Integrand size = 29, antiderivative size = 154 \[ \int \frac {1}{x^4 \left (a c+b c x^3+d \sqrt {a+b x^3}\right )} \, dx=-\frac {a c-d \sqrt {a+b x^3}}{3 a \left (a c^2-d^2\right ) x^3}-\frac {b d \left (3 a c^2-d^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{3 a^{3/2} \left (a c^2-d^2\right )^2}-\frac {b c^3 \log (x)}{\left (a c^2-d^2\right )^2}+\frac {2 b c^3 \log \left (d+c \sqrt {a+b x^3}\right )}{3 \left (a c^2-d^2\right )^2} \]
-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
Time = 0.21 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x^4 \left (a c+b c x^3+d \sqrt {a+b x^3}\right )} \, dx=\frac {b d \left (-3 a c^2+d^2\right ) x^3 \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )+\sqrt {a} \left (-\left (\left (a c^2-d^2\right ) \left (a c-d \sqrt {a+b x^3}\right )\right )-a b c^3 x^3 \log \left (b x^3\right )+2 a b c^3 x^3 \log \left (d+c \sqrt {a+b x^3}\right )\right )}{3 a^{3/2} \left (-a c^2+d^2\right )^2 x^3} \]
(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)
Time = 0.55 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.21, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2586, 7267, 496, 25, 657, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {1}{x^4 \left (d \sqrt {a+b x^3}+a c+b c x^3\right )} \, dx\) |
| \(\Big \downarrow \) 2586 |
| \(\displaystyle \frac {1}{3} \int \frac {1}{x^6 \left (b c x^3+a c+d \sqrt {b x^3+a}\right )}dx^3\) |
| \(\Big \downarrow \) 7267 |
| \(\displaystyle \frac {2}{3} b \int \frac {1}{\left (a-x^6\right )^2 \left (\sqrt {b x^3+a} c+d\right )}d\sqrt {b x^3+a}\) |
| \(\Big \downarrow \) 496 |
| \(\displaystyle \frac {2}{3} b \left (\frac {a c-d \sqrt {a+b x^3}}{2 a \left (a-x^6\right ) \left (a c^2-d^2\right )}-\frac {\int -\frac {2 a c^2-d \sqrt {b x^3+a} c-d^2}{\left (a-x^6\right ) \left (\sqrt {b x^3+a} c+d\right )}d\sqrt {b x^3+a}}{2 a \left (a c^2-d^2\right )}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2}{3} b \left (\frac {\int \frac {2 a c^2-d \sqrt {b x^3+a} c-d^2}{\left (a-x^6\right ) \left (\sqrt {b x^3+a} c+d\right )}d\sqrt {b x^3+a}}{2 a \left (a c^2-d^2\right )}+\frac {a c-d \sqrt {a+b x^3}}{2 a \left (a-x^6\right ) \left (a c^2-d^2\right )}\right )\) |
| \(\Big \downarrow \) 657 |
| \(\displaystyle \frac {2}{3} b \left (\frac {\int \left (\frac {2 a c^4}{\left (a c^2-d^2\right ) \left (\sqrt {b x^3+a} c+d\right )}+\frac {2 a \sqrt {b x^3+a} c^3-3 a d c^2+d^3}{\left (a c^2-d^2\right ) \left (a-x^6\right )}\right )d\sqrt {b x^3+a}}{2 a \left (a c^2-d^2\right )}+\frac {a c-d \sqrt {a+b x^3}}{2 a \left (a-x^6\right ) \left (a c^2-d^2\right )}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2}{3} b \left (\frac {-\frac {d \left (3 a c^2-d^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{\sqrt {a} \left (a c^2-d^2\right )}+\frac {2 a c^3 \log \left (c \sqrt {a+b x^3}+d\right )}{a c^2-d^2}-\frac {a c^3 \log \left (a-x^6\right )}{a c^2-d^2}}{2 a \left (a c^2-d^2\right )}+\frac {a c-d \sqrt {a+b x^3}}{2 a \left (a-x^6\right ) \left (a c^2-d^2\right )}\right )\) |
(2*b*((a*c - d*Sqrt[a + b*x^3])/(2*a*(a*c^2 - d^2)*(a - x^6)) + (-((d*(3*a *c^2 - d^2)*ArcTanh[Sqrt[a + b*x^3]/Sqrt[a]])/(Sqrt[a]*(a*c^2 - d^2))) - ( a*c^3*Log[a - x^6])/(a*c^2 - d^2) + (2*a*c^3*Log[d + c*Sqrt[a + b*x^3]])/( a*c^2 - d^2))/(2*a*(a*c^2 - d^2))))/3
3.6.56.3.1 Defintions of rubi rules used
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-(a*d + b*c*x))*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1)*(b*c^2 + a*d^2))), x] + Simp[1/(2*a*(p + 1)*(b*c^2 + a*d^2)) Int[(c + d*x)^n*(a + b*x^2)^(p + 1)*Simp[b*c^2*(2*p + 3) + a*d^2*(n + 2*p + 3) + b*c*d*(n + 2 *p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[p, -1] && IntQuad raticQ[a, 0, b, c, d, n, p, x]
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
Int[(x_)^(m_.)/((c_) + (d_.)*(x_)^(n_) + (e_.)*Sqrt[(a_) + (b_.)*(x_)^(n_)] ), x_Symbol] :> Simp[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]
Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si mp[lst[[2]]*lst[[4]] Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x ] /; !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
Leaf count of result is larger than twice the leaf count of optimal. \(404\) vs. \(2(138)=276\).
