Integrand size = 24, antiderivative size = 158 \[ \int \frac {1}{x^4 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=-\frac {d (b c-3 a d)}{3 a c^2 (b c-a d) \sqrt {c+d x^3}}-\frac {1}{3 a c x^3 \sqrt {c+d x^3}}+\frac {(2 b c+3 a d) \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^2 c^{5/2}}-\frac {2 b^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 a^2 (b c-a d)^{3/2}} \]
1/3*(3*a*d+2*b*c)*arctanh((d*x^3+c)^(1/2)/c^(1/2))/a^2/c^(5/2)-2/3*b^(5/2) *arctanh(b^(1/2)*(d*x^3+c)^(1/2)/(-a*d+b*c)^(1/2))/a^2/(-a*d+b*c)^(3/2)-1/ 3*d*(-3*a*d+b*c)/a/c^2/(-a*d+b*c)/(d*x^3+c)^(1/2)-1/3/a/c/x^3/(d*x^3+c)^(1 /2)
Time = 0.68 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x^4 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {\frac {a \left (-b c \left (c+d x^3\right )+a d \left (c+3 d x^3\right )\right )}{c^2 (b c-a d) x^3 \sqrt {c+d x^3}}-\frac {2 b^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {-b c+a d}}\right )}{(-b c+a d)^{3/2}}+\frac {(2 b c+3 a d) \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{c^{5/2}}}{3 a^2} \]
((a*(-(b*c*(c + d*x^3)) + a*d*(c + 3*d*x^3)))/(c^2*(b*c - a*d)*x^3*Sqrt[c + d*x^3]) - (2*b^(5/2)*ArcTan[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[-(b*c) + a*d] ])/(-(b*c) + a*d)^(3/2) + ((2*b*c + 3*a*d)*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c] ])/c^(5/2))/(3*a^2)
Time = 0.36 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.19, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {948, 114, 27, 169, 27, 174, 73, 221}
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.
\(\displaystyle \int \frac {1}{x^4 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int \frac {1}{x^6 \left (b x^3+a\right ) \left (d x^3+c\right )^{3/2}}dx^3\) |
\(\Big \downarrow \) 114 |
\(\displaystyle \frac {1}{3} \left (-\frac {\int \frac {3 b d x^3+2 b c+3 a d}{2 x^3 \left (b x^3+a\right ) \left (d x^3+c\right )^{3/2}}dx^3}{a c}-\frac {1}{a c x^3 \sqrt {c+d x^3}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \left (-\frac {\int \frac {3 b d x^3+2 b c+3 a d}{x^3 \left (b x^3+a\right ) \left (d x^3+c\right )^{3/2}}dx^3}{2 a c}-\frac {1}{a c x^3 \sqrt {c+d x^3}}\right )\) |
\(\Big \downarrow \) 169 |
\(\displaystyle \frac {1}{3} \left (-\frac {\frac {2 d (b c-3 a d)}{c \sqrt {c+d x^3} (b c-a d)}-\frac {2 \int -\frac {b d (b c-3 a d) x^3+(b c-a d) (2 b c+3 a d)}{2 x^3 \left (b x^3+a\right ) \sqrt {d x^3+c}}dx^3}{c (b c-a d)}}{2 a c}-\frac {1}{a c x^3 \sqrt {c+d x^3}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \left (-\frac {\frac {\int \frac {b d (b c-3 a d) x^3+(b c-a d) (2 b c+3 a d)}{x^3 \left (b x^3+a\right ) \sqrt {d x^3+c}}dx^3}{c (b c-a d)}+\frac {2 d (b c-3 a d)}{c \sqrt {c+d x^3} (b c-a d)}}{2 a c}-\frac {1}{a c x^3 \sqrt {c+d x^3}}\right )\) |
\(\Big \downarrow \) 174 |
\(\displaystyle \frac {1}{3} \left (-\frac {\frac {\frac {(b c-a d) (3 a d+2 b c) \int \frac {1}{x^3 \sqrt {d x^3+c}}dx^3}{a}-\frac {2 b^3 c^2 \int \frac {1}{\left (b x^3+a\right ) \sqrt {d x^3+c}}dx^3}{a}}{c (b c-a d)}+\frac {2 d (b c-3 a d)}{c \sqrt {c+d x^3} (b c-a d)}}{2 a c}-\frac {1}{a c x^3 \sqrt {c+d x^3}}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{3} \left (-\frac {\frac {\frac {2 (b c-a d) (3 a d+2 b c) \int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{a d}-\frac {4 b^3 c^2 \int \frac {1}{\frac {b x^6}{d}+a-\frac {b c}{d}}d\sqrt {d x^3+c}}{a d}}{c (b c-a d)}+\frac {2 d (b c-3 a d)}{c \sqrt {c+d x^3} (b c-a d)}}{2 a c}-\frac {1}{a c x^3 \sqrt {c+d x^3}}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{3} \left (-\frac {\frac {\frac {4 b^{5/2} c^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{a \sqrt {b c-a d}}-\frac {2 (b c-a d) (3 a d+2 b c) \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{a \sqrt {c}}}{c (b c-a d)}+\frac {2 d (b c-3 a d)}{c \sqrt {c+d x^3} (b c-a d)}}{2 a c}-\frac {1}{a c x^3 \sqrt {c+d x^3}}\right )\) |
(-(1/(a*c*x^3*Sqrt[c + d*x^3])) - ((2*d*(b*c - 3*a*d))/(c*(b*c - a*d)*Sqrt [c + d*x^3]) + ((-2*(b*c - a*d)*(2*b*c + 3*a*d)*ArcTanh[Sqrt[c + d*x^3]/Sq rt[c]])/(a*Sqrt[c]) + (4*b^(5/2)*c^2*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqr t[b*c - a*d]])/(a*Sqrt[b*c - a*d]))/(c*(b*c - a*d)))/(2*a*c))/3
3.4.92.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[1/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] && (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n *(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[ 2*m, 2*n, 2*p]
Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* ((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d) Int[(e + f*x)^ p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d) Int[(e + f*x)^p/(c + d *x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Time = 4.56 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.91
method | result | size |
risch | \(-\frac {\sqrt {d \,x^{3}+c}}{3 c^{2} a \,x^{3}}-\frac {-\frac {2 \left (3 a d +2 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 a \sqrt {c}}+\frac {4 a \,d^{2}}{3 \left (a d -b c \right ) \sqrt {d \,x^{3}+c}}+\frac {4 b^{3} c^{2} \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 \left (a d -b c \right ) a \sqrt {\left (a d -b c \right ) b}}}{2 a \,c^{2}}\) | \(144\) |
pseudoelliptic | \(\frac {d^{2} \left (-\frac {2 b^{3} \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right ) a^{2} d^{2} \sqrt {\left (a d -b c \right ) b}}-\frac {-3 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right ) a d \,x^{3}-2 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right ) b c \,x^{3}+\sqrt {d \,x^{3}+c}\, a \sqrt {c}}{x^{3} c^{\frac {5}{2}} a^{2} d^{2}}-\frac {2}{\left (a d -b c \right ) c^{2} \sqrt {d \,x^{3}+c}}\right )}{3}\) | \(159\) |
default | \(\frac {-\frac {\sqrt {d \,x^{3}+c}}{3 c^{2} x^{3}}-\frac {2 d}{3 c^{2} \sqrt {\left (x^{3}+\frac {c}{d}\right ) d}}+\frac {d \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{c^{\frac {5}{2}}}}{a}-\frac {b \left (\frac {2}{3 c \sqrt {\left (x^{3}+\frac {c}{d}\right ) d}}-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 c^{\frac {3}{2}}}\right )}{a^{2}}-\frac {2 b^{2} \left (b \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right ) \sqrt {d \,x^{3}+c}+\sqrt {\left (a d -b c \right ) b}\right )}{a^{2} \sqrt {\left (a d -b c \right ) b}\, \sqrt {d \,x^{3}+c}\, \left (3 a d -3 b c \right )}\) | \(194\) |
elliptic | \(\text {Expression too large to display}\) | \(1682\) |
-1/3/c^2/a*(d*x^3+c)^(1/2)/x^3-1/2/a/c^2*(-2/3*(3*a*d+2*b*c)/a*arctanh((d* x^3+c)^(1/2)/c^(1/2))/c^(1/2)+4/3*a*d^2/(a*d-b*c)/(d*x^3+c)^(1/2)+4/3*b^3* c^2/(a*d-b*c)/a/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x^3+c)^(1/2)/((a*d-b*c)*b) ^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 270 vs. \(2 (130) = 260\).
