3.3.68 \(\int \frac {x^8}{(a+b x^3) (c+d x^3)^{3/2}} \, dx\)

Optimal. Leaf size=107 \[ -\frac {2 a^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{3/2} (b c-a d)^{3/2}}+\frac {2 c^2}{3 d^2 \sqrt {c+d x^3} (b c-a d)}+\frac {2 \sqrt {c+d x^3}}{3 b d^2} \]

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Rubi [A]  time = 0.12, antiderivative size = 107, 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, 87, 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^{3/2} (b c-a d)^{3/2}}+\frac {2 c^2}{3 d^2 \sqrt {c+d x^3} (b c-a d)}+\frac {2 \sqrt {c+d x^3}}{3 b d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^8/((a + b*x^3)*(c + d*x^3)^(3/2)),x]

[Out]

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

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 87

Int[(((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegr
and[(e + f*x)^FractionalPart[p], ((c + d*x)^n*(e + f*x)^IntegerPart[p])/(a + b*x), x], x] /; FreeQ[{a, b, c, d
, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]

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

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Mathematica [C]  time = 0.08, size = 100, normalized size = 0.93 \begin {gather*} \frac {2 \left (-a^2 d^2 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {b \left (d x^3+c\right )}{b c-a d}\right )+a^2 d^2+a b d \left (c+d x^3\right )+b^2 (-c) \left (2 c+d x^3\right )\right )}{3 b^2 d^2 \sqrt {c+d x^3} (a d-b c)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^8/((a + b*x^3)*(c + d*x^3)^(3/2)),x]

[Out]

(2*(a^2*d^2 + a*b*d*(c + d*x^3) - b^2*c*(2*c + d*x^3) - a^2*d^2*Hypergeometric2F1[-1/2, 1, 1/2, (b*(c + d*x^3)
)/(b*c - a*d)]))/(3*b^2*d^2*(-(b*c) + a*d)*Sqrt[c + d*x^3])

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IntegrateAlgebraic [A]  time = 0.21, size = 124, normalized size = 1.16 \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^{3/2} (a d-b c)^{3/2}}-\frac {2 \left (-a c d-a d^2 x^3+2 b c^2+b c d x^3\right )}{3 b d^2 \sqrt {c+d x^3} (a d-b c)} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^8/((a + b*x^3)*(c + d*x^3)^(3/2)),x]

[Out]

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

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fricas [B]  time = 0.80, size = 440, normalized size = 4.11 \begin {gather*} \left [-\frac {{\left (a^{2} d^{3} x^{3} + a^{2} c d^{2}\right )} \sqrt {b^{2} c - a b d} \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^{3} - 3 \, a b^{2} c^{2} d + a^{2} b c d^{2} + {\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{3}\right )} \sqrt {d x^{3} + c}}{3 \, {\left (b^{4} c^{3} d^{2} - 2 \, a b^{3} c^{2} d^{3} + a^{2} b^{2} c d^{4} + {\left (b^{4} c^{2} d^{3} - 2 \, a b^{3} c d^{4} + a^{2} b^{2} d^{5}\right )} x^{3}\right )}}, \frac {2 \, {\left ({\left (a^{2} d^{3} x^{3} + a^{2} c d^{2}\right )} \sqrt {-b^{2} c + a b d} \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^{3} - 3 \, a b^{2} c^{2} d + a^{2} b c d^{2} + {\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{3}\right )} \sqrt {d x^{3} + c}\right )}}{3 \, {\left (b^{4} c^{3} d^{2} - 2 \, a b^{3} c^{2} d^{3} + a^{2} b^{2} c d^{4} + {\left (b^{4} c^{2} d^{3} - 2 \, a b^{3} c d^{4} + a^{2} b^{2} d^{5}\right )} x^{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)^(3/2),x, algorithm="fricas")

[Out]

[-1/3*((a^2*d^3*x^3 + a^2*c*d^2)*sqrt(b^2*c - a*b*d)*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^3 - 3*a*b^2*c^2*d + a^2*b*c*d^2 + (b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)
*x^3)*sqrt(d*x^3 + c))/(b^4*c^3*d^2 - 2*a*b^3*c^2*d^3 + a^2*b^2*c*d^4 + (b^4*c^2*d^3 - 2*a*b^3*c*d^4 + a^2*b^2
*d^5)*x^3), 2/3*((a^2*d^3*x^3 + a^2*c*d^2)*sqrt(-b^2*c + a*b*d)*arctan(sqrt(d*x^3 + c)*sqrt(-b^2*c + a*b*d)/(b
*d*x^3 + b*c)) + (2*b^3*c^3 - 3*a*b^2*c^2*d + a^2*b*c*d^2 + (b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*x^3)*sqrt(
d*x^3 + c))/(b^4*c^3*d^2 - 2*a*b^3*c^2*d^3 + a^2*b^2*c*d^4 + (b^4*c^2*d^3 - 2*a*b^3*c*d^4 + a^2*b^2*d^5)*x^3)]

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giac [A]  time = 0.18, size = 103, normalized size = 0.96 \begin {gather*} \frac {2 \, a^{2} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{3 \, {\left (b^{2} c - a b d\right )} \sqrt {-b^{2} c + a b d}} + \frac {2 \, c^{2}}{3 \, {\left (b c d^{2} - a d^{3}\right )} \sqrt {d x^{3} + c}} + \frac {2 \, \sqrt {d x^{3} + c}}{3 \, b d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.34, size = 527, normalized size = 4.93 \begin {gather*} \frac {\left (-\frac {i b \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 d^{2} \left (-a d +b c \right ) \left (a d -b c \right ) \sqrt {d \,x^{3}+c}}-\frac {2}{3 \left (a d -b c \right ) \sqrt {\left (x^{3}+\frac {c}{d}\right ) d}}\right ) a^{2}}{b^{2}}+\frac {\left (\frac {2 c}{3 \sqrt {\left (x^{3}+\frac {c}{d}\right ) d}\, d^{2}}+\frac {2 \sqrt {d \,x^{3}+c}}{3 d^{2}}\right ) b +\frac {2 a}{3 \sqrt {d \,x^{3}+c}\, 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)^(3/2),x)

[Out]

1/b^2*(b*(2/3/((x^3+c/d)*d)^(1/2)*c/d^2+2/3*(d*x^3+c)^(1/2)/d^2)+2/3*a/d/(d*x^3+c)^(1/2))+a^2/b^2*(-2/3/(a*d-b
*c)/((x^3+c/d)*d)^(1/2)-1/3*I*b/d^2*2^(1/2)*sum(1/(-a*d+b*c)/(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)))

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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)^(3/2),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?

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mupad [B]  time = 6.46, size = 115, normalized size = 1.07 \begin {gather*} \frac {2\,\sqrt {d\,x^3+c}}{3\,b\,d^2}-\frac {2\,c^2}{3\,d^2\,\sqrt {d\,x^3+c}\,\left (a\,d-b\,c\right )}+\frac {a^2\,\ln \left (\frac {a\,d-2\,b\,c-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^{3/2}\,{\left (a\,d-b\,c\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{8}}{\left (a + b x^{3}\right ) \left (c + d x^{3}\right )^{\frac {3}{2}}}\, 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)**(3/2),x)

[Out]

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

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