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

Optimal. Leaf size=84 \[ -\frac {15 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{7/2}}+\frac {5 x^{3/2}}{4 b^2 (a-b x)}-\frac {x^{5/2}}{2 b (a-b x)^2}+\frac {15 \sqrt {x}}{4 b^3} \]

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 84, 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, 208} \begin {gather*} \frac {5 x^{3/2}}{4 b^2 (a-b x)}-\frac {15 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{7/2}}-\frac {x^{5/2}}{2 b (a-b x)^2}+\frac {15 \sqrt {x}}{4 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/(-a + b*x)^3,x]

[Out]

(15*Sqrt[x])/(4*b^3) - x^(5/2)/(2*b*(a - b*x)^2) + (5*x^(3/2))/(4*b^2*(a - b*x)) - (15*Sqrt[a]*ArcTanh[(Sqrt[b
]*Sqrt[x])/Sqrt[a]])/(4*b^(7/2))

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 208

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

Rubi steps

\begin {align*} \int \frac {x^{5/2}}{(-a+b x)^3} \, dx &=-\frac {x^{5/2}}{2 b (a-b x)^2}+\frac {5 \int \frac {x^{3/2}}{(-a+b x)^2} \, dx}{4 b}\\ &=-\frac {x^{5/2}}{2 b (a-b x)^2}+\frac {5 x^{3/2}}{4 b^2 (a-b x)}+\frac {15 \int \frac {\sqrt {x}}{-a+b x} \, dx}{8 b^2}\\ &=\frac {15 \sqrt {x}}{4 b^3}-\frac {x^{5/2}}{2 b (a-b x)^2}+\frac {5 x^{3/2}}{4 b^2 (a-b x)}+\frac {(15 a) \int \frac {1}{\sqrt {x} (-a+b x)} \, dx}{8 b^3}\\ &=\frac {15 \sqrt {x}}{4 b^3}-\frac {x^{5/2}}{2 b (a-b x)^2}+\frac {5 x^{3/2}}{4 b^2 (a-b x)}+\frac {(15 a) \operatorname {Subst}\left (\int \frac {1}{-a+b x^2} \, dx,x,\sqrt {x}\right )}{4 b^3}\\ &=\frac {15 \sqrt {x}}{4 b^3}-\frac {x^{5/2}}{2 b (a-b x)^2}+\frac {5 x^{3/2}}{4 b^2 (a-b x)}-\frac {15 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.00, size = 26, normalized size = 0.31 \begin {gather*} -\frac {2 x^{7/2} \, _2F_1\left (3,\frac {7}{2};\frac {9}{2};\frac {b x}{a}\right )}{7 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/(-a + b*x)^3,x]

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.12, size = 78, normalized size = 0.93 \begin {gather*} \frac {15 a^2 \sqrt {x}-25 a b x^{3/2}+8 b^2 x^{5/2}}{4 b^3 (b x-a)^2}-\frac {15 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(15*a^2*Sqrt[x] - 25*a*b*x^(3/2) + 8*b^2*x^(5/2))/(4*b^3*(-a + b*x)^2) - (15*Sqrt[a]*ArcTanh[(Sqrt[b]*Sqrt[x])
/Sqrt[a]])/(4*b^(7/2))

________________________________________________________________________________________

fricas [A]  time = 0.65, size = 199, normalized size = 2.37 \begin {gather*} \left [\frac {15 \, {\left (b^{2} x^{2} - 2 \, a b x + a^{2}\right )} \sqrt {\frac {a}{b}} \log \left (\frac {b x - 2 \, b \sqrt {x} \sqrt {\frac {a}{b}} + a}{b x - a}\right ) + 2 \, {\left (8 \, b^{2} x^{2} - 25 \, a b x + 15 \, a^{2}\right )} \sqrt {x}}{8 \, {\left (b^{5} x^{2} - 2 \, a b^{4} x + a^{2} b^{3}\right )}}, \frac {15 \, {\left (b^{2} x^{2} - 2 \, a b x + a^{2}\right )} \sqrt {-\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {-\frac {a}{b}}}{a}\right ) + {\left (8 \, b^{2} x^{2} - 25 \, a b x + 15 \, a^{2}\right )} \sqrt {x}}{4 \, {\left (b^{5} x^{2} - 2 \, a b^{4} x + a^{2} b^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(15*(b^2*x^2 - 2*a*b*x + a^2)*sqrt(a/b)*log((b*x - 2*b*sqrt(x)*sqrt(a/b) + a)/(b*x - a)) + 2*(8*b^2*x^2 -
25*a*b*x + 15*a^2)*sqrt(x))/(b^5*x^2 - 2*a*b^4*x + a^2*b^3), 1/4*(15*(b^2*x^2 - 2*a*b*x + a^2)*sqrt(-a/b)*arc
tan(b*sqrt(x)*sqrt(-a/b)/a) + (8*b^2*x^2 - 25*a*b*x + 15*a^2)*sqrt(x))/(b^5*x^2 - 2*a*b^4*x + a^2*b^3)]

