3.3.63 \(\int \frac {x^8}{(a+b x^3) \sqrt {c+d x^3}} \, dx\)

Optimal. Leaf size=104 \[ -\frac {2 a^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{5/2} \sqrt {b c-a d}}-\frac {2 \sqrt {c+d x^3} (a d+b c)}{3 b^2 d^2}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 b d^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {446, 88, 63, 208} \begin {gather*} -\frac {2 a^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{5/2} \sqrt {b c-a d}}-\frac {2 \sqrt {c+d x^3} (a d+b c)}{3 b^2 d^2}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 b d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^8/((a + b*x^3)*Sqrt[c + d*x^3]),x]

[Out]

(-2*(b*c + a*d)*Sqrt[c + d*x^3])/(3*b^2*d^2) + (2*(c + d*x^3)^(3/2))/(9*b*d^2) - (2*a^2*ArcTanh[(Sqrt[b]*Sqrt[
c + d*x^3])/Sqrt[b*c - a*d]])/(3*b^(5/2)*Sqrt[b*c - a*d])

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 208

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

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {x^8}{\left (a+b x^3\right ) \sqrt {c+d x^3}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^2}{(a+b x) \sqrt {c+d x}} \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (\frac {-b c-a d}{b^2 d \sqrt {c+d x}}+\frac {a^2}{b^2 (a+b x) \sqrt {c+d x}}+\frac {\sqrt {c+d x}}{b d}\right ) \, dx,x,x^3\right )\\ &=-\frac {2 (b c+a d) \sqrt {c+d x^3}}{3 b^2 d^2}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 b d^2}+\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^3\right )}{3 b^2}\\ &=-\frac {2 (b c+a d) \sqrt {c+d x^3}}{3 b^2 d^2}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 b d^2}+\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x^3}\right )}{3 b^2 d}\\ &=-\frac {2 (b c+a d) \sqrt {c+d x^3}}{3 b^2 d^2}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 b d^2}-\frac {2 a^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{5/2} \sqrt {b c-a d}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.22, size = 91, normalized size = 0.88 \begin {gather*} \frac {2 \sqrt {c+d x^3} \left (-3 a d-2 b c+b d x^3\right )}{9 b^2 d^2}-\frac {2 a^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{5/2} \sqrt {b c-a d}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^8/((a + b*x^3)*Sqrt[c + d*x^3]),x]

[Out]

(2*Sqrt[c + d*x^3]*(-2*b*c - 3*a*d + b*d*x^3))/(9*b^2*d^2) - (2*a^2*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c
- a*d]])/(3*b^(5/2)*Sqrt[b*c - a*d])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.19, size = 102, normalized size = 0.98 \begin {gather*} -\frac {2 a^2 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3} \sqrt {a d-b c}}{b c-a d}\right )}{3 b^{5/2} \sqrt {a d-b c}}-\frac {2 \sqrt {c+d x^3} \left (3 a d+2 b c-b d x^3\right )}{9 b^2 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^8/((a + b*x^3)*Sqrt[c + d*x^3]),x]

[Out]

(-2*Sqrt[c + d*x^3]*(2*b*c + 3*a*d - b*d*x^3))/(9*b^2*d^2) - (2*a^2*ArcTan[(Sqrt[b]*Sqrt[-(b*c) + a*d]*Sqrt[c
+ d*x^3])/(b*c - a*d)])/(3*b^(5/2)*Sqrt[-(b*c) + a*d])

