\(\int \frac {(a^2+2 a b x^3+b^2 x^6)^{3/2}}{x^3} \, dx\) [35]

Optimal result

Integrand size = 26, antiderivative size = 163 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^3} \, dx=-\frac {a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 x^2 \left (a+b x^3\right )}+\frac {3 a^2 b x \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac {3 a b^2 x^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 \left (a+b x^3\right )}+\frac {b^3 x^7 \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 \left (a+b x^3\right )} \]

[Out]

-1/2*a^3*((b*x^3+a)^2)^(1/2)/x^2/(b*x^3+a)+3*a^2*b*x*((b*x^3+a)^2)^(1/2)/(b*x^3+a)+3/4*a*b^2*x^4*((b*x^3+a)^2)
^(1/2)/(b*x^3+a)+1/7*b^3*x^7*((b*x^3+a)^2)^(1/2)/(b*x^3+a)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 163, 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, 276} \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^3} \, dx=\frac {3 a^2 b x \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac {3 a b^2 x^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 \left (a+b x^3\right )}+\frac {b^3 x^7 \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 \left (a+b x^3\right )}-\frac {a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 x^2 \left (a+b x^3\right )} \]

[In]

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

[Out]

-1/2*(a^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(x^2*(a + b*x^3)) + (3*a^2*b*x*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(a
+ b*x^3) + (3*a*b^2*x^4*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(4*(a + b*x^3)) + (b^3*x^7*Sqrt[a^2 + 2*a*b*x^3 + b^2
*x^6])/(7*(a + b*x^3))

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

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^3} \, dx}{b^2 \left (a b+b^2 x^3\right )} \\ & = \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \left (3 a^2 b^4+\frac {a^3 b^3}{x^3}+3 a b^5 x^3+b^6 x^6\right ) \, dx}{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}}{2 x^2 \left (a+b x^3\right )}+\frac {3 a^2 b x \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac {3 a b^2 x^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 \left (a+b x^3\right )}+\frac {b^3 x^7 \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 \left (a+b x^3\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.37 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^3} \, dx=\frac {\sqrt {\left (a+b x^3\right )^2} \left (-14 a^3+84 a^2 b x^3+21 a b^2 x^6+4 b^3 x^9\right )}{28 x^2 \left (a+b x^3\right )} \]

[In]

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

[Out]

(Sqrt[(a + b*x^3)^2]*(-14*a^3 + 84*a^2*b*x^3 + 21*a*b^2*x^6 + 4*b^3*x^9))/(28*x^2*(a + b*x^3))

Maple [A] (verified)

Time = 3.13 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.36

method	result	size
gosper	\(-\frac {\left (-4 b^{3} x^{9}-21 b^{2} x^{6} a -84 a^{2} b \,x^{3}+14 a^{3}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}}}{28 x^{2} \left (b \,x^{3}+a \right )^{3}}\)	\(58\)
default	\(-\frac {\left (-4 b^{3} x^{9}-21 b^{2} x^{6} a -84 a^{2} b \,x^{3}+14 a^{3}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}}}{28 x^{2} \left (b \,x^{3}+a \right )^{3}}\)	\(58\)
risch	\(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, b \left (\frac {1}{7} b^{2} x^{7}+\frac {3}{4} a b \,x^{4}+3 a^{2} x \right )}{b \,x^{3}+a}-\frac {a^{3} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{2 x^{2} \left (b \,x^{3}+a \right )}\)	\(74\)

[In]

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

[Out]

-1/28*(-4*b^3*x^9-21*a*b^2*x^6-84*a^2*b*x^3+14*a^3)*((b*x^3+a)^2)^(3/2)/x^2/(b*x^3+a)^3

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.23 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^3} \, dx=\frac {4 \, b^{3} x^{9} + 21 \, a b^{2} x^{6} + 84 \, a^{2} b x^{3} - 14 \, a^{3}}{28 \, x^{2}} \]

[In]

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

[Out]

1/28*(4*b^3*x^9 + 21*a*b^2*x^6 + 84*a^2*b*x^3 - 14*a^3)/x^2

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.23 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^3} \, dx=\frac {4 \, b^{3} x^{9} + 21 \, a b^{2} x^{6} + 84 \, a^{2} b x^{3} - 14 \, a^{3}}{28 \, x^{2}} \]

[In]

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

[Out]

1/28*(4*b^3*x^9 + 21*a*b^2*x^6 + 84*a^2*b*x^3 - 14*a^3)/x^2

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.40 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^3} \, dx=\frac {1}{7} \, b^{3} x^{7} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {3}{4} \, a b^{2} x^{4} \mathrm {sgn}\left (b x^{3} + a\right ) + 3 \, a^{2} b x \mathrm {sgn}\left (b x^{3} + a\right ) - \frac {a^{3} \mathrm {sgn}\left (b x^{3} + a\right )}{2 \, x^{2}} \]

[In]

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

[Out]

1/7*b^3*x^7*sgn(b*x^3 + a) + 3/4*a*b^2*x^4*sgn(b*x^3 + a) + 3*a^2*b*x*sgn(b*x^3 + a) - 1/2*a^3*sgn(b*x^3 + a)/
x^2

Mupad [F(-1)]

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

[In]

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

[Out]

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