\(\int x^{11} \sqrt {a+b x^3+c x^6} \, dx\) [186]

Optimal result

Integrand size = 20, antiderivative size = 171 \[ \int x^{11} \sqrt {a+b x^3+c x^6} \, dx=-\frac {b \left (7 b^2-12 a c\right ) \left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{384 c^4}+\frac {x^6 \left (a+b x^3+c x^6\right )^{3/2}}{15 c}+\frac {\left (35 b^2-32 a c-42 b c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{720 c^3}+\frac {b \left (7 b^2-12 a c\right ) \left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{768 c^{9/2}} \]

[Out]

1/15*x^6*(c*x^6+b*x^3+a)^(3/2)/c+1/720*(-42*b*c*x^3-32*a*c+35*b^2)*(c*x^6+b*x^3+a)^(3/2)/c^3+1/768*b*(-12*a*c+
7*b^2)*(-4*a*c+b^2)*arctanh(1/2*(2*c*x^3+b)/c^(1/2)/(c*x^6+b*x^3+a)^(1/2))/c^(9/2)-1/384*b*(-12*a*c+7*b^2)*(2*
c*x^3+b)*(c*x^6+b*x^3+a)^(1/2)/c^4

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1371, 756, 793, 626, 635, 212} \[ \int x^{11} \sqrt {a+b x^3+c x^6} \, dx=\frac {b \left (7 b^2-12 a c\right ) \left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{768 c^{9/2}}-\frac {b \left (7 b^2-12 a c\right ) \left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{384 c^4}+\frac {\left (-32 a c+35 b^2-42 b c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{720 c^3}+\frac {x^6 \left (a+b x^3+c x^6\right )^{3/2}}{15 c} \]

[In]

Int[x^11*Sqrt[a + b*x^3 + c*x^6],x]

[Out]

-1/384*(b*(7*b^2 - 12*a*c)*(b + 2*c*x^3)*Sqrt[a + b*x^3 + c*x^6])/c^4 + (x^6*(a + b*x^3 + c*x^6)^(3/2))/(15*c)
+ ((35*b^2 - 32*a*c - 42*b*c*x^3)*(a + b*x^3 + c*x^6)^(3/2))/(720*c^3) + (b*(7*b^2 - 12*a*c)*(b^2 - 4*a*c)*Ar
cTanh[(b + 2*c*x^3)/(2*Sqrt[c]*Sqrt[a + b*x^3 + c*x^6])])/(768*c^(9/2))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 626

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x)*((a + b*x + c*x^2)^p/(2*c*(2*p + 1
))), x] - Dist[p*((b^2 - 4*a*c)/(2*c*(2*p + 1))), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 635

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 756

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*
((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2
*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]*(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]

Rule 793

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x))*((a + b*x + c*x^2)^(p + 1)/(2*c^2*(p + 1)*(2*p +
3))), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(
a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rule 1371

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && NeQ[
b^2 - 4*a*c, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int x^3 \sqrt {a+b x+c x^2} \, dx,x,x^3\right ) \\ & = \frac {x^6 \left (a+b x^3+c x^6\right )^{3/2}}{15 c}+\frac {\text {Subst}\left (\int x \left (-2 a-\frac {7 b x}{2}\right ) \sqrt {a+b x+c x^2} \, dx,x,x^3\right )}{15 c} \\ & = \frac {x^6 \left (a+b x^3+c x^6\right )^{3/2}}{15 c}+\frac {\left (35 b^2-32 a c-42 b c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{720 c^3}-\frac {\left (b \left (7 b^2-12 a c\right )\right ) \text {Subst}\left (\int \sqrt {a+b x+c x^2} \, dx,x,x^3\right )}{96 c^3} \\ & = -\frac {b \left (7 b^2-12 a c\right ) \left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{384 c^4}+\frac {x^6 \left (a+b x^3+c x^6\right )^{3/2}}{15 c}+\frac {\left (35 b^2-32 a c-42 b c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{720 c^3}+\frac {\left (b \left (7 b^2-12 a c\right ) \left (b^2-4 a c\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x+c x^2}} \, dx,x,x^3\right )}{768 c^4} \\ & = -\frac {b \left (7 b^2-12 a c\right ) \left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{384 c^4}+\frac {x^6 \left (a+b x^3+c x^6\right )^{3/2}}{15 c}+\frac {\left (35 b^2-32 a c-42 b c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{720 c^3}+\frac {\left (b \left (7 b^2-12 a c\right ) \left (b^2-4 a c\right )\right ) \text {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x^3}{\sqrt {a+b x^3+c x^6}}\right )}{384 c^4} \\ & = -\frac {b \left (7 b^2-12 a c\right ) \left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{384 c^4}+\frac {x^6 \left (a+b x^3+c x^6\right )^{3/2}}{15 c}+\frac {\left (35 b^2-32 a c-42 b c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{720 c^3}+\frac {b \left (7 b^2-12 a c\right ) \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{768 c^{9/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.97 \[ \int x^{11} \sqrt {a+b x^3+c x^6} \, dx=\frac {\sqrt {a+b x^3+c x^6} \left (-105 b^4+70 b^3 c x^3+4 b^2 c \left (115 a-14 c x^6\right )+8 b c^2 x^3 \left (-29 a+6 c x^6\right )+128 c^2 \left (-2 a^2+a c x^6+3 c^2 x^{12}\right )\right )}{5760 c^4}-\frac {\left (7 b^5-40 a b^3 c+48 a^2 b c^2\right ) \log \left (c^4 \left (b+2 c x^3-2 \sqrt {c} \sqrt {a+b x^3+c x^6}\right )\right )}{768 c^{9/2}} \]

