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

Optimal. Leaf size=93 \[ -\frac {15}{4} a^2 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )-\frac {2 (a-b x)^{5/2}}{\sqrt {x}}-\frac {5}{2} b \sqrt {x} (a-b x)^{3/2}-\frac {15}{4} a b \sqrt {x} \sqrt {a-b x} \]

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 93, 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 = {47, 50, 63, 217, 203} \begin {gather*} -\frac {15}{4} a^2 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )-\frac {2 (a-b x)^{5/2}}{\sqrt {x}}-\frac {5}{2} b \sqrt {x} (a-b x)^{3/2}-\frac {15}{4} a b \sqrt {x} \sqrt {a-b x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a - b*x)^(5/2)/x^(3/2),x]

[Out]

(-15*a*b*Sqrt[x]*Sqrt[a - b*x])/4 - (5*b*Sqrt[x]*(a - b*x)^(3/2))/2 - (2*(a - b*x)^(5/2))/Sqrt[x] - (15*a^2*Sq
rt[b]*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a - b*x]])/4

Rule 47

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 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 \frac {(a-b x)^{5/2}}{x^{3/2}} \, dx &=-\frac {2 (a-b x)^{5/2}}{\sqrt {x}}-(5 b) \int \frac {(a-b x)^{3/2}}{\sqrt {x}} \, dx\\ &=-\frac {5}{2} b \sqrt {x} (a-b x)^{3/2}-\frac {2 (a-b x)^{5/2}}{\sqrt {x}}-\frac {1}{4} (15 a b) \int \frac {\sqrt {a-b x}}{\sqrt {x}} \, dx\\ &=-\frac {15}{4} a b \sqrt {x} \sqrt {a-b x}-\frac {5}{2} b \sqrt {x} (a-b x)^{3/2}-\frac {2 (a-b x)^{5/2}}{\sqrt {x}}-\frac {1}{8} \left (15 a^2 b\right ) \int \frac {1}{\sqrt {x} \sqrt {a-b x}} \, dx\\ &=-\frac {15}{4} a b \sqrt {x} \sqrt {a-b x}-\frac {5}{2} b \sqrt {x} (a-b x)^{3/2}-\frac {2 (a-b x)^{5/2}}{\sqrt {x}}-\frac {1}{4} \left (15 a^2 b\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a-b x^2}} \, dx,x,\sqrt {x}\right )\\ &=-\frac {15}{4} a b \sqrt {x} \sqrt {a-b x}-\frac {5}{2} b \sqrt {x} (a-b x)^{3/2}-\frac {2 (a-b x)^{5/2}}{\sqrt {x}}-\frac {1}{4} \left (15 a^2 b\right ) \operatorname {Subst}\left (\int \frac {1}{1+b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a-b x}}\right )\\ &=-\frac {15}{4} a b \sqrt {x} \sqrt {a-b x}-\frac {5}{2} b \sqrt {x} (a-b x)^{3/2}-\frac {2 (a-b x)^{5/2}}{\sqrt {x}}-\frac {15}{4} a^2 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.01, size = 49, normalized size = 0.53 \begin {gather*} -\frac {2 a^2 \sqrt {a-b x} \, _2F_1\left (-\frac {5}{2},-\frac {1}{2};\frac {1}{2};\frac {b x}{a}\right )}{\sqrt {x} \sqrt {1-\frac {b x}{a}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a - b*x)^(5/2)/x^(3/2),x]

[Out]

(-2*a^2*Sqrt[a - b*x]*Hypergeometric2F1[-5/2, -1/2, 1/2, (b*x)/a])/(Sqrt[x]*Sqrt[1 - (b*x)/a])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.16, size = 79, normalized size = 0.85 \begin {gather*} \frac {\sqrt {a-b x} \left (-8 a^2-9 a b x+2 b^2 x^2\right )}{4 \sqrt {x}}-\frac {15}{4} a^2 \sqrt {-b} \log \left (\sqrt {a-b x}-\sqrt {-b} \sqrt {x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a - b*x]*(-8*a^2 - 9*a*b*x + 2*b^2*x^2))/(4*Sqrt[x]) - (15*a^2*Sqrt[-b]*Log[-(Sqrt[-b]*Sqrt[x]) + Sqrt[a
- b*x]])/4

