\(\int \frac {(a^2+2 a b x^3+b^2 x^6)^{5/2}}{x} \, dx\) [64]

Optimal result

Integrand size = 26, antiderivative size = 251 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x} \, dx=\frac {5 a^4 b x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 \left (a+b x^3\right )}+\frac {5 a^3 b^2 x^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 \left (a+b x^3\right )}+\frac {10 a^2 b^3 x^9 \sqrt {a^2+2 a b x^3+b^2 x^6}}{9 \left (a+b x^3\right )}+\frac {5 a b^4 x^{12} \sqrt {a^2+2 a b x^3+b^2 x^6}}{12 \left (a+b x^3\right )}+\frac {b^5 x^{15} \sqrt {a^2+2 a b x^3+b^2 x^6}}{15 \left (a+b x^3\right )}+\frac {a^5 \sqrt {a^2+2 a b x^3+b^2 x^6} \log (x)}{a+b x^3} \]

[Out]

5/3*a^4*b*x^3*((b*x^3+a)^2)^(1/2)/(b*x^3+a)+5/3*a^3*b^2*x^6*((b*x^3+a)^2)^(1/2)/(b*x^3+a)+10/9*a^2*b^3*x^9*((b
*x^3+a)^2)^(1/2)/(b*x^3+a)+5/12*a*b^4*x^12*((b*x^3+a)^2)^(1/2)/(b*x^3+a)+1/15*b^5*x^15*((b*x^3+a)^2)^(1/2)/(b*
x^3+a)+a^5*ln(x)*((b*x^3+a)^2)^(1/2)/(b*x^3+a)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 251, 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} \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x} \, dx=\frac {b^5 x^{15} \sqrt {a^2+2 a b x^3+b^2 x^6}}{15 \left (a+b x^3\right )}+\frac {5 a b^4 x^{12} \sqrt {a^2+2 a b x^3+b^2 x^6}}{12 \left (a+b x^3\right )}+\frac {10 a^2 b^3 x^9 \sqrt {a^2+2 a b x^3+b^2 x^6}}{9 \left (a+b x^3\right )}+\frac {a^5 \log (x) \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac {5 a^4 b x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 \left (a+b x^3\right )}+\frac {5 a^3 b^2 x^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 \left (a+b x^3\right )} \]

[In]

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

[Out]

(5*a^4*b*x^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(3*(a + b*x^3)) + (5*a^3*b^2*x^6*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]
)/(3*(a + b*x^3)) + (10*a^2*b^3*x^9*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(9*(a + b*x^3)) + (5*a*b^4*x^12*Sqrt[a^2
+ 2*a*b*x^3 + b^2*x^6])/(12*(a + b*x^3)) + (b^5*x^15*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(15*(a + b*x^3)) + (a^5*
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 )^5}{x} \, dx}{b^4 \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 )^5}{x} \, dx,x,x^3\right )}{3 b^4 \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 (5 a^4 b^6+\frac {a^5 b^5}{x}+10 a^3 b^7 x+10 a^2 b^8 x^2+5 a b^9 x^3+b^{10} x^4\right ) \, dx,x,x^3\right )}{3 b^4 \left (a b+b^2 x^3\right )} \\ & = \frac {5 a^4 b x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 \left (a+b x^3\right )}+\frac {5 a^3 b^2 x^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 \left (a+b x^3\right )}+\frac {10 a^2 b^3 x^9 \sqrt {a^2+2 a b x^3+b^2 x^6}}{9 \left (a+b x^3\right )}+\frac {5 a b^4 x^{12} \sqrt {a^2+2 a b x^3+b^2 x^6}}{12 \left (a+b x^3\right )}+\frac {b^5 x^{15} \sqrt {a^2+2 a b x^3+b^2 x^6}}{15 \left (a+b x^3\right )}+\frac {a^5 \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 = 82, normalized size of antiderivative = 0.33 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x} \, dx=\frac {\sqrt {\left (a+b x^3\right )^2} \left (b x^3 \left (300 a^4+300 a^3 b x^3+200 a^2 b^2 x^6+75 a b^3 x^9+12 b^4 x^{12}\right )+180 a^5 \log (x)\right )}{180 \left (a+b x^3\right )} \]

[In]

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

[Out]

