3.6.56 \(\int \frac {1}{x^4 (a c+b c x^3+d \sqrt {a+b x^3})} \, dx\) [556]

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

[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 = 0.21, 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 = {2186, 755, 815, 649, 212, 266} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(a*c - d*Sqrt[a + b*x^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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[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 755

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 815

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 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^4 \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^2 \left (a c+b c x+d \sqrt {a+b x}\right )} \, dx,x,x^3\right )\\ &=\frac {1}{3} (2 b) \text {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 \text {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 \text {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 \text {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 ) \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 )^2}-\frac {\left (b d \left (3 a c^2-d^2\right )\right ) \text {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 = 0.20, size = 139, normalized size = 0.90 \begin {gather*} \frac {b d \left (-3 a c^2+d^2\right ) x^3 \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )+\sqrt {a} \left (-\left (\left (a c^2-d^2\right ) \left (a c-d \sqrt {a+b x^3}\right )\right )-a b c^3 x^3 \log \left (b x^3\right )+2 a b c^3 x^3 \log \left (d+c \sqrt {a+b x^3}\right )\right )}{3 a^{3/2} \left (-a c^2+d^2\right )^2 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.21, size = 774, normalized size = 5.03

method	result	size
default	\(\frac {a \,c^{5} b \ln \left (b \,c^{2} x^{3}+c^{2} a -d^{2}\right )}{3 \left (c^{2} a -d^{2}\right )^{2} d^{2}}-\frac {c}{3 \left (c^{2} a -d^{2}\right ) x^{3}}-\frac {2 b \,c^{3} \ln \left (x \right )}{\left (c^{2} a -d^{2}\right )^{2}}+\frac {c b \ln \left (x \right ) d^{2}}{a \left (c^{2} a -d^{2}\right )^{2}}-\frac {b \,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 {b c \ln \left (x \right )}{a \left (c^{2} a -d^{2}\right )}-d \left (-\frac {2 b \sqrt {b \,x^{3}+a}}{3 a^{2} d^{2}}+\frac {b^{2} c^{6} \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 )^{2} d^{2}}-\frac {b \left (2 c^{2} a -d^{2}\right ) \left (\frac {2 \sqrt {b \,x^{3}+a}}{3}-\frac {2 \sqrt {a}\, \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{3}\right )}{a^{2} \left (c^{2} a -d^{2}\right )^{2}}+\frac {-\frac {\sqrt {b \,x^{3}+a}}{3 x^{3}}-\frac {b \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{3 \sqrt {a}}}{a \left (c^{2} a -d^{2}\right )}\right )\)	\(774\)
elliptic	\(\text {Expression too large to display}\)	\(1862\)

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,method=_RETURNVERBOSE)

[Out]

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

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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^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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Fricas [A]
time = 0.45, size = 445, normalized size = 2.89 \begin {gather*} \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 \left (x\right ) - 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 \left (x\right ) - 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 ] \end {gather*}

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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{4} \left (a c + b c x^{3} + d \sqrt {a + b x^{3}}\right )}\, dx \end {gather*}

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]

Integral(1/(x**4*(a*c + b*c*x**3 + d*sqrt(a + b*x**3))), x)

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Giac [A]
time = 4.20, size = 211, normalized size = 1.37 \begin {gather*} \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}} \end {gather*}

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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Mupad [B]
time = 5.46, size = 248, normalized size = 1.61 \begin {gather*} \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 \left (x\right )}{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} \end {gather*}

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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