\(\int x^5 \sqrt {a+b x^3+c x^6} \, dx\) [188]

Optimal result

Integrand size = 20, antiderivative size = 108 \[ \int x^5 \sqrt {a+b x^3+c x^6} \, dx=-\frac {b \left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{24 c^2}+\frac {\left (a+b x^3+c x^6\right )^{3/2}}{9 c}+\frac {b \left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{48 c^{5/2}} \]

[Out]

1/9*(c*x^6+b*x^3+a)^(3/2)/c+1/48*b*(-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^(5/2)
-1/24*b*(2*c*x^3+b)*(c*x^6+b*x^3+a)^(1/2)/c^2

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1371, 654, 626, 635, 212} \[ \int x^5 \sqrt {a+b x^3+c x^6} \, dx=\frac {b \left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{48 c^{5/2}}-\frac {b \left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{24 c^2}+\frac {\left (a+b x^3+c x^6\right )^{3/2}}{9 c} \]

[In]

Int[x^5*Sqrt[a + b*x^3 + c*x^6],x]

[Out]

-1/24*(b*(b + 2*c*x^3)*Sqrt[a + b*x^3 + c*x^6])/c^2 + (a + b*x^3 + c*x^6)^(3/2)/(9*c) + (b*(b^2 - 4*a*c)*ArcTa
nh[(b + 2*c*x^3)/(2*Sqrt[c]*Sqrt[a + b*x^3 + c*x^6])])/(48*c^(5/2))

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 654

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*((a + b*x + c*x^2)^(p +
1)/(2*c*(p + 1))), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[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 \sqrt {a+b x+c x^2} \, dx,x,x^3\right ) \\ & = \frac {\left (a+b x^3+c x^6\right )^{3/2}}{9 c}-\frac {b \text {Subst}\left (\int \sqrt {a+b x+c x^2} \, dx,x,x^3\right )}{6 c} \\ & = -\frac {b \left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{24 c^2}+\frac {\left (a+b x^3+c x^6\right )^{3/2}}{9 c}+\frac {\left (b \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 )}{48 c^2} \\ & = -\frac {b \left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{24 c^2}+\frac {\left (a+b x^3+c x^6\right )^{3/2}}{9 c}+\frac {\left (b \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 )}{24 c^2} \\ & = -\frac {b \left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{24 c^2}+\frac {\left (a+b x^3+c x^6\right )^{3/2}}{9 c}+\frac {b \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 )}{48 c^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.94 \[ \int x^5 \sqrt {a+b x^3+c x^6} \, dx=\frac {\sqrt {a+b x^3+c x^6} \left (-3 b^2+2 b c x^3+8 c \left (a+c x^6\right )\right )}{72 c^2}-\frac {\left (b^3-4 a b c\right ) \log \left (c^2 \left (b+2 c x^3-2 \sqrt {c} \sqrt {a+b x^3+c x^6}\right )\right )}{48 c^{5/2}} \]

[In]

Integrate[x^5*Sqrt[a + b*x^3 + c*x^6],x]

[Out]

(Sqrt[a + b*x^3 + c*x^6]*(-3*b^2 + 2*b*c*x^3 + 8*c*(a + c*x^6)))/(72*c^2) - ((b^3 - 4*a*b*c)*Log[c^2*(b + 2*c*
x^3 - 2*Sqrt[c]*Sqrt[a + b*x^3 + c*x^6])])/(48*c^(5/2))

Maple [F]

\[\int x^{5} \sqrt {c \,x^{6}+b \,x^{3}+a}d x\]

[In]

int(x^5*(c*x^6+b*x^3+a)^(1/2),x)

[Out]

int(x^5*(c*x^6+b*x^3+a)^(1/2),x)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 237, normalized size of antiderivative = 2.19 \[ \int x^5 \sqrt {a+b x^3+c x^6} \, dx=\left [-\frac {3 \, {\left (b^{3} - 4 \, a b c\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{6} - 8 \, b c x^{3} - b^{2} + 4 \, \sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, {\left (8 \, c^{3} x^{6} + 2 \, b c^{2} x^{3} - 3 \, b^{2} c + 8 \, a c^{2}\right )} \sqrt {c x^{6} + b x^{3} + a}}{288 \, c^{3}}, -\frac {3 \, {\left (b^{3} - 4 \, a b c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{6} + b c x^{3} + a c\right )}}\right ) - 2 \, {\left (8 \, c^{3} x^{6} + 2 \, b c^{2} x^{3} - 3 \, b^{2} c + 8 \, a c^{2}\right )} \sqrt {c x^{6} + b x^{3} + a}}{144 \, c^{3}}\right ] \]

[In]

integrate(x^5*(c*x^6+b*x^3+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/288*(3*(b^3 - 4*a*b*c)*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*(8*c^3*x^6 + 2*b*c^2*x^3 - 3*b^2*c + 8*a*c^2)*sqrt(c*x^6 + b*x^3 + a))/c^3, -1/144*(3*(b^
3 - 4*a*b*c)*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
*(8*c^3*x^6 + 2*b*c^2*x^3 - 3*b^2*c + 8*a*c^2)*sqrt(c*x^6 + b*x^3 + a))/c^3]

Sympy [F]

\[ \int x^5 \sqrt {a+b x^3+c x^6} \, dx=\int x^{5} \sqrt {a + b x^{3} + c x^{6}}\, dx \]

[In]

integrate(x**5*(c*x**6+b*x**3+a)**(1/2),x)

[Out]

Integral(x**5*sqrt(a + b*x**3 + c*x**6), x)

Maxima [F(-2)]

Exception generated. \[ \int x^5 \sqrt {a+b x^3+c x^6} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(x^5*(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 [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.89 \[ \int x^5 \sqrt {a+b x^3+c x^6} \, dx=\frac {1}{72} \, \sqrt {c x^{6} + b x^{3} + a} {\left (2 \, {\left (4 \, x^{3} + \frac {b}{c}\right )} x^{3} - \frac {3 \, b^{2} - 8 \, a c}{c^{2}}\right )} - \frac {{\left (b^{3} - 4 \, a b c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{3} - \sqrt {c x^{6} + b x^{3} + a}\right )} \sqrt {c} + b \right |}\right )}{48 \, c^{\frac {5}{2}}} \]

[In]

integrate(x^5*(c*x^6+b*x^3+a)^(1/2),x, algorithm="giac")

[Out]

1/72*sqrt(c*x^6 + b*x^3 + a)*(2*(4*x^3 + b/c)*x^3 - (3*b^2 - 8*a*c)/c^2) - 1/48*(b^3 - 4*a*b*c)*log(abs(2*(sqr
t(c)*x^3 - sqrt(c*x^6 + b*x^3 + a))*sqrt(c) + b))/c^(5/2)

Mupad [B] (verification not implemented)

Time = 8.39 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.81 \[ \int x^5 \sqrt {a+b x^3+c x^6} \, dx=\frac {\left (8\,c\,\left (c\,x^6+a\right )-3\,b^2+2\,b\,c\,x^3\right )\,\sqrt {c\,x^6+b\,x^3+a}}{72\,c^2}+\frac {\ln \left (2\,\sqrt {c\,x^6+b\,x^3+a}+\frac {2\,c\,x^3+b}{\sqrt {c}}\right )\,\left (b^3-4\,a\,b\,c\right )}{48\,c^{5/2}} \]

[In]

int(x^5*(a + b*x^3 + c*x^6)^(1/2),x)

[Out]

((8*c*(a + c*x^6) - 3*b^2 + 2*b*c*x^3)*(a + b*x^3 + c*x^6)^(1/2))/(72*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))/(48*c^(5/2))