3.5.88 \(\int \frac {1}{x^{5/2} (-a+b x)^3} \, dx\)

Optimal. Leaf size=97 \[ -\frac {35 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{9/2}}+\frac {35 b}{4 a^4 \sqrt {x}}+\frac {35}{12 a^3 x^{3/2}}-\frac {7}{4 a^2 x^{3/2} (a-b x)}-\frac {1}{2 a x^{3/2} (a-b x)^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {51, 63, 208} \begin {gather*} -\frac {35 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{9/2}}-\frac {7}{4 a^2 x^{3/2} (a-b x)}+\frac {35 b}{4 a^4 \sqrt {x}}+\frac {35}{12 a^3 x^{3/2}}-\frac {1}{2 a x^{3/2} (a-b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^(5/2)*(-a + b*x)^3),x]

[Out]

35/(12*a^3*x^(3/2)) + (35*b)/(4*a^4*Sqrt[x]) - 1/(2*a*x^(3/2)*(a - b*x)^2) - 7/(4*a^2*x^(3/2)*(a - b*x)) - (35
*b^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*a^(9/2))

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 {1}{x^{5/2} (-a+b x)^3} \, dx &=-\frac {1}{2 a x^{3/2} (a-b x)^2}-\frac {7 \int \frac {1}{x^{5/2} (-a+b x)^2} \, dx}{4 a}\\ &=-\frac {1}{2 a x^{3/2} (a-b x)^2}-\frac {7}{4 a^2 x^{3/2} (a-b x)}+\frac {35 \int \frac {1}{x^{5/2} (-a+b x)} \, dx}{8 a^2}\\ &=\frac {35}{12 a^3 x^{3/2}}-\frac {1}{2 a x^{3/2} (a-b x)^2}-\frac {7}{4 a^2 x^{3/2} (a-b x)}+\frac {(35 b) \int \frac {1}{x^{3/2} (-a+b x)} \, dx}{8 a^3}\\ &=\frac {35}{12 a^3 x^{3/2}}+\frac {35 b}{4 a^4 \sqrt {x}}-\frac {1}{2 a x^{3/2} (a-b x)^2}-\frac {7}{4 a^2 x^{3/2} (a-b x)}+\frac {\left (35 b^2\right ) \int \frac {1}{\sqrt {x} (-a+b x)} \, dx}{8 a^4}\\ &=\frac {35}{12 a^3 x^{3/2}}+\frac {35 b}{4 a^4 \sqrt {x}}-\frac {1}{2 a x^{3/2} (a-b x)^2}-\frac {7}{4 a^2 x^{3/2} (a-b x)}+\frac {\left (35 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{-a+b x^2} \, dx,x,\sqrt {x}\right )}{4 a^4}\\ &=\frac {35}{12 a^3 x^{3/2}}+\frac {35 b}{4 a^4 \sqrt {x}}-\frac {1}{2 a x^{3/2} (a-b x)^2}-\frac {7}{4 a^2 x^{3/2} (a-b x)}-\frac {35 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{9/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.00, size = 26, normalized size = 0.27 \begin {gather*} \frac {2 \, _2F_1\left (-\frac {3}{2},3;-\frac {1}{2};\frac {b x}{a}\right )}{3 a^3 x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(5/2)*(-a + b*x)^3),x]

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.12, size = 82, normalized size = 0.85 \begin {gather*} \frac {8 a^3+56 a^2 b x-175 a b^2 x^2+105 b^3 x^3}{12 a^4 x^{3/2} (a-b x)^2}-\frac {35 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^(5/2)*(-a + b*x)^3),x]

[Out]

(8*a^3 + 56*a^2*b*x - 175*a*b^2*x^2 + 105*b^3*x^3)/(12*a^4*x^(3/2)*(a - b*x)^2) - (35*b^(3/2)*ArcTanh[(Sqrt[b]
*Sqrt[x])/Sqrt[a]])/(4*a^(9/2))

