3.556 \(\int \frac {1}{x^4 (a c+b c x^3+d \sqrt {a+b x^3})} \, dx\)

Optimal. Leaf size=154 \[ -\frac {b d \left (3 a c^2-d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{3 a^{3/2} \left (a c^2-d^2\right )^2}-\frac {a c-d \sqrt {a+b x^3}}{3 a x^3 \left (a c^2-d^2\right )}+\frac {2 b c^3 \log \left (c \sqrt {a+b x^3}+d\right )}{3 \left (a c^2-d^2\right )^2}-\frac {b c^3 \log (x)}{\left (a c^2-d^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.30, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2155, 741, 801, 635, 206, 260} \[ -\frac {b d \left (3 a c^2-d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{3 a^{3/2} \left (a c^2-d^2\right )^2}-\frac {a c-d \sqrt {a+b x^3}}{3 a x^3 \left (a c^2-d^2\right )}+\frac {2 b c^3 \log \left (c \sqrt {a+b x^3}+d\right )}{3 \left (a c^2-d^2\right )^2}-\frac {b c^3 \log (x)}{\left (a c^2-d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a*c + b*c*x^3 + d*Sqrt[a + b*x^3])),x]

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[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 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

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^4 \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^2 \left (a c+b c x+d \sqrt {a+b x}\right )} \, dx,x,x^3\right )\\ &=\frac {1}{3} (2 b) \operatorname {Subst}\left (\int \frac {1}{(d+c x) \left (a-x^2\right )^2} \, dx,x,\sqrt {a+b x^3}\right )\\ &=-\frac {a c-d \sqrt {a+b x^3}}{3 a \left (a c^2-d^2\right ) x^3}-\frac {b \operatorname {Subst}\left (\int \frac {-2 a c^2+d^2+c d x}{(d+c x) \left (a-x^2\right )} \, dx,x,\sqrt {a+b x^3}\right )}{3 a \left (a c^2-d^2\right )}\\ &=-\frac {a c-d \sqrt {a+b x^3}}{3 a \left (a c^2-d^2\right ) x^3}-\frac {b \operatorname {Subst}\left (\int \left (-\frac {2 a c^4}{\left (a c^2-d^2\right ) (d+c x)}+\frac {3 a c^2 d-d^3-2 a c^3 x}{\left (a c^2-d^2\right ) \left (a-x^2\right )}\right ) \, dx,x,\sqrt {a+b x^3}\right )}{3 a \left (a c^2-d^2\right )}\\ &=-\frac {a c-d \sqrt {a+b x^3}}{3 a \left (a c^2-d^2\right ) x^3}+\frac {2 b c^3 \log \left (d+c \sqrt {a+b x^3}\right )}{3 \left (a c^2-d^2\right )^2}-\frac {b \operatorname {Subst}\left (\int \frac {3 a c^2 d-d^3-2 a c^3 x}{a-x^2} \, dx,x,\sqrt {a+b x^3}\right )}{3 a \left (a c^2-d^2\right )^2}\\ &=-\frac {a c-d \sqrt {a+b x^3}}{3 a \left (a c^2-d^2\right ) x^3}+\frac {2 b c^3 \log \left (d+c \sqrt {a+b x^3}\right )}{3 \left (a c^2-d^2\right )^2}+\frac {\left (2 b c^3\right ) \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 )^2}-\frac {\left (b d \left (3 a c^2-d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a-x^2} \, dx,x,\sqrt {a+b x^3}\right )}{3 a \left (a c^2-d^2\right )^2}\\ &=-\frac {a c-d \sqrt {a+b x^3}}{3 a \left (a c^2-d^2\right ) x^3}-\frac {b d \left (3 a c^2-d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{3 a^{3/2} \left (a c^2-d^2\right )^2}-\frac {b c^3 \log (x)}{\left (a c^2-d^2\right )^2}+\frac {2 b c^3 \log \left (d+c \sqrt {a+b x^3}\right )}{3 \left (a c^2-d^2\right )^2}\\ \end {align*}