Time = 0.45 (sec) , antiderivative size = 1120, normalized size of antiderivative = 7.09 \[ \int \frac {1}{x^4 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\left [-\frac {2 \, {\left (b^{2} c^{3} d x^{6} + b^{2} c^{4} x^{3}\right )} \sqrt {\frac {b}{b c - a d}} \log \left (\frac {b d x^{3} + 2 \, b c - a d + 2 \, \sqrt {d x^{3} + c} {\left (b c - a d\right )} \sqrt {\frac {b}{b c - a d}}}{b x^{3} + a}\right ) - {\left ({\left (2 \, b^{2} c^{2} d + a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{6} + {\left (2 \, b^{2} c^{3} + a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x^{3}\right )} \sqrt {c} \log \left (\frac {d x^{3} + 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) + 2 \, {\left (a b c^{3} - a^{2} c^{2} d + {\left (a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x^{3}\right )} \sqrt {d x^{3} + c}}{6 \, {\left ({\left (a^{2} b c^{4} d - a^{3} c^{3} d^{2}\right )} x^{6} + {\left (a^{2} b c^{5} - a^{3} c^{4} d\right )} x^{3}\right )}}, -\frac {4 \, {\left (b^{2} c^{3} d x^{6} + b^{2} c^{4} x^{3}\right )} \sqrt {-\frac {b}{b c - a d}} \arctan \left (-\frac {\sqrt {d x^{3} + c} {\left (b c - a d\right )} \sqrt {-\frac {b}{b c - a d}}}{b d x^{3} + b c}\right ) - {\left ({\left (2 \, b^{2} c^{2} d + a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{6} + {\left (2 \, b^{2} c^{3} + a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x^{3}\right )} \sqrt {c} \log \left (\frac {d x^{3} + 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) + 2 \, {\left (a b c^{3} - a^{2} c^{2} d + {\left (a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x^{3}\right )} \sqrt {d x^{3} + c}}{6 \, {\left ({\left (a^{2} b c^{4} d - a^{3} c^{3} d^{2}\right )} x^{6} + {\left (a^{2} b c^{5} - a^{3} c^{4} d\right )} x^{3}\right )}}, -\frac {{\left ({\left (2 \, b^{2} c^{2} d + a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{6} + {\left (2 \, b^{2} c^{3} + a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) + {\left (b^{2} c^{3} d x^{6} + b^{2} c^{4} x^{3}\right )} \sqrt {\frac {b}{b c - a d}} \log \left (\frac {b d x^{3} + 2 \, b c - a d + 2 \, \sqrt {d x^{3} + c} {\left (b c - a d\right )} \sqrt {\frac {b}{b c - a d}}}{b x^{3} + a}\right ) + {\left (a b c^{3} - a^{2} c^{2} d + {\left (a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x^{3}\right )} \sqrt {d x^{3} + c}}{3 \, {\left ({\left (a^{2} b c^{4} d - a^{3} c^{3} d^{2}\right )} x^{6} + {\left (a^{2} b c^{5} - a^{3} c^{4} d\right )} x^{3}\right )}}, -\frac {2 \, {\left (b^{2} c^{3} d x^{6} + b^{2} c^{4} x^{3}\right )} \sqrt {-\frac {b}{b c - a d}} \arctan \left (-\frac {\sqrt {d x^{3} + c} {\left (b c - a d\right )} \sqrt {-\frac {b}{b c - a d}}}{b d x^{3} + b c}\right ) + {\left ({\left (2 \, b^{2} c^{2} d + a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{6} + {\left (2 \, b^{2} c^{3} + a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) + {\left (a b c^{3} - a^{2} c^{2} d + {\left (a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x^{3}\right )} \sqrt {d x^{3} + c}}{3 \, {\left ({\left (a^{2} b c^{4} d - a^{3} c^{3} d^{2}\right )} x^{6} + {\left (a^{2} b c^{5} - a^{3} c^{4} d\right )} x^{3}\right )}}\right ] \]
