3.6.55 \(\int \frac {1}{x (a c+b c x^3+d \sqrt {a+b x^3})} \, dx\) [555]

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

[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)

________________________________________________________________________________________

Rubi [A]
time = 0.15, 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 = {2186, 720, 31, 649, 213, 266} \begin {gather*} -\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} \end {gather*}

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 213

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

Rule 266

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 649

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 720

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 2186

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

________________________________________________________________________________________

Mathematica [A]
time = 0.07, size = 69, normalized size = 0.74 \begin {gather*} \frac {\frac {2 d \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{\sqrt {a}}+c \log \left (b x^3\right )-2 c \log \left (d+c \sqrt {a+b x^3}\right )}{3 a c^2-3 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.24, size = 621, normalized size = 6.68

method	result	size
default	\(-\frac {a \,c^{3} \ln \left (b \,c^{2} x^{3}+c^{2} a -d^{2}\right )}{3 \left (c^{2} a -d^{2}\right ) d^{2}}+\frac {c \ln \left (x \right )}{c^{2} a -d^{2}}+\frac {c \ln \left (b \,c^{2} x^{3}+c^{2} a -d^{2}\right )}{3 d^{2}}-d \left (\frac {2 \sqrt {b \,x^{3}+a}}{3 a \,d^{2}}-\frac {b \,c^{4} \left (\frac {2 \sqrt {b \,x^{3}+a}}{3 c^{2} b}+\frac {i \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (b \,c^{2} \textit {\_Z}^{3}+c^{2} a -d^{2}\right )}{\sum }\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {2}\, \sqrt {\frac {i b \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 )}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {b \left (x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}\right )}{-3 \left (-a \,b^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i b \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 )}{2 \left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \left (i \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \underline {\hspace {1.25 ex}}\alpha b -i \left (-a \,b^{2}\right )^{\frac {2}{3}} \sqrt {3}+2 \underline {\hspace {1.25 ex}}\alpha ^{2} b^{2}-\left (-a \,b^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha b -\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 {c^{2} \left (2 i \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \underline {\hspace {1.25 ex}}\alpha ^{2} b -i \left (-a \,b^{2}\right )^{\frac {2}{3}} \sqrt {3}\, \underline {\hspace {1.25 ex}}\alpha +i \sqrt {3}\, a b -3 \left (-a \,b^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha -3 a b \right )}{2 b \,d^{2}}, \sqrt {\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{b \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 )}}\right )}{2 \sqrt {b \,x^{3}+a}}\right )}{3 b^{3} c^{2}}\right )}{\left (c^{2} a -d^{2}\right ) d^{2}}+\frac {\frac {2 \sqrt {b \,x^{3}+a}}{3}-\frac {2 \sqrt {a}\, \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{3}}{a \left (c^{2} a -d^{2}\right )}\right )\)	\(621\)
elliptic	\(\text {Expression too large to display}\)	\(1768\)

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,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

________________________________________________________________________________________

Fricas [A]
time = 0.38, size = 232, normalized size = 2.49 \begin {gather*} \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 \left (x\right ) - \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 \left (x\right ) + 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 ] \end {gather*}

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)]

________________________________________________________________________________________

Sympy [A]
time = 4.23, size = 97, normalized size = 1.04 \begin {gather*} - \frac {2 c^{2} \left (\begin {cases} \frac {\sqrt {a + b x^{3}}}{d} & \text {for}\: c = 0 \\\frac {\log {\left (c \sqrt {a + b x^{3}} + d \right )}}{c} & \text {otherwise} \end {cases}\right )}{3 \left (a c^{2} - d^{2}\right )} - \frac {2 \left (- \frac {c \log {\left (- b x^{3} \right )}}{2} + \frac {d \operatorname {atan}{\left (\frac {\sqrt {a + b x^{3}}}{\sqrt {- a}} \right )}}{\sqrt {- a}}\right )}{3 \left (a c^{2} - d^{2}\right )} \end {gather*}

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]

-2*c**2*Piecewise((sqrt(a + b*x**3)/d, Eq(c, 0)), (log(c*sqrt(a + b*x**3) + d)/c, True))/(3*(a*c**2 - d**2)) -
2*(-c*log(-b*x**3)/2 + d*atan(sqrt(a + b*x**3)/sqrt(-a))/sqrt(-a))/(3*(a*c**2 - d**2))

________________________________________________________________________________________

Giac [A]
time = 3.73, size = 94, normalized size = 1.01 \begin {gather*} -\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}} \end {gather*}

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))

________________________________________________________________________________________

Mupad [B]
time = 4.31, size = 156, normalized size = 1.68 \begin {gather*} \frac {c\,\ln \left (x\right )}{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 )} \end {gather*}

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))

________________________________________________________________________________________