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Mathematica [A]  time = 0.67, size = 307, normalized size = 1.99 \[ \frac {\sqrt {a} \left (-a^2 c^3 \sqrt {a+b x^3}+a^2 c^2 d+2 a b c^3 x^3 \sqrt {a+b x^3} \tanh ^{-1}\left (\frac {c \sqrt {a+b x^3}}{d}\right )-3 a b c^3 x^3 \log (x) \sqrt {a+b x^3}+b d x^3 \sqrt {\frac {b x^3}{a}+1} \left (a c^2-d^2\right ) \tanh ^{-1}\left (\sqrt {\frac {b x^3}{a}+1}\right )+a b c^2 d x^3+a b c^3 x^3 \sqrt {a+b x^3} \log \left (a c^2+b c^2 x^3-d^2\right )+a c d^2 \sqrt {a+b x^3}-a d^3-b d^3 x^3\right )-2 b d x^3 \sqrt {a+b x^3} \left (2 a c^2-d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{3 a^{3/2} x^3 \sqrt {a+b x^3} \left (d^2-a c^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(a*c + b*c*x^3 + d*Sqrt[a + b*x^3])),x]

[Out]

(-2*b*d*(2*a*c^2 - d^2)*x^3*Sqrt[a + b*x^3]*ArcTanh[Sqrt[a + b*x^3]/Sqrt[a]] + Sqrt[a]*(a^2*c^2*d - a*d^3 + a*
b*c^2*d*x^3 - b*d^3*x^3 - a^2*c^3*Sqrt[a + b*x^3] + a*c*d^2*Sqrt[a + b*x^3] + 2*a*b*c^3*x^3*Sqrt[a + b*x^3]*Ar
cTanh[(c*Sqrt[a + b*x^3])/d] + b*d*(a*c^2 - d^2)*x^3*Sqrt[1 + (b*x^3)/a]*ArcTanh[Sqrt[1 + (b*x^3)/a]] - 3*a*b*
c^3*x^3*Sqrt[a + b*x^3]*Log[x] + a*b*c^3*x^3*Sqrt[a + b*x^3]*Log[a*c^2 - d^2 + b*c^2*x^3]))/(3*a^(3/2)*(-(a*c^
2) + d^2)^2*x^3*Sqrt[a + b*x^3])

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fricas [A]  time = 0.55, size = 445, normalized size = 2.89 \[ \left [\frac {2 \, a^{2} b c^{3} x^{3} \log \left (b c^{2} x^{3} + a c^{2} - d^{2}\right ) + 2 \, a^{2} b c^{3} x^{3} \log \left (\sqrt {b x^{3} + a} c + d\right ) - 2 \, a^{2} b c^{3} x^{3} \log \left (\sqrt {b x^{3} + a} c - d\right ) - 6 \, a^{2} b c^{3} x^{3} \log \relax (x) - 2 \, a^{3} c^{3} - {\left (3 \, a b c^{2} d - b d^{3}\right )} \sqrt {a} x^{3} \log \left (\frac {b x^{3} + 2 \, \sqrt {b x^{3} + a} \sqrt {a} + 2 \, a}{x^{3}}\right ) + 2 \, a^{2} c d^{2} + 2 \, {\left (a^{2} c^{2} d - a d^{3}\right )} \sqrt {b x^{3} + a}}{6 \, {\left (a^{4} c^{4} - 2 \, a^{3} c^{2} d^{2} + a^{2} d^{4}\right )} x^{3}}, \frac {a^{2} b c^{3} x^{3} \log \left (b c^{2} x^{3} + a c^{2} - d^{2}\right ) + a^{2} b c^{3} x^{3} \log \left (\sqrt {b x^{3} + a} c + d\right ) - a^{2} b c^{3} x^{3} \log \left (\sqrt {b x^{3} + a} c - d\right ) - 3 \, a^{2} b c^{3} x^{3} \log \relax (x) - a^{3} c^{3} + {\left (3 \, a b c^{2} d - b d^{3}\right )} \sqrt {-a} x^{3} \arctan \left (\frac {\sqrt {b x^{3} + a} \sqrt {-a}}{a}\right ) + a^{2} c d^{2} + {\left (a^{2} c^{2} d - a d^{3}\right )} \sqrt {b x^{3} + a}}{3 \, {\left (a^{4} c^{4} - 2 \, a^{3} c^{2} d^{2} + a^{2} d^{4}\right )} x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(a*c+b*c*x^3+d*(b*x^3+a)^(1/2)),x, algorithm="fricas")

