3.5.88 \(\int \frac {1}{x^{5/2} (-a+b x)^3} \, dx\) [488]

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

[Out]

35/12/a^3/x^(3/2)-1/2/a/x^(3/2)/(-b*x+a)^2-7/4/a^2/x^(3/2)/(-b*x+a)-35/4*b^(3/2)*arctanh(b^(1/2)*x^(1/2)/a^(1/
2))/a^(9/2)+35/4*b/a^4/x^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {44, 53, 65, 214} \begin {gather*} -\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} \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 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 53

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 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*} \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 ) \text {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*}

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Mathematica [A]
time = 0.11, 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]

Integrate[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))

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Maple [A]
time = 0.13, size = 68, normalized size = 0.70

method	result	size
risch	\(\frac {6 b x +\frac {2 a}{3}}{a^{4} x^{\frac {3}{2}}}+\frac {b^{2} \left (\frac {\frac {11 b \,x^{\frac {3}{2}}}{4}-\frac {13 a \sqrt {x}}{4}}{\left (b x -a \right )^{2}}-\frac {35 \arctanh \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}}\right )}{a^{4}}\)	\(66\)
derivativedivides	\(\frac {2}{3 a^{3} x^{\frac {3}{2}}}+\frac {6 b}{a^{4} \sqrt {x}}-\frac {2 b^{2} \left (\frac {-\frac {11 b \,x^{\frac {3}{2}}}{8}+\frac {13 a \sqrt {x}}{8}}{\left (-b x +a \right )^{2}}+\frac {35 \arctanh \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{4}}\)	\(68\)
default	\(\frac {2}{3 a^{3} x^{\frac {3}{2}}}+\frac {6 b}{a^{4} \sqrt {x}}-\frac {2 b^{2} \left (\frac {-\frac {11 b \,x^{\frac {3}{2}}}{8}+\frac {13 a \sqrt {x}}{8}}{\left (-b x +a \right )^{2}}+\frac {35 \arctanh \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{4}}\)	\(68\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3/a^3/x^(3/2)+6*b/a^4/x^(1/2)-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)*arctan
h(b*x^(1/2)/(a*b)^(1/2)))

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Maxima [A]
time = 0.51, 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)

