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

Optimal. Leaf size=87 \[ -\frac {15 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{7/2}}+\frac {15 b^2}{4 a^3 \sqrt {a+b x}}+\frac {5 b}{4 a^2 x \sqrt {a+b x}}-\frac {1}{2 a x^2 \sqrt {a+b x}} \]

[Out]

-15/4*b^2*arctanh((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 = 85, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {51, 63, 208} \[ -\frac {15 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{7/2}}-\frac {5 \sqrt {a+b x}}{2 a^2 x^2}+\frac {15 b \sqrt {a+b x}}{4 a^3 x}+\frac {2}{a x^2 \sqrt {a+b x}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*x)^(3/2)),x]

[Out]

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

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^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) \operatorname {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 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 33, normalized size = 0.38 \[ \frac {2 b^2 \, _2F_1\left (-\frac {1}{2},3;\frac {1}{2};\frac {b x}{a}+1\right )}{a^3 \sqrt {a+b x}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a + b*x)^(3/2)),x]

[Out]

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

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fricas [A]  time = 0.47, size = 189, normalized size = 2.17 \[ \left [\frac {15 \, {\left (b^{3} x^{3} + a b^{2} x^{2}\right )} \sqrt {a} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (15 \, a b^{2} x^{2} + 5 \, a^{2} b x - 2 \, a^{3}\right )} \sqrt {b x + a}}{8 \, {\left (a^{4} b x^{3} + a^{5} x^{2}\right )}}, \frac {15 \, {\left (b^{3} x^{3} + a b^{2} x^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (15 \, a b^{2} x^{2} + 5 \, a^{2} b x - 2 \, a^{3}\right )} \sqrt {b x + a}}{4 \, {\left (a^{4} b x^{3} + a^{5} x^{2}\right )}}\right ] \]

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)/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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giac [A]  time = 1.05, size = 80, normalized size = 0.92 \[ \frac {15 \, b^{2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a^{3}} + \frac {2 \, b^{2}}{\sqrt {b x + a} a^{3}} + \frac {7 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{2} - 9 \, \sqrt {b x + a} a b^{2}}{4 \, a^{3} b^{2} x^{2}} \]

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

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maple [A]  time = 0.01, size = 67, normalized size = 0.77 \[ 2 \left (\frac {1}{\sqrt {b x +a}\, a^{3}}+\frac {-\frac {15 \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 \sqrt {a}}+\frac {-\frac {9 \sqrt {b x +a}\, a}{8}+\frac {7 \left (b x +a \right )^{\frac {3}{2}}}{8}}{b^{2} x^{2}}}{a^{3}}\right ) b^{2} \]

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

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maxima [A]  time = 3.06, size = 108, normalized size = 1.24 \[ \frac {15 \, {\left (b x + a\right )}^{2} b^{2} - 25 \, {\left (b x + a\right )} a b^{2} + 8 \, a^{2} b^{2}}{4 \, {\left ({\left (b x + a\right )}^{\frac {5}{2}} a^{3} - 2 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} + \sqrt {b x + a} a^{5}\right )}} + \frac {15 \, b^{2} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{8 \, a^{\frac {7}{2}}} \]

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) + 15/8*b^2*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/a^(7/2)

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mupad [B]  time = 0.06, size = 90, normalized size = 1.03 \[ \frac {\frac {2\,b^2}{a}+\frac {15\,b^2\,{\left (a+b\,x\right )}^2}{4\,a^3}-\frac {25\,b^2\,\left (a+b\,x\right )}{4\,a^2}}{{\left (a+b\,x\right )}^{5/2}-2\,a\,{\left (a+b\,x\right )}^{3/2}+a^2\,\sqrt {a+b\,x}}-\frac {15\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )}{4\,a^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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sympy [A]  time = 5.98, size = 107, normalized size = 1.23 \[ - \frac {1}{2 a \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {5 \sqrt {b}}{4 a^{2} x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {15 b^{\frac {3}{2}}}{4 a^{3} \sqrt {x} \sqrt {\frac {a}{b x} + 1}} - \frac {15 b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4 a^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/(4*a**(7/2))

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