3.6.54 \(\int \frac {(a-b x)^{5/2}}{\sqrt {x}} \, dx\) [554]

Optimal. Leaf size=96 \[ \frac {5}{8} a^2 \sqrt {x} \sqrt {a-b x}+\frac {5}{12} a \sqrt {x} (a-b x)^{3/2}+\frac {1}{3} \sqrt {x} (a-b x)^{5/2}+\frac {5 a^3 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )}{8 \sqrt {b}} \]

[Out]

5/8*a^3*arctan(b^(1/2)*x^(1/2)/(-b*x+a)^(1/2))/b^(1/2)+5/12*a*(-b*x+a)^(3/2)*x^(1/2)+1/3*(-b*x+a)^(5/2)*x^(1/2
)+5/8*a^2*x^(1/2)*(-b*x+a)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {52, 65, 223, 209} \begin {gather*} \frac {5 a^3 \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )}{8 \sqrt {b}}+\frac {5}{8} a^2 \sqrt {x} \sqrt {a-b x}+\frac {5}{12} a \sqrt {x} (a-b x)^{3/2}+\frac {1}{3} \sqrt {x} (a-b x)^{5/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a - b*x)^(5/2)/Sqrt[x],x]

[Out]

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

Rule 52

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 65

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 209

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

Rule 223

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

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Mathematica [A]
time = 0.13, size = 79, normalized size = 0.82 \begin {gather*} \frac {1}{24} \sqrt {x} \sqrt {a-b x} \left (33 a^2-26 a b x+8 b^2 x^2\right )-\frac {5 a^3 \log \left (-\sqrt {-b} \sqrt {x}+\sqrt {a-b x}\right )}{8 \sqrt {-b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a - b*x)^(5/2)/Sqrt[x],x]

[Out]

(Sqrt[x]*Sqrt[a - b*x]*(33*a^2 - 26*a*b*x + 8*b^2*x^2))/24 - (5*a^3*Log[-(Sqrt[-b]*Sqrt[x]) + Sqrt[a - b*x]])/
(8*Sqrt[-b])

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Maple [A]
time = 0.10, size = 100, normalized size = 1.04

method	result	size
risch	\(\frac {\left (8 x^{2} b^{2}-26 a b x +33 a^{2}\right ) \sqrt {x}\, \sqrt {-b x +a}}{24}+\frac {5 a^{3} \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {a}{2 b}\right )}{\sqrt {-x^{2} b +a x}}\right ) \sqrt {x \left (-b x +a \right )}}{16 \sqrt {b}\, \sqrt {x}\, \sqrt {-b x +a}}\)	\(88\)
default	\(\frac {\left (-b x +a \right )^{\frac {5}{2}} \sqrt {x}}{3}+\frac {5 a \left (\frac {\left (-b x +a \right )^{\frac {3}{2}} \sqrt {x}}{2}+\frac {3 a \left (\sqrt {x}\, \sqrt {-b x +a}+\frac {a \sqrt {x \left (-b x +a \right )}\, \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {a}{2 b}\right )}{\sqrt {-x^{2} b +a x}}\right )}{2 \sqrt {-b x +a}\, \sqrt {x}\, \sqrt {b}}\right )}{4}\right )}{6}\)	\(100\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(-b*x+a)^(5/2)*x^(1/2)+5/6*a*(1/2*(-b*x+a)^(3/2)*x^(1/2)+3/4*a*(x^(1/2)*(-b*x+a)^(1/2)+1/2*a*(x*(-b*x+a))^
(1/2)/(-b*x+a)^(1/2)/x^(1/2)/b^(1/2)*arctan(b^(1/2)*(x-1/2*a/b)/(-b*x^2+a*x)^(1/2))))

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Maxima [A]
time = 0.51, size = 130, normalized size = 1.35 \begin {gather*} -\frac {5 \, a^{3} \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right )}{8 \, \sqrt {b}} + \frac {\frac {15 \, \sqrt {-b x + a} a^{3} b^{2}}{\sqrt {x}} + \frac {40 \, {\left (-b x + a\right )}^{\frac {3}{2}} a^{3} b}{x^{\frac {3}{2}}} + \frac {33 \, {\left (-b x + a\right )}^{\frac {5}{2}} a^{3}}{x^{\frac {5}{2}}}}{24 \, {\left (b^{3} - \frac {3 \, {\left (b x - a\right )} b^{2}}{x} + \frac {3 \, {\left (b x - a\right )}^{2} b}{x^{2}} - \frac {{\left (b x - a\right )}^{3}}{x^{3}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/8*a^3*arctan(sqrt(-b*x + a)/(sqrt(b)*sqrt(x)))/sqrt(b) + 1/24*(15*sqrt(-b*x + a)*a^3*b^2/sqrt(x) + 40*(-b*x
+ a)^(3/2)*a^3*b/x^(3/2) + 33*(-b*x + a)^(5/2)*a^3/x^(5/2))/(b^3 - 3*(b*x - a)*b^2/x + 3*(b*x - a)^2*b/x^2 -
(b*x - a)^3/x^3)

