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

Optimal. Leaf size=121 \[ \frac {5 a^4 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )}{64 b^{3/2}}-\frac {5 a^3 \sqrt {x} \sqrt {a-b x}}{64 b}+\frac {5}{32} a^2 x^{3/2} \sqrt {a-b x}+\frac {5}{24} a x^{3/2} (a-b x)^{3/2}+\frac {1}{4} x^{3/2} (a-b x)^{5/2} \]

[Out]

5/24*a*x^(3/2)*(-b*x+a)^(3/2)+1/4*x^(3/2)*(-b*x+a)^(5/2)+5/64*a^4*arctan(b^(1/2)*x^(1/2)/(-b*x+a)^(1/2))/b^(3/
2)+5/32*a^2*x^(3/2)*(-b*x+a)^(1/2)-5/64*a^3*x^(1/2)*(-b*x+a)^(1/2)/b

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Rubi [A]  time = 0.04, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {50, 63, 217, 203} \[ \frac {5 a^4 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )}{64 b^{3/2}}+\frac {5}{32} a^2 x^{3/2} \sqrt {a-b x}-\frac {5 a^3 \sqrt {x} \sqrt {a-b x}}{64 b}+\frac {5}{24} a x^{3/2} (a-b x)^{3/2}+\frac {1}{4} x^{3/2} (a-b x)^{5/2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]*(a - b*x)^(5/2),x]

[Out]

(-5*a^3*Sqrt[x]*Sqrt[a - b*x])/(64*b) + (5*a^2*x^(3/2)*Sqrt[a - b*x])/32 + (5*a*x^(3/2)*(a - b*x)^(3/2))/24 +
(x^(3/2)*(a - b*x)^(5/2))/4 + (5*a^4*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a - b*x]])/(64*b^(3/2))

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 \sqrt {x} (a-b x)^{5/2} \, dx &=\frac {1}{4} x^{3/2} (a-b x)^{5/2}+\frac {1}{8} (5 a) \int \sqrt {x} (a-b x)^{3/2} \, dx\\ &=\frac {5}{24} a x^{3/2} (a-b x)^{3/2}+\frac {1}{4} x^{3/2} (a-b x)^{5/2}+\frac {1}{16} \left (5 a^2\right ) \int \sqrt {x} \sqrt {a-b x} \, dx\\ &=\frac {5}{32} a^2 x^{3/2} \sqrt {a-b x}+\frac {5}{24} a x^{3/2} (a-b x)^{3/2}+\frac {1}{4} x^{3/2} (a-b x)^{5/2}+\frac {1}{64} \left (5 a^3\right ) \int \frac {\sqrt {x}}{\sqrt {a-b x}} \, dx\\ &=-\frac {5 a^3 \sqrt {x} \sqrt {a-b x}}{64 b}+\frac {5}{32} a^2 x^{3/2} \sqrt {a-b x}+\frac {5}{24} a x^{3/2} (a-b x)^{3/2}+\frac {1}{4} x^{3/2} (a-b x)^{5/2}+\frac {\left (5 a^4\right ) \int \frac {1}{\sqrt {x} \sqrt {a-b x}} \, dx}{128 b}\\ &=-\frac {5 a^3 \sqrt {x} \sqrt {a-b x}}{64 b}+\frac {5}{32} a^2 x^{3/2} \sqrt {a-b x}+\frac {5}{24} a x^{3/2} (a-b x)^{3/2}+\frac {1}{4} x^{3/2} (a-b x)^{5/2}+\frac {\left (5 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a-b x^2}} \, dx,x,\sqrt {x}\right )}{64 b}\\ &=-\frac {5 a^3 \sqrt {x} \sqrt {a-b x}}{64 b}+\frac {5}{32} a^2 x^{3/2} \sqrt {a-b x}+\frac {5}{24} a x^{3/2} (a-b x)^{3/2}+\frac {1}{4} x^{3/2} (a-b x)^{5/2}+\frac {\left (5 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{1+b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a-b x}}\right )}{64 b}\\ &=-\frac {5 a^3 \sqrt {x} \sqrt {a-b x}}{64 b}+\frac {5}{32} a^2 x^{3/2} \sqrt {a-b x}+\frac {5}{24} a x^{3/2} (a-b x)^{3/2}+\frac {1}{4} x^{3/2} (a-b x)^{5/2}+\frac {5 a^4 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )}{64 b^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 98, normalized size = 0.81 \[ \frac {\sqrt {a-b x} \left (\frac {15 a^{7/2} \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {1-\frac {b x}{a}}}+\sqrt {b} \sqrt {x} \left (-15 a^3+118 a^2 b x-136 a b^2 x^2+48 b^3 x^3\right )\right )}{192 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]*(a - b*x)^(5/2),x]

