\(\int \frac {(a^2+2 a b x^3+b^2 x^6)^{3/2}}{x} \, dx\) [33]

Optimal result

Integrand size = 26, antiderivative size = 160 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x} \, dx=\frac {a^2 b x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac {a b^2 x^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}+\frac {b^3 x^9 \sqrt {a^2+2 a b x^3+b^2 x^6}}{9 \left (a+b x^3\right )}+\frac {a^3 \sqrt {a^2+2 a b x^3+b^2 x^6} \log (x)}{a+b x^3} \]

[Out]

a^2*b*x^3*((b*x^3+a)^2)^(1/2)/(b*x^3+a)+1/2*a*b^2*x^6*((b*x^3+a)^2)^(1/2)/(b*x^3+a)+1/9*b^3*x^9*((b*x^3+a)^2)^
(1/2)/(b*x^3+a)+a^3*ln(x)*((b*x^3+a)^2)^(1/2)/(b*x^3+a)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1369, 272, 45} \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x} \, dx=\frac {a b^2 x^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}+\frac {a^2 b x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac {b^3 x^9 \sqrt {a^2+2 a b x^3+b^2 x^6}}{9 \left (a+b x^3\right )}+\frac {a^3 \log (x) \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3} \]

[In]

Int[(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)/x,x]

[Out]

(a^2*b*x^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(a + b*x^3) + (a*b^2*x^6*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(2*(a +
b*x^3)) + (b^3*x^9*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(9*(a + b*x^3)) + (a^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]*Log
[x])/(a + b*x^3)

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

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]]

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 {\left (a b+b^2 x^3\right )^3}{x} \, dx}{b^2 \left (a b+b^2 x^3\right )} \\ & = \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \text {Subst}\left (\int \frac {\left (a b+b^2 x\right )^3}{x} \, dx,x,x^3\right )}{3 b^2 \left (a b+b^2 x^3\right )} \\ & = \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \text {Subst}\left (\int \left (3 a^2 b^4+\frac {a^3 b^3}{x}+3 a b^5 x+b^6 x^2\right ) \, dx,x,x^3\right )}{3 b^2 \left (a b+b^2 x^3\right )} \\ & = \frac {a^2 b x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac {a b^2 x^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}+\frac {b^3 x^9 \sqrt {a^2+2 a b x^3+b^2 x^6}}{9 \left (a+b x^3\right )}+\frac {a^3 \sqrt {a^2+2 a b x^3+b^2 x^6} \log (x)}{a+b x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.38 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x} \, dx=\frac {\sqrt {\left (a+b x^3\right )^2} \left (b x^3 \left (18 a^2+9 a b x^3+2 b^2 x^6\right )+18 a^3 \log (x)\right )}{18 \left (a+b x^3\right )} \]

[In]

Integrate[(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)/x,x]

[Out]

(Sqrt[(a + b*x^3)^2]*(b*x^3*(18*a^2 + 9*a*b*x^3 + 2*b^2*x^6) + 18*a^3*Log[x]))/(18*(a + b*x^3))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.08 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.34

method	result	size
pseudoelliptic	\(\frac {\operatorname {csgn}\left (b \,x^{3}+a \right ) \left (2 b^{3} x^{9}+9 b^{2} x^{6} a +18 a^{2} b \,x^{3}+6 a^{3} \ln \left (b \,x^{3}\right )+11 a^{3}\right )}{18}\)	\(54\)
default	\(\frac {{\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}} \left (2 b^{3} x^{9}+9 b^{2} x^{6} a +18 a^{2} b \,x^{3}+18 a^{3} \ln \left (x \right )\right )}{18 \left (b \,x^{3}+a \right )^{3}}\)	\(57\)
risch	\(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, b \left (\frac {1}{9} b^{2} x^{9}+\frac {1}{2} a b \,x^{6}+a^{2} x^{3}\right )}{b \,x^{3}+a}+\frac {a^{3} \ln \left (x \right ) \sqrt {\left (b \,x^{3}+a \right )^{2}}}{b \,x^{3}+a}\)	\(73\)

[In]

int((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x,x,method=_RETURNVERBOSE)

[Out]

1/18*csgn(b*x^3+a)*(2*b^3*x^9+9*b^2*x^6*a+18*a^2*b*x^3+6*a^3*ln(b*x^3)+11*a^3)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.20 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x} \, dx=\frac {1}{9} \, b^{3} x^{9} + \frac {1}{2} \, a b^{2} x^{6} + a^{2} b x^{3} + a^{3} \log \left (x\right ) \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x,x, algorithm="fricas")

[Out]

1/9*b^3*x^9 + 1/2*a*b^2*x^6 + a^2*b*x^3 + a^3*log(x)

Sympy [F]

\[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x} \, dx=\int \frac {\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {3}{2}}}{x}\, dx \]

[In]

integrate((b**2*x**6+2*a*b*x**3+a**2)**(3/2)/x,x)

[Out]

Integral(((a + b*x**3)**2)**(3/2)/x, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x} \, dx=\frac {1}{6} \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} a b x^{3} + \frac {1}{3} \, \left (-1\right )^{2 \, b^{2} x^{3} + 2 \, a b} a^{3} \log \left (2 \, b^{2} x^{3} + 2 \, a b\right ) - \frac {1}{3} \, \left (-1\right )^{2 \, a b x^{3} + 2 \, a^{2}} a^{3} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{x^{2} {\left | x \right |}}\right ) + \frac {1}{2} \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} a^{2} + \frac {1}{9} \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x,x, algorithm="maxima")

[Out]

1/6*sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*a*b*x^3 + 1/3*(-1)^(2*b^2*x^3 + 2*a*b)*a^3*log(2*b^2*x^3 + 2*a*b) - 1/3*(-
1)^(2*a*b*x^3 + 2*a^2)*a^3*log(2*a*b*x/abs(x) + 2*a^2/(x^2*abs(x))) + 1/2*sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*a^2
+ 1/9*(b^2*x^6 + 2*a*b*x^3 + a^2)^(3/2)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.41 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x} \, dx=\frac {1}{9} \, b^{3} x^{9} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{2} \, a b^{2} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + a^{2} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + a^{3} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x^{3} + a\right ) \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x,x, algorithm="giac")

[Out]

1/9*b^3*x^9*sgn(b*x^3 + a) + 1/2*a*b^2*x^6*sgn(b*x^3 + a) + a^2*b*x^3*sgn(b*x^3 + a) + a^3*log(abs(x))*sgn(b*x
^3 + a)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{3/2}}{x} \,d x \]

[In]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2)/x,x)

[Out]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2)/x, x)