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

Optimal. Leaf size=146 \[ -\frac{3 a^4 \sqrt{x} \sqrt{a-b x}}{128 b^2}+\frac{3 a^5 \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a-b x}}\right )}{128 b^{5/2}}-\frac{a^3 x^{3/2} \sqrt{a-b x}}{64 b}+\frac{1}{16} a^2 x^{5/2} \sqrt{a-b x}+\frac{1}{8} a x^{5/2} (a-b x)^{3/2}+\frac{1}{5} x^{5/2} (a-b x)^{5/2} \]

[Out]

(-3*a^4*Sqrt[x]*Sqrt[a - b*x])/(128*b^2) - (a^3*x^(3/2)*Sqrt[a - b*x])/(64*b) + (a^2*x^(5/2)*Sqrt[a - b*x])/16
+ (a*x^(5/2)*(a - b*x)^(3/2))/8 + (x^(5/2)*(a - b*x)^(5/2))/5 + (3*a^5*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a - b*x]
])/(128*b^(5/2))

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Rubi [A]  time = 0.0505688, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {50, 63, 217, 203} \[ -\frac{3 a^4 \sqrt{x} \sqrt{a-b x}}{128 b^2}+\frac{3 a^5 \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a-b x}}\right )}{128 b^{5/2}}-\frac{a^3 x^{3/2} \sqrt{a-b x}}{64 b}+\frac{1}{16} a^2 x^{5/2} \sqrt{a-b x}+\frac{1}{8} a x^{5/2} (a-b x)^{3/2}+\frac{1}{5} x^{5/2} (a-b x)^{5/2} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)*(a - b*x)^(5/2),x]

[Out]

(-3*a^4*Sqrt[x]*Sqrt[a - b*x])/(128*b^2) - (a^3*x^(3/2)*Sqrt[a - b*x])/(64*b) + (a^2*x^(5/2)*Sqrt[a - b*x])/16
+ (a*x^(5/2)*(a - b*x)^(3/2))/8 + (x^(5/2)*(a - b*x)^(5/2))/5 + (3*a^5*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a - b*x]
])/(128*b^(5/2))

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 x^{3/2} (a-b x)^{5/2} \, dx &=\frac{1}{5} x^{5/2} (a-b x)^{5/2}+\frac{1}{2} a \int x^{3/2} (a-b x)^{3/2} \, dx\\ &=\frac{1}{8} a x^{5/2} (a-b x)^{3/2}+\frac{1}{5} x^{5/2} (a-b x)^{5/2}+\frac{1}{16} \left (3 a^2\right ) \int x^{3/2} \sqrt{a-b x} \, dx\\ &=\frac{1}{16} a^2 x^{5/2} \sqrt{a-b x}+\frac{1}{8} a x^{5/2} (a-b x)^{3/2}+\frac{1}{5} x^{5/2} (a-b x)^{5/2}+\frac{1}{32} a^3 \int \frac{x^{3/2}}{\sqrt{a-b x}} \, dx\\ &=-\frac{a^3 x^{3/2} \sqrt{a-b x}}{64 b}+\frac{1}{16} a^2 x^{5/2} \sqrt{a-b x}+\frac{1}{8} a x^{5/2} (a-b x)^{3/2}+\frac{1}{5} x^{5/2} (a-b x)^{5/2}+\frac{\left (3 a^4\right ) \int \frac{\sqrt{x}}{\sqrt{a-b x}} \, dx}{128 b}\\ &=-\frac{3 a^4 \sqrt{x} \sqrt{a-b x}}{128 b^2}-\frac{a^3 x^{3/2} \sqrt{a-b x}}{64 b}+\frac{1}{16} a^2 x^{5/2} \sqrt{a-b x}+\frac{1}{8} a x^{5/2} (a-b x)^{3/2}+\frac{1}{5} x^{5/2} (a-b x)^{5/2}+\frac{\left (3 a^5\right ) \int \frac{1}{\sqrt{x} \sqrt{a-b x}} \, dx}{256 b^2}\\ &=-\frac{3 a^4 \sqrt{x} \sqrt{a-b x}}{128 b^2}-\frac{a^3 x^{3/2} \sqrt{a-b x}}{64 b}+\frac{1}{16} a^2 x^{5/2} \sqrt{a-b x}+\frac{1}{8} a x^{5/2} (a-b x)^{3/2}+\frac{1}{5} x^{5/2} (a-b x)^{5/2}+\frac{\left (3 a^5\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a-b x^2}} \, dx,x,\sqrt{x}\right )}{128 b^2}\\ &=-\frac{3 a^4 \sqrt{x} \sqrt{a-b x}}{128 b^2}-\frac{a^3 x^{3/2} \sqrt{a-b x}}{64 b}+\frac{1}{16} a^2 x^{5/2} \sqrt{a-b x}+\frac{1}{8} a x^{5/2} (a-b x)^{3/2}+\frac{1}{5} x^{5/2} (a-b x)^{5/2}+\frac{\left (3 a^5\right ) \operatorname{Subst}\left (\int \frac{1}{1+b x^2} \, dx,x,\frac{\sqrt{x}}{\sqrt{a-b x}}\right )}{128 b^2}\\ &=-\frac{3 a^4 \sqrt{x} \sqrt{a-b x}}{128 b^2}-\frac{a^3 x^{3/2} \sqrt{a-b x}}{64 b}+\frac{1}{16} a^2 x^{5/2} \sqrt{a-b x}+\frac{1}{8} a x^{5/2} (a-b x)^{3/2}+\frac{1}{5} x^{5/2} (a-b x)^{5/2}+\frac{3 a^5 \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a-b x}}\right )}{128 b^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.142884, size = 109, normalized size = 0.75 \[ \frac{\sqrt{a-b x} \left (\sqrt{b} \sqrt{x} \left (248 a^2 b^2 x^2-10 a^3 b x-15 a^4-336 a b^3 x^3+128 b^4 x^4\right )+\frac{15 a^{9/2} \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{\sqrt{1-\frac{b x}{a}}}\right )}{640 b^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)*(a - b*x)^(5/2),x]

