3.2205 \(\int \frac {1}{(a+b \sqrt {x})^2 x^3} \, dx\)

Optimal. Leaf size=95 \[ -\frac {10 b^4 \log \left (a+b \sqrt {x}\right )}{a^6}+\frac {5 b^4 \log (x)}{a^6}+\frac {2 b^4}{a^5 \left (a+b \sqrt {x}\right )}+\frac {8 b^3}{a^5 \sqrt {x}}-\frac {3 b^2}{a^4 x}+\frac {4 b}{3 a^3 x^{3/2}}-\frac {1}{2 a^2 x^2} \]

[Out]

-1/2/a^2/x^2+4/3*b/a^3/x^(3/2)-3*b^2/a^4/x+5*b^4*ln(x)/a^6-10*b^4*ln(a+b*x^(1/2))/a^6+8*b^3/a^5/x^(1/2)+2*b^4/
a^5/(a+b*x^(1/2))

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Rubi [A]  time = 0.06, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {266, 44} \[ \frac {2 b^4}{a^5 \left (a+b \sqrt {x}\right )}+\frac {8 b^3}{a^5 \sqrt {x}}-\frac {3 b^2}{a^4 x}-\frac {10 b^4 \log \left (a+b \sqrt {x}\right )}{a^6}+\frac {5 b^4 \log (x)}{a^6}+\frac {4 b}{3 a^3 x^{3/2}}-\frac {1}{2 a^2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*Sqrt[x])^2*x^3),x]

[Out]

(2*b^4)/(a^5*(a + b*Sqrt[x])) - 1/(2*a^2*x^2) + (4*b)/(3*a^3*x^(3/2)) - (3*b^2)/(a^4*x) + (8*b^3)/(a^5*Sqrt[x]
) - (10*b^4*Log[a + b*Sqrt[x]])/a^6 + (5*b^4*Log[x])/a^6

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {1}{\left (a+b \sqrt {x}\right )^2 x^3} \, dx &=2 \operatorname {Subst}\left (\int \frac {1}{x^5 (a+b x)^2} \, dx,x,\sqrt {x}\right )\\ &=2 \operatorname {Subst}\left (\int \left (\frac {1}{a^2 x^5}-\frac {2 b}{a^3 x^4}+\frac {3 b^2}{a^4 x^3}-\frac {4 b^3}{a^5 x^2}+\frac {5 b^4}{a^6 x}-\frac {b^5}{a^5 (a+b x)^2}-\frac {5 b^5}{a^6 (a+b x)}\right ) \, dx,x,\sqrt {x}\right )\\ &=\frac {2 b^4}{a^5 \left (a+b \sqrt {x}\right )}-\frac {1}{2 a^2 x^2}+\frac {4 b}{3 a^3 x^{3/2}}-\frac {3 b^2}{a^4 x}+\frac {8 b^3}{a^5 \sqrt {x}}-\frac {10 b^4 \log \left (a+b \sqrt {x}\right )}{a^6}+\frac {5 b^4 \log (x)}{a^6}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 91, normalized size = 0.96 \[ \frac {\frac {a \left (-3 a^4+5 a^3 b \sqrt {x}-10 a^2 b^2 x+30 a b^3 x^{3/2}+60 b^4 x^2\right )}{x^2 \left (a+b \sqrt {x}\right )}-60 b^4 \log \left (a+b \sqrt {x}\right )+30 b^4 \log (x)}{6 a^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(-3*a^4 + 5*a^3*b*Sqrt[x] - 10*a^2*b^2*x + 30*a*b^3*x^(3/2) + 60*b^4*x^2))/((a + b*Sqrt[x])*x^2) - 60*b^4*
Log[a + b*Sqrt[x]] + 30*b^4*Log[x])/(6*a^6)