[In]

Integrate[x^11*Sqrt[a + b*x^3 + c*x^6],x]

[Out]

(Sqrt[a + b*x^3 + c*x^6]*(-105*b^4 + 70*b^3*c*x^3 + 4*b^2*c*(115*a - 14*c*x^6) + 8*b*c^2*x^3*(-29*a + 6*c*x^6)
+ 128*c^2*(-2*a^2 + a*c*x^6 + 3*c^2*x^12)))/(5760*c^4) - ((7*b^5 - 40*a*b^3*c + 48*a^2*b*c^2)*Log[c^4*(b + 2*
c*x^3 - 2*Sqrt[c]*Sqrt[a + b*x^3 + c*x^6])])/(768*c^(9/2))

Maple [F]

\[\int x^{11} \sqrt {c \,x^{6}+b \,x^{3}+a}d x\]

[In]

int(x^11*(c*x^6+b*x^3+a)^(1/2),x)

[Out]

int(x^11*(c*x^6+b*x^3+a)^(1/2),x)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 367, normalized size of antiderivative = 2.15 \[ \int x^{11} \sqrt {a+b x^3+c x^6} \, dx=\left [\frac {15 \, {\left (7 \, b^{5} - 40 \, a b^{3} c + 48 \, a^{2} b c^{2}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{6} - 8 \, b c x^{3} - b^{2} - 4 \, \sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (384 \, c^{5} x^{12} + 48 \, b c^{4} x^{9} - 8 \, {\left (7 \, b^{2} c^{3} - 16 \, a c^{4}\right )} x^{6} - 105 \, b^{4} c + 460 \, a b^{2} c^{2} - 256 \, a^{2} c^{3} + 2 \, {\left (35 \, b^{3} c^{2} - 116 \, a b c^{3}\right )} x^{3}\right )} \sqrt {c x^{6} + b x^{3} + a}}{23040 \, c^{5}}, -\frac {15 \, {\left (7 \, b^{5} - 40 \, a b^{3} c + 48 \, a^{2} b c^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{6} + b c x^{3} + a c\right )}}\right ) - 2 \, {\left (384 \, c^{5} x^{12} + 48 \, b c^{4} x^{9} - 8 \, {\left (7 \, b^{2} c^{3} - 16 \, a c^{4}\right )} x^{6} - 105 \, b^{4} c + 460 \, a b^{2} c^{2} - 256 \, a^{2} c^{3} + 2 \, {\left (35 \, b^{3} c^{2} - 116 \, a b c^{3}\right )} x^{3}\right )} \sqrt {c x^{6} + b x^{3} + a}}{11520 \, c^{5}}\right ] \]

[In]

integrate(x^11*(c*x^6+b*x^3+a)^(1/2),x, algorithm="fricas")

[Out]

[1/23040*(15*(7*b^5 - 40*a*b^3*c + 48*a^2*b*c^2)*sqrt(c)*log(-8*c^2*x^6 - 8*b*c*x^3 - b^2 - 4*sqrt(c*x^6 + b*x
^3 + a)*(2*c*x^3 + b)*sqrt(c) - 4*a*c) + 4*(384*c^5*x^12 + 48*b*c^4*x^9 - 8*(7*b^2*c^3 - 16*a*c^4)*x^6 - 105*b
^4*c + 460*a*b^2*c^2 - 256*a^2*c^3 + 2*(35*b^3*c^2 - 116*a*b*c^3)*x^3)*sqrt(c*x^6 + b*x^3 + a))/c^5, -1/11520*
(15*(7*b^5 - 40*a*b^3*c + 48*a^2*b*c^2)*sqrt(-c)*arctan(1/2*sqrt(c*x^6 + b*x^3 + a)*(2*c*x^3 + b)*sqrt(-c)/(c^
2*x^6 + b*c*x^3 + a*c)) - 2*(384*c^5*x^12 + 48*b*c^4*x^9 - 8*(7*b^2*c^3 - 16*a*c^4)*x^6 - 105*b^4*c + 460*a*b^
2*c^2 - 256*a^2*c^3 + 2*(35*b^3*c^2 - 116*a*b*c^3)*x^3)*sqrt(c*x^6 + b*x^3 + a))/c^5]