________________________________________________________________________________________

fricas [A]  time = 0.87, size = 289, normalized size = 2.78 \begin {gather*} \left [\frac {3 \, \sqrt {b^{2} c - a b d} a^{2} d^{2} \log \left (\frac {b d x^{3} + 2 \, b c - a d - 2 \, \sqrt {d x^{3} + c} \sqrt {b^{2} c - a b d}}{b x^{3} + a}\right ) - 2 \, {\left (2 \, b^{3} c^{2} + a b^{2} c d - 3 \, a^{2} b d^{2} - {\left (b^{3} c d - a b^{2} d^{2}\right )} x^{3}\right )} \sqrt {d x^{3} + c}}{9 \, {\left (b^{4} c d^{2} - a b^{3} d^{3}\right )}}, \frac {2 \, {\left (3 \, \sqrt {-b^{2} c + a b d} a^{2} d^{2} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-b^{2} c + a b d}}{b d x^{3} + b c}\right ) - {\left (2 \, b^{3} c^{2} + a b^{2} c d - 3 \, a^{2} b d^{2} - {\left (b^{3} c d - a b^{2} d^{2}\right )} x^{3}\right )} \sqrt {d x^{3} + c}\right )}}{9 \, {\left (b^{4} c d^{2} - a b^{3} d^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^3+a)/(d*x^3+c)^(1/2),x, algorithm="fricas")

[Out]

[1/9*(3*sqrt(b^2*c - a*b*d)*a^2*d^2*log((b*d*x^3 + 2*b*c - a*d - 2*sqrt(d*x^3 + c)*sqrt(b^2*c - a*b*d))/(b*x^3
+ a)) - 2*(2*b^3*c^2 + a*b^2*c*d - 3*a^2*b*d^2 - (b^3*c*d - a*b^2*d^2)*x^3)*sqrt(d*x^3 + c))/(b^4*c*d^2 - a*b
^3*d^3), 2/9*(3*sqrt(-b^2*c + a*b*d)*a^2*d^2*arctan(sqrt(d*x^3 + c)*sqrt(-b^2*c + a*b*d)/(b*d*x^3 + b*c)) - (2
*b^3*c^2 + a*b^2*c*d - 3*a^2*b*d^2 - (b^3*c*d - a*b^2*d^2)*x^3)*sqrt(d*x^3 + c))/(b^4*c*d^2 - a*b^3*d^3)]

________________________________________________________________________________________

giac [A]  time = 0.17, size = 106, normalized size = 1.02 \begin {gather*} \frac {2 \, a^{2} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{3 \, \sqrt {-b^{2} c + a b d} b^{2}} + \frac {2 \, {\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} b^{2} d^{4} - 3 \, \sqrt {d x^{3} + c} b^{2} c d^{4} - 3 \, \sqrt {d x^{3} + c} a b d^{5}\right )}}{9 \, b^{3} d^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^3+a)/(d*x^3+c)^(1/2),x, algorithm="giac")

[Out]

2/3*a^2*arctan(sqrt(d*x^3 + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*b^2) + 2/9*((d*x^3 + c)^(3/2)*b^2
*d^4 - 3*sqrt(d*x^3 + c)*b^2*c*d^4 - 3*sqrt(d*x^3 + c)*a*b*d^5)/(b^3*d^6)

________________________________________________________________________________________

maple [C]  time = 0.34, size = 488, normalized size = 4.69 \begin {gather*} -\frac {i a^{2} \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (2 x +\frac {-i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {\left (x -\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{-3 \left (-c \,d^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i \left (2 x +\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{2 \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \left (2 \RootOf \left (\textit {\_Z}^{3} b +a \right )^{2} d^{2}+i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (\textit {\_Z}^{3} b +a \right ) d -\left (-c \,d^{2}\right )^{\frac {1}{3}} \RootOf \left (\textit {\_Z}^{3} b +a \right ) d -i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {2}{3}}-\left (-c \,d^{2}\right )^{\frac {2}{3}}\right ) \EllipticPi \left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}-\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) \sqrt {3}\, d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}}{3}, \frac {\left (2 i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (\textit {\_Z}^{3} b +a \right )^{2} d +i \sqrt {3}\, c d -3 c d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \sqrt {3}\, \RootOf \left (\textit {\_Z}^{3} b +a \right )-3 \left (-c \,d^{2}\right )^{\frac {2}{3}} \RootOf \left (\textit {\_Z}^{3} b +a \right )\right ) b}{2 \left (a d -b c \right ) d}, \sqrt {\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{\left (-\frac {3 \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}+\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) d}}\right )}{3 b^{2} d^{2} \left (a d -b c \right ) \sqrt {d \,x^{3}+c}}+\frac {\left (\frac {2 \sqrt {d \,x^{3}+c}\, x^{3}}{9 d}-\frac {4 \sqrt {d \,x^{3}+c}\, c}{9 d^{2}}\right ) b -\frac {2 \sqrt {d \,x^{3}+c}\, a}{3 d}}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(b*x^3+a)/(d*x^3+c)^(1/2),x)

