3.555 \(\int \frac {1}{x (a c+b c x^3+d \sqrt {a+b x^3})} \, dx\)

Optimal. Leaf size=93 \[ -\frac {2 c \log \left (c \sqrt {a+b x^3}+d\right )}{3 \left (a c^2-d^2\right )}+\frac {2 d \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{3 \sqrt {a} \left (a c^2-d^2\right )}+\frac {c \log (x)}{a c^2-d^2} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2155, 706, 31, 635, 207, 260} \[ -\frac {2 c \log \left (c \sqrt {a+b x^3}+d\right )}{3 \left (a c^2-d^2\right )}+\frac {2 d \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{3 \sqrt {a} \left (a c^2-d^2\right )}+\frac {c \log (x)}{a c^2-d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 706

Int[1/(((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 + a*e^2), Int[1/(d + e*x), x],
x] + Dist[1/(c*d^2 + a*e^2), Int[(c*d - c*e*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a
*e^2, 0]

Rule 2155

Int[(x_)^(m_.)/((c_) + (d_.)*(x_)^(n_) + (e_.)*Sqrt[(a_) + (b_.)*(x_)^(n_)]), x_Symbol] :> Dist[1/n, Subst[Int
[x^((m + 1)/n - 1)/(c + d*x + e*Sqrt[a + b*x]), x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && EqQ[b*c
- a*d, 0] && IntegerQ[(m + 1)/n]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 107, normalized size = 1.15 \[ \frac {\left (\sqrt {a} c-d\right ) \log \left (\sqrt {a}-\sqrt {a+b x^3}\right )+\left (\sqrt {a} c+d\right ) \log \left (\sqrt {a+b x^3}+\sqrt {a}\right )-2 \sqrt {a} c \log \left (c \sqrt {a+b x^3}+d\right )}{3 \sqrt {a} \left (a c^2-d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.47, size = 232, normalized size = 2.49 \[ \left [-\frac {a c \log \left (b c^{2} x^{3} + a c^{2} - d^{2}\right ) + a c \log \left (\sqrt {b x^{3} + a} c + d\right ) - a c \log \left (\sqrt {b x^{3} + a} c - d\right ) - 3 \, a c \log \relax (x) - \sqrt {a} d \log \left (\frac {b x^{3} + 2 \, \sqrt {b x^{3} + a} \sqrt {a} + 2 \, a}{x^{3}}\right )}{3 \, {\left (a^{2} c^{2} - a d^{2}\right )}}, -\frac {a c \log \left (b c^{2} x^{3} + a c^{2} - d^{2}\right ) + a c \log \left (\sqrt {b x^{3} + a} c + d\right ) - a c \log \left (\sqrt {b x^{3} + a} c - d\right ) - 3 \, a c \log \relax (x) + 2 \, \sqrt {-a} d \arctan \left (\frac {\sqrt {b x^{3} + a} \sqrt {-a}}{a}\right )}{3 \, {\left (a^{2} c^{2} - a d^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.35, size = 94, normalized size = 1.01 \[ -\frac {2 \, c^{2} \log \left ({\left | \sqrt {b x^{3} + a} c + d \right |}\right )}{3 \, {\left (a c^{3} - c d^{2}\right )}} + \frac {c \log \left (b x^{3}\right )}{3 \, {\left (a c^{2} - d^{2}\right )}} - \frac {2 \, d \arctan \left (\frac {\sqrt {b x^{3} + a}}{\sqrt {-a}}\right )}{3 \, {\left (a c^{2} - d^{2}\right )} \sqrt {-a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.06, size = 636, normalized size = 6.84 \[ -\frac {a \,c^{3} \ln \left (b \,c^{2} x^{3}+a \,c^{2}-d^{2}\right )}{3 \left (a \,c^{2}-d^{2}\right ) d^{2}}+\frac {c \ln \relax (x )}{a \,c^{2}-d^{2}}+\frac {2 d \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{3 \left (a \,c^{2}-d^{2}\right ) \sqrt {a}}+\frac {2 \sqrt {b \,x^{3}+a}\, c^{2}}{3 \left (a \,c^{2}-d^{2}\right ) d}-\frac {2 \sqrt {b \,x^{3}+a}\, d}{3 \left (a \,c^{2}-d^{2}\right ) a}+\frac {i c^{2} \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (2 x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}-i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}\right ) b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {\left (x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}\right ) b}{-3 \left (-a \,b^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i \left (2 x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}\right ) b}{2 \left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \left (2 \RootOf \left (b \,c^{2} \textit {\_Z}^{3}+a \,c^{2}-d^{2}\right )^{2} b^{2}+i \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (b \,c^{2} \textit {\_Z}^{3}+a \,c^{2}-d^{2}\right ) b -\left (-a \,b^{2}\right )^{\frac {1}{3}} \RootOf \left (b \,c^{2} \textit {\_Z}^{3}+a \,c^{2}-d^{2}\right ) b -i \left (-a \,b^{2}\right )^{\frac {2}{3}} \sqrt {3}-\left (-a \,b^{2}\right )^{\frac {2}{3}}\right ) \EllipticPi \left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}}{3}, -\frac {\left (2 i \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (b \,c^{2} \textit {\_Z}^{3}+a \,c^{2}-d^{2}\right )^{2} b +i \sqrt {3}\, a b -3 a b -i \left (-a \,b^{2}\right )^{\frac {2}{3}} \sqrt {3}\, \RootOf \left (b \,c^{2} \textit {\_Z}^{3}+a \,c^{2}-d^{2}\right )-3 \left (-a \,b^{2}\right )^{\frac {2}{3}} \RootOf \left (b \,c^{2} \textit {\_Z}^{3}+a \,c^{2}-d^{2}\right )\right ) c^{2}}{2 b \,d^{2}}, \sqrt {\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{\left (-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) b}}\right )}{3 \left (a \,c^{2}-d^{2}\right ) b^{2} d \sqrt {b \,x^{3}+a}}+\frac {c \ln \left (b \,c^{2} x^{3}+a \,c^{2}-d^{2}\right )}{3 d^{2}}-\frac {2 \sqrt {b \,x^{3}+a}}{3 a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.31, size = 156, normalized size = 1.68 \[ \frac {c\,\ln \relax (x)}{a\,c^2-d^2}+\frac {c\,\ln \left (\frac {d-c\,\sqrt {b\,x^3+a}}{d+c\,\sqrt {b\,x^3+a}}\right )}{3\,\left (a\,c^2-d^2\right )}-\frac {c\,\ln \left (b\,c^2\,x^3+a\,c^2-d^2\right )}{3\,a\,c^2-3\,d^2}+\frac {d\,\ln \left (\frac {\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )\,{\left (\sqrt {b\,x^3+a}+\sqrt {a}\right )}^3}{x^6}\right )}{3\,\sqrt {a}\,\left (a\,c^2-d^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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