Sympy [F]

\[ \int x^{11} \sqrt {a+b x^3+c x^6} \, dx=\int x^{11} \sqrt {a + b x^{3} + c x^{6}}\, dx \]

[In]

integrate(x**11*(c*x**6+b*x**3+a)**(1/2),x)

[Out]

Integral(x**11*sqrt(a + b*x**3 + c*x**6), x)

Maxima [F(-2)]

Exception generated. \[ \int x^{11} \sqrt {a+b x^3+c x^6} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(x^11*(c*x^6+b*x^3+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta

Giac [F]

\[ \int x^{11} \sqrt {a+b x^3+c x^6} \, dx=\int { \sqrt {c x^{6} + b x^{3} + a} x^{11} \,d x } \]

[In]

integrate(x^11*(c*x^6+b*x^3+a)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(c*x^6 + b*x^3 + a)*x^11, x)

Mupad [B] (verification not implemented)

Time = 8.69 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.84 \[ \int x^{11} \sqrt {a+b x^3+c x^6} \, dx=\frac {x^6\,{\left (c\,x^6+b\,x^3+a\right )}^{3/2}}{15\,c}+\frac {7\,b\,\left (\frac {a\,\left (\left (\frac {b}{4\,c}+\frac {x^3}{2}\right )\,\sqrt {c\,x^6+b\,x^3+a}+\frac {\ln \left (\sqrt {c\,x^6+b\,x^3+a}+\frac {c\,x^3+\frac {b}{2}}{\sqrt {c}}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}\right )}{4\,c}-\frac {x^3\,{\left (c\,x^6+b\,x^3+a\right )}^{3/2}}{4\,c}+\frac {5\,b\,\left (\frac {\left (8\,c\,\left (c\,x^6+a\right )-3\,b^2+2\,b\,c\,x^3\right )\,\sqrt {c\,x^6+b\,x^3+a}}{24\,c^2}+\frac {\ln \left (2\,\sqrt {c\,x^6+b\,x^3+a}+\frac {2\,c\,x^3+b}{\sqrt {c}}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}\right )}{8\,c}\right )}{30\,c}-\frac {2\,a\,\left (\frac {\left (8\,c\,\left (c\,x^6+a\right )-3\,b^2+2\,b\,c\,x^3\right )\,\sqrt {c\,x^6+b\,x^3+a}}{24\,c^2}+\frac {\ln \left (2\,\sqrt {c\,x^6+b\,x^3+a}+\frac {2\,c\,x^3+b}{\sqrt {c}}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}\right )}{15\,c} \]

[In]

int(x^11*(a + b*x^3 + c*x^6)^(1/2),x)

[Out]

(x^6*(a + b*x^3 + c*x^6)^(3/2))/(15*c) + (7*b*((a*((b/(4*c) + x^3/2)*(a + b*x^3 + c*x^6)^(1/2) + (log((a + b*x
^3 + c*x^6)^(1/2) + (b/2 + c*x^3)/c^(1/2))*(a*c - b^2/4))/(2*c^(3/2))))/(4*c) - (x^3*(a + b*x^3 + c*x^6)^(3/2)
)/(4*c) + (5*b*(((8*c*(a + c*x^6) - 3*b^2 + 2*b*c*x^3)*(a + b*x^3 + c*x^6)^(1/2))/(24*c^2) + (log(2*(a + b*x^3
+ c*x^6)^(1/2) + (b + 2*c*x^3)/c^(1/2))*(b^3 - 4*a*b*c))/(16*c^(5/2))))/(8*c)))/(30*c) - (2*a*(((8*c*(a + c*x
^6) - 3*b^2 + 2*b*c*x^3)*(a + b*x^3 + c*x^6)^(1/2))/(24*c^2) + (log(2*(a + b*x^3 + c*x^6)^(1/2) + (b + 2*c*x^3
)/c^(1/2))*(b^3 - 4*a*b*c))/(16*c^(5/2))))/(15*c)