________________________________________________________________________________________

fricas [A]  time = 1.30, size = 137, normalized size = 1.47 \begin {gather*} \left [\frac {15 \, a^{2} \sqrt {-b} x \log \left (-2 \, b x + 2 \, \sqrt {-b x + a} \sqrt {-b} \sqrt {x} + a\right ) + 2 \, {\left (2 \, b^{2} x^{2} - 9 \, a b x - 8 \, a^{2}\right )} \sqrt {-b x + a} \sqrt {x}}{8 \, x}, \frac {15 \, a^{2} \sqrt {b} x \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right ) + {\left (2 \, b^{2} x^{2} - 9 \, a b x - 8 \, a^{2}\right )} \sqrt {-b x + a} \sqrt {x}}{4 \, x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(15*a^2*sqrt(-b)*x*log(-2*b*x + 2*sqrt(-b*x + a)*sqrt(-b)*sqrt(x) + a) + 2*(2*b^2*x^2 - 9*a*b*x - 8*a^2)*
sqrt(-b*x + a)*sqrt(x))/x, 1/4*(15*a^2*sqrt(b)*x*arctan(sqrt(-b*x + a)/(sqrt(b)*sqrt(x))) + (2*b^2*x^2 - 9*a*b
*x - 8*a^2)*sqrt(-b*x + a)*sqrt(x))/x]

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

maple [F]  time = 0.03, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (-b x +a \right )^{\frac {5}{2}}}{x^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 2.99, size = 112, normalized size = 1.20 \begin {gather*} \frac {15}{4} \, a^{2} \sqrt {b} \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right ) - \frac {2 \, \sqrt {-b x + a} a^{2}}{\sqrt {x}} - \frac {\frac {7 \, \sqrt {-b x + a} a^{2} b^{2}}{\sqrt {x}} + \frac {9 \, {\left (-b x + a\right )}^{\frac {3}{2}} a^{2} b}{x^{\frac {3}{2}}}}{4 \, {\left (b^{2} - \frac {2 \, {\left (b x - a\right )} b}{x} + \frac {{\left (b x - a\right )}^{2}}{x^{2}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15/4*a^2*sqrt(b)*arctan(sqrt(-b*x + a)/(sqrt(b)*sqrt(x))) - 2*sqrt(-b*x + a)*a^2/sqrt(x) - 1/4*(7*sqrt(-b*x +
a)*a^2*b^2/sqrt(x) + 9*(-b*x + a)^(3/2)*a^2*b/x^(3/2))/(b^2 - 2*(b*x - a)*b/x + (b*x - a)^2/x^2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a - b*x)^(5/2)/x^(3/2),x)

[Out]

int((a - b*x)^(5/2)/x^(3/2), x)

________________________________________________________________________________________

sympy [A]  time = 6.22, size = 267, normalized size = 2.87 \begin {gather*} \begin {cases} \frac {2 i a^{\frac {5}{2}}}{\sqrt {x} \sqrt {-1 + \frac {b x}{a}}} + \frac {i a^{\frac {3}{2}} b \sqrt {x}}{4 \sqrt {-1 + \frac {b x}{a}}} - \frac {11 i \sqrt {a} b^{2} x^{\frac {3}{2}}}{4 \sqrt {-1 + \frac {b x}{a}}} + \frac {15 i a^{2} \sqrt {b} \operatorname {acosh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{4} + \frac {i b^{3} x^{\frac {5}{2}}}{2 \sqrt {a} \sqrt {-1 + \frac {b x}{a}}} & \text {for}\: \left |{\frac {b x}{a}}\right | > 1 \\- \frac {2 a^{\frac {5}{2}}}{\sqrt {x} \sqrt {1 - \frac {b x}{a}}} - \frac {a^{\frac {3}{2}} b \sqrt {x}}{4 \sqrt {1 - \frac {b x}{a}}} + \frac {11 \sqrt {a} b^{2} x^{\frac {3}{2}}}{4 \sqrt {1 - \frac {b x}{a}}} - \frac {15 a^{2} \sqrt {b} \operatorname {asin}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{4} - \frac {b^{3} x^{\frac {5}{2}}}{2 \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((-b*x+a)**(5/2)/x**(3/2),x)

[Out]

Piecewise((2*I*a**(5/2)/(sqrt(x)*sqrt(-1 + b*x/a)) + I*a**(3/2)*b*sqrt(x)/(4*sqrt(-1 + b*x/a)) - 11*I*sqrt(a)*
b**2*x**(3/2)/(4*sqrt(-1 + b*x/a)) + 15*I*a**2*sqrt(b)*acosh(sqrt(b)*sqrt(x)/sqrt(a))/4 + I*b**3*x**(5/2)/(2*s
qrt(a)*sqrt(-1 + b*x/a)), Abs(b*x/a) > 1), (-2*a**(5/2)/(sqrt(x)*sqrt(1 - b*x/a)) - a**(3/2)*b*sqrt(x)/(4*sqrt
(1 - b*x/a)) + 11*sqrt(a)*b**2*x**(3/2)/(4*sqrt(1 - b*x/a)) - 15*a**2*sqrt(b)*asin(sqrt(b)*sqrt(x)/sqrt(a))/4
- b**3*x**(5/2)/(2*sqrt(a)*sqrt(1 - b*x/a)), True))

________________________________________________________________________________________