[Out]

1/b^2*(b*(2/9*(d*x^3+c)^(1/2)/d*x^3-4/9*(d*x^3+c)^(1/2)*c/d^2)-2/3*a/d*(d*x^3+c)^(1/2))-1/3*I*a^2/b^2/d^2*2^(1
/2)*sum(1/(a*d-b*c)*(-c*d^2)^(1/3)*(1/2*I*(2*x+(-I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3))/d)/(-c*d^2)^(1/3)*d)
^(1/2)*((x-(-c*d^2)^(1/3)/d)/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^2)^(1/3))*d)^(1/2)*(-1/2*I*(2*x+(I*3^(1/2)*(-c
*d^2)^(1/3)+(-c*d^2)^(1/3))/d)/(-c*d^2)^(1/3)*d)^(1/2)/(d*x^3+c)^(1/2)*(2*_alpha^2*d^2+I*(-c*d^2)^(1/3)*3^(1/2
)*_alpha*d-(-c*d^2)^(1/3)*_alpha*d-I*3^(1/2)*(-c*d^2)^(2/3)-(-c*d^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2*(
-c*d^2)^(1/3)/d-1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2),1/2*(2*I*(-c*d^2)^(1/3)*3^(1/2
)*_alpha^2*d+I*3^(1/2)*c*d-3*c*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha-3*(-c*d^2)^(2/3)*_alpha)/(a*d-b*c)*b/d,(I*3^(
1/2)*(-c*d^2)^(1/3)/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)/d)^(1/2)),_alpha=RootOf(_Z^3*b+a))

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^3+a)/(d*x^3+c)^(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(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c positive or negative?

________________________________________________________________________________________

mupad [B]  time = 5.43, size = 121, normalized size = 1.16 \begin {gather*} \frac {2\,x^3\,\sqrt {d\,x^3+c}}{9\,b\,d}-\frac {\left (\frac {2\,a}{b^2}+\frac {4\,c}{3\,b\,d}\right )\,\sqrt {d\,x^3+c}}{3\,d}+\frac {a^2\,\ln \left (\frac {2\,b\,c-a\,d+b\,d\,x^3+\sqrt {b}\,\sqrt {d\,x^3+c}\,\sqrt {a\,d-b\,c}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,1{}\mathrm {i}}{3\,b^{5/2}\,\sqrt {a\,d-b\,c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/((a + b*x^3)*(c + d*x^3)^(1/2)),x)

[Out]

(2*x^3*(c + d*x^3)^(1/2))/(9*b*d) - (((2*a)/b^2 + (4*c)/(3*b*d))*(c + d*x^3)^(1/2))/(3*d) + (a^2*log((2*b*c -
a*d + b^(1/2)*(c + d*x^3)^(1/2)*(a*d - b*c)^(1/2)*2i + b*d*x^3)/(a + b*x^3))*1i)/(3*b^(5/2)*(a*d - b*c)^(1/2))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{8}}{\left (a + b x^{3}\right ) \sqrt {c + d x^{3}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8/(b*x**3+a)/(d*x**3+c)**(1/2),x)

[Out]

Integral(x**8/((a + b*x**3)*sqrt(c + d*x**3)), x)

________________________________________________________________________________________