3.15 \(\int \frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{x^4} \, dx\)

Optimal. Leaf size=75 \[ \frac {b \log (x) \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}-\frac {a \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 x^3 \left (a+b x^3\right )} \]

[Out]

-1/3*a*((b*x^3+a)^2)^(1/2)/x^3/(b*x^3+a)+b*ln(x)*((b*x^3+a)^2)^(1/2)/(b*x^3+a)

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Rubi [A]  time = 0.02, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1355, 14} \[ \frac {b \log (x) \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}-\frac {a \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 x^3 \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]/x^4,x]

[Out]

-(a*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(3*x^3*(a + b*x^3)) + (b*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]*Log[x])/(a + b*x
^3)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 1355

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^
(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^n)^(2*p), x], x] /; Fr
eeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 39, normalized size = 0.52 \[ -\frac {\sqrt {\left (a+b x^3\right )^2} \left (a-3 b x^3 \log (x)\right )}{3 x^3 \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]/x^4,x]

[Out]

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

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fricas [A]  time = 0.64, size = 17, normalized size = 0.23 \[ \frac {3 \, b x^{3} \log \relax (x) - a}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x^3+a)^2)^(1/2)/x^4,x, algorithm="fricas")

[Out]

1/3*(3*b*x^3*log(x) - a)/x^3

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giac [A]  time = 0.35, size = 43, normalized size = 0.57 \[ b \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x^{3} + a\right ) - \frac {b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + a \mathrm {sgn}\left (b x^{3} + a\right )}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x^3+a)^2)^(1/2)/x^4,x, algorithm="giac")

[Out]

b*log(abs(x))*sgn(b*x^3 + a) - 1/3*(b*x^3*sgn(b*x^3 + a) + a*sgn(b*x^3 + a))/x^3

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maple [A]  time = 0.01, size = 38, normalized size = 0.51 \[ \frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (3 b \,x^{3} \ln \relax (x )-a \right )}{3 \left (b \,x^{3}+a \right ) x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x^3+a)^2)^(1/2)/x^4,x)

[Out]

1/3*((b*x^3+a)^2)^(1/2)*(3*b*ln(x)*x^3-a)/(b*x^3+a)/x^3

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maxima [A]  time = 0.46, size = 99, normalized size = 1.32 \[ \frac {1}{3} \, \left (-1\right )^{2 \, b^{2} x^{3} + 2 \, a b} b \log \left (2 \, b^{2} x^{3} + 2 \, a b\right ) - \frac {1}{3} \, \left (-1\right )^{2 \, a b x^{3} + 2 \, a^{2}} b \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{x^{2} {\left | x \right |}}\right ) - \frac {\sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}}}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x^3+a)^2)^(1/2)/x^4,x, algorithm="maxima")

[Out]

1/3*(-1)^(2*b^2*x^3 + 2*a*b)*b*log(2*b^2*x^3 + 2*a*b) - 1/3*(-1)^(2*a*b*x^3 + 2*a^2)*b*log(2*a*b*x/abs(x) + 2*
a^2/(x^2*abs(x))) - 1/3*sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)/x^3

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mupad [B]  time = 1.38, size = 112, normalized size = 1.49 \[ \frac {\ln \left (a\,b+\sqrt {{\left (b\,x^3+a\right )}^2}\,\sqrt {b^2}+b^2\,x^3\right )\,\sqrt {b^2}}{3}-\frac {\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{3\,x^3}-\frac {a\,b\,\ln \left (a\,b+\frac {a^2}{x^3}+\frac {\sqrt {a^2}\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{x^3}\right )}{3\,\sqrt {a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^3)^2)^(1/2)/x^4,x)

[Out]

(log(a*b + ((a + b*x^3)^2)^(1/2)*(b^2)^(1/2) + b^2*x^3)*(b^2)^(1/2))/3 - (a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2)/(3*
x^3) - (a*b*log(a*b + a^2/x^3 + ((a^2)^(1/2)*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/x^3))/(3*(a^2)^(1/2))

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sympy [A]  time = 0.16, size = 10, normalized size = 0.13 \[ - \frac {a}{3 x^{3}} + b \log {\relax (x )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x**3+a)**2)**(1/2)/x**4,x)

[Out]

-a/(3*x**3) + b*log(x)

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