3.6.56 \(\int \frac {(a-b x)^{5/2}}{x^{5/2}} \, dx\) [556]

Optimal. Leaf size=90 \[ 5 b^2 \sqrt {x} \sqrt {a-b x}+\frac {10 b (a-b x)^{3/2}}{3 \sqrt {x}}-\frac {2 (a-b x)^{5/2}}{3 x^{3/2}}+5 a b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right ) \]

[Out]

-2/3*(-b*x+a)^(5/2)/x^(3/2)+5*a*b^(3/2)*arctan(b^(1/2)*x^(1/2)/(-b*x+a)^(1/2))+10/3*b*(-b*x+a)^(3/2)/x^(1/2)+5
*b^2*x^(1/2)*(-b*x+a)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {49, 52, 65, 223, 209} \begin {gather*} 5 a b^{3/2} \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )+5 b^2 \sqrt {x} \sqrt {a-b x}-\frac {2 (a-b x)^{5/2}}{3 x^{3/2}}+\frac {10 b (a-b x)^{3/2}}{3 \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a - b*x)^(5/2)/x^(5/2),x]

[Out]

5*b^2*Sqrt[x]*Sqrt[a - b*x] + (10*b*(a - b*x)^(3/2))/(3*Sqrt[x]) - (2*(a - b*x)^(5/2))/(3*x^(3/2)) + 5*a*b^(3/
2)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a - b*x]]

Rule 49

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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}}{x^{5/2}} \, dx &=-\frac {2 (a-b x)^{5/2}}{3 x^{3/2}}-\frac {1}{3} (5 b) \int \frac {(a-b x)^{3/2}}{x^{3/2}} \, dx\\ &=\frac {10 b (a-b x)^{3/2}}{3 \sqrt {x}}-\frac {2 (a-b x)^{5/2}}{3 x^{3/2}}+\left (5 b^2\right ) \int \frac {\sqrt {a-b x}}{\sqrt {x}} \, dx\\ &=5 b^2 \sqrt {x} \sqrt {a-b x}+\frac {10 b (a-b x)^{3/2}}{3 \sqrt {x}}-\frac {2 (a-b x)^{5/2}}{3 x^{3/2}}+\frac {1}{2} \left (5 a b^2\right ) \int \frac {1}{\sqrt {x} \sqrt {a-b x}} \, dx\\ &=5 b^2 \sqrt {x} \sqrt {a-b x}+\frac {10 b (a-b x)^{3/2}}{3 \sqrt {x}}-\frac {2 (a-b x)^{5/2}}{3 x^{3/2}}+\left (5 a b^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a-b x^2}} \, dx,x,\sqrt {x}\right )\\ &=5 b^2 \sqrt {x} \sqrt {a-b x}+\frac {10 b (a-b x)^{3/2}}{3 \sqrt {x}}-\frac {2 (a-b x)^{5/2}}{3 x^{3/2}}+\left (5 a b^2\right ) \text {Subst}\left (\int \frac {1}{1+b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a-b x}}\right )\\ &=5 b^2 \sqrt {x} \sqrt {a-b x}+\frac {10 b (a-b x)^{3/2}}{3 \sqrt {x}}-\frac {2 (a-b x)^{5/2}}{3 x^{3/2}}+5 a b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 76, normalized size = 0.84 \begin {gather*} \frac {\sqrt {a-b x} \left (-2 a^2+14 a b x+3 b^2 x^2\right )}{3 x^{3/2}}+5 a \sqrt {-b} b \log \left (-\sqrt {-b} \sqrt {x}+\sqrt {a-b x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a - b*x)^(5/2)/x^(5/2),x]

[Out]

(Sqrt[a - b*x]*(-2*a^2 + 14*a*b*x + 3*b^2*x^2))/(3*x^(3/2)) + 5*a*Sqrt[-b]*b*Log[-(Sqrt[-b]*Sqrt[x]) + Sqrt[a
- b*x]]

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Maple [A]
time = 0.12, size = 86, normalized size = 0.96

method	result	size
risch	\(-\frac {\sqrt {-b x +a}\, \left (-3 x^{2} b^{2}-14 a b x +2 a^{2}\right )}{3 x^{\frac {3}{2}}}+\frac {5 a \,b^{\frac {3}{2}} \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {a}{2 b}\right )}{\sqrt {-x^{2} b +a x}}\right ) \sqrt {x \left (-b x +a \right )}}{2 \sqrt {x}\, \sqrt {-b x +a}}\)	\(86\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(-b*x+a)^(1/2)*(-3*b^2*x^2-14*a*b*x+2*a^2)/x^(3/2)+5/2*a*b^(3/2)*arctan(b^(1/2)*(x-1/2*a/b)/(-b*x^2+a*x)^
(1/2))*(x*(-b*x+a))^(1/2)/x^(1/2)/(-b*x+a)^(1/2)

