3.391 \(\int \frac {1}{x (a+b x^3) (c+d x^3)^{3/2}} \, dx\)

Optimal. Leaf size=114 \[ \frac {2 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 a (b c-a d)^{3/2}}-\frac {2 d}{3 c \sqrt {c+d x^3} (b c-a d)}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a c^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {446, 85, 156, 63, 208} \[ \frac {2 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 a (b c-a d)^{3/2}}-\frac {2 d}{3 c \sqrt {c+d x^3} (b c-a d)}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a c^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d)/(3*c*(b*c - a*d)*Sqrt[c + d*x^3]) - (2*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(3*a*c^(3/2)) + (2*b^(3/2)*Arc
Tanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]])/(3*a*(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 85

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[(f*(e + f*x)^(p +
1))/((p + 1)*(b*e - a*f)*(d*e - c*f)), x] + Dist[1/((b*e - a*f)*(d*e - c*f)), Int[((b*d*e - b*c*f - a*d*f - b
*d*f*x)*(e + f*x)^(p + 1))/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[p, -1]

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, 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 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 {1}{x \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{x (a+b x) (c+d x)^{3/2}} \, dx,x,x^3\right )\\ &=-\frac {2 d}{3 c (b c-a d) \sqrt {c+d x^3}}+\frac {\operatorname {Subst}\left (\int \frac {b c-a d-b d x}{x (a+b x) \sqrt {c+d x}} \, dx,x,x^3\right )}{3 c (b c-a d)}\\ &=-\frac {2 d}{3 c (b c-a d) \sqrt {c+d x^3}}+\frac {\operatorname {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^3\right )}{3 a c}-\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^3\right )}{3 a (b c-a d)}\\ &=-\frac {2 d}{3 c (b c-a d) \sqrt {c+d x^3}}+\frac {2 \operatorname {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^3}\right )}{3 a c d}-\frac {\left (2 b^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 a d (b c-a d)}\\ &=-\frac {2 d}{3 c (b c-a d) \sqrt {c+d x^3}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a c^{3/2}}+\frac {2 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 a (b c-a d)^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 89, normalized size = 0.78 \[ -\frac {2 \left (b c \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {b \left (d x^3+c\right )}{b c-a d}\right )+(a d-b c) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {d x^3}{c}+1\right )\right )}{3 a c \sqrt {c+d x^3} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.19, size = 790, normalized size = 6.93 \[ \left [-\frac {2 \, \sqrt {d x^{3} + c} a c d + {\left (b c^{2} d x^{3} + b c^{3}\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 ) - {\left ({\left (b c d - a d^{2}\right )} x^{3} + b c^{2} - a c d\right )} \sqrt {c} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right )}{3 \, {\left (a b c^{4} - a^{2} c^{3} d + {\left (a b c^{3} d - a^{2} c^{2} d^{2}\right )} x^{3}\right )}}, -\frac {2 \, \sqrt {d x^{3} + c} a c d - 2 \, {\left (b c^{2} d x^{3} + b c^{3}\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 ) - {\left ({\left (b c d - a d^{2}\right )} x^{3} + b c^{2} - a c d\right )} \sqrt {c} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right )}{3 \, {\left (a b c^{4} - a^{2} c^{3} d + {\left (a b c^{3} d - a^{2} c^{2} d^{2}\right )} x^{3}\right )}}, -\frac {2 \, \sqrt {d x^{3} + c} a c d - 2 \, {\left ({\left (b c d - a d^{2}\right )} x^{3} + b c^{2} - a c d\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) + {\left (b c^{2} d x^{3} + b c^{3}\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 )}{3 \, {\left (a b c^{4} - a^{2} c^{3} d + {\left (a b c^{3} d - a^{2} c^{2} d^{2}\right )} x^{3}\right )}}, -\frac {2 \, {\left (\sqrt {d x^{3} + c} a c d - {\left (b c^{2} d x^{3} + b c^{3}\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 ) - {\left ({\left (b c d - a d^{2}\right )} x^{3} + b c^{2} - a c d\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right )\right )}}{3 \, {\left (a b c^{4} - a^{2} c^{3} d + {\left (a b c^{3} d - a^{2} c^{2} d^{2}\right )} x^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 111, normalized size = 0.97 \[ -\frac {2 \, b^{2} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{3 \, {\left (a b c - a^{2} d\right )} \sqrt {-b^{2} c + a b d}} - \frac {2 \, d}{3 \, \sqrt {d x^{3} + c} {\left (b c^{2} - a c d\right )}} + \frac {2 \, \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{3 \, a \sqrt {-c} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.29, size = 512, normalized size = 4.49 \[ -\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 ) b}{a}+\frac {-\frac {2 \arctanh \left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 c^{\frac {3}{2}}}+\frac {2}{3 \sqrt {\left (x^{3}+\frac {c}{d}\right ) d}\, c}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{3} + a\right )} {\left (d x^{3} + c\right )}^{\frac {3}{2}} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 8.44, size = 139, normalized size = 1.22 \[ \frac {\ln \left (\frac {{\left (\sqrt {d\,x^3+c}-\sqrt {c}\right )}^3\,\left (\sqrt {d\,x^3+c}+\sqrt {c}\right )}{x^6}\right )}{3\,a\,c^{3/2}}+\frac {2\,d}{3\,c\,\sqrt {d\,x^3+c}\,\left (a\,d-b\,c\right )}+\frac {b^{3/2}\,\ln \left (\frac {2\,b\,c-a\,d+b\,d\,x^3+\sqrt {b}\,\sqrt {d\,x^3+c}\,\sqrt {a\,d-b\,c}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,1{}\mathrm {i}}{3\,a\,{\left (a\,d-b\,c\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log((((c + d*x^3)^(1/2) - c^(1/2))^3*((c + d*x^3)^(1/2) + c^(1/2)))/x^6)/(3*a*c^(3/2)) + (2*d)/(3*c*(c + d*x^3
)^(1/2)*(a*d - b*c)) + (b^(3/2)*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*a*(a*d - b*c)^(3/2))

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sympy [A]  time = 24.63, size = 104, normalized size = 0.91 \[ \frac {2 d}{3 c \sqrt {c + d x^{3}} \left (a d - b c\right )} + \frac {2 b \operatorname {atan}{\left (\frac {\sqrt {c + d x^{3}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{3 a \sqrt {\frac {a d - b c}{b}} \left (a d - b c\right )} + \frac {2 \operatorname {atan}{\left (\frac {\sqrt {c + d x^{3}}}{\sqrt {- c}} \right )}}{3 a c \sqrt {- c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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