{5}{2}}}\sqrt{-bx+a}}-{\frac{5\,a}{12\,{b}^{2}}{x}^{{\frac{3}{2}}}\sqrt{-bx+a}}-{\frac{5\,{a}^{2}}{8\,{b}^{3}}\sqrt{x}\sqrt{-bx+a}}+{\frac{5\,{a}^{3}}{16}\sqrt{x \left ( -bx+a \right ) }\arctan \left ({\sqrt{b} \left ( x-{\frac{a}{2\,b}} \right ){\frac{1}{\sqrt{-b{x}^{2}+ax}}}} \right ){b}^{-{\frac{7}{2}}}{\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{-bx+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*x^(5/2)*(-b*x+a)^(1/2)/b-5/12*a*x^(3/2)*(-b*x+a)^(1/2)/b^2-5/8*a^2*x^(1/2)*(-b*x+a)^(1/2)/b^3+5/16/b^(7/2
)*a^3*(x*(-b*x+a))^(1/2)/x^(1/2)/(-b*x+a)^(1/2)*arctan(b^(1/2)*(x-1/2/b*a)/(-b*x^2+a*x)^(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(x^(5/2)/(-b*x+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.15587, size = 375, normalized size = 3.57 \begin{align*} \left [-\frac{15 \, a^{3} \sqrt{-b} \log \left (-2 \, b x + 2 \, \sqrt{-b x + a} \sqrt{-b} \sqrt{x} + a\right ) + 2 \,{\left (8 \, b^{3} x^{2} + 10 \, a b^{2} x + 15 \, a^{2} b\right )} \sqrt{-b x + a} \sqrt{x}}{48 \, b^{4}}, -\frac{15 \, a^{3} \sqrt{b} \arctan \left (\frac{\sqrt{-b x + a}}{\sqrt{b} \sqrt{x}}\right ) +{\left (8 \, b^{3} x^{2} + 10 \, a b^{2} x + 15 \, a^{2} b\right )} \sqrt{-b x + a} \sqrt{x}}{24 \, b^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/48*(15*a^3*sqrt(-b)*log(-2*b*x + 2*sqrt(-b*x + a)*sqrt(-b)*sqrt(x) + a) + 2*(8*b^3*x^2 + 10*a*b^2*x + 15*a
^2*b)*sqrt(-b*x + a)*sqrt(x))/b^4, -1/24*(15*a^3*sqrt(b)*arctan(sqrt(-b*x + a)/(sqrt(b)*sqrt(x))) + (8*b^3*x^2
+ 10*a*b^2*x + 15*a^2*b)*sqrt(-b*x + a)*sqrt(x))/b^4]

________________________________________________________________________________________

Sympy [A]  time = 15.0901, size = 272, normalized size = 2.59 \begin{align*} \begin{cases} \frac{5 i a^{\frac{5}{2}} \sqrt{x}}{8 b^{3} \sqrt{-1 + \frac{b x}{a}}} - \frac{5 i a^{\frac{3}{2}} x^{\frac{3}{2}}}{24 b^{2} \sqrt{-1 + \frac{b x}{a}}} - \frac{i \sqrt{a} x^{\frac{5}{2}}}{12 b \sqrt{-1 + \frac{b x}{a}}} - \frac{5 i a^{3} \operatorname{acosh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{8 b^{\frac{7}{2}}} - \frac{i x^{\frac{7}{2}}}{3 \sqrt{a} \sqrt{-1 + \frac{b x}{a}}} & \text{for}\: \frac{\left |{b x}\right |}{\left |{a}\right |} > 1 \\- \frac{5 a^{\frac{5}{2}} \sqrt{x}}{8 b^{3} \sqrt{1 - \frac{b x}{a}}} + \frac{5 a^{\frac{3}{2}} x^{\frac{3}{2}}}{24 b^{2} \sqrt{1 - \frac{b x}{a}}} + \frac{\sqrt{a} x^{\frac{5}{2}}}{12 b \sqrt{1 - \frac{b x}{a}}} + \frac{5 a^{3} \operatorname{asin}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{8 b^{\frac{7}{2}}} + \frac{x^{\frac{7}{2}}}{3 \sqrt{a} \sqrt{1 - \frac{b x}{a}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((5*I*a**(5/2)*sqrt(x)/(8*b**3*sqrt(-1 + b*x/a)) - 5*I*a**(3/2)*x**(3/2)/(24*b**2*sqrt(-1 + b*x/a)) -
I*sqrt(a)*x**(5/2)/(12*b*sqrt(-1 + b*x/a)) - 5*I*a**3*acosh(sqrt(b)*sqrt(x)/sqrt(a))/(8*b**(7/2)) - I*x**(7/2
)/(3*sqrt(a)*sqrt(-1 + b*x/a)), Abs(b*x)/Abs(a) > 1), (-5*a**(5/2)*sqrt(x)/(8*b**3*sqrt(1 - b*x/a)) + 5*a**(3/
2)*x**(3/2)/(24*b**2*sqrt(1 - b*x/a)) + sqrt(a)*x**(5/2)/(12*b*sqrt(1 - b*x/a)) + 5*a**3*asin(sqrt(b)*sqrt(x)/
sqrt(a))/(8*b**(7/2)) + x**(7/2)/(3*sqrt(a)*sqrt(1 - b*x/a)), True))

________________________________________________________________________________________

Giac [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)^(1/2),x, algorithm="giac")

[Out]

Timed out