________________________________________________________________________________________

giac [A]  time = 0.95, size = 63, normalized size = 0.75 \begin {gather*} \frac {15 \, a \arctan \left (\frac {b \sqrt {x}}{\sqrt {-a b}}\right )}{4 \, \sqrt {-a b} b^{3}} + \frac {2 \, \sqrt {x}}{b^{3}} - \frac {9 \, a b x^{\frac {3}{2}} - 7 \, a^{2} \sqrt {x}}{4 \, {\left (b x - a\right )}^{2} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.01, size = 58, normalized size = 0.69 \begin {gather*} \frac {2 \left (-\frac {15 \arctanh \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}+\frac {-\frac {9 b \,x^{\frac {3}{2}}}{8}+\frac {7 a \sqrt {x}}{8}}{\left (b x -a \right )^{2}}\right ) a}{b^{3}}+\frac {2 \sqrt {x}}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 3.00, size = 90, normalized size = 1.07 \begin {gather*} -\frac {9 \, a b x^{\frac {3}{2}} - 7 \, a^{2} \sqrt {x}}{4 \, {\left (b^{5} x^{2} - 2 \, a b^{4} x + a^{2} b^{3}\right )}} + \frac {15 \, a \log \left (\frac {b \sqrt {x} - \sqrt {a b}}{b \sqrt {x} + \sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{3}} + \frac {2 \, \sqrt {x}}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(9*a*b*x^(3/2) - 7*a^2*sqrt(x))/(b^5*x^2 - 2*a*b^4*x + a^2*b^3) + 15/8*a*log((b*sqrt(x) - sqrt(a*b))/(b*s
qrt(x) + sqrt(a*b)))/(sqrt(a*b)*b^3) + 2*sqrt(x)/b^3

________________________________________________________________________________________

mupad [B]  time = 0.06, size = 69, normalized size = 0.82 \begin {gather*} \frac {\frac {7\,a^2\,\sqrt {x}}{4}-\frac {9\,a\,b\,x^{3/2}}{4}}{a^2\,b^3-2\,a\,b^4\,x+b^5\,x^2}+\frac {2\,\sqrt {x}}{b^3}-\frac {15\,\sqrt {a}\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{4\,b^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-x^(5/2)/(a - b*x)^3,x)

[Out]

((7*a^2*x^(1/2))/4 - (9*a*b*x^(3/2))/4)/(a^2*b^3 + b^5*x^2 - 2*a*b^4*x) + (2*x^(1/2))/b^3 - (15*a^(1/2)*atanh(
(b^(1/2)*x^(1/2))/a^(1/2)))/(4*b^(7/2))

