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

Optimal. Leaf size=149 \[ \frac {3 a^5 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )}{128 b^{7/2}}-\frac {3 a^4 \sqrt {x} \sqrt {a-b x}}{128 b^3}-\frac {a^3 x^{3/2} \sqrt {a-b x}}{64 b^2}-\frac {a^2 x^{5/2} \sqrt {a-b x}}{80 b}+\frac {3}{40} a x^{7/2} \sqrt {a-b x}+\frac {1}{5} x^{7/2} (a-b x)^{3/2} \]

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Rubi [A]  time = 0.05, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {50, 63, 217, 203} \begin {gather*} -\frac {a^3 x^{3/2} \sqrt {a-b x}}{64 b^2}-\frac {3 a^4 \sqrt {x} \sqrt {a-b x}}{128 b^3}+\frac {3 a^5 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )}{128 b^{7/2}}-\frac {a^2 x^{5/2} \sqrt {a-b x}}{80 b}+\frac {3}{40} a x^{7/2} \sqrt {a-b x}+\frac {1}{5} x^{7/2} (a-b x)^{3/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(5/2)*(a - b*x)^(3/2),x]

[Out]

(-3*a^4*Sqrt[x]*Sqrt[a - b*x])/(128*b^3) - (a^3*x^(3/2)*Sqrt[a - b*x])/(64*b^2) - (a^2*x^(5/2)*Sqrt[a - b*x])/
(80*b) + (3*a*x^(7/2)*Sqrt[a - b*x])/40 + (x^(7/2)*(a - b*x)^(3/2))/5 + (3*a^5*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a
- b*x]])/(128*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 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])

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]

Rubi steps

\begin {align*} \int x^{5/2} (a-b x)^{3/2} \, dx &=\frac {1}{5} x^{7/2} (a-b x)^{3/2}+\frac {1}{10} (3 a) \int x^{5/2} \sqrt {a-b x} \, dx\\ &=\frac {3}{40} a x^{7/2} \sqrt {a-b x}+\frac {1}{5} x^{7/2} (a-b x)^{3/2}+\frac {1}{80} \left (3 a^2\right ) \int \frac {x^{5/2}}{\sqrt {a-b x}} \, dx\\ &=-\frac {a^2 x^{5/2} \sqrt {a-b x}}{80 b}+\frac {3}{40} a x^{7/2} \sqrt {a-b x}+\frac {1}{5} x^{7/2} (a-b x)^{3/2}+\frac {a^3 \int \frac {x^{3/2}}{\sqrt {a-b x}} \, dx}{32 b}\\ &=-\frac {a^3 x^{3/2} \sqrt {a-b x}}{64 b^2}-\frac {a^2 x^{5/2} \sqrt {a-b x}}{80 b}+\frac {3}{40} a x^{7/2} \sqrt {a-b x}+\frac {1}{5} x^{7/2} (a-b x)^{3/2}+\frac {\left (3 a^4\right ) \int \frac {\sqrt {x}}{\sqrt {a-b x}} \, dx}{128 b^2}\\ &=-\frac {3 a^4 \sqrt {x} \sqrt {a-b x}}{128 b^3}-\frac {a^3 x^{3/2} \sqrt {a-b x}}{64 b^2}-\frac {a^2 x^{5/2} \sqrt {a-b x}}{80 b}+\frac {3}{40} a x^{7/2} \sqrt {a-b x}+\frac {1}{5} x^{7/2} (a-b x)^{3/2}+\frac {\left (3 a^5\right ) \int \frac {1}{\sqrt {x} \sqrt {a-b x}} \, dx}{256 b^3}\\ &=-\frac {3 a^4 \sqrt {x} \sqrt {a-b x}}{128 b^3}-\frac {a^3 x^{3/2} \sqrt {a-b x}}{64 b^2}-\frac {a^2 x^{5/2} \sqrt {a-b x}}{80 b}+\frac {3}{40} a x^{7/2} \sqrt {a-b x}+\frac {1}{5} x^{7/2} (a-b x)^{3/2}+\frac {\left (3 a^5\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a-b x^2}} \, dx,x,\sqrt {x}\right )}{128 b^3}\\ &=-\frac {3 a^4 \sqrt {x} \sqrt {a-b x}}{128 b^3}-\frac {a^3 x^{3/2} \sqrt {a-b x}}{64 b^2}-\frac {a^2 x^{5/2} \sqrt {a-b x}}{80 b}+\frac {3}{40} a x^{7/2} \sqrt {a-b x}+\frac {1}{5} x^{7/2} (a-b x)^{3/2}+\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^3}\\ &=-\frac {3 a^4 \sqrt {x} \sqrt {a-b x}}{128 b^3}-\frac {a^3 x^{3/2} \sqrt {a-b x}}{64 b^2}-\frac {a^2 x^{5/2} \sqrt {a-b x}}{80 b}+\frac {3}{40} a x^{7/2} \sqrt {a-b x}+\frac {1}{5} x^{7/2} (a-b x)^{3/2}+\frac {3 a^5 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )}{128 b^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 110, normalized size = 0.74 \begin {gather*} \frac {\sqrt {a-b x} \left (\frac {15 a^{9/2} \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {1-\frac {b x}{a}}}-\sqrt {b} \sqrt {x} \left (15 a^4+10 a^3 b x+8 a^2 b^2 x^2-176 a b^3 x^3+128 b^4 x^4\right )\right )}{640 b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)*(a - b*x)^(3/2),x]