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fricas [A]  time = 1.00, size = 133, normalized size = 1.40 \[ -\frac {30 \, a^{2} b^{4} x^{2} - 15 \, a^{4} b^{2} x - 3 \, a^{6} + 60 \, {\left (b^{6} x^{3} - a^{2} b^{4} x^{2}\right )} \log \left (b \sqrt {x} + a\right ) - 60 \, {\left (b^{6} x^{3} - a^{2} b^{4} x^{2}\right )} \log \left (\sqrt {x}\right ) - 4 \, {\left (15 \, a b^{5} x^{2} - 10 \, a^{3} b^{3} x - 2 \, a^{5} b\right )} \sqrt {x}}{6 \, {\left (a^{6} b^{2} x^{3} - a^{8} x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(30*a^2*b^4*x^2 - 15*a^4*b^2*x - 3*a^6 + 60*(b^6*x^3 - a^2*b^4*x^2)*log(b*sqrt(x) + a) - 60*(b^6*x^3 - a^
2*b^4*x^2)*log(sqrt(x)) - 4*(15*a*b^5*x^2 - 10*a^3*b^3*x - 2*a^5*b)*sqrt(x))/(a^6*b^2*x^3 - a^8*x^2)

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giac [A]  time = 0.16, size = 90, normalized size = 0.95 \[ -\frac {10 \, b^{4} \log \left ({\left | b \sqrt {x} + a \right |}\right )}{a^{6}} + \frac {5 \, b^{4} \log \left ({\left | x \right |}\right )}{a^{6}} + \frac {60 \, a b^{4} x^{2} + 30 \, a^{2} b^{3} x^{\frac {3}{2}} - 10 \, a^{3} b^{2} x + 5 \, a^{4} b \sqrt {x} - 3 \, a^{5}}{6 \, {\left (b \sqrt {x} + a\right )} a^{6} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-10*b^4*log(abs(b*sqrt(x) + a))/a^6 + 5*b^4*log(abs(x))/a^6 + 1/6*(60*a*b^4*x^2 + 30*a^2*b^3*x^(3/2) - 10*a^3*
b^2*x + 5*a^4*b*sqrt(x) - 3*a^5)/((b*sqrt(x) + a)*a^6*x^2)

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maple [A]  time = 0.01, size = 84, normalized size = 0.88 \[ \frac {2 b^{4}}{\left (b \sqrt {x}+a \right ) a^{5}}+\frac {5 b^{4} \ln \relax (x )}{a^{6}}-\frac {10 b^{4} \ln \left (b \sqrt {x}+a \right )}{a^{6}}+\frac {8 b^{3}}{a^{5} \sqrt {x}}-\frac {3 b^{2}}{a^{4} x}+\frac {4 b}{3 a^{3} x^{\frac {3}{2}}}-\frac {1}{2 a^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/a^2/x^2+4/3*b/a^3/x^(3/2)-3*b^2/a^4/x+5*b^4*ln(x)/a^6-10*b^4*ln(b*x^(1/2)+a)/a^6+8*b^3/a^5/x^(1/2)+2*b^4/
a^5/(b*x^(1/2)+a)

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maxima [A]  time = 0.91, size = 88, normalized size = 0.93 \[ \frac {60 \, b^{4} x^{2} + 30 \, a b^{3} x^{\frac {3}{2}} - 10 \, a^{2} b^{2} x + 5 \, a^{3} b \sqrt {x} - 3 \, a^{4}}{6 \, {\left (a^{5} b x^{\frac {5}{2}} + a^{6} x^{2}\right )}} - \frac {10 \, b^{4} \log \left (b \sqrt {x} + a\right )}{a^{6}} + \frac {5 \, b^{4} \log \relax (x)}{a^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(60*b^4*x^2 + 30*a*b^3*x^(3/2) - 10*a^2*b^2*x + 5*a^3*b*sqrt(x) - 3*a^4)/(a^5*b*x^(5/2) + a^6*x^2) - 10*b^
4*log(b*sqrt(x) + a)/a^6 + 5*b^4*log(x)/a^6

