\(\int \frac {(a^2+2 a b x^3+b^2 x^6)^{3/2}}{x^4} \, dx\) [36]

Optimal result

Integrand size = 26, antiderivative size = 161 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^4} \, dx=-\frac {a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 x^3 \left (a+b x^3\right )}+\frac {a b^2 x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac {b^3 x^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}{6 \left (a+b x^3\right )}+\frac {3 a^2 b \sqrt {a^2+2 a b x^3+b^2 x^6} \log (x)}{a+b x^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 161, 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} \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^4} \, dx=\frac {a b^2 x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac {3 a^2 b \log (x) \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac {b^3 x^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}{6 \left (a+b x^3\right )}-\frac {a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 x^3 \left (a+b x^3\right )} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.39 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^4} \, dx=\frac {\sqrt {\left (a+b x^3\right )^2} \left (-2 a^3+6 a b^2 x^6+b^3 x^9+18 a^2 b x^3 \log (x)\right )}{6 x^3 \left (a+b x^3\right )} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.37

method	result	size
default	\(\frac {{\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}} \left (b^{3} x^{9}+6 b^{2} x^{6} a +18 a^{2} b \ln \left (x \right ) x^{3}-2 a^{3}\right )}{6 x^{3} \left (b \,x^{3}+a \right )^{3}}\)	\(59\)
pseudoelliptic	\(-\frac {\operatorname {csgn}\left (b \,x^{3}+a \right ) \left (-\frac {b^{3} x^{9}}{2}-3 b^{2} x^{6} a -3 \ln \left (b \,x^{3}\right ) a^{2} b \,x^{3}-\frac {5 a^{2} b \,x^{3}}{2}+a^{3}\right )}{3 x^{3}}\)	\(59\)
risch	\(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, b \left (b \,x^{3}+3 a \right )^{2}}{6 b \,x^{3}+6 a}-\frac {a^{3} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{3 x^{3} \left (b \,x^{3}+a \right )}+\frac {3 a^{2} b \ln \left (x \right ) \sqrt {\left (b \,x^{3}+a \right )^{2}}}{b \,x^{3}+a}\)	\(92\)

[In]

int((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^4} \, dx=\frac {b^{3} x^{9} + 6 \, a b^{2} x^{6} + 18 \, a^{2} b x^{3} \log \left (x\right ) - 2 \, a^{3}}{6 \, x^{3}} \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^4,x, algorithm="fricas")

[Out]

1/6*(b^3*x^9 + 6*a*b^2*x^6 + 18*a^2*b*x^3*log(x) - 2*a^3)/x^3

Sympy [F]

\[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^4} \, dx=\int \frac {\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {3}{2}}}{x^{4}}\, dx \]

[In]

integrate((b**2*x**6+2*a*b*x**3+a**2)**(3/2)/x**4,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.97 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^4} \, dx=\frac {1}{2} \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} b^{2} x^{3} + \left (-1\right )^{2 \, b^{2} x^{3} + 2 \, a b} a^{2} b \log \left (2 \, b^{2} x^{3} + 2 \, a b\right ) - \left (-1\right )^{2 \, a b x^{3} + 2 \, a^{2}} a^{2} b \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{x^{2} {\left | x \right |}}\right ) + \frac {3}{2} \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} a b - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}}}{3 \, x^{3}} \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^4,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.53 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^4} \, dx=\frac {1}{6} \, b^{3} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + a b^{2} x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + 3 \, a^{2} b \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x^{3} + a\right ) - \frac {3 \, a^{2} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + a^{3} \mathrm {sgn}\left (b x^{3} + a\right )}{3 \, x^{3}} \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^4,x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^4} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{3/2}}{x^4} \,d x \]

[In]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2)/x^4,x)

[Out]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2)/x^4, x)