[Out]

(Sqrt[a - b*x]*(Sqrt[b]*Sqrt[x]*(-15*a^3 + 118*a^2*b*x - 136*a*b^2*x^2 + 48*b^3*x^3) + (15*a^(7/2)*ArcSin[(Sqr
t[b]*Sqrt[x])/Sqrt[a]])/Sqrt[1 - (b*x)/a]))/(192*b^(3/2))

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fricas [A]  time = 0.45, size = 164, normalized size = 1.36 \[ \left [-\frac {15 \, a^{4} \sqrt {-b} \log \left (-2 \, b x + 2 \, \sqrt {-b x + a} \sqrt {-b} \sqrt {x} + a\right ) - 2 \, {\left (48 \, b^{4} x^{3} - 136 \, a b^{3} x^{2} + 118 \, a^{2} b^{2} x - 15 \, a^{3} b\right )} \sqrt {-b x + a} \sqrt {x}}{384 \, b^{2}}, -\frac {15 \, a^{4} \sqrt {b} \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right ) - {\left (48 \, b^{4} x^{3} - 136 \, a b^{3} x^{2} + 118 \, a^{2} b^{2} x - 15 \, a^{3} b\right )} \sqrt {-b x + a} \sqrt {x}}{192 \, b^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/384*(15*a^4*sqrt(-b)*log(-2*b*x + 2*sqrt(-b*x + a)*sqrt(-b)*sqrt(x) + a) - 2*(48*b^4*x^3 - 136*a*b^3*x^2 +
118*a^2*b^2*x - 15*a^3*b)*sqrt(-b*x + a)*sqrt(x))/b^2, -1/192*(15*a^4*sqrt(b)*arctan(sqrt(-b*x + a)/(sqrt(b)*
sqrt(x))) - (48*b^4*x^3 - 136*a*b^3*x^2 + 118*a^2*b^2*x - 15*a^3*b)*sqrt(-b*x + a)*sqrt(x))/b^2]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [A]  time = 0.01, size = 118, normalized size = 0.98 \[ \frac {5 \sqrt {-b x +a}\, a^{2} x^{\frac {3}{2}}}{32}+\frac {5 \sqrt {\left (-b x +a \right ) x}\, a^{4} \arctan \left (\frac {\left (x -\frac {a}{2 b}\right ) \sqrt {b}}{\sqrt {-b \,x^{2}+a x}}\right )}{128 \sqrt {-b x +a}\, b^{\frac {3}{2}} \sqrt {x}}-\frac {5 \sqrt {-b x +a}\, a^{3} \sqrt {x}}{64 b}+\frac {5 \left (-b x +a \right )^{\frac {3}{2}} a \,x^{\frac {3}{2}}}{24}+\frac {\left (-b x +a \right )^{\frac {5}{2}} x^{\frac {3}{2}}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^(3/2)*(-b*x+a)^(5/2)+5/24*a*x^(3/2)*(-b*x+a)^(3/2)+5/32*a^2*x^(3/2)*(-b*x+a)^(1/2)-5/64*a^3*x^(1/2)*(-b*
x+a)^(1/2)/b+5/128*a^4/b^(3/2)*((-b*x+a)*x)^(1/2)/x^(1/2)/(-b*x+a)^(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 = 3.05, size = 168, normalized size = 1.39 \[ -\frac {5 \, a^{4} \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right )}{64 \, b^{\frac {3}{2}}} + \frac {\frac {15 \, \sqrt {-b x + a} a^{4} b^{3}}{\sqrt {x}} + \frac {55 \, {\left (-b x + a\right )}^{\frac {3}{2}} a^{4} b^{2}}{x^{\frac {3}{2}}} + \frac {73 \, {\left (-b x + a\right )}^{\frac {5}{2}} a^{4} b}{x^{\frac {5}{2}}} - \frac {15 \, {\left (-b x + a\right )}^{\frac {7}{2}} a^{4}}{x^{\frac {7}{2}}}}{192 \, {\left (b^{5} - \frac {4 \, {\left (b x - a\right )} b^{4}}{x} + \frac {6 \, {\left (b x - a\right )}^{2} b^{3}}{x^{2}} - \frac {4 \, {\left (b x - a\right )}^{3} b^{2}}{x^{3}} + \frac {{\left (b x - a\right )}^{4} b}{x^{4}}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/64*a^4*arctan(sqrt(-b*x + a)/(sqrt(b)*sqrt(x)))/b^(3/2) + 1/192*(15*sqrt(-b*x + a)*a^4*b^3/sqrt(x) + 55*(-b
*x + a)^(3/2)*a^4*b^2/x^(3/2) + 73*(-b*x + a)^(5/2)*a^4*b/x^(5/2) - 15*(-b*x + a)^(7/2)*a^4/x^(7/2))/(b^5 - 4*
(b*x - a)*b^4/x + 6*(b*x - a)^2*b^3/x^2 - 4*(b*x - a)^3*b^2/x^3 + (b*x - a)^4*b/x^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)*(a - b*x)^(5/2),x)