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Maxima [A]
time = 0.50, size = 84, normalized size = 0.93 \begin {gather*} -5 \, a b^{\frac {3}{2}} \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right ) + \frac {4 \, \sqrt {-b x + a} a b}{\sqrt {x}} + \frac {\sqrt {-b x + a} a b^{2}}{{\left (b - \frac {b x - a}{x}\right )} \sqrt {x}} - \frac {2 \, {\left (-b x + a\right )}^{\frac {3}{2}} a}{3 \, x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.60, size = 139, normalized size = 1.54 \begin {gather*} \left [\frac {15 \, a \sqrt {-b} b x^{2} \log \left (-2 \, b x - 2 \, \sqrt {-b x + a} \sqrt {-b} \sqrt {x} + a\right ) + 2 \, {\left (3 \, b^{2} x^{2} + 14 \, a b x - 2 \, a^{2}\right )} \sqrt {-b x + a} \sqrt {x}}{6 \, x^{2}}, -\frac {15 \, a b^{\frac {3}{2}} x^{2} \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right ) - {\left (3 \, b^{2} x^{2} + 14 \, a b x - 2 \, a^{2}\right )} \sqrt {-b x + a} \sqrt {x}}{3 \, x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(15*a*sqrt(-b)*b*x^2*log(-2*b*x - 2*sqrt(-b*x + a)*sqrt(-b)*sqrt(x) + a) + 2*(3*b^2*x^2 + 14*a*b*x - 2*a^
2)*sqrt(-b*x + a)*sqrt(x))/x^2, -1/3*(15*a*b^(3/2)*x^2*arctan(sqrt(-b*x + a)/(sqrt(b)*sqrt(x))) - (3*b^2*x^2 +
14*a*b*x - 2*a^2)*sqrt(-b*x + a)*sqrt(x))/x^2]

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Sympy [C] Result contains complex when optimal does not.
time = 3.58, size = 245, normalized size = 2.72 \begin {gather*} \begin {cases} - \frac {2 a^{2} \sqrt {b} \sqrt {\frac {a}{b x} - 1}}{3 x} + \frac {14 a b^{\frac {3}{2}} \sqrt {\frac {a}{b x} - 1}}{3} - 5 i a b^{\frac {3}{2}} \log {\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )} + \frac {5 i a b^{\frac {3}{2}} \log {\left (\frac {a}{b x} \right )}}{2} + 5 a b^{\frac {3}{2}} \operatorname {asin}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )} + b^{\frac {5}{2}} x \sqrt {\frac {a}{b x} - 1} & \text {for}\: \left |{\frac {a}{b x}}\right | > 1 \\- \frac {2 i a^{2} \sqrt {b} \sqrt {- \frac {a}{b x} + 1}}{3 x} + \frac {14 i a b^{\frac {3}{2}} \sqrt {- \frac {a}{b x} + 1}}{3} + \frac {5 i a b^{\frac {3}{2}} \log {\left (\frac {a}{b x} \right )}}{2} - 5 i a b^{\frac {3}{2}} \log {\left (\sqrt {- \frac {a}{b x} + 1} + 1 \right )} + i b^{\frac {5}{2}} x \sqrt {- \frac {a}{b x} + 1} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-2*a**2*sqrt(b)*sqrt(a/(b*x) - 1)/(3*x) + 14*a*b**(3/2)*sqrt(a/(b*x) - 1)/3 - 5*I*a*b**(3/2)*log(sq
rt(a)/(sqrt(b)*sqrt(x))) + 5*I*a*b**(3/2)*log(a/(b*x))/2 + 5*a*b**(3/2)*asin(sqrt(b)*sqrt(x)/sqrt(a)) + b**(5/
2)*x*sqrt(a/(b*x) - 1), Abs(a/(b*x)) > 1), (-2*I*a**2*sqrt(b)*sqrt(-a/(b*x) + 1)/(3*x) + 14*I*a*b**(3/2)*sqrt(
-a/(b*x) + 1)/3 + 5*I*a*b**(3/2)*log(a/(b*x))/2 - 5*I*a*b**(3/2)*log(sqrt(-a/(b*x) + 1) + 1) + I*b**(5/2)*x*sq
rt(-a/(b*x) + 1), 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^(5/2),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}}{x^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a - b*x)^(5/2)/x^(5/2),x)

[Out]

int((a - b*x)^(5/2)/x^(5/2), x)

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