[Out]

(Sqrt[a - b*x]*(-(Sqrt[b]*Sqrt[x]*(15*a^4 + 10*a^3*b*x + 8*a^2*b^2*x^2 - 176*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^(7/2))

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IntegrateAlgebraic [A]  time = 0.18, size = 117, normalized size = 0.79 \begin {gather*} \frac {3 a^5 \sqrt {-b} \log \left (\sqrt {a-b x}-\sqrt {-b} \sqrt {x}\right )}{128 b^4}+\frac {\sqrt {a-b x} \left (-15 a^4 \sqrt {x}-10 a^3 b x^{3/2}-8 a^2 b^2 x^{5/2}+176 a b^3 x^{7/2}-128 b^4 x^{9/2}\right )}{640 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(5/2)*(a - b*x)^(3/2),x]

[Out]

(Sqrt[a - b*x]*(-15*a^4*Sqrt[x] - 10*a^3*b*x^(3/2) - 8*a^2*b^2*x^(5/2) + 176*a*b^3*x^(7/2) - 128*b^4*x^(9/2)))
/(640*b^3) + (3*a^5*Sqrt[-b]*Log[-(Sqrt[-b]*Sqrt[x]) + Sqrt[a - b*x]])/(128*b^4)

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fricas [A]  time = 1.09, size = 185, normalized size = 1.24 \begin {gather*} \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} - 176 \, a b^{4} x^{3} + 8 \, 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^{4}}, -\frac {15 \, a^{5} \sqrt {b} \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right ) + {\left (128 \, b^{5} x^{4} - 176 \, a b^{4} x^{3} + 8 \, 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^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(-b*x+a)^(3/2),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 - 176*a*b^4*x^3
+ 8*a^2*b^3*x^2 + 10*a^3*b^2*x + 15*a^4*b)*sqrt(-b*x + a)*sqrt(x))/b^4, -1/640*(15*a^5*sqrt(b)*arctan(sqrt(-b
*x + a)/(sqrt(b)*sqrt(x))) + (128*b^5*x^4 - 176*a*b^4*x^3 + 8*a^2*b^3*x^2 + 10*a^3*b^2*x + 15*a^4*b)*sqrt(-b*x
+ a)*sqrt(x))/b^4]

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [A]  time = 0.01, size = 146, normalized size = 0.98 \begin {gather*} \frac {3 \sqrt {\left (-b x +a \right ) x}\, a^{5} \arctan \left (\frac {\left (x -\frac {a}{2 b}\right ) \sqrt {b}}{\sqrt {-b \,x^{2}+a x}}\right )}{256 \sqrt {-b x +a}\, b^{\frac {7}{2}} \sqrt {x}}+\frac {3 \sqrt {-b x +a}\, a^{4} \sqrt {x}}{128 b^{3}}-\frac {\left (-b x +a \right )^{\frac {5}{2}} x^{\frac {5}{2}}}{5 b}+\frac {\left (-b x +a \right )^{\frac {3}{2}} a^{3} \sqrt {x}}{64 b^{3}}-\frac {\left (-b x +a \right )^{\frac {5}{2}} a \,x^{\frac {3}{2}}}{8 b^{2}}-\frac {\left (-b x +a \right )^{\frac {5}{2}} a^{2} \sqrt {x}}{16 b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5/b*x^(5/2)*(-b*x+a)^(5/2)-1/8*a/b^2*x^(3/2)*(-b*x+a)^(5/2)-1/16*a^2/b^3*x^(1/2)*(-b*x+a)^(5/2)+1/64*a^3/b^
3*(-b*x+a)^(3/2)*x^(1/2)+3/128*a^4*x^(1/2)*(-b*x+a)^(1/2)/b^3+3/256*a^5/b^(7/2)*((-b*x+a)*x)^(1/2)/(-b*x+a)^(1
/2)/x^(1/2)*arctan((x-1/2*a/b)/(-b*x^2+a*x)^(1/2)*b^(1/2))