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Fricas [A]
time = 0.57, size = 142, normalized size = 1.48 \begin {gather*} \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} - 26 \, a b^{2} x + 33 \, a^{2} b\right )} \sqrt {-b x + a} \sqrt {x}}{48 \, b}, -\frac {15 \, a^{3} \sqrt {b} \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right ) - {\left (8 \, b^{3} x^{2} - 26 \, a b^{2} x + 33 \, a^{2} b\right )} \sqrt {-b x + a} \sqrt {x}}{24 \, b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/48*(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 - 26*a*b^2*x + 33*a
^2*b)*sqrt(-b*x + a)*sqrt(x))/b, -1/24*(15*a^3*sqrt(b)*arctan(sqrt(-b*x + a)/(sqrt(b)*sqrt(x))) - (8*b^3*x^2 -
26*a*b^2*x + 33*a^2*b)*sqrt(-b*x + a)*sqrt(x))/b]

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Sympy [C] Result contains complex when optimal does not.
time = 3.91, size = 246, normalized size = 2.56 \begin {gather*} \begin {cases} - \frac {11 i a^{\frac {5}{2}} \sqrt {x}}{8 \sqrt {-1 + \frac {b x}{a}}} + \frac {59 i a^{\frac {3}{2}} b x^{\frac {3}{2}}}{24 \sqrt {-1 + \frac {b x}{a}}} - \frac {17 i \sqrt {a} b^{2} x^{\frac {5}{2}}}{12 \sqrt {-1 + \frac {b x}{a}}} - \frac {5 i a^{3} \operatorname {acosh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{8 \sqrt {b}} + \frac {i b^{3} x^{\frac {7}{2}}}{3 \sqrt {a} \sqrt {-1 + \frac {b x}{a}}} & \text {for}\: \left |{\frac {b x}{a}}\right | > 1 \\\frac {11 a^{\frac {5}{2}} \sqrt {x} \sqrt {1 - \frac {b x}{a}}}{8} - \frac {13 a^{\frac {3}{2}} b x^{\frac {3}{2}} \sqrt {1 - \frac {b x}{a}}}{12} + \frac {\sqrt {a} b^{2} x^{\frac {5}{2}} \sqrt {1 - \frac {b x}{a}}}{3} + \frac {5 a^{3} \operatorname {asin}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{8 \sqrt {b}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-11*I*a**(5/2)*sqrt(x)/(8*sqrt(-1 + b*x/a)) + 59*I*a**(3/2)*b*x**(3/2)/(24*sqrt(-1 + b*x/a)) - 17*I
*sqrt(a)*b**2*x**(5/2)/(12*sqrt(-1 + b*x/a)) - 5*I*a**3*acosh(sqrt(b)*sqrt(x)/sqrt(a))/(8*sqrt(b)) + I*b**3*x*
*(7/2)/(3*sqrt(a)*sqrt(-1 + b*x/a)), Abs(b*x/a) > 1), (11*a**(5/2)*sqrt(x)*sqrt(1 - b*x/a)/8 - 13*a**(3/2)*b*x
**(3/2)*sqrt(1 - b*x/a)/12 + sqrt(a)*b**2*x**(5/2)*sqrt(1 - b*x/a)/3 + 5*a**3*asin(sqrt(b)*sqrt(x)/sqrt(a))/(8
*sqrt(b)), True))

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Warning, choosing root of [1,0,%%%{-4,[1
,0,0]%%%}+%%%{4,[0,1,1]%%%}+%%%{4,[0,1,0]%%%}+%%%{-4,[0,0,1]%%%},0,%%%{6,[2,0,0]%%%}+%%%{-12,[1,1,1]%%%}+%%%{-
4,[1,1,0]%%%}+%%%{4,[

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a - b*x)^(5/2)/x^(1/2),x)

[Out]

int((a - b*x)^(5/2)/x^(1/2), x)

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