[Out]

(Sqrt[a - b*x]*(Sqrt[b]*Sqrt[x]*(-15*a^4 - 10*a^3*b*x + 248*a^2*b^2*x^2 - 336*a*b^3*x^3 + 128*b^4*x^4) + (15*a
^(9/2)*ArcSin[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/Sqrt[1 - (b*x)/a]))/(640*b^(5/2))

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Maple [A]  time = 0.005, size = 146, normalized size = 1. \begin{align*} -{\frac{1}{5\,b}{x}^{{\frac{3}{2}}} \left ( -bx+a \right ) ^{{\frac{7}{2}}}}-{\frac{3\,a}{40\,{b}^{2}}\sqrt{x} \left ( -bx+a \right ) ^{{\frac{7}{2}}}}+{\frac{{a}^{2}}{80\,{b}^{2}} \left ( -bx+a \right ) ^{{\frac{5}{2}}}\sqrt{x}}+{\frac{{a}^{3}}{64\,{b}^{2}} \left ( -bx+a \right ) ^{{\frac{3}{2}}}\sqrt{x}}+{\frac{3\,{a}^{4}}{128\,{b}^{2}}\sqrt{x}\sqrt{-bx+a}}+{\frac{3\,{a}^{5}}{256}\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{5}{2}}}{\frac{1}{\sqrt{-bx+a}}}{\frac{1}{\sqrt{x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(-b*x+a)^(5/2),x)

[Out]

-1/5/b*x^(3/2)*(-b*x+a)^(7/2)-3/40/b^2*a*x^(1/2)*(-b*x+a)^(7/2)+1/80/b^2*a^2*(-b*x+a)^(5/2)*x^(1/2)+1/64/b^2*a
^3*(-b*x+a)^(3/2)*x^(1/2)+3/128*a^4*x^(1/2)*(-b*x+a)^(1/2)/b^2+3/256/b^(5/2)*a^5*(x*(-b*x+a))^(1/2)/(-b*x+a)^(
1/2)/x^(1/2)*arctan(b^(1/2)*(x-1/2/b*a)/(-b*x^2+a*x)^(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^(3/2)*(-b*x+a)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.90761, size = 482, normalized size = 3.3 \begin{align*} \left [-\frac{15 \, a^{5} \sqrt{-b} \log \left (-2 \, b x + 2 \, \sqrt{-b x + a} \sqrt{-b} \sqrt{x} + a\right ) - 2 \,{\left (128 \, b^{5} x^{4} - 336 \, a b^{4} x^{3} + 248 \, a^{2} b^{3} x^{2} - 10 \, a^{3} b^{2} x - 15 \, a^{4} b\right )} \sqrt{-b x + a} \sqrt{x}}{1280 \, b^{3}}, -\frac{15 \, a^{5} \sqrt{b} \arctan \left (\frac{\sqrt{-b x + a}}{\sqrt{b} \sqrt{x}}\right ) -{\left (128 \, b^{5} x^{4} - 336 \, a b^{4} x^{3} + 248 \, a^{2} b^{3} x^{2} - 10 \, a^{3} b^{2} x - 15 \, a^{4} b\right )} \sqrt{-b x + a} \sqrt{x}}{640 \, b^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/1280*(15*a^5*sqrt(-b)*log(-2*b*x + 2*sqrt(-b*x + a)*sqrt(-b)*sqrt(x) + a) - 2*(128*b^5*x^4 - 336*a*b^4*x^3
+ 248*a^2*b^3*x^2 - 10*a^3*b^2*x - 15*a^4*b)*sqrt(-b*x + a)*sqrt(x))/b^3, -1/640*(15*a^5*sqrt(b)*arctan(sqrt(
-b*x + a)/(sqrt(b)*sqrt(x))) - (128*b^5*x^4 - 336*a*b^4*x^3 + 248*a^2*b^3*x^2 - 10*a^3*b^2*x - 15*a^4*b)*sqrt(
-b*x + a)*sqrt(x))/b^3]