________________________________________________________________________________________

sympy [A]  time = 53.45, size = 756, normalized size = 9.00 \begin {gather*} \begin {cases} \tilde {\infty } \sqrt {x} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 \sqrt {x}}{b^{3}} & \text {for}\: a = 0 \\- \frac {2 x^{\frac {7}{2}}}{7 a^{3}} & \text {for}\: b = 0 \\\frac {30 a^{\frac {5}{2}} b \sqrt {x} \sqrt {\frac {1}{b}}}{8 a^{\frac {5}{2}} b^{4} \sqrt {\frac {1}{b}} - 16 a^{\frac {3}{2}} b^{5} x \sqrt {\frac {1}{b}} + 8 \sqrt {a} b^{6} x^{2} \sqrt {\frac {1}{b}}} - \frac {50 a^{\frac {3}{2}} b^{2} x^{\frac {3}{2}} \sqrt {\frac {1}{b}}}{8 a^{\frac {5}{2}} b^{4} \sqrt {\frac {1}{b}} - 16 a^{\frac {3}{2}} b^{5} x \sqrt {\frac {1}{b}} + 8 \sqrt {a} b^{6} x^{2} \sqrt {\frac {1}{b}}} + \frac {16 \sqrt {a} b^{3} x^{\frac {5}{2}} \sqrt {\frac {1}{b}}}{8 a^{\frac {5}{2}} b^{4} \sqrt {\frac {1}{b}} - 16 a^{\frac {3}{2}} b^{5} x \sqrt {\frac {1}{b}} + 8 \sqrt {a} b^{6} x^{2} \sqrt {\frac {1}{b}}} + \frac {15 a^{3} \log {\left (- \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{\frac {5}{2}} b^{4} \sqrt {\frac {1}{b}} - 16 a^{\frac {3}{2}} b^{5} x \sqrt {\frac {1}{b}} + 8 \sqrt {a} b^{6} x^{2} \sqrt {\frac {1}{b}}} - \frac {15 a^{3} \log {\left (\sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{\frac {5}{2}} b^{4} \sqrt {\frac {1}{b}} - 16 a^{\frac {3}{2}} b^{5} x \sqrt {\frac {1}{b}} + 8 \sqrt {a} b^{6} x^{2} \sqrt {\frac {1}{b}}} - \frac {30 a^{2} b x \log {\left (- \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{\frac {5}{2}} b^{4} \sqrt {\frac {1}{b}} - 16 a^{\frac {3}{2}} b^{5} x \sqrt {\frac {1}{b}} + 8 \sqrt {a} b^{6} x^{2} \sqrt {\frac {1}{b}}} + \frac {30 a^{2} b x \log {\left (\sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{\frac {5}{2}} b^{4} \sqrt {\frac {1}{b}} - 16 a^{\frac {3}{2}} b^{5} x \sqrt {\frac {1}{b}} + 8 \sqrt {a} b^{6} x^{2} \sqrt {\frac {1}{b}}} + \frac {15 a b^{2} x^{2} \log {\left (- \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{\frac {5}{2}} b^{4} \sqrt {\frac {1}{b}} - 16 a^{\frac {3}{2}} b^{5} x \sqrt {\frac {1}{b}} + 8 \sqrt {a} b^{6} x^{2} \sqrt {\frac {1}{b}}} - \frac {15 a b^{2} x^{2} \log {\left (\sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{\frac {5}{2}} b^{4} \sqrt {\frac {1}{b}} - 16 a^{\frac {3}{2}} b^{5} x \sqrt {\frac {1}{b}} + 8 \sqrt {a} b^{6} x^{2} \sqrt {\frac {1}{b}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*sqrt(x), Eq(a, 0) & Eq(b, 0)), (2*sqrt(x)/b**3, Eq(a, 0)), (-2*x**(7/2)/(7*a**3), Eq(b, 0)), (3
0*a**(5/2)*b*sqrt(x)*sqrt(1/b)/(8*a**(5/2)*b**4*sqrt(1/b) - 16*a**(3/2)*b**5*x*sqrt(1/b) + 8*sqrt(a)*b**6*x**2
*sqrt(1/b)) - 50*a**(3/2)*b**2*x**(3/2)*sqrt(1/b)/(8*a**(5/2)*b**4*sqrt(1/b) - 16*a**(3/2)*b**5*x*sqrt(1/b) +
8*sqrt(a)*b**6*x**2*sqrt(1/b)) + 16*sqrt(a)*b**3*x**(5/2)*sqrt(1/b)/(8*a**(5/2)*b**4*sqrt(1/b) - 16*a**(3/2)*b
**5*x*sqrt(1/b) + 8*sqrt(a)*b**6*x**2*sqrt(1/b)) + 15*a**3*log(-sqrt(a)*sqrt(1/b) + sqrt(x))/(8*a**(5/2)*b**4*
sqrt(1/b) - 16*a**(3/2)*b**5*x*sqrt(1/b) + 8*sqrt(a)*b**6*x**2*sqrt(1/b)) - 15*a**3*log(sqrt(a)*sqrt(1/b) + sq
rt(x))/(8*a**(5/2)*b**4*sqrt(1/b) - 16*a**(3/2)*b**5*x*sqrt(1/b) + 8*sqrt(a)*b**6*x**2*sqrt(1/b)) - 30*a**2*b*
x*log(-sqrt(a)*sqrt(1/b) + sqrt(x))/(8*a**(5/2)*b**4*sqrt(1/b) - 16*a**(3/2)*b**5*x*sqrt(1/b) + 8*sqrt(a)*b**6
*x**2*sqrt(1/b)) + 30*a**2*b*x*log(sqrt(a)*sqrt(1/b) + sqrt(x))/(8*a**(5/2)*b**4*sqrt(1/b) - 16*a**(3/2)*b**5*
x*sqrt(1/b) + 8*sqrt(a)*b**6*x**2*sqrt(1/b)) + 15*a*b**2*x**2*log(-sqrt(a)*sqrt(1/b) + sqrt(x))/(8*a**(5/2)*b*
*4*sqrt(1/b) - 16*a**(3/2)*b**5*x*sqrt(1/b) + 8*sqrt(a)*b**6*x**2*sqrt(1/b)) - 15*a*b**2*x**2*log(sqrt(a)*sqrt
(1/b) + sqrt(x))/(8*a**(5/2)*b**4*sqrt(1/b) - 16*a**(3/2)*b**5*x*sqrt(1/b) + 8*sqrt(a)*b**6*x**2*sqrt(1/b)), T
rue))

________________________________________________________________________________________