(Sqrt[(a + b*x^3)^2]*(b*x^3*(300*a^4 + 300*a^3*b*x^3 + 200*a^2*b^2*x^6 + 75*a*b^3*x^9 + 12*b^4*x^12) + 180*a^5
*Log[x]))/(180*(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 = 75, normalized size of antiderivative = 0.30

method	result	size
pseudoelliptic	\(\frac {\operatorname {csgn}\left (b \,x^{3}+a \right ) \left (\frac {b^{5} x^{15}}{5}+\frac {5 a \,b^{4} x^{12}}{4}+\frac {10 a^{2} b^{3} x^{9}}{3}+5 a^{3} b^{2} x^{6}+5 a^{4} b \,x^{3}+a^{5} \ln \left (b \,x^{3}\right )+\frac {137 a^{5}}{60}\right )}{3}\)	\(75\)
default	\(\frac {{\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}} \left (12 b^{5} x^{15}+75 a \,b^{4} x^{12}+200 a^{2} b^{3} x^{9}+300 a^{3} b^{2} x^{6}+300 a^{4} b \,x^{3}+180 a^{5} \ln \left (x \right )\right )}{180 \left (b \,x^{3}+a \right )^{5}}\)	\(79\)
risch	\(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, b \left (\frac {1}{15} b^{4} x^{15}+\frac {5}{12} a \,b^{3} x^{12}+\frac {10}{9} a^{2} b^{2} x^{9}+\frac {5}{3} a^{3} b \,x^{6}+\frac {5}{3} a^{4} x^{3}\right )}{b \,x^{3}+a}+\frac {a^{5} \ln \left (x \right ) \sqrt {\left (b \,x^{3}+a \right )^{2}}}{b \,x^{3}+a}\)	\(96\)

[In]

int((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x,x,method=_RETURNVERBOSE)

[Out]

1/3*csgn(b*x^3+a)*(1/5*b^5*x^15+5/4*a*b^4*x^12+10/3*a^2*b^3*x^9+5*a^3*b^2*x^6+5*a^4*b*x^3+a^5*ln(b*x^3)+137/60
*a^5)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x} \, dx=\frac {1}{15} \, b^{5} x^{15} + \frac {5}{12} \, a b^{4} x^{12} + \frac {10}{9} \, a^{2} b^{3} x^{9} + \frac {5}{3} \, a^{3} b^{2} x^{6} + \frac {5}{3} \, a^{4} b x^{3} + a^{5} \log \left (x\right ) \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x,x, algorithm="fricas")

[Out]

1/15*b^5*x^15 + 5/12*a*b^4*x^12 + 10/9*a^2*b^3*x^9 + 5/3*a^3*b^2*x^6 + 5/3*a^4*b*x^3 + a^5*log(x)

Sympy [F]

\[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x} \, dx=\int \frac {\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}}{x}\, dx \]

[In]

integrate((b**2*x**6+2*a*b*x**3+a**2)**(5/2)/x,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.82 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x} \, dx=\frac {1}{6} \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} a^{3} b x^{3} + \frac {1}{3} \, \left (-1\right )^{2 \, b^{2} x^{3} + 2 \, a b} a^{5} \log \left (2 \, b^{2} x^{3} + 2 \, a b\right ) - \frac {1}{3} \, \left (-1\right )^{2 \, a b x^{3} + 2 \, a^{2}} a^{5} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{x^{2} {\left | x \right |}}\right ) + \frac {1}{12} \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} a b x^{3} + \frac {1}{2} \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} a^{4} + \frac {7}{36} \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} a^{2} + \frac {1}{15} \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x,x, algorithm="maxima")

[Out]

1/6*sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*a^3*b*x^3 + 1/3*(-1)^(2*b^2*x^3 + 2*a*b)*a^5*log(2*b^2*x^3 + 2*a*b) - 1/3*
(-1)^(2*a*b*x^3 + 2*a^2)*a^5*log(2*a*b*x/abs(x) + 2*a^2/(x^2*abs(x))) + 1/12*(b^2*x^6 + 2*a*b*x^3 + a^2)^(3/2)
*a*b*x^3 + 1/2*sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*a^4 + 7/36*(b^2*x^6 + 2*a*b*x^3 + a^2)^(3/2)*a^2 + 1/15*(b^2*x^
6 + 2*a*b*x^3 + a^2)^(5/2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.41 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x} \, dx=\frac {1}{15} \, b^{5} x^{15} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{12} \, a b^{4} x^{12} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {10}{9} \, a^{2} b^{3} x^{9} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{3} \, a^{3} b^{2} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{3} \, a^{4} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + a^{5} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x^{3} + a\right ) \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x,x, algorithm="giac")

[Out]

1/15*b^5*x^15*sgn(b*x^3 + a) + 5/12*a*b^4*x^12*sgn(b*x^3 + a) + 10/9*a^2*b^3*x^9*sgn(b*x^3 + a) + 5/3*a^3*b^2*
x^6*sgn(b*x^3 + a) + 5/3*a^4*b*x^3*sgn(b*x^3 + a) + a^5*log(abs(x))*sgn(b*x^3 + a)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{5/2}}{x} \,d x \]

[In]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2)/x,x)

[Out]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2)/x, x)