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Sympy [A]  time = 52.7216, size = 381, normalized size = 2.61 \begin{align*} \begin{cases} \frac{3 i a^{\frac{9}{2}} \sqrt{x}}{128 b^{2} \sqrt{-1 + \frac{b x}{a}}} - \frac{i a^{\frac{7}{2}} x^{\frac{3}{2}}}{128 b \sqrt{-1 + \frac{b x}{a}}} - \frac{129 i a^{\frac{5}{2}} x^{\frac{5}{2}}}{320 \sqrt{-1 + \frac{b x}{a}}} + \frac{73 i a^{\frac{3}{2}} b x^{\frac{7}{2}}}{80 \sqrt{-1 + \frac{b x}{a}}} - \frac{29 i \sqrt{a} b^{2} x^{\frac{9}{2}}}{40 \sqrt{-1 + \frac{b x}{a}}} - \frac{3 i a^{5} \operatorname{acosh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{128 b^{\frac{5}{2}}} + \frac{i b^{3} x^{\frac{11}{2}}}{5 \sqrt{a} \sqrt{-1 + \frac{b x}{a}}} & \text{for}\: \frac{\left |{b x}\right |}{\left |{a}\right |} > 1 \\- \frac{3 a^{\frac{9}{2}} \sqrt{x}}{128 b^{2} \sqrt{1 - \frac{b x}{a}}} + \frac{a^{\frac{7}{2}} x^{\frac{3}{2}}}{128 b \sqrt{1 - \frac{b x}{a}}} + \frac{129 a^{\frac{5}{2}} x^{\frac{5}{2}}}{320 \sqrt{1 - \frac{b x}{a}}} - \frac{73 a^{\frac{3}{2}} b x^{\frac{7}{2}}}{80 \sqrt{1 - \frac{b x}{a}}} + \frac{29 \sqrt{a} b^{2} x^{\frac{9}{2}}}{40 \sqrt{1 - \frac{b x}{a}}} + \frac{3 a^{5} \operatorname{asin}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{128 b^{\frac{5}{2}}} - \frac{b^{3} x^{\frac{11}{2}}}{5 \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**(3/2)*(-b*x+a)**(5/2),x)

[Out]

Piecewise((3*I*a**(9/2)*sqrt(x)/(128*b**2*sqrt(-1 + b*x/a)) - I*a**(7/2)*x**(3/2)/(128*b*sqrt(-1 + b*x/a)) - 1
29*I*a**(5/2)*x**(5/2)/(320*sqrt(-1 + b*x/a)) + 73*I*a**(3/2)*b*x**(7/2)/(80*sqrt(-1 + b*x/a)) - 29*I*sqrt(a)*
b**2*x**(9/2)/(40*sqrt(-1 + b*x/a)) - 3*I*a**5*acosh(sqrt(b)*sqrt(x)/sqrt(a))/(128*b**(5/2)) + I*b**3*x**(11/2
)/(5*sqrt(a)*sqrt(-1 + b*x/a)), Abs(b*x)/Abs(a) > 1), (-3*a**(9/2)*sqrt(x)/(128*b**2*sqrt(1 - b*x/a)) + a**(7/
2)*x**(3/2)/(128*b*sqrt(1 - b*x/a)) + 129*a**(5/2)*x**(5/2)/(320*sqrt(1 - b*x/a)) - 73*a**(3/2)*b*x**(7/2)/(80
*sqrt(1 - b*x/a)) + 29*sqrt(a)*b**2*x**(9/2)/(40*sqrt(1 - b*x/a)) + 3*a**5*asin(sqrt(b)*sqrt(x)/sqrt(a))/(128*
b**(5/2)) - b**3*x**(11/2)/(5*sqrt(a)*sqrt(1 - b*x/a)), True))

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

[Out]

Timed out