________________________________________________________________________________________

fricas [A]  time = 1.01, size = 249, normalized size = 2.57 \begin {gather*} \left [\frac {105 \, {\left (b^{3} x^{4} - 2 \, a b^{2} x^{3} + a^{2} b x^{2}\right )} \sqrt {\frac {b}{a}} \log \left (\frac {b x - 2 \, a \sqrt {x} \sqrt {\frac {b}{a}} + a}{b x - a}\right ) + 2 \, {\left (105 \, b^{3} x^{3} - 175 \, a b^{2} x^{2} + 56 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {x}}{24 \, {\left (a^{4} b^{2} x^{4} - 2 \, a^{5} b x^{3} + a^{6} x^{2}\right )}}, \frac {105 \, {\left (b^{3} x^{4} - 2 \, a b^{2} x^{3} + a^{2} b x^{2}\right )} \sqrt {-\frac {b}{a}} \arctan \left (\frac {a \sqrt {-\frac {b}{a}}}{b \sqrt {x}}\right ) + {\left (105 \, b^{3} x^{3} - 175 \, a b^{2} x^{2} + 56 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {x}}{12 \, {\left (a^{4} b^{2} x^{4} - 2 \, a^{5} b x^{3} + a^{6} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(b*x-a)^3,x, algorithm="fricas")

[Out]

[1/24*(105*(b^3*x^4 - 2*a*b^2*x^3 + a^2*b*x^2)*sqrt(b/a)*log((b*x - 2*a*sqrt(x)*sqrt(b/a) + a)/(b*x - a)) + 2*
(105*b^3*x^3 - 175*a*b^2*x^2 + 56*a^2*b*x + 8*a^3)*sqrt(x))/(a^4*b^2*x^4 - 2*a^5*b*x^3 + a^6*x^2), 1/12*(105*(
b^3*x^4 - 2*a*b^2*x^3 + a^2*b*x^2)*sqrt(-b/a)*arctan(a*sqrt(-b/a)/(b*sqrt(x))) + (105*b^3*x^3 - 175*a*b^2*x^2
+ 56*a^2*b*x + 8*a^3)*sqrt(x))/(a^4*b^2*x^4 - 2*a^5*b*x^3 + a^6*x^2)]

________________________________________________________________________________________

giac [A]  time = 1.04, size = 73, normalized size = 0.75 \begin {gather*} \frac {35 \, b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {-a b}}\right )}{4 \, \sqrt {-a b} a^{4}} + \frac {2 \, {\left (9 \, b x + a\right )}}{3 \, a^{4} x^{\frac {3}{2}}} + \frac {11 \, b^{3} x^{\frac {3}{2}} - 13 \, a b^{2} \sqrt {x}}{4 \, {\left (b x - a\right )}^{2} a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(b*x-a)^3,x, algorithm="giac")

[Out]

35/4*b^2*arctan(b*sqrt(x)/sqrt(-a*b))/(sqrt(-a*b)*a^4) + 2/3*(9*b*x + a)/(a^4*x^(3/2)) + 1/4*(11*b^3*x^(3/2) -
13*a*b^2*sqrt(x))/((b*x - a)^2*a^4)

________________________________________________________________________________________

maple [A]  time = 0.02, size = 69, normalized size = 0.71 \begin {gather*} \frac {2 \left (-\frac {35 \arctanh \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}+\frac {\frac {11 b \,x^{\frac {3}{2}}}{8}-\frac {13 a \sqrt {x}}{8}}{\left (b x -a \right )^{2}}\right ) b^{2}}{a^{4}}+\frac {6 b}{a^{4} \sqrt {x}}+\frac {2}{3 a^{3} x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(5/2)/(b*x-a)^3,x)

[Out]

2/a^4*b^2*((11/8*b*x^(3/2)-13/8*a*x^(1/2))/(b*x-a)^2-35/8/(a*b)^(1/2)*arctanh(1/(a*b)^(1/2)*b*x^(1/2)))+2/3/a^
3/x^(3/2)+6/a^4*b/x^(1/2)

________________________________________________________________________________________

maxima [A]  time = 2.95, size = 103, normalized size = 1.06 \begin {gather*} \frac {105 \, b^{3} x^{3} - 175 \, a b^{2} x^{2} + 56 \, a^{2} b x + 8 \, a^{3}}{12 \, {\left (a^{4} b^{2} x^{\frac {7}{2}} - 2 \, a^{5} b x^{\frac {5}{2}} + a^{6} x^{\frac {3}{2}}\right )}} + \frac {35 \, b^{2} \log \left (\frac {b \sqrt {x} - \sqrt {a b}}{b \sqrt {x} + \sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(b*x-a)^3,x, algorithm="maxima")