[Out]

[1/6*(2*a^2*b*c^3*x^3*log(b*c^2*x^3 + a*c^2 - d^2) + 2*a^2*b*c^3*x^3*log(sqrt(b*x^3 + a)*c + d) - 2*a^2*b*c^3*
x^3*log(sqrt(b*x^3 + a)*c - d) - 6*a^2*b*c^3*x^3*log(x) - 2*a^3*c^3 - (3*a*b*c^2*d - b*d^3)*sqrt(a)*x^3*log((b
*x^3 + 2*sqrt(b*x^3 + a)*sqrt(a) + 2*a)/x^3) + 2*a^2*c*d^2 + 2*(a^2*c^2*d - a*d^3)*sqrt(b*x^3 + a))/((a^4*c^4
- 2*a^3*c^2*d^2 + a^2*d^4)*x^3), 1/3*(a^2*b*c^3*x^3*log(b*c^2*x^3 + a*c^2 - d^2) + a^2*b*c^3*x^3*log(sqrt(b*x^
3 + a)*c + d) - a^2*b*c^3*x^3*log(sqrt(b*x^3 + a)*c - d) - 3*a^2*b*c^3*x^3*log(x) - a^3*c^3 + (3*a*b*c^2*d - b
*d^3)*sqrt(-a)*x^3*arctan(sqrt(b*x^3 + a)*sqrt(-a)/a) + a^2*c*d^2 + (a^2*c^2*d - a*d^3)*sqrt(b*x^3 + a))/((a^4
*c^4 - 2*a^3*c^2*d^2 + a^2*d^4)*x^3)]

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giac [A]  time = 0.34, size = 211, normalized size = 1.37 \[ \frac {2 \, b c^{4} \log \left ({\left | \sqrt {b x^{3} + a} c + d \right |}\right )}{3 \, {\left (a^{2} c^{5} - 2 \, a c^{3} d^{2} + c d^{4}\right )}} - \frac {b c^{3} \log \left (-b x^{3}\right )}{3 \, {\left (a^{2} c^{4} - 2 \, a c^{2} d^{2} + d^{4}\right )}} + \frac {{\left (3 \, a b c^{2} d - b d^{3}\right )} \arctan \left (\frac {\sqrt {b x^{3} + a}}{\sqrt {-a}}\right )}{3 \, {\left (a^{3} c^{4} - 2 \, a^{2} c^{2} d^{2} + a d^{4}\right )} \sqrt {-a}} - \frac {a^{2} b c^{3} - a b c d^{2} - {\left (a b c^{2} d - b d^{3}\right )} \sqrt {b x^{3} + a}}{3 \, {\left (a c^{2} - d^{2}\right )}^{2} a b x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(a*c+b*c*x^3+d*(b*x^3+a)^(1/2)),x, algorithm="giac")

[Out]

2/3*b*c^4*log(abs(sqrt(b*x^3 + a)*c + d))/(a^2*c^5 - 2*a*c^3*d^2 + c*d^4) - 1/3*b*c^3*log(-b*x^3)/(a^2*c^4 - 2
*a*c^2*d^2 + d^4) + 1/3*(3*a*b*c^2*d - b*d^3)*arctan(sqrt(b*x^3 + a)/sqrt(-a))/((a^3*c^4 - 2*a^2*c^2*d^2 + a*d
^4)*sqrt(-a)) - 1/3*(a^2*b*c^3 - a*b*c*d^2 - (a*b*c^2*d - b*d^3)*sqrt(b*x^3 + a))/((a*c^2 - d^2)^2*a*b*x^3)