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maxima [A]  time = 3.12, size = 207, normalized size = 1.39 \begin {gather*} -\frac {3 \, a^{5} \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right )}{128 \, b^{\frac {7}{2}}} + \frac {\frac {15 \, \sqrt {-b x + a} a^{5} b^{4}}{\sqrt {x}} + \frac {70 \, {\left (-b x + a\right )}^{\frac {3}{2}} a^{5} b^{3}}{x^{\frac {3}{2}}} - \frac {128 \, {\left (-b x + a\right )}^{\frac {5}{2}} a^{5} b^{2}}{x^{\frac {5}{2}}} - \frac {70 \, {\left (-b x + a\right )}^{\frac {7}{2}} a^{5} b}{x^{\frac {7}{2}}} - \frac {15 \, {\left (-b x + a\right )}^{\frac {9}{2}} a^{5}}{x^{\frac {9}{2}}}}{640 \, {\left (b^{8} - \frac {5 \, {\left (b x - a\right )} b^{7}}{x} + \frac {10 \, {\left (b x - a\right )}^{2} b^{6}}{x^{2}} - \frac {10 \, {\left (b x - a\right )}^{3} b^{5}}{x^{3}} + \frac {5 \, {\left (b x - a\right )}^{4} b^{4}}{x^{4}} - \frac {{\left (b x - a\right )}^{5} b^{3}}{x^{5}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/128*a^5*arctan(sqrt(-b*x + a)/(sqrt(b)*sqrt(x)))/b^(7/2) + 1/640*(15*sqrt(-b*x + a)*a^5*b^4/sqrt(x) + 70*(-
b*x + a)^(3/2)*a^5*b^3/x^(3/2) - 128*(-b*x + a)^(5/2)*a^5*b^2/x^(5/2) - 70*(-b*x + a)^(7/2)*a^5*b/x^(7/2) - 15
*(-b*x + a)^(9/2)*a^5/x^(9/2))/(b^8 - 5*(b*x - a)*b^7/x + 10*(b*x - a)^2*b^6/x^2 - 10*(b*x - a)^3*b^5/x^3 + 5*
(b*x - a)^4*b^4/x^4 - (b*x - a)^5*b^3/x^5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^{5/2}\,{\left (a-b\,x\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(a - b*x)^(3/2),x)

[Out]

int(x^(5/2)*(a - b*x)^(3/2), x)

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sympy [A]  time = 17.69, size = 376, normalized size = 2.52 \begin {gather*} \begin {cases} \frac {3 i a^{\frac {9}{2}} \sqrt {x}}{128 b^{3} \sqrt {-1 + \frac {b x}{a}}} - \frac {i a^{\frac {7}{2}} x^{\frac {3}{2}}}{128 b^{2} \sqrt {-1 + \frac {b x}{a}}} - \frac {i a^{\frac {5}{2}} x^{\frac {5}{2}}}{320 b \sqrt {-1 + \frac {b x}{a}}} - \frac {23 i a^{\frac {3}{2}} x^{\frac {7}{2}}}{80 \sqrt {-1 + \frac {b x}{a}}} + \frac {19 i \sqrt {a} b 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 {7}{2}}} - \frac {i b^{2} x^{\frac {11}{2}}}{5 \sqrt {a} \sqrt {-1 + \frac {b x}{a}}} & \text {for}\: \left |{\frac {b x}{a}}\right | > 1 \\- \frac {3 a^{\frac {9}{2}} \sqrt {x}}{128 b^{3} \sqrt {1 - \frac {b x}{a}}} + \frac {a^{\frac {7}{2}} x^{\frac {3}{2}}}{128 b^{2} \sqrt {1 - \frac {b x}{a}}} + \frac {a^{\frac {5}{2}} x^{\frac {5}{2}}}{320 b \sqrt {1 - \frac {b x}{a}}} + \frac {23 a^{\frac {3}{2}} x^{\frac {7}{2}}}{80 \sqrt {1 - \frac {b x}{a}}} - \frac {19 \sqrt {a} b 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 {7}{2}}} + \frac {b^{2} x^{\frac {11}{2}}}{5 \sqrt {a} \sqrt {1 - \frac {b x}{a}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((3*I*a**(9/2)*sqrt(x)/(128*b**3*sqrt(-1 + b*x/a)) - I*a**(7/2)*x**(3/2)/(128*b**2*sqrt(-1 + b*x/a))
- I*a**(5/2)*x**(5/2)/(320*b*sqrt(-1 + b*x/a)) - 23*I*a**(3/2)*x**(7/2)/(80*sqrt(-1 + b*x/a)) + 19*I*sqrt(a)*b
*x**(9/2)/(40*sqrt(-1 + b*x/a)) - 3*I*a**5*acosh(sqrt(b)*sqrt(x)/sqrt(a))/(128*b**(7/2)) - I*b**2*x**(11/2)/(5
*sqrt(a)*sqrt(-1 + b*x/a)), Abs(b*x/a) > 1), (-3*a**(9/2)*sqrt(x)/(128*b**3*sqrt(1 - b*x/a)) + a**(7/2)*x**(3/
2)/(128*b**2*sqrt(1 - b*x/a)) + a**(5/2)*x**(5/2)/(320*b*sqrt(1 - b*x/a)) + 23*a**(3/2)*x**(7/2)/(80*sqrt(1 -
b*x/a)) - 19*sqrt(a)*b*x**(9/2)/(40*sqrt(1 - b*x/a)) + 3*a**5*asin(sqrt(b)*sqrt(x)/sqrt(a))/(128*b**(7/2)) + b
**2*x**(11/2)/(5*sqrt(a)*sqrt(1 - b*x/a)), True))

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