\(\int \frac {x^{5/2}}{(-a+b x)^3} \, dx\) [483]

Optimal result

Integrand size = 15, antiderivative size = 84 \[ \int \frac {x^{5/2}}{(-a+b x)^3} \, dx=\frac {15 \sqrt {x}}{4 b^3}-\frac {x^{5/2}}{2 b (a-b x)^2}+\frac {5 x^{3/2}}{4 b^2 (a-b x)}-\frac {15 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{7/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {43, 52, 65, 214} \[ \int \frac {x^{5/2}}{(-a+b x)^3} \, dx=-\frac {15 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{7/2}}+\frac {5 x^{3/2}}{4 b^2 (a-b x)}-\frac {x^{5/2}}{2 b (a-b x)^2}+\frac {15 \sqrt {x}}{4 b^3} \]

[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 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

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 214

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

Rubi steps \begin{align*} \text {integral}& = -\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) \text {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 [A] (verified)

Time = 0.12 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.85 \[ \int \frac {x^{5/2}}{(-a+b x)^3} \, dx=\frac {\sqrt {x} \left (15 a^2-25 a b x+8 b^2 x^2\right )}{4 b^3 (a-b x)^2}-\frac {15 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{7/2}} \]

[In]

Integrate[x^(5/2)/(-a + b*x)^3,x]

[Out]

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

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.68

method	result	size
derivativedivides	\(\frac {2 \sqrt {x}}{b^{3}}-\frac {2 a \left (\frac {\frac {9 b \,x^{\frac {3}{2}}}{8}-\frac {7 a \sqrt {x}}{8}}{\left (-b x +a \right )^{2}}+\frac {15 \,\operatorname {arctanh}\left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{b^{3}}\)	\(57\)
default	\(\frac {2 \sqrt {x}}{b^{3}}-\frac {2 a \left (\frac {\frac {9 b \,x^{\frac {3}{2}}}{8}-\frac {7 a \sqrt {x}}{8}}{\left (-b x +a \right )^{2}}+\frac {15 \,\operatorname {arctanh}\left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{b^{3}}\)	\(57\)
risch	\(\frac {2 \sqrt {x}}{b^{3}}+\frac {a \left (\frac {-\frac {9 b \,x^{\frac {3}{2}}}{4}+\frac {7 a \sqrt {x}}{4}}{\left (b x -a \right )^{2}}-\frac {15 \,\operatorname {arctanh}\left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}}\right )}{b^{3}}\)	\(58\)

[In]

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

[Out]

2*x^(1/2)/b^3-2/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(b*x^(1/2)/(a*b)^(1/2)
))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 199, normalized size of antiderivative = 2.37 \[ \int \frac {x^{5/2}}{(-a+b x)^3} \, dx=\left [\frac {15 \, {\left (b^{2} x^{2} - 2 \, a b x + a^{2}\right )} \sqrt {\frac {a}{b}} \log \left (\frac {b x - 2 \, b \sqrt {x} \sqrt {\frac {a}{b}} + a}{b x - a}\right ) + 2 \, {\left (8 \, b^{2} x^{2} - 25 \, a b x + 15 \, a^{2}\right )} \sqrt {x}}{8 \, {\left (b^{5} x^{2} - 2 \, a b^{4} x + a^{2} b^{3}\right )}}, \frac {15 \, {\left (b^{2} x^{2} - 2 \, a b x + a^{2}\right )} \sqrt {-\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {-\frac {a}{b}}}{a}\right ) + {\left (8 \, b^{2} x^{2} - 25 \, a b x + 15 \, a^{2}\right )} \sqrt {x}}{4 \, {\left (b^{5} x^{2} - 2 \, a b^{4} x + a^{2} b^{3}\right )}}\right ] \]

[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)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 624 vs. \(2 (73) = 146\).