[Out]

1/12*(105*b^3*x^3 - 175*a*b^2*x^2 + 56*a^2*b*x + 8*a^3)/(a^4*b^2*x^(7/2) - 2*a^5*b*x^(5/2) + a^6*x^(3/2)) + 35
/8*b^2*log((b*sqrt(x) - sqrt(a*b))/(b*sqrt(x) + sqrt(a*b)))/(sqrt(a*b)*a^4)

________________________________________________________________________________________

mupad [B]  time = 0.17, size = 80, normalized size = 0.82 \begin {gather*} \frac {\frac {2}{3\,a}-\frac {175\,b^2\,x^2}{12\,a^3}+\frac {35\,b^3\,x^3}{4\,a^4}+\frac {14\,b\,x}{3\,a^2}}{a^2\,x^{3/2}+b^2\,x^{7/2}-2\,a\,b\,x^{5/2}}-\frac {35\,b^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{4\,a^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(x^(5/2)*(a - b*x)^3),x)

[Out]

(2/(3*a) - (175*b^2*x^2)/(12*a^3) + (35*b^3*x^3)/(4*a^4) + (14*b*x)/(3*a^2))/(a^2*x^(3/2) + b^2*x^(7/2) - 2*a*
b*x^(5/2)) - (35*b^(3/2)*atanh((b^(1/2)*x^(1/2))/a^(1/2)))/(4*a^(9/2))

________________________________________________________________________________________

sympy [A]  time = 138.88, size = 892, normalized size = 9.20 \begin {gather*} \begin {cases} \frac {\tilde {\infty }}{x^{\frac {9}{2}}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2}{3 a^{3} x^{\frac {3}{2}}} & \text {for}\: b = 0 \\- \frac {2}{9 b^{3} x^{\frac {9}{2}}} & \text {for}\: a = 0 \\\frac {16 a^{\frac {7}{2}} \sqrt {\frac {1}{b}}}{24 a^{\frac {13}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} - 48 a^{\frac {11}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}} + 24 a^{\frac {9}{2}} b^{2} x^{\frac {7}{2}} \sqrt {\frac {1}{b}}} + \frac {112 a^{\frac {5}{2}} b x \sqrt {\frac {1}{b}}}{24 a^{\frac {13}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} - 48 a^{\frac {11}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}} + 24 a^{\frac {9}{2}} b^{2} x^{\frac {7}{2}} \sqrt {\frac {1}{b}}} - \frac {350 a^{\frac {3}{2}} b^{2} x^{2} \sqrt {\frac {1}{b}}}{24 a^{\frac {13}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} - 48 a^{\frac {11}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}} + 24 a^{\frac {9}{2}} b^{2} x^{\frac {7}{2}} \sqrt {\frac {1}{b}}} + \frac {210 \sqrt {a} b^{3} x^{3} \sqrt {\frac {1}{b}}}{24 a^{\frac {13}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} - 48 a^{\frac {11}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}} + 24 a^{\frac {9}{2}} b^{2} x^{\frac {7}{2}} \sqrt {\frac {1}{b}}} + \frac {105 a^{2} b x^{\frac {3}{2}} \log {\left (- \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{24 a^{\frac {13}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} - 48 a^{\frac {11}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}} + 24 a^{\frac {9}{2}} b^{2} x^{\frac {7}{2}} \sqrt {\frac {1}{b}}} - \frac {105 a^{2} b x^{\frac {3}{2}} \log {\left (\sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{24 a^{\frac {13}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} - 48 a^{\frac {11}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}} + 24 a^{\frac {9}{2}} b^{2} x^{\frac {7}{2}} \sqrt {\frac {1}{b}}} - \frac {210 a b^{2} x^{\frac {5}{2}} \log {\left (- \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{24 a^{\frac {13}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} - 48 a^{\frac {11}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}} + 24 a^{\frac {9}{2}} b^{2} x^{\frac {7}{2}} \sqrt {\frac {1}{b}}} + \frac {210 a b^{2} x^{\frac {5}{2}} \log {\left (\sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{24 a^{\frac {13}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} - 48 a^{\frac {11}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}} + 24 a^{\frac {9}{2}} b^{2} x^{\frac {7}{2}} \sqrt {\frac {1}{b}}} + \frac {105 b^{3} x^{\frac {7}{2}} \log {\left (- \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{24 a^{\frac {13}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} - 48 a^{\frac {11}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}} + 24 a^{\frac {9}{2}} b^{2} x^{\frac {7}{2}} \sqrt {\frac {1}{b}}} - \frac {105 b^{3} x^{\frac {7}{2}} \log {\left (\sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{24 a^{\frac {13}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} - 48 a^{\frac {11}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}} + 24 a^{\frac {9}{2}} b^{2} x^{\frac {7}{2}} \sqrt {\frac {1}{b}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(5/2)/(b*x-a)**3,x)