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mupad [B]  time = 1.17, size = 81, normalized size = 0.85 \[ \frac {\frac {5\,b\,\sqrt {x}}{6\,a^2}-\frac {1}{2\,a}-\frac {5\,b^2\,x}{3\,a^3}+\frac {10\,b^4\,x^2}{a^5}+\frac {5\,b^3\,x^{3/2}}{a^4}}{a\,x^2+b\,x^{5/2}}-\frac {20\,b^4\,\mathrm {atanh}\left (\frac {2\,b\,\sqrt {x}}{a}+1\right )}{a^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a + b*x^(1/2))^2),x)

[Out]

((5*b*x^(1/2))/(6*a^2) - 1/(2*a) - (5*b^2*x)/(3*a^3) + (10*b^4*x^2)/a^5 + (5*b^3*x^(3/2))/a^4)/(a*x^2 + b*x^(5
/2)) - (20*b^4*atanh((2*b*x^(1/2))/a + 1))/a^6

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sympy [A]  time = 4.07, size = 333, normalized size = 3.51 \[ \begin {cases} \frac {\tilde {\infty }}{x^{3}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {1}{3 b^{2} x^{3}} & \text {for}\: a = 0 \\- \frac {1}{2 a^{2} x^{2}} & \text {for}\: b = 0 \\- \frac {3 a^{5} \sqrt {x}}{6 a^{7} x^{\frac {5}{2}} + 6 a^{6} b x^{3}} + \frac {5 a^{4} b x}{6 a^{7} x^{\frac {5}{2}} + 6 a^{6} b x^{3}} - \frac {10 a^{3} b^{2} x^{\frac {3}{2}}}{6 a^{7} x^{\frac {5}{2}} + 6 a^{6} b x^{3}} + \frac {30 a^{2} b^{3} x^{2}}{6 a^{7} x^{\frac {5}{2}} + 6 a^{6} b x^{3}} + \frac {30 a b^{4} x^{\frac {5}{2}} \log {\relax (x )}}{6 a^{7} x^{\frac {5}{2}} + 6 a^{6} b x^{3}} - \frac {60 a b^{4} x^{\frac {5}{2}} \log {\left (\frac {a}{b} + \sqrt {x} \right )}}{6 a^{7} x^{\frac {5}{2}} + 6 a^{6} b x^{3}} + \frac {60 a b^{4} x^{\frac {5}{2}}}{6 a^{7} x^{\frac {5}{2}} + 6 a^{6} b x^{3}} + \frac {30 b^{5} x^{3} \log {\relax (x )}}{6 a^{7} x^{\frac {5}{2}} + 6 a^{6} b x^{3}} - \frac {60 b^{5} x^{3} \log {\left (\frac {a}{b} + \sqrt {x} \right )}}{6 a^{7} x^{\frac {5}{2}} + 6 a^{6} b x^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo/x**3, Eq(a, 0) & Eq(b, 0)), (-1/(3*b**2*x**3), Eq(a, 0)), (-1/(2*a**2*x**2), Eq(b, 0)), (-3*a**
5*sqrt(x)/(6*a**7*x**(5/2) + 6*a**6*b*x**3) + 5*a**4*b*x/(6*a**7*x**(5/2) + 6*a**6*b*x**3) - 10*a**3*b**2*x**(
3/2)/(6*a**7*x**(5/2) + 6*a**6*b*x**3) + 30*a**2*b**3*x**2/(6*a**7*x**(5/2) + 6*a**6*b*x**3) + 30*a*b**4*x**(5
/2)*log(x)/(6*a**7*x**(5/2) + 6*a**6*b*x**3) - 60*a*b**4*x**(5/2)*log(a/b + sqrt(x))/(6*a**7*x**(5/2) + 6*a**6
*b*x**3) + 60*a*b**4*x**(5/2)/(6*a**7*x**(5/2) + 6*a**6*b*x**3) + 30*b**5*x**3*log(x)/(6*a**7*x**(5/2) + 6*a**
6*b*x**3) - 60*b**5*x**3*log(a/b + sqrt(x))/(6*a**7*x**(5/2) + 6*a**6*b*x**3), True))

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