[Out]

int(x^(1/2)*(a - b*x)^(5/2), x)

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sympy [A]  time = 9.81, size = 326, normalized size = 2.69 \[ \begin {cases} \frac {5 i a^{\frac {7}{2}} \sqrt {x}}{64 b \sqrt {-1 + \frac {b x}{a}}} - \frac {133 i a^{\frac {5}{2}} x^{\frac {3}{2}}}{192 \sqrt {-1 + \frac {b x}{a}}} + \frac {127 i a^{\frac {3}{2}} b x^{\frac {5}{2}}}{96 \sqrt {-1 + \frac {b x}{a}}} - \frac {23 i \sqrt {a} b^{2} x^{\frac {7}{2}}}{24 \sqrt {-1 + \frac {b x}{a}}} - \frac {5 i a^{4} \operatorname {acosh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{64 b^{\frac {3}{2}}} + \frac {i b^{3} x^{\frac {9}{2}}}{4 \sqrt {a} \sqrt {-1 + \frac {b x}{a}}} & \text {for}\: \left |{\frac {b x}{a}}\right | > 1 \\- \frac {5 a^{\frac {7}{2}} \sqrt {x}}{64 b \sqrt {1 - \frac {b x}{a}}} + \frac {133 a^{\frac {5}{2}} x^{\frac {3}{2}}}{192 \sqrt {1 - \frac {b x}{a}}} - \frac {127 a^{\frac {3}{2}} b x^{\frac {5}{2}}}{96 \sqrt {1 - \frac {b x}{a}}} + \frac {23 \sqrt {a} b^{2} x^{\frac {7}{2}}}{24 \sqrt {1 - \frac {b x}{a}}} + \frac {5 a^{4} \operatorname {asin}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{64 b^{\frac {3}{2}}} - \frac {b^{3} x^{\frac {9}{2}}}{4 \sqrt {a} \sqrt {1 - \frac {b x}{a}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((5*I*a**(7/2)*sqrt(x)/(64*b*sqrt(-1 + b*x/a)) - 133*I*a**(5/2)*x**(3/2)/(192*sqrt(-1 + b*x/a)) + 127
*I*a**(3/2)*b*x**(5/2)/(96*sqrt(-1 + b*x/a)) - 23*I*sqrt(a)*b**2*x**(7/2)/(24*sqrt(-1 + b*x/a)) - 5*I*a**4*aco
sh(sqrt(b)*sqrt(x)/sqrt(a))/(64*b**(3/2)) + I*b**3*x**(9/2)/(4*sqrt(a)*sqrt(-1 + b*x/a)), Abs(b*x/a) > 1), (-5
*a**(7/2)*sqrt(x)/(64*b*sqrt(1 - b*x/a)) + 133*a**(5/2)*x**(3/2)/(192*sqrt(1 - b*x/a)) - 127*a**(3/2)*b*x**(5/
2)/(96*sqrt(1 - b*x/a)) + 23*sqrt(a)*b**2*x**(7/2)/(24*sqrt(1 - b*x/a)) + 5*a**4*asin(sqrt(b)*sqrt(x)/sqrt(a))
/(64*b**(3/2)) - b**3*x**(9/2)/(4*sqrt(a)*sqrt(1 - b*x/a)), True))

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