3.4.64 \(\int \frac {1}{x^3 (-a+b x)^{3/2}} \, dx\) [364]

Optimal. Leaf size=95 \[ -\frac {15 b^2}{4 a^3 \sqrt {-a+b x}}+\frac {1}{2 a x^2 \sqrt {-a+b x}}+\frac {5 b}{4 a^2 x \sqrt {-a+b x}}-\frac {15 b^2 \tan ^{-1}\left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right )}{4 a^{7/2}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 95, 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 = {44, 53, 65, 211} \begin {gather*} -\frac {15 b^2 \text {ArcTan}\left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )}{4 a^{7/2}}-\frac {15 b^2}{4 a^3 \sqrt {b x-a}}+\frac {5 b}{4 a^2 x \sqrt {b x-a}}+\frac {1}{2 a x^2 \sqrt {b x-a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(-a + b*x)^(3/2)),x]

[Out]

(-15*b^2)/(4*a^3*Sqrt[-a + b*x]) + 1/(2*a*x^2*Sqrt[-a + b*x]) + (5*b)/(4*a^2*x*Sqrt[-a + b*x]) - (15*b^2*ArcTa
n[Sqrt[-a + b*x]/Sqrt[a]])/(4*a^(7/2))

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 211

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

Rubi steps

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

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Mathematica [A]
time = 0.10, size = 71, normalized size = 0.75 \begin {gather*} \frac {2 a^2+5 a b x-15 b^2 x^2}{4 a^3 x^2 \sqrt {-a+b x}}-\frac {15 b^2 \tan ^{-1}\left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right )}{4 a^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(-a + b*x)^(3/2)),x]

[Out]

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

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Maple [A]
time = 0.10, size = 77, normalized size = 0.81

method	result	size
risch	\(\frac {\left (-b x +a \right ) \left (7 b x +2 a \right )}{4 a^{3} x^{2} \sqrt {b x -a}}-\frac {2 b^{2}}{a^{3} \sqrt {b x -a}}-\frac {15 b^{2} \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{4 a^{\frac {7}{2}}}\)	\(72\)
derivativedivides	\(2 b^{2} \left (-\frac {\frac {\frac {7 \left (b x -a \right )^{\frac {3}{2}}}{8}+\frac {9 a \sqrt {b x -a}}{8}}{b^{2} x^{2}}+\frac {15 \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{8 \sqrt {a}}}{a^{3}}-\frac {1}{a^{3} \sqrt {b x -a}}\right )\)	\(77\)
default	\(2 b^{2} \left (-\frac {\frac {\frac {7 \left (b x -a \right )^{\frac {3}{2}}}{8}+\frac {9 a \sqrt {b x -a}}{8}}{b^{2} x^{2}}+\frac {15 \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{8 \sqrt {a}}}{a^{3}}-\frac {1}{a^{3} \sqrt {b x -a}}\right )\)	\(77\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.48, size = 104, normalized size = 1.09 \begin {gather*} -\frac {15 \, {\left (b x - a\right )}^{2} b^{2} + 25 \, {\left (b x - a\right )} a b^{2} + 8 \, a^{2} b^{2}}{4 \, {\left ({\left (b x - a\right )}^{\frac {5}{2}} a^{3} + 2 \, {\left (b x - a\right )}^{\frac {3}{2}} a^{4} + \sqrt {b x - a} a^{5}\right )}} - \frac {15 \, b^{2} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right )}{4 \, a^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(15*(b*x - a)^2*b^2 + 25*(b*x - a)*a*b^2 + 8*a^2*b^2)/((b*x - a)^(5/2)*a^3 + 2*(b*x - a)^(3/2)*a^4 + sqrt
(b*x - a)*a^5) - 15/4*b^2*arctan(sqrt(b*x - a)/sqrt(a))/a^(7/2)

