\(\int \frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{x^{11}} \, dx\) [22]

Optimal result

Integrand size = 26, antiderivative size = 79 \[ \int \frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{x^{11}} \, dx=-\frac {a \sqrt {a^2+2 a b x^3+b^2 x^6}}{10 x^{10} \left (a+b x^3\right )}-\frac {b \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 x^7 \left (a+b x^3\right )} \]

[Out]

-1/10*a*((b*x^3+a)^2)^(1/2)/x^10/(b*x^3+a)-1/7*b*((b*x^3+a)^2)^(1/2)/x^7/(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} \[ \int \frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{x^{11}} \, dx=-\frac {a \sqrt {a^2+2 a b x^3+b^2 x^6}}{10 x^{10} \left (a+b x^3\right )}-\frac {b \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 x^7 \left (a+b x^3\right )} \]

[In]

Int[Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]/x^11,x]

[Out]

-1/10*(a*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(x^10*(a + b*x^3)) - (b*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(7*x^7*(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^{11}} \, 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^{11}}+\frac {b^2}{x^8}\right ) \, dx}{a b+b^2 x^3} \\ & = -\frac {a \sqrt {a^2+2 a b x^3+b^2 x^6}}{10 x^{10} \left (a+b x^3\right )}-\frac {b \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 x^7 \left (a+b x^3\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.49 \[ \int \frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{x^{11}} \, dx=-\frac {\sqrt {\left (a+b x^3\right )^2} \left (7 a+10 b x^3\right )}{70 x^{10} \left (a+b x^3\right )} \]

[In]

Integrate[Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]/x^11,x]

[Out]

-1/70*(Sqrt[(a + b*x^3)^2]*(7*a + 10*b*x^3))/(x^10*(a + b*x^3))

Maple [A] (verified)

Time = 11.83 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.44

method	result	size
risch	\(\frac {\left (-\frac {b \,x^{3}}{7}-\frac {a}{10}\right ) \sqrt {\left (b \,x^{3}+a \right )^{2}}}{x^{10} \left (b \,x^{3}+a \right )}\)	\(35\)
gosper	\(-\frac {\left (10 b \,x^{3}+7 a \right ) \sqrt {\left (b \,x^{3}+a \right )^{2}}}{70 x^{10} \left (b \,x^{3}+a \right )}\)	\(36\)
default	\(-\frac {\left (10 b \,x^{3}+7 a \right ) \sqrt {\left (b \,x^{3}+a \right )^{2}}}{70 x^{10} \left (b \,x^{3}+a \right )}\)	\(36\)

[In]

int(((b*x^3+a)^2)^(1/2)/x^11,x,method=_RETURNVERBOSE)

[Out]

1/x^10*(-1/7*b*x^3-1/10*a)/(b*x^3+a)*((b*x^3+a)^2)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.19 \[ \int \frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{x^{11}} \, dx=-\frac {10 \, b x^{3} + 7 \, a}{70 \, x^{10}} \]

[In]

integrate(((b*x^3+a)^2)^(1/2)/x^11,x, algorithm="fricas")

[Out]

-1/70*(10*b*x^3 + 7*a)/x^10

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{x^{11}} \, dx=\text {Timed out} \]

[In]

integrate(((b*x**3+a)**2)**(1/2)/x**11,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.19 \[ \int \frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{x^{11}} \, dx=-\frac {10 \, b x^{3} + 7 \, a}{70 \, x^{10}} \]

[In]

integrate(((b*x^3+a)^2)^(1/2)/x^11,x, algorithm="maxima")

[Out]

-1/70*(10*b*x^3 + 7*a)/x^10

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.39 \[ \int \frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{x^{11}} \, dx=-\frac {10 \, b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + 7 \, a \mathrm {sgn}\left (b x^{3} + a\right )}{70 \, x^{10}} \]

[In]

integrate(((b*x^3+a)^2)^(1/2)/x^11,x, algorithm="giac")

[Out]

-1/70*(10*b*x^3*sgn(b*x^3 + a) + 7*a*sgn(b*x^3 + a))/x^10

Mupad [B] (verification not implemented)

Time = 8.21 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.44 \[ \int \frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{x^{11}} \, dx=-\frac {\left (10\,b\,x^3+7\,a\right )\,\sqrt {{\left (b\,x^3+a\right )}^2}}{70\,x^{10}\,\left (b\,x^3+a\right )} \]

[In]

int(((a + b*x^3)^2)^(1/2)/x^11,x)

[Out]

-((7*a + 10*b*x^3)*((a + b*x^3)^2)^(1/2))/(70*x^10*(a + b*x^3))