[Out]

Piecewise((zoo/x**(9/2), Eq(a, 0) & Eq(b, 0)), (2/(3*a**3*x**(3/2)), Eq(b, 0)), (-2/(9*b**3*x**(9/2)), Eq(a, 0
)), (16*a**(7/2)*sqrt(1/b)/(24*a**(13/2)*x**(3/2)*sqrt(1/b) - 48*a**(11/2)*b*x**(5/2)*sqrt(1/b) + 24*a**(9/2)*
b**2*x**(7/2)*sqrt(1/b)) + 112*a**(5/2)*b*x*sqrt(1/b)/(24*a**(13/2)*x**(3/2)*sqrt(1/b) - 48*a**(11/2)*b*x**(5/
2)*sqrt(1/b) + 24*a**(9/2)*b**2*x**(7/2)*sqrt(1/b)) - 350*a**(3/2)*b**2*x**2*sqrt(1/b)/(24*a**(13/2)*x**(3/2)*
sqrt(1/b) - 48*a**(11/2)*b*x**(5/2)*sqrt(1/b) + 24*a**(9/2)*b**2*x**(7/2)*sqrt(1/b)) + 210*sqrt(a)*b**3*x**3*s
qrt(1/b)/(24*a**(13/2)*x**(3/2)*sqrt(1/b) - 48*a**(11/2)*b*x**(5/2)*sqrt(1/b) + 24*a**(9/2)*b**2*x**(7/2)*sqrt
(1/b)) + 105*a**2*b*x**(3/2)*log(-sqrt(a)*sqrt(1/b) + sqrt(x))/(24*a**(13/2)*x**(3/2)*sqrt(1/b) - 48*a**(11/2)
*b*x**(5/2)*sqrt(1/b) + 24*a**(9/2)*b**2*x**(7/2)*sqrt(1/b)) - 105*a**2*b*x**(3/2)*log(sqrt(a)*sqrt(1/b) + sqr
t(x))/(24*a**(13/2)*x**(3/2)*sqrt(1/b) - 48*a**(11/2)*b*x**(5/2)*sqrt(1/b) + 24*a**(9/2)*b**2*x**(7/2)*sqrt(1/
b)) - 210*a*b**2*x**(5/2)*log(-sqrt(a)*sqrt(1/b) + sqrt(x))/(24*a**(13/2)*x**(3/2)*sqrt(1/b) - 48*a**(11/2)*b*
x**(5/2)*sqrt(1/b) + 24*a**(9/2)*b**2*x**(7/2)*sqrt(1/b)) + 210*a*b**2*x**(5/2)*log(sqrt(a)*sqrt(1/b) + sqrt(x
))/(24*a**(13/2)*x**(3/2)*sqrt(1/b) - 48*a**(11/2)*b*x**(5/2)*sqrt(1/b) + 24*a**(9/2)*b**2*x**(7/2)*sqrt(1/b))
+ 105*b**3*x**(7/2)*log(-sqrt(a)*sqrt(1/b) + sqrt(x))/(24*a**(13/2)*x**(3/2)*sqrt(1/b) - 48*a**(11/2)*b*x**(5
/2)*sqrt(1/b) + 24*a**(9/2)*b**2*x**(7/2)*sqrt(1/b)) - 105*b**3*x**(7/2)*log(sqrt(a)*sqrt(1/b) + sqrt(x))/(24*
a**(13/2)*x**(3/2)*sqrt(1/b) - 48*a**(11/2)*b*x**(5/2)*sqrt(1/b) + 24*a**(9/2)*b**2*x**(7/2)*sqrt(1/b)), True)
)

________________________________________________________________________________________