\(\int \frac {(a^2+2 a b x^3+b^2 x^6)^{3/2}}{x^{10}} \, dx\) [42]

Optimal result

Integrand size = 26, antiderivative size = 161 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{10}} \, dx=-\frac {a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{9 x^9 \left (a+b x^3\right )}-\frac {a^2 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 x^6 \left (a+b x^3\right )}-\frac {a b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{x^3 \left (a+b x^3\right )}+\frac {b^3 \sqrt {a^2+2 a b x^3+b^2 x^6} \log (x)}{a+b x^3} \]

[Out]

-1/9*a^3*((b*x^3+a)^2)^(1/2)/x^9/(b*x^3+a)-1/2*a^2*b*((b*x^3+a)^2)^(1/2)/x^6/(b*x^3+a)-a*b^2*((b*x^3+a)^2)^(1/
2)/x^3/(b*x^3+a)+b^3*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} \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{10}} \, dx=-\frac {a^2 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 x^6 \left (a+b x^3\right )}-\frac {a b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{x^3 \left (a+b x^3\right )}+\frac {b^3 \log (x) \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}-\frac {a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{9 x^9 \left (a+b x^3\right )} \]

[In]

Int[(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)/x^10,x]

[Out]

-1/9*(a^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(x^9*(a + b*x^3)) - (a^2*b*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(2*x^6*
(a + b*x^3)) - (a*b^2*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(x^3*(a + b*x^3)) + (b^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^{10}} \, 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^4} \, 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 (\frac {a^3 b^3}{x^4}+\frac {3 a^2 b^4}{x^3}+\frac {3 a b^5}{x^2}+\frac {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}}{9 x^9 \left (a+b x^3\right )}-\frac {a^2 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 x^6 \left (a+b x^3\right )}-\frac {a b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{x^3 \left (a+b x^3\right )}+\frac {b^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 = 0.31 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.65 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{10}} \, dx=\frac {2 a^3 \sqrt {a^2}+9 \left (a^2\right )^{3/2} b x^3+18 a \sqrt {a^2} b^2 x^6-2 a^3 \sqrt {\left (a+b x^3\right )^2}-7 a^2 b x^3 \sqrt {\left (a+b x^3\right )^2}-11 a b^2 x^6 \sqrt {\left (a+b x^3\right )^2}-12 a b^3 x^9 \text {arctanh}\left (\frac {b x^3}{\sqrt {a^2}-\sqrt {\left (a+b x^3\right )^2}}\right )-12 \sqrt {a^2} b^3 x^9 \log \left (x^3\right )+6 \sqrt {a^2} b^3 x^9 \log \left (a \left (\sqrt {a^2}-b x^3-\sqrt {\left (a+b x^3\right )^2}\right )\right )+6 \sqrt {a^2} b^3 x^9 \log \left (a \left (\sqrt {a^2}+b x^3-\sqrt {\left (a+b x^3\right )^2}\right )\right )}{36 a x^9} \]

[In]

Integrate[(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)/x^10,x]

[Out]

(2*a^3*Sqrt[a^2] + 9*(a^2)^(3/2)*b*x^3 + 18*a*Sqrt[a^2]*b^2*x^6 - 2*a^3*Sqrt[(a + b*x^3)^2] - 7*a^2*b*x^3*Sqrt
[(a + b*x^3)^2] - 11*a*b^2*x^6*Sqrt[(a + b*x^3)^2] - 12*a*b^3*x^9*ArcTanh[(b*x^3)/(Sqrt[a^2] - Sqrt[(a + b*x^3
)^2])] - 12*Sqrt[a^2]*b^3*x^9*Log[x^3] + 6*Sqrt[a^2]*b^3*x^9*Log[a*(Sqrt[a^2] - b*x^3 - Sqrt[(a + b*x^3)^2])]
+ 6*Sqrt[a^2]*b^3*x^9*Log[a*(Sqrt[a^2] + b*x^3 - Sqrt[(a + b*x^3)^2])])/(36*a*x^9)

