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

Optimal. Leaf size=77 \[ \frac {2}{3 \sqrt {c+d x^3} (b c-a d)}-\frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 (b c-a d)^{3/2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {444, 51, 63, 208} \begin {gather*} \frac {2}{3 \sqrt {c+d x^3} (b c-a d)}-\frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 (b c-a d)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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 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 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rubi steps

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

________________________________________________________________________________________

Mathematica [C]  time = 0.03, size = 52, normalized size = 0.68 \begin {gather*} -\frac {2 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {b \left (d x^3+c\right )}{b c-a d}\right )}{3 \sqrt {c+d x^3} (a d-b c)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.09, size = 109, normalized size = 1.42 \begin {gather*} \frac {\frac {2}{3}-\frac {2 \sqrt {b} \sqrt {c+d x^3} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3} \sqrt {a d-b c}}{b c-a d}\right )}{3 \sqrt {a d-b c}}}{b c \sqrt {c+d x^3}-a d \sqrt {c+d x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.70, size = 236, normalized size = 3.06 \begin {gather*} \left [-\frac {{\left (d x^{3} + c\right )} \sqrt {\frac {b}{b c - a d}} \log \left (\frac {b d x^{3} + 2 \, b c - a d + 2 \, \sqrt {d x^{3} + c} {\left (b c - a d\right )} \sqrt {\frac {b}{b c - a d}}}{b x^{3} + a}\right ) - 2 \, \sqrt {d x^{3} + c}}{3 \, {\left ({\left (b c d - a d^{2}\right )} x^{3} + b c^{2} - a c d\right )}}, -\frac {2 \, {\left ({\left (d x^{3} + c\right )} \sqrt {-\frac {b}{b c - a d}} \arctan \left (-\frac {\sqrt {d x^{3} + c} {\left (b c - a d\right )} \sqrt {-\frac {b}{b c - a d}}}{b d x^{3} + b c}\right ) - \sqrt {d x^{3} + c}\right )}}{3 \, {\left ({\left (b c d - a d^{2}\right )} x^{3} + b c^{2} - a c d\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.19, size = 73, normalized size = 0.95 \begin {gather*} \frac {2 \, b \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} {\left (b c - a d\right )}} + \frac {2}{3 \, \sqrt {d x^{3} + c} {\left (b c - a d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 0.23, size = 463, normalized size = 6.01 \begin {gather*} -\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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))

________________________________________________________________________________________

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^2/(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?

________________________________________________________________________________________

mupad [B]  time = 5.85, size = 89, normalized size = 1.16 \begin {gather*} -\frac {2}{3\,\sqrt {d\,x^3+c}\,\left (a\,d-b\,c\right )}+\frac {\sqrt {b}\,\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\,{\left (a\,d-b\,c\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 32.78, size = 66, normalized size = 0.86 \begin {gather*} - \frac {2}{3 \sqrt {c + d x^{3}} \left (a d - b c\right )} - \frac {2 \operatorname {atan}{\left (\frac {\sqrt {c + d x^{3}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{3 \sqrt {\frac {a d - b c}{b}} \left (a d - b c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________