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maple [C]  time = 0.07, size = 863, normalized size = 5.60 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(a*c+b*c*x^3+d*(b*x^3+a)^(1/2)),x)

[Out]

-1/3*c/(a*c^2-d^2)/x^3-2/(a*c^2-d^2)^2*b*c^3*ln(x)+1/(a*c^2-d^2)^2/a*b*c*d^2*ln(x)+1/3*a*c^5*b/(a*c^2-d^2)^2/d
^2*ln(b*c^2*x^3+a*c^2-d^2)+1/(a*c^2-d^2)/a*b*c*ln(x)-1/3*b*c^3/(a*c^2-d^2)/d^2*ln(b*c^2*x^3+a*c^2-d^2)+2/3*b/d
/a^2*(b*x^3+a)^(1/2)+1/3*d/a/(a*c^2-d^2)*(b*x^3+a)^(1/2)/x^3+1/3*d/a^(3/2)/(a*c^2-d^2)*b*arctanh((b*x^3+a)^(1/
2)/a^(1/2))+4/3*d*b/a/(a*c^2-d^2)^2*(b*x^3+a)^(1/2)*c^2-2/3*b/a^2/(a*c^2-d^2)^2*(b*x^3+a)^(1/2)*d^3-4/3*d*b/a^
(1/2)/(a*c^2-d^2)^2*arctanh((b*x^3+a)^(1/2)/a^(1/2))*c^2+2/3*b/a^(3/2)/(a*c^2-d^2)^2*arctanh((b*x^3+a)^(1/2)/a
^(1/2))*d^3-2/3*b*c^4/(a*c^2-d^2)^2/d*(b*x^3+a)^(1/2)-1/3*I/b*c^4/(a*c^2-d^2)^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 \[ \int \frac {1}{{\left (b c x^{3} + a c + \sqrt {b x^{3} + a} d\right )} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(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^4), x)

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mupad [B]  time = 5.46, size = 248, normalized size = 1.61 \[ \frac {b\,c^3\,\ln \left (b\,c^2\,x^3+a\,c^2-d^2\right )}{3\,a^2\,c^4-6\,a\,c^2\,d^2+3\,d^4}-\frac {b\,c^3\,\ln \relax (x)}{a^2\,c^4-2\,a\,c^2\,d^2+d^4}-\frac {c}{3\,x^3\,\left (a\,c^2-d^2\right )}+\frac {b\,c^3\,\ln \left (\frac {d+c\,\sqrt {b\,x^3+a}}{d-c\,\sqrt {b\,x^3+a}}\right )}{3\,{\left (a\,c^2-d^2\right )}^2}+\frac {d\,\sqrt {b\,x^3+a}}{3\,a\,x^3\,\left (a\,c^2-d^2\right )}+\frac {b\,d\,\ln \left (\frac {{\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )}^3\,\left (\sqrt {b\,x^3+a}+\sqrt {a}\right )}{x^6}\right )\,\left (3\,a\,c^2-d^2\right )}{6\,a^{3/2}\,{\left (a\,c^2-d^2\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(a*c + d*(a + b*x^3)^(1/2) + b*c*x^3)),x)

[Out]

(b*c^3*log(a*c^2 - d^2 + b*c^2*x^3))/(3*d^4 + 3*a^2*c^4 - 6*a*c^2*d^2) - (b*c^3*log(x))/(d^4 + a^2*c^4 - 2*a*c
^2*d^2) - c/(3*x^3*(a*c^2 - d^2)) + (b*c^3*log((d + c*(a + b*x^3)^(1/2))/(d - c*(a + b*x^3)^(1/2))))/(3*(a*c^2
- d^2)^2) + (d*(a + b*x^3)^(1/2))/(3*a*x^3*(a*c^2 - d^2)) + (b*d*log((((a + b*x^3)^(1/2) - a^(1/2))^3*((a + b
*x^3)^(1/2) + a^(1/2)))/x^6)*(3*a*c^2 - d^2))/(6*a^(3/2)*(a*c^2 - d^2)^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**4/(a*c+b*c*x**3+d*(b*x**3+a)**(1/2)),x)

[Out]

Timed out

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