\(\int \frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{x^7} \, dx\) [18]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [C] (warning: unable to verify)
   Fricas [A] (verification not implemented)
   Sympy [F(-1)]
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 26, antiderivative size = 79 \[ \int \frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{x^7} \, dx=-\frac {a \sqrt {a^2+2 a b x^3+b^2 x^6}}{6 x^6 \left (a+b x^3\right )}-\frac {b \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 x^3 \left (a+b x^3\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1369, 14} \[ \int \frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{x^7} \, dx=-\frac {a \sqrt {a^2+2 a b x^3+b^2 x^6}}{6 x^6 \left (a+b x^3\right )}-\frac {b \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 x^3 \left (a+b x^3\right )} \]

[In]

Int[Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]/x^7,x]

[Out]

-1/6*(a*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(x^6*(a + b*x^3)) - (b*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(3*x^3*(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 1369

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*} \text {integral}& = \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {a b+b^2 x^3}{x^7} \, 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^7}+\frac {b^2}{x^4}\right ) \, dx}{a b+b^2 x^3} \\ & = -\frac {a \sqrt {a^2+2 a b x^3+b^2 x^6}}{6 x^6 \left (a+b x^3\right )}-\frac {b \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 x^3 \left (a+b x^3\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.47 \[ \int \frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{x^7} \, dx=-\frac {\sqrt {\left (a+b x^3\right )^2} \left (a+2 b x^3\right )}{6 x^6 \left (a+b x^3\right )} \]

[In]

Integrate[Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]/x^7,x]

[Out]

-1/6*(Sqrt[(a + b*x^3)^2]*(a + 2*b*x^3))/(x^6*(a + b*x^3))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 2.

Time = 0.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.28

method result size
pseudoelliptic \(-\frac {\operatorname {csgn}\left (b \,x^{3}+a \right ) \left (2 b \,x^{3}+a \right )}{6 x^{6}}\) \(22\)
gosper \(-\frac {\left (2 b \,x^{3}+a \right ) \sqrt {\left (b \,x^{3}+a \right )^{2}}}{6 x^{6} \left (b \,x^{3}+a \right )}\) \(34\)
default \(-\frac {\left (2 b \,x^{3}+a \right ) \sqrt {\left (b \,x^{3}+a \right )^{2}}}{6 x^{6} \left (b \,x^{3}+a \right )}\) \(34\)
risch \(\frac {\left (-\frac {b \,x^{3}}{3}-\frac {a}{6}\right ) \sqrt {\left (b \,x^{3}+a \right )^{2}}}{x^{6} \left (b \,x^{3}+a \right )}\) \(35\)

[In]

int(((b*x^3+a)^2)^(1/2)/x^7,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.16 \[ \int \frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{x^7} \, dx=-\frac {2 \, b x^{3} + a}{6 \, x^{6}} \]

[In]

integrate(((b*x^3+a)^2)^(1/2)/x^7,x, algorithm="fricas")

[Out]

-1/6*(2*b*x^3 + a)/x^6

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{x^7} \, dx=\text {Timed out} \]

[In]

integrate(((b*x**3+a)**2)**(1/2)/x**7,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.09 \[ \int \frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{x^7} \, dx=\frac {\sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} b^{2}}{6 \, a^{2}} + \frac {\sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} b}{6 \, a x^{3}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}}}{6 \, a^{2} x^{6}} \]

[In]

integrate(((b*x^3+a)^2)^(1/2)/x^7,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.38 \[ \int \frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{x^7} \, dx=-\frac {2 \, b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + a \mathrm {sgn}\left (b x^{3} + a\right )}{6 \, x^{6}} \]

[In]

integrate(((b*x^3+a)^2)^(1/2)/x^7,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.17 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.42 \[ \int \frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{x^7} \, dx=-\frac {\left (2\,b\,x^3+a\right )\,\sqrt {{\left (b\,x^3+a\right )}^2}}{6\,x^6\,\left (b\,x^3+a\right )} \]

[In]

int(((a + b*x^3)^2)^(1/2)/x^7,x)

[Out]

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