\(\int \frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{x^5} \, dx\) [16]

Optimal result

Integrand size = 26, antiderivative size = 77 \[ \int \frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{x^5} \, dx=-\frac {a \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 x^4 \left (a+b x^3\right )}-\frac {b \sqrt {a^2+2 a b x^3+b^2 x^6}}{x \left (a+b x^3\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 77, 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} \[ \int \frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{x^5} \, dx=-\frac {b \sqrt {a^2+2 a b x^3+b^2 x^6}}{x \left (a+b x^3\right )}-\frac {a \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 x^4 \left (a+b x^3\right )} \]

[In]

Int[Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]/x^5,x]

[Out]

-1/4*(a*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(x^4*(a + b*x^3)) - (b*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(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 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^5} \, 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^5}+\frac {b^2}{x^2}\right ) \, dx}{a b+b^2 x^3} \\ & = -\frac {a \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 x^4 \left (a+b x^3\right )}-\frac {b \sqrt {a^2+2 a b x^3+b^2 x^6}}{x \left (a+b x^3\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.48 \[ \int \frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{x^5} \, dx=-\frac {\sqrt {\left (a+b x^3\right )^2} \left (a+4 b x^3\right )}{4 x^4 \left (a+b x^3\right )} \]

[In]

Integrate[Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]/x^5,x]

[Out]

-1/4*(Sqrt[(a + b*x^3)^2]*(a + 4*b*x^3))/(x^4*(a + b*x^3))

Maple [A] (verified)

Time = 4.52 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.44

method	result	size
gosper	\(-\frac {\left (4 b \,x^{3}+a \right ) \sqrt {\left (b \,x^{3}+a \right )^{2}}}{4 \left (b \,x^{3}+a \right ) x^{4}}\)	\(34\)
default	\(-\frac {\left (4 b \,x^{3}+a \right ) \sqrt {\left (b \,x^{3}+a \right )^{2}}}{4 \left (b \,x^{3}+a \right ) x^{4}}\)	\(34\)
risch	\(\frac {\left (-b \,x^{3}-\frac {a}{4}\right ) \sqrt {\left (b \,x^{3}+a \right )^{2}}}{x^{4} \left (b \,x^{3}+a \right )}\)	\(35\)

[In]

int(((b*x^3+a)^2)^(1/2)/x^5,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.17 \[ \int \frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{x^5} \, dx=-\frac {4 \, b x^{3} + a}{4 \, x^{4}} \]

[In]

integrate(((b*x^3+a)^2)^(1/2)/x^5,x, algorithm="fricas")

[Out]

-1/4*(4*b*x^3 + a)/x^4

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{x^5} \, dx=\text {Timed out} \]

[In]

integrate(((b*x**3+a)**2)**(1/2)/x**5,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.17 \[ \int \frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{x^5} \, dx=-\frac {4 \, b x^{3} + a}{4 \, x^{4}} \]

[In]

integrate(((b*x^3+a)^2)^(1/2)/x^5,x, algorithm="maxima")

[Out]

-1/4*(4*b*x^3 + a)/x^4

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.39 \[ \int \frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{x^5} \, dx=-\frac {4 \, b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + a \mathrm {sgn}\left (b x^{3} + a\right )}{4 \, x^{4}} \]

[In]

integrate(((b*x^3+a)^2)^(1/2)/x^5,x, algorithm="giac")

[Out]

-1/4*(4*b*x^3*sgn(b*x^3 + a) + a*sgn(b*x^3 + a))/x^4

Mupad [B] (verification not implemented)

Time = 8.23 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.43 \[ \int \frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{x^5} \, dx=-\frac {\left (4\,b\,x^3+a\right )\,\sqrt {{\left (b\,x^3+a\right )}^2}}{4\,x^4\,\left (b\,x^3+a\right )} \]

[In]

int(((a + b*x^3)^2)^(1/2)/x^5,x)

[Out]

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