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

Optimal. Leaf size=86 \[ \frac {15}{4} b^2 \sqrt {b x-a}-\frac {15}{4} \sqrt {a} b^2 \tan ^{-1}\left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )-\frac {(b x-a)^{5/2}}{2 x^2}-\frac {5 b (b x-a)^{3/2}}{4 x} \]

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {47, 50, 63, 205} \begin {gather*} \frac {15}{4} b^2 \sqrt {b x-a}-\frac {15}{4} \sqrt {a} b^2 \tan ^{-1}\left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )-\frac {(b x-a)^{5/2}}{2 x^2}-\frac {5 b (b x-a)^{3/2}}{4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-a + b*x)^(5/2)/x^3,x]

[Out]

(15*b^2*Sqrt[-a + b*x])/4 - (5*b*(-a + b*x)^(3/2))/(4*x) - (-a + b*x)^(5/2)/(2*x^2) - (15*Sqrt[a]*b^2*ArcTan[S
qrt[-a + b*x]/Sqrt[a]])/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 205

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [C]  time = 0.01, size = 38, normalized size = 0.44 \begin {gather*} \frac {2 b^2 (b x-a)^{7/2} \, _2F_1\left (3,\frac {7}{2};\frac {9}{2};1-\frac {b x}{a}\right )}{7 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-a + b*x)^(5/2)/x^3,x]

[Out]

(2*b^2*(-a + b*x)^(7/2)*Hypergeometric2F1[3, 7/2, 9/2, 1 - (b*x)/a])/(7*a^3)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.09, size = 76, normalized size = 0.88 \begin {gather*} \frac {\sqrt {b x-a} \left (15 a^2+25 a (b x-a)+8 (b x-a)^2\right )}{4 x^2}-\frac {15}{4} \sqrt {a} b^2 \tan ^{-1}\left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[-a + b*x]*(15*a^2 + 25*a*(-a + b*x) + 8*(-a + b*x)^2))/(4*x^2) - (15*Sqrt[a]*b^2*ArcTan[Sqrt[-a + b*x]/S
qrt[a]])/4

________________________________________________________________________________________

fricas [A]  time = 0.90, size = 139, normalized size = 1.62 \begin {gather*} \left [\frac {15 \, \sqrt {-a} b^{2} x^{2} \log \left (\frac {b x - 2 \, \sqrt {b x - a} \sqrt {-a} - 2 \, a}{x}\right ) + 2 \, {\left (8 \, b^{2} x^{2} + 9 \, a b x - 2 \, a^{2}\right )} \sqrt {b x - a}}{8 \, x^{2}}, -\frac {15 \, \sqrt {a} b^{2} x^{2} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) - {\left (8 \, b^{2} x^{2} + 9 \, a b x - 2 \, a^{2}\right )} \sqrt {b x - a}}{4 \, x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 1.04, size = 83, normalized size = 0.97 \begin {gather*} -\frac {15 \, \sqrt {a} b^{3} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) - 8 \, \sqrt {b x - a} b^{3} - \frac {9 \, {\left (b x - a\right )}^{\frac {3}{2}} a b^{3} + 7 \, \sqrt {b x - a} a^{2} b^{3}}{b^{2} x^{2}}}{4 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(15*sqrt(a)*b^3*arctan(sqrt(b*x - a)/sqrt(a)) - 8*sqrt(b*x - a)*b^3 - (9*(b*x - a)^(3/2)*a*b^3 + 7*sqrt(b
*x - a)*a^2*b^3)/(b^2*x^2))/b

________________________________________________________________________________________

maple [A]  time = 0.01, size = 70, normalized size = 0.81 \begin {gather*} -\frac {15 \sqrt {a}\, b^{2} \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{4}+2 \sqrt {b x -a}\, b^{2}+\frac {7 \sqrt {b x -a}\, a^{2}}{4 x^{2}}+\frac {9 \left (b x -a \right )^{\frac {3}{2}} a}{4 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 2.92, size = 97, normalized size = 1.13 \begin {gather*} -\frac {15}{4} \, \sqrt {a} b^{2} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) + 2 \, \sqrt {b x - a} b^{2} + \frac {9 \, {\left (b x - a\right )}^{\frac {3}{2}} a b^{2} + 7 \, \sqrt {b x - a} a^{2} b^{2}}{4 \, {\left ({\left (b x - a\right )}^{2} + 2 \, {\left (b x - a\right )} a + a^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.09, size = 69, normalized size = 0.80 \begin {gather*} 2\,b^2\,\sqrt {b\,x-a}-\frac {15\,\sqrt {a}\,b^2\,\mathrm {atan}\left (\frac {\sqrt {b\,x-a}}{\sqrt {a}}\right )}{4}+\frac {9\,a\,{\left (b\,x-a\right )}^{3/2}}{4\,x^2}+\frac {7\,a^2\,\sqrt {b\,x-a}}{4\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x - a)^(5/2)/x^3,x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 3.93, size = 267, normalized size = 3.10 \begin {gather*} \begin {cases} - \frac {15 i \sqrt {a} b^{2} \operatorname {acosh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4} - \frac {i a^{3}}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} - 1}} + \frac {11 i a^{2} \sqrt {b}}{4 x^{\frac {3}{2}} \sqrt {\frac {a}{b x} - 1}} - \frac {i a b^{\frac {3}{2}}}{4 \sqrt {x} \sqrt {\frac {a}{b x} - 1}} - \frac {2 i b^{\frac {5}{2}} \sqrt {x}}{\sqrt {\frac {a}{b x} - 1}} & \text {for}\: \left |{\frac {a}{b x}}\right | > 1 \\\frac {15 \sqrt {a} b^{2} \operatorname {asin}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4} + \frac {a^{3}}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {- \frac {a}{b x} + 1}} - \frac {11 a^{2} \sqrt {b}}{4 x^{\frac {3}{2}} \sqrt {- \frac {a}{b x} + 1}} + \frac {a b^{\frac {3}{2}}}{4 \sqrt {x} \sqrt {- \frac {a}{b x} + 1}} + \frac {2 b^{\frac {5}{2}} \sqrt {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**3,x)

[Out]

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

________________________________________________________________________________________