Time = 0.97 (sec) , antiderivative size = 405, normalized size of antiderivative = 2.63
| method | result | size |
| default | \(a c \left (-\frac {b \ln \left (b \,x^{3}+a \right )}{3 a^{2} d^{2}}-\frac {1}{3 a \left (a \,c^{2}-d^{2}\right ) x^{3}}-\frac {b \left (2 a \,c^{2}-d^{2}\right ) \ln \left (x \right )}{a^{2} \left (a \,c^{2}-d^{2}\right )^{2}}+\frac {b \,c^{4} \ln \left (b \,c^{2} x^{3}+a \,c^{2}-d^{2}\right )}{3 \left (a \,c^{2}-d^{2}\right )^{2} d^{2}}\right )+b c \left (\frac {\ln \left (b \,x^{3}+a \right )}{3 a \,d^{2}}+\frac {\ln \left (x \right )}{a \left (a \,c^{2}-d^{2}\right )}-\frac {c^{2} \ln \left (b \,c^{2} x^{3}+a \,c^{2}-d^{2}\right )}{3 \left (a \,c^{2}-d^{2}\right ) d^{2}}\right )-d \left (\frac {-\frac {\sqrt {b \,x^{3}+a}}{3 x^{3}}-\frac {b \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{3 \sqrt {a}}}{a \left (a \,c^{2}-d^{2}\right )}-\frac {2 b \sqrt {b \,x^{3}+a}}{3 a^{2} d^{2}}-\frac {b \left (2 a \,c^{2}-d^{2}\right ) \left (\frac {2 \sqrt {b \,x^{3}+a}}{3}-\frac {2 \sqrt {a}\, \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{3}\right )}{a^{2} \left (a \,c^{2}-d^{2}\right )^{2}}+\frac {b \,c^{3} \left (d \ln \left (c \sqrt {b \,x^{3}+a}-d \right )-d \ln \left (d +c \sqrt {b \,x^{3}+a}\right )+2 c \sqrt {b \,x^{3}+a}\right )}{3 \left (a \,c^{2}-d^{2}\right )^{2} d^{2}}\right )\) | \(405\) |
| elliptic | \(\text {Expression too large to display}\) | \(1862\) |
a*c*(-1/3*b/a^2/d^2*ln(b*x^3+a)-1/3/a/(a*c^2-d^2)/x^3-b*(2*a*c^2-d^2)/a^2/ (a*c^2-d^2)^2*ln(x)+1/3*b*c^4/(a*c^2-d^2)^2/d^2*ln(b*c^2*x^3+a*c^2-d^2))+b *c*(1/3/a/d^2*ln(b*x^3+a)+1/a/(a*c^2-d^2)*ln(x)-1/3*c^2/(a*c^2-d^2)/d^2*ln (b*c^2*x^3+a*c^2-d^2))-d*(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))-2/3*b/a^2/d^2*(b*x^3+a)^(1/2)-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)))+1/3*b*c^3/(a*c^2-d^2)^2/d^2*(d*ln(c*(b*x^3+a)^(1/ 2)-d)-d*ln(d+c*(b*x^3+a)^(1/2))+2*c*(b*x^3+a)^(1/2)))
Time = 0.37 (sec) , antiderivative size = 445, normalized size of antiderivative = 2.89 \[ \int \frac {1}{x^4 \left (a c+b c x^3+d \sqrt {a+b x^3}\right )} \, dx=\left [\frac {2 \, a^{2} b c^{3} x^{3} \log \left (b c^{2} x^{3} + a c^{2} - d^{2}\right ) + 2 \, a^{2} b c^{3} x^{3} \log \left (\sqrt {b x^{3} + a} c + d\right ) - 2 \, a^{2} b c^{3} x^{3} \log \left (\sqrt {b x^{3} + a} c - d\right ) - 6 \, a^{2} b c^{3} x^{3} \log \left (x\right ) - 2 \, a^{3} c^{3} - {\left (3 \, a b c^{2} d - b d^{3}\right )} \sqrt {a} x^{3} \log \left (\frac {b x^{3} + 2 \, \sqrt {b x^{3} + a} \sqrt {a} + 2 \, a}{x^{3}}\right ) + 2 \, a^{2} c d^{2} + 2 \, {\left (a^{2} c^{2} d - a d^{3}\right )} \sqrt {b x^{3} + a}}{6 \, {\left (a^{4} c^{4} - 2 \, a^{3} c^{2} d^{2} + a^{2} d^{4}\right )} x^{3}}, \frac {a^{2} b c^{3} x^{3} \log \left (b c^{2} x^{3} + a c^{2} - d^{2}\right ) + a^{2} b c^{3} x^{3} \log \left (\sqrt {b x^{3} + a} c + d\right ) - a^{2} b c^{3} x^{3} \log \left (\sqrt {b x^{3} + a} c - d\right ) - 3 \, a^{2} b c^{3} x^{3} \log \left (x\right ) - a^{3} c^{3} + {\left (3 \, a b c^{2} d - b d^{3}\right )} \sqrt {-a} x^{3} \arctan \left (\frac {\sqrt {b x^{3} + a} \sqrt {-a}}{a}\right ) + a^{2} c d^{2} + {\left (a^{2} c^{2} d - a d^{3}\right )} \sqrt {b x^{3} + a}}{3 \, {\left (a^{4} c^{4} - 2 \, a^{3} c^{2} d^{2} + a^{2} d^{4}\right )} x^{3}}\right ] \]
[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(s qrt(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)]
\[ \int \frac {1}{x^4 \left (a c+b c x^3+d \sqrt {a+b x^3}\right )} \, dx=\int \frac {1}{x^{4} \left (a c + b c x^{3} + d \sqrt {a + b x^{3}}\right )}\, dx \]
\[ \int \frac {1}{x^4 \left (a c+b c x^3+d \sqrt {a+b x^3}\right )} \, dx=\int { \frac {1}{{\left (b c x^{3} + a c + \sqrt {b x^{3} + a} d\right )} x^{4}} \,d x } \]
Time = 0.32 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.37 \[ \int \frac {1}{x^4 \left (a c+b c x^3+d \sqrt {a+b x^3}\right )} \, dx=\frac {2 \, b c^{4} \log \left ({\left | \sqrt {b x^{3} + a} c + d \right |}\right )}{3 \, {\left (a^{2} c^{5} - 2 \, a c^{3} d^{2} + c d^{4}\right )}} - \frac {b c^{3} \log \left (-b x^{3}\right )}{3 \, {\left (a^{2} c^{4} - 2 \, a c^{2} d^{2} + d^{4}\right )}} + \frac {{\left (3 \, a b c^{2} d - b d^{3}\right )} \arctan \left (\frac {\sqrt {b x^{3} + a}}{\sqrt {-a}}\right )}{3 \, {\left (a^{3} c^{4} - 2 \, a^{2} c^{2} d^{2} + a d^{4}\right )} \sqrt {-a}} - \frac {a^{2} b c^{3} - a b c d^{2} - {\left (a b c^{2} d - b d^{3}\right )} \sqrt {b x^{3} + a}}{3 \, {\left (a c^{2} - d^{2}\right )}^{2} a b x^{3}} \]
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)
Time = 18.90 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.61 \[ \int \frac {1}{x^4 \left (a c+b c x^3+d \sqrt {a+b x^3}\right )} \, dx=\frac {b\,c^3\,\ln \left (b\,c^2\,x^3+a\,c^2-d^2\right )}{3\,a^2\,c^4-6\,a\,c^2\,d^2+3\,d^4}-\frac {b\,c^3\,\ln \left (x\right )}{a^2\,c^4-2\,a\,c^2\,d^2+d^4}-\frac {c}{3\,x^3\,\left (a\,c^2-d^2\right )}+\frac {b\,c^3\,\ln \left (\frac {d+c\,\sqrt {b\,x^3+a}}{d-c\,\sqrt {b\,x^3+a}}\right )}{3\,{\left (a\,c^2-d^2\right )}^2}+\frac {d\,\sqrt {b\,x^3+a}}{3\,a\,x^3\,\left (a\,c^2-d^2\right )}+\frac {b\,d\,\ln \left (\frac {{\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )}^3\,\left (\sqrt {b\,x^3+a}+\sqrt {a}\right )}{x^6}\right )\,\left (3\,a\,c^2-d^2\right )}{6\,a^{3/2}\,{\left (a\,c^2-d^2\right )}^2} \]
(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)