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Fricas [A]
time = 0.43, 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)]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 799 vs. \(2 (88) = 176\).
time = 94.62, size = 799, normalized size = 8.24 \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^{3} \sqrt {\frac {a}{b}}}{24 a^{6} x^{\frac {3}{2}} \sqrt {\frac {a}{b}} - 48 a^{5} b x^{\frac {5}{2}} \sqrt {\frac {a}{b}} + 24 a^{4} b^{2} x^{\frac {7}{2}} \sqrt {\frac {a}{b}}} + \frac {105 a^{2} b x^{\frac {3}{2}} \log {\left (\sqrt {x} - \sqrt {\frac {a}{b}} \right )}}{24 a^{6} x^{\frac {3}{2}} \sqrt {\frac {a}{b}} - 48 a^{5} b x^{\frac {5}{2}} \sqrt {\frac {a}{b}} + 24 a^{4} b^{2} x^{\frac {7}{2}} \sqrt {\frac {a}{b}}} - \frac {105 a^{2} b x^{\frac {3}{2}} \log {\left (\sqrt {x} + \sqrt {\frac {a}{b}} \right )}}{24 a^{6} x^{\frac {3}{2}} \sqrt {\frac {a}{b}} - 48 a^{5} b x^{\frac {5}{2}} \sqrt {\frac {a}{b}} + 24 a^{4} b^{2} x^{\frac {7}{2}} \sqrt {\frac {a}{b}}} + \frac {112 a^{2} b x \sqrt {\frac {a}{b}}}{24 a^{6} x^{\frac {3}{2}} \sqrt {\frac {a}{b}} - 48 a^{5} b x^{\frac {5}{2}} \sqrt {\frac {a}{b}} + 24 a^{4} b^{2} x^{\frac {7}{2}} \sqrt {\frac {a}{b}}} - \frac {210 a b^{2} x^{\frac {5}{2}} \log {\left (\sqrt {x} - \sqrt {\frac {a}{b}} \right )}}{24 a^{6} x^{\frac {3}{2}} \sqrt {\frac {a}{b}} - 48 a^{5} b x^{\frac {5}{2}} \sqrt {\frac {a}{b}} + 24 a^{4} b^{2} x^{\frac {7}{2}} \sqrt {\frac {a}{b}}} + \frac {210 a b^{2} x^{\frac {5}{2}} \log {\left (\sqrt {x} + \sqrt {\frac {a}{b}} \right )}}{24 a^{6} x^{\frac {3}{2}} \sqrt {\frac {a}{b}} - 48 a^{5} b x^{\frac {5}{2}} \sqrt {\frac {a}{b}} + 24 a^{4} b^{2} x^{\frac {7}{2}} \sqrt {\frac {a}{b}}} - \frac {350 a b^{2} x^{2} \sqrt {\frac {a}{b}}}{24 a^{6} x^{\frac {3}{2}} \sqrt {\frac {a}{b}} - 48 a^{5} b x^{\frac {5}{2}} \sqrt {\frac {a}{b}} + 24 a^{4} b^{2} x^{\frac {7}{2}} \sqrt {\frac {a}{b}}} + \frac {105 b^{3} x^{\frac {7}{2}} \log {\left (\sqrt {x} - \sqrt {\frac {a}{b}} \right )}}{24 a^{6} x^{\frac {3}{2}} \sqrt {\frac {a}{b}} - 48 a^{5} b x^{\frac {5}{2}} \sqrt {\frac {a}{b}} + 24 a^{4} b^{2} x^{\frac {7}{2}} \sqrt {\frac {a}{b}}} - \frac {105 b^{3} x^{\frac {7}{2}} \log {\left (\sqrt {x} + \sqrt {\frac {a}{b}} \right )}}{24 a^{6} x^{\frac {3}{2}} \sqrt {\frac {a}{b}} - 48 a^{5} b x^{\frac {5}{2}} \sqrt {\frac {a}{b}} + 24 a^{4} b^{2} x^{\frac {7}{2}} \sqrt {\frac {a}{b}}} + \frac {210 b^{3} x^{3} \sqrt {\frac {a}{b}}}{24 a^{6} x^{\frac {3}{2}} \sqrt {\frac {a}{b}} - 48 a^{5} b x^{\frac {5}{2}} \sqrt {\frac {a}{b}} + 24 a^{4} b^{2} x^{\frac {7}{2}} \sqrt {\frac {a}{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**3*sqrt(a/b)/(24*a**6*x**(3/2)*sqrt(a/b) - 48*a**5*b*x**(5/2)*sqrt(a/b) + 24*a**4*b**2*x**(7/2)*sqrt
(a/b)) + 105*a**2*b*x**(3/2)*log(sqrt(x) - sqrt(a/b))/(24*a**6*x**(3/2)*sqrt(a/b) - 48*a**5*b*x**(5/2)*sqrt(a/
b) + 24*a**4*b**2*x**(7/2)*sqrt(a/b)) - 105*a**2*b*x**(3/2)*log(sqrt(x) + sqrt(a/b))/(24*a**6*x**(3/2)*sqrt(a/
b) - 48*a**5*b*x**(5/2)*sqrt(a/b) + 24*a**4*b**2*x**(7/2)*sqrt(a/b)) + 112*a**2*b*x*sqrt(a/b)/(24*a**6*x**(3/2
)*sqrt(a/b) - 48*a**5*b*x**(5/2)*sqrt(a/b) + 24*a**4*b**2*x**(7/2)*sqrt(a/b)) - 210*a*b**2*x**(5/2)*log(sqrt(x
) - sqrt(a/b))/(24*a**6*x**(3/2)*sqrt(a/b) - 48*a**5*b*x**(5/2)*sqrt(a/b) + 24*a**4*b**2*x**(7/2)*sqrt(a/b)) +
210*a*b**2*x**(5/2)*log(sqrt(x) + sqrt(a/b))/(24*a**6*x**(3/2)*sqrt(a/b) - 48*a**5*b*x**(5/2)*sqrt(a/b) + 24*
a**4*b**2*x**(7/2)*sqrt(a/b)) - 350*a*b**2*x**2*sqrt(a/b)/(24*a**6*x**(3/2)*sqrt(a/b) - 48*a**5*b*x**(5/2)*sqr
t(a/b) + 24*a**4*b**2*x**(7/2)*sqrt(a/b)) + 105*b**3*x**(7/2)*log(sqrt(x) - sqrt(a/b))/(24*a**6*x**(3/2)*sqrt(
a/b) - 48*a**5*b*x**(5/2)*sqrt(a/b) + 24*a**4*b**2*x**(7/2)*sqrt(a/b)) - 105*b**3*x**(7/2)*log(sqrt(x) + sqrt(
a/b))/(24*a**6*x**(3/2)*sqrt(a/b) - 48*a**5*b*x**(5/2)*sqrt(a/b) + 24*a**4*b**2*x**(7/2)*sqrt(a/b)) + 210*b**3
*x**3*sqrt(a/b)/(24*a**6*x**(3/2)*sqrt(a/b) - 48*a**5*b*x**(5/2)*sqrt(a/b) + 24*a**4*b**2*x**(7/2)*sqrt(a/b)),
True))

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Giac [A]
time = 1.46, 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)

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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))

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