Maple [C] (warning: unable to verify)

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

Time = 0.09 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.32

method	result	size
pseudoelliptic	\(\frac {\operatorname {csgn}\left (b \,x^{3}+a \right ) \left (6 \ln \left (b \,x^{3}\right ) b^{3} x^{9}-18 b^{2} x^{6} a -9 a^{2} b \,x^{3}-2 a^{3}\right )}{18 x^{9}}\)	\(52\)
default	\(\frac {{\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}} \left (18 b^{3} \ln \left (x \right ) x^{9}-18 b^{2} x^{6} a -9 a^{2} b \,x^{3}-2 a^{3}\right )}{18 \left (b \,x^{3}+a \right )^{3} x^{9}}\)	\(60\)
risch	\(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (-b^{2} x^{6} a -\frac {1}{2} a^{2} b \,x^{3}-\frac {1}{9} a^{3}\right )}{\left (b \,x^{3}+a \right ) x^{9}}+\frac {b^{3} \ln \left (x \right ) \sqrt {\left (b \,x^{3}+a \right )^{2}}}{b \,x^{3}+a}\)	\(76\)

[In]

int((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^10,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{10}} \, dx=\frac {18 \, b^{3} x^{9} \log \left (x\right ) - 18 \, a b^{2} x^{6} - 9 \, a^{2} b x^{3} - 2 \, a^{3}}{18 \, x^{9}} \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^10,x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

integrate((b**2*x**6+2*a*b*x**3+a**2)**(3/2)/x**10,x)

[Out]

Integral(((a + b*x**3)**2)**(3/2)/x**10, x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (113) = 226\).

Time = 0.22 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.57 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{10}} \, dx=\frac {\sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} b^{4} x^{3}}{6 \, a^{2}} + \frac {1}{3} \, \left (-1\right )^{2 \, b^{2} x^{3} + 2 \, a b} b^{3} \log \left (2 \, b^{2} x^{3} + 2 \, a b\right ) - \frac {1}{3} \, \left (-1\right )^{2 \, a b x^{3} + 2 \, a^{2}} b^{3} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{x^{2} {\left | x \right |}}\right ) + \frac {\sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} b^{3}}{2 \, a} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} b^{3}}{18 \, a^{3}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} b^{2}}{6 \, a^{2} x^{3}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b}{18 \, a^{3} x^{6}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}}}{9 \, a^{2} x^{9}} \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^10,x, algorithm="maxima")

[Out]

1/6*sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*b^4*x^3/a^2 + 1/3*(-1)^(2*b^2*x^3 + 2*a*b)*b^3*log(2*b^2*x^3 + 2*a*b) - 1/
3*(-1)^(2*a*b*x^3 + 2*a^2)*b^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)*
b^3/a - 1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(3/2)*b^3/a^3 - 1/6*(b^2*x^6 + 2*a*b*x^3 + a^2)^(3/2)*b^2/(a^2*x^3) +
1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*b/(a^3*x^6) - 1/9*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)/(a^2*x^9)

Giac [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.53 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{10}} \, dx=b^{3} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x^{3} + a\right ) - \frac {11 \, b^{3} x^{9} \mathrm {sgn}\left (b x^{3} + a\right ) + 18 \, a b^{2} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + 9 \, a^{2} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + 2 \, a^{3} \mathrm {sgn}\left (b x^{3} + a\right )}{18 \, x^{9}} \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^10,x, algorithm="giac")

[Out]

b^3*log(abs(x))*sgn(b*x^3 + a) - 1/18*(11*b^3*x^9*sgn(b*x^3 + a) + 18*a*b^2*x^6*sgn(b*x^3 + a) + 9*a^2*b*x^3*s
gn(b*x^3 + a) + 2*a^3*sgn(b*x^3 + a))/x^9

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{10}} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{3/2}}{x^{10}} \,d x \]

[In]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2)/x^10,x)

[Out]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2)/x^10, x)