\(\int x^{11} (a^2+2 a b x^3+b^2 x^6)^{5/2} \, dx\) [52]

Optimal result

Integrand size = 26, antiderivative size = 160 \[ \int x^{11} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=-\frac {a^3 \left (a+b x^3\right )^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{18 b^4}+\frac {a^2 \left (a+b x^3\right )^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 b^4}-\frac {a \left (a+b x^3\right )^7 \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 b^4}+\frac {\left (a+b x^3\right )^8 \sqrt {a^2+2 a b x^3+b^2 x^6}}{27 b^4} \]

[Out]

-1/18*a^3*(b*x^3+a)^5*((b*x^3+a)^2)^(1/2)/b^4+1/7*a^2*(b*x^3+a)^6*((b*x^3+a)^2)^(1/2)/b^4-1/8*a*(b*x^3+a)^7*((
b*x^3+a)^2)^(1/2)/b^4+1/27*(b*x^3+a)^8*((b*x^3+a)^2)^(1/2)/b^4

Rubi [A] (verified)

Time = 0.08 (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} \[ \int x^{11} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \left (a+b x^3\right )^8}{27 b^4}-\frac {a \sqrt {a^2+2 a b x^3+b^2 x^6} \left (a+b x^3\right )^7}{8 b^4}+\frac {a^2 \sqrt {a^2+2 a b x^3+b^2 x^6} \left (a+b x^3\right )^6}{7 b^4}-\frac {a^3 \sqrt {a^2+2 a b x^3+b^2 x^6} \left (a+b x^3\right )^5}{18 b^4} \]

[In]

Int[x^11*(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2),x]

[Out]

-1/18*(a^3*(a + b*x^3)^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/b^4 + (a^2*(a + b*x^3)^6*Sqrt[a^2 + 2*a*b*x^3 + b^2*
x^6])/(7*b^4) - (a*(a + b*x^3)^7*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(8*b^4) + ((a + b*x^3)^8*Sqrt[a^2 + 2*a*b*x^
3 + b^2*x^6])/(27*b^4)

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 x^{11} \left (a b+b^2 x^3\right )^5 \, 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 x^3 \left (a b+b^2 x\right )^5 \, 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 (-\frac {a^3 \left (a b+b^2 x\right )^5}{b^3}+\frac {3 a^2 \left (a b+b^2 x\right )^6}{b^4}-\frac {3 a \left (a b+b^2 x\right )^7}{b^5}+\frac {\left (a b+b^2 x\right )^8}{b^6}\right ) \, dx,x,x^3\right )}{3 b^4 \left (a b+b^2 x^3\right )} \\ & = -\frac {a^3 \left (a+b x^3\right )^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{18 b^4}+\frac {a^2 \left (a+b x^3\right )^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 b^4}-\frac {a \left (a+b x^3\right )^7 \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 b^4}+\frac {\left (a+b x^3\right )^8 \sqrt {a^2+2 a b x^3+b^2 x^6}}{27 b^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.81 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.84 \[ \int x^{11} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {x^{12} \left (126 a^5+504 a^4 b x^3+840 a^3 b^2 x^6+720 a^2 b^3 x^9+315 a b^4 x^{12}+56 b^5 x^{15}\right ) \left (\sqrt {a^2} b x^3+a \left (\sqrt {a^2}-\sqrt {\left (a+b x^3\right )^2}\right )\right )}{1512 \left (-a^2-a b x^3+\sqrt {a^2} \sqrt {\left (a+b x^3\right )^2}\right )} \]

[In]

Integrate[x^11*(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2),x]

[Out]

(x^12*(126*a^5 + 504*a^4*b*x^3 + 840*a^3*b^2*x^6 + 720*a^2*b^3*x^9 + 315*a*b^4*x^12 + 56*b^5*x^15)*(Sqrt[a^2]*
b*x^3 + a*(Sqrt[a^2] - Sqrt[(a + b*x^3)^2])))/(1512*(-a^2 - a*b*x^3 + Sqrt[a^2]*Sqrt[(a + b*x^3)^2]))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 2.

Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.33

method	result	size
pseudoelliptic	\(-\frac {\operatorname {csgn}\left (b \,x^{3}+a \right ) \left (b \,x^{3}+a \right )^{6} \left (-56 b^{3} x^{9}+21 b^{2} x^{6} a -6 a^{2} b \,x^{3}+a^{3}\right )}{1512 b^{4}}\)	\(53\)
gosper	\(\frac {x^{12} \left (56 b^{5} x^{15}+315 a \,b^{4} x^{12}+720 a^{2} b^{3} x^{9}+840 a^{3} b^{2} x^{6}+504 a^{4} b \,x^{3}+126 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{1512 \left (b \,x^{3}+a \right )^{5}}\)	\(80\)
default	\(\frac {x^{12} \left (56 b^{5} x^{15}+315 a \,b^{4} x^{12}+720 a^{2} b^{3} x^{9}+840 a^{3} b^{2} x^{6}+504 a^{4} b \,x^{3}+126 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{1512 \left (b \,x^{3}+a \right )^{5}}\)	\(80\)
risch	\(\frac {5 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, b^{4} a \,x^{24}}{24 \left (b \,x^{3}+a \right )}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, b^{5} x^{27}}{27 b \,x^{3}+27 a}+\frac {5 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{3} b^{2} x^{18}}{9 \left (b \,x^{3}+a \right )}+\frac {10 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{2} b^{3} x^{21}}{21 \left (b \,x^{3}+a \right )}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{5} x^{12}}{12 b \,x^{3}+12 a}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, b \,a^{4} x^{15}}{3 b \,x^{3}+3 a}\)	\(178\)

[In]

int(x^11*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/1512*csgn(b*x^3+a)*(b*x^3+a)^6*(-56*b^3*x^9+21*a*b^2*x^6-6*a^2*b*x^3+a^3)/b^4

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.36 \[ \int x^{11} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{27} \, b^{5} x^{27} + \frac {5}{24} \, a b^{4} x^{24} + \frac {10}{21} \, a^{2} b^{3} x^{21} + \frac {5}{9} \, a^{3} b^{2} x^{18} + \frac {1}{3} \, a^{4} b x^{15} + \frac {1}{12} \, a^{5} x^{12} \]

[In]

integrate(x^11*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/27*b^5*x^27 + 5/24*a*b^4*x^24 + 10/21*a^2*b^3*x^21 + 5/9*a^3*b^2*x^18 + 1/3*a^4*b*x^15 + 1/12*a^5*x^12

Sympy [F]

\[ \int x^{11} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\int x^{11} \left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}\, dx \]

[In]

integrate(x**11*(b**2*x**6+2*a*b*x**3+a**2)**(5/2),x)

[Out]

Integral(x**11*((a + b*x**3)**2)**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.91 \[ \int x^{11} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}} x^{6}}{27 \, b^{2}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} a^{3} x^{3}}{18 \, b^{3}} - \frac {11 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}} a x^{3}}{216 \, b^{3}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} a^{4}}{18 \, b^{4}} + \frac {83 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}} a^{2}}{1512 \, b^{4}} \]

[In]

integrate(x^11*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="maxima")

[Out]

1/27*(b^2*x^6 + 2*a*b*x^3 + a^2)^(7/2)*x^6/b^2 - 1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*a^3*x^3/b^3 - 11/216*(
b^2*x^6 + 2*a*b*x^3 + a^2)^(7/2)*a*x^3/b^3 - 1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*a^4/b^4 + 83/1512*(b^2*x^6
+ 2*a*b*x^3 + a^2)^(7/2)*a^2/b^4

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.66 \[ \int x^{11} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{27} \, b^{5} x^{27} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{24} \, a b^{4} x^{24} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {10}{21} \, a^{2} b^{3} x^{21} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{9} \, a^{3} b^{2} x^{18} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{3} \, a^{4} b x^{15} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{12} \, a^{5} x^{12} \mathrm {sgn}\left (b x^{3} + a\right ) \]

[In]

integrate(x^11*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="giac")

[Out]

1/27*b^5*x^27*sgn(b*x^3 + a) + 5/24*a*b^4*x^24*sgn(b*x^3 + a) + 10/21*a^2*b^3*x^21*sgn(b*x^3 + a) + 5/9*a^3*b^
2*x^18*sgn(b*x^3 + a) + 1/3*a^4*b*x^15*sgn(b*x^3 + a) + 1/12*a^5*x^12*sgn(b*x^3 + a)

Mupad [F(-1)]

Timed out. \[ \int x^{11} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\int x^{11}\,{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{5/2} \,d x \]

[In]

int(x^11*(a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2),x)

[Out]

int(x^11*(a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2), x)