Time = 22.94 (sec) , antiderivative size = 624, normalized size of antiderivative = 7.43 \[ \int \frac {x^{5/2}}{(-a+b x)^3} \, dx=\begin {cases} \tilde {\infty } \sqrt {x} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {2 x^{\frac {7}{2}}}{7 a^{3}} & \text {for}\: b = 0 \\\frac {2 \sqrt {x}}{b^{3}} & \text {for}\: a = 0 \\\frac {15 a^{3} \log {\left (\sqrt {x} - \sqrt {\frac {a}{b}} \right )}}{8 a^{2} b^{4} \sqrt {\frac {a}{b}} - 16 a b^{5} x \sqrt {\frac {a}{b}} + 8 b^{6} x^{2} \sqrt {\frac {a}{b}}} - \frac {15 a^{3} \log {\left (\sqrt {x} + \sqrt {\frac {a}{b}} \right )}}{8 a^{2} b^{4} \sqrt {\frac {a}{b}} - 16 a b^{5} x \sqrt {\frac {a}{b}} + 8 b^{6} x^{2} \sqrt {\frac {a}{b}}} + \frac {30 a^{2} b \sqrt {x} \sqrt {\frac {a}{b}}}{8 a^{2} b^{4} \sqrt {\frac {a}{b}} - 16 a b^{5} x \sqrt {\frac {a}{b}} + 8 b^{6} x^{2} \sqrt {\frac {a}{b}}} - \frac {30 a^{2} b x \log {\left (\sqrt {x} - \sqrt {\frac {a}{b}} \right )}}{8 a^{2} b^{4} \sqrt {\frac {a}{b}} - 16 a b^{5} x \sqrt {\frac {a}{b}} + 8 b^{6} x^{2} \sqrt {\frac {a}{b}}} + \frac {30 a^{2} b x \log {\left (\sqrt {x} + \sqrt {\frac {a}{b}} \right )}}{8 a^{2} b^{4} \sqrt {\frac {a}{b}} - 16 a b^{5} x \sqrt {\frac {a}{b}} + 8 b^{6} x^{2} \sqrt {\frac {a}{b}}} - \frac {50 a b^{2} x^{\frac {3}{2}} \sqrt {\frac {a}{b}}}{8 a^{2} b^{4} \sqrt {\frac {a}{b}} - 16 a b^{5} x \sqrt {\frac {a}{b}} + 8 b^{6} x^{2} \sqrt {\frac {a}{b}}} + \frac {15 a b^{2} x^{2} \log {\left (\sqrt {x} - \sqrt {\frac {a}{b}} \right )}}{8 a^{2} b^{4} \sqrt {\frac {a}{b}} - 16 a b^{5} x \sqrt {\frac {a}{b}} + 8 b^{6} x^{2} \sqrt {\frac {a}{b}}} - \frac {15 a b^{2} x^{2} \log {\left (\sqrt {x} + \sqrt {\frac {a}{b}} \right )}}{8 a^{2} b^{4} \sqrt {\frac {a}{b}} - 16 a b^{5} x \sqrt {\frac {a}{b}} + 8 b^{6} x^{2} \sqrt {\frac {a}{b}}} + \frac {16 b^{3} x^{\frac {5}{2}} \sqrt {\frac {a}{b}}}{8 a^{2} b^{4} \sqrt {\frac {a}{b}} - 16 a b^{5} x \sqrt {\frac {a}{b}} + 8 b^{6} x^{2} \sqrt {\frac {a}{b}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((zoo*sqrt(x), Eq(a, 0) & Eq(b, 0)), (-2*x**(7/2)/(7*a**3), Eq(b, 0)), (2*sqrt(x)/b**3, Eq(a, 0)), (1
5*a**3*log(sqrt(x) - sqrt(a/b))/(8*a**2*b**4*sqrt(a/b) - 16*a*b**5*x*sqrt(a/b) + 8*b**6*x**2*sqrt(a/b)) - 15*a
**3*log(sqrt(x) + sqrt(a/b))/(8*a**2*b**4*sqrt(a/b) - 16*a*b**5*x*sqrt(a/b) + 8*b**6*x**2*sqrt(a/b)) + 30*a**2
*b*sqrt(x)*sqrt(a/b)/(8*a**2*b**4*sqrt(a/b) - 16*a*b**5*x*sqrt(a/b) + 8*b**6*x**2*sqrt(a/b)) - 30*a**2*b*x*log
(sqrt(x) - sqrt(a/b))/(8*a**2*b**4*sqrt(a/b) - 16*a*b**5*x*sqrt(a/b) + 8*b**6*x**2*sqrt(a/b)) + 30*a**2*b*x*lo
g(sqrt(x) + sqrt(a/b))/(8*a**2*b**4*sqrt(a/b) - 16*a*b**5*x*sqrt(a/b) + 8*b**6*x**2*sqrt(a/b)) - 50*a*b**2*x**
(3/2)*sqrt(a/b)/(8*a**2*b**4*sqrt(a/b) - 16*a*b**5*x*sqrt(a/b) + 8*b**6*x**2*sqrt(a/b)) + 15*a*b**2*x**2*log(s
qrt(x) - sqrt(a/b))/(8*a**2*b**4*sqrt(a/b) - 16*a*b**5*x*sqrt(a/b) + 8*b**6*x**2*sqrt(a/b)) - 15*a*b**2*x**2*l
og(sqrt(x) + sqrt(a/b))/(8*a**2*b**4*sqrt(a/b) - 16*a*b**5*x*sqrt(a/b) + 8*b**6*x**2*sqrt(a/b)) + 16*b**3*x**(
5/2)*sqrt(a/b)/(8*a**2*b**4*sqrt(a/b) - 16*a*b**5*x*sqrt(a/b) + 8*b**6*x**2*sqrt(a/b)), True))

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.07 \[ \int \frac {x^{5/2}}{(-a+b x)^3} \, dx=-\frac {9 \, a b x^{\frac {3}{2}} - 7 \, a^{2} \sqrt {x}}{4 \, {\left (b^{5} x^{2} - 2 \, a b^{4} x + a^{2} b^{3}\right )}} + \frac {15 \, a \log \left (\frac {b \sqrt {x} - \sqrt {a b}}{b \sqrt {x} + \sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{3}} + \frac {2 \, \sqrt {x}}{b^{3}} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.75 \[ \int \frac {x^{5/2}}{(-a+b x)^3} \, dx=\frac {15 \, a \arctan \left (\frac {b \sqrt {x}}{\sqrt {-a b}}\right )}{4 \, \sqrt {-a b} b^{3}} + \frac {2 \, \sqrt {x}}{b^{3}} - \frac {9 \, a b x^{\frac {3}{2}} - 7 \, a^{2} \sqrt {x}}{4 \, {\left (b x - a\right )}^{2} b^{3}} \]

[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)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.82 \[ \int \frac {x^{5/2}}{(-a+b x)^3} \, dx=\frac {\frac {7\,a^2\,\sqrt {x}}{4}-\frac {9\,a\,b\,x^{3/2}}{4}}{a^2\,b^3-2\,a\,b^4\,x+b^5\,x^2}+\frac {2\,\sqrt {x}}{b^3}-\frac {15\,\sqrt {a}\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{4\,b^{7/2}} \]

[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))