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Fricas [A]
time = 1.03, size = 198, normalized size = 2.08 \begin {gather*} \left [-\frac {15 \, {\left (b^{3} x^{3} - a b^{2} x^{2}\right )} \sqrt {-a} \log \left (\frac {b x + 2 \, \sqrt {b x - a} \sqrt {-a} - 2 \, a}{x}\right ) + 2 \, {\left (15 \, a b^{2} x^{2} - 5 \, a^{2} b x - 2 \, a^{3}\right )} \sqrt {b x - a}}{8 \, {\left (a^{4} b x^{3} - a^{5} x^{2}\right )}}, -\frac {15 \, {\left (b^{3} x^{3} - a b^{2} x^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) + {\left (15 \, a b^{2} x^{2} - 5 \, a^{2} b x - 2 \, a^{3}\right )} \sqrt {b x - a}}{4 \, {\left (a^{4} b x^{3} - a^{5} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 3.98, size = 226, normalized size = 2.38 \begin {gather*} \begin {cases} - \frac {i}{2 a \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} - 1}} - \frac {5 i \sqrt {b}}{4 a^{2} x^{\frac {3}{2}} \sqrt {\frac {a}{b x} - 1}} + \frac {15 i b^{\frac {3}{2}}}{4 a^{3} \sqrt {x} \sqrt {\frac {a}{b x} - 1}} - \frac {15 i b^{2} \operatorname {acosh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4 a^{\frac {7}{2}}} & \text {for}\: \left |{\frac {a}{b x}}\right | > 1 \\\frac {1}{2 a \sqrt {b} x^{\frac {5}{2}} \sqrt {- \frac {a}{b x} + 1}} + \frac {5 \sqrt {b}}{4 a^{2} x^{\frac {3}{2}} \sqrt {- \frac {a}{b x} + 1}} - \frac {15 b^{\frac {3}{2}}}{4 a^{3} \sqrt {x} \sqrt {- \frac {a}{b x} + 1}} + \frac {15 b^{2} \operatorname {asin}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4 a^{\frac {7}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-I/(2*a*sqrt(b)*x**(5/2)*sqrt(a/(b*x) - 1)) - 5*I*sqrt(b)/(4*a**2*x**(3/2)*sqrt(a/(b*x) - 1)) + 15*
I*b**(3/2)/(4*a**3*sqrt(x)*sqrt(a/(b*x) - 1)) - 15*I*b**2*acosh(sqrt(a)/(sqrt(b)*sqrt(x)))/(4*a**(7/2)), Abs(a
/(b*x)) > 1), (1/(2*a*sqrt(b)*x**(5/2)*sqrt(-a/(b*x) + 1)) + 5*sqrt(b)/(4*a**2*x**(3/2)*sqrt(-a/(b*x) + 1)) -
15*b**(3/2)/(4*a**3*sqrt(x)*sqrt(-a/(b*x) + 1)) + 15*b**2*asin(sqrt(a)/(sqrt(b)*sqrt(x)))/(4*a**(7/2)), True))

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Giac [A]
time = 1.51, size = 81, normalized size = 0.85 \begin {gather*} -\frac {15 \, b^{2} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right )}{4 \, a^{\frac {7}{2}}} - \frac {2 \, b^{2}}{\sqrt {b x - a} a^{3}} - \frac {7 \, {\left (b x - a\right )}^{\frac {3}{2}} b^{2} + 9 \, \sqrt {b x - a} a b^{2}}{4 \, a^{3} b^{2} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.13, size = 101, normalized size = 1.06 \begin {gather*} -\frac {\frac {2\,b^2}{a}+\frac {15\,b^2\,{\left (a-b\,x\right )}^2}{4\,a^3}-\frac {25\,b^2\,\left (a-b\,x\right )}{4\,a^2}}{2\,a\,{\left (b\,x-a\right )}^{3/2}+{\left (b\,x-a\right )}^{5/2}+a^2\,\sqrt {b\,x-a}}-\frac {15\,b^2\,\mathrm {atan}\left (\frac {\sqrt {b\,x-a}}{\sqrt {a}}\right )}{4\,a^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(b*x - a)^(3/2)),x)

[Out]

- ((2*b^2)/a + (15*b^2*(a - b*x)^2)/(4*a^3) - (25*b^2*(a - b*x))/(4*a^2))/(2*a*(b*x - a)^(3/2) + (b*x - a)^(5/
2) + a^2*(b*x - a)^(1/2)) - (15*b^2*atan((b*x - a)^(1/2)/a^(1/2)))/(4*a^(7/2))

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