[-1/6*(2*(b^2*c^3*d*x^6 + b^2*c^4*x^3)*sqrt(b/(b*c - a*d))*log((b*d*x^3 + 2*b*c - a*d + 2*sqrt(d*x^3 + c)*(b*c - a*d)*sqrt(b/(b*c - a*d)))/(b*x^3 + a)) - ((2*b^2*c^2*d + a*b*c*d^2 - 3*a^2*d^3)*x^6 + (2*b^2*c^3 + a*b*c^2*d - 3*a^2*c*d^2)*x^3)*sqrt(c)*log((d*x^3 + 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/ x^3) + 2*(a*b*c^3 - a^2*c^2*d + (a*b*c^2*d - 3*a^2*c*d^2)*x^3)*sqrt(d*x^3 + c))/((a^2*b*c^4*d - a^3*c^3*d^2)*x^6 + (a^2*b*c^5 - a^3*c^4*d)*x^3), -1/ 6*(4*(b^2*c^3*d*x^6 + b^2*c^4*x^3)*sqrt(-b/(b*c - a*d))*arctan(-sqrt(d*x^3 + c)*(b*c - a*d)*sqrt(-b/(b*c - a*d))/(b*d*x^3 + b*c)) - ((2*b^2*c^2*d + a*b*c*d^2 - 3*a^2*d^3)*x^6 + (2*b^2*c^3 + a*b*c^2*d - 3*a^2*c*d^2)*x^3)*sq rt(c)*log((d*x^3 + 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^3) + 2*(a*b*c^3 - a^ 2*c^2*d + (a*b*c^2*d - 3*a^2*c*d^2)*x^3)*sqrt(d*x^3 + c))/((a^2*b*c^4*d - a^3*c^3*d^2)*x^6 + (a^2*b*c^5 - a^3*c^4*d)*x^3), -1/3*(((2*b^2*c^2*d + a*b *c*d^2 - 3*a^2*d^3)*x^6 + (2*b^2*c^3 + a*b*c^2*d - 3*a^2*c*d^2)*x^3)*sqrt( -c)*arctan(sqrt(d*x^3 + c)*sqrt(-c)/c) + (b^2*c^3*d*x^6 + b^2*c^4*x^3)*sqr t(b/(b*c - a*d))*log((b*d*x^3 + 2*b*c - a*d + 2*sqrt(d*x^3 + c)*(b*c - a*d )*sqrt(b/(b*c - a*d)))/(b*x^3 + a)) + (a*b*c^3 - a^2*c^2*d + (a*b*c^2*d - 3*a^2*c*d^2)*x^3)*sqrt(d*x^3 + c))/((a^2*b*c^4*d - a^3*c^3*d^2)*x^6 + (a^2 *b*c^5 - a^3*c^4*d)*x^3), -1/3*(2*(b^2*c^3*d*x^6 + b^2*c^4*x^3)*sqrt(-b/(b *c - a*d))*arctan(-sqrt(d*x^3 + c)*(b*c - a*d)*sqrt(-b/(b*c - a*d))/(b*d*x ^3 + b*c)) + ((2*b^2*c^2*d + a*b*c*d^2 - 3*a^2*d^3)*x^6 + (2*b^2*c^3 + ...
\[ \int \frac {1}{x^4 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {1}{x^{4} \left (a + b x^{3}\right ) \left (c + d x^{3}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {1}{x^4 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )} {\left (d x^{3} + c\right )}^{\frac {3}{2}} x^{4}} \,d x } \]
Time = 0.29 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.09 \[ \int \frac {1}{x^4 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {2 \, b^{3} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{3 \, {\left (a^{2} b c - a^{3} d\right )} \sqrt {-b^{2} c + a b d}} - \frac {{\left (d x^{3} + c\right )} b c d - 3 \, {\left (d x^{3} + c\right )} a d^{2} + 2 \, a c d^{2}}{3 \, {\left (a b c^{3} - a^{2} c^{2} d\right )} {\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} - \sqrt {d x^{3} + c} c\right )}} - \frac {{\left (2 \, b c + 3 \, a d\right )} \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{3 \, a^{2} \sqrt {-c} c^{2}} \]
2/3*b^3*arctan(sqrt(d*x^3 + c)*b/sqrt(-b^2*c + a*b*d))/((a^2*b*c - a^3*d)* sqrt(-b^2*c + a*b*d)) - 1/3*((d*x^3 + c)*b*c*d - 3*(d*x^3 + c)*a*d^2 + 2*a *c*d^2)/((a*b*c^3 - a^2*c^2*d)*((d*x^3 + c)^(3/2) - sqrt(d*x^3 + c)*c)) - 1/3*(2*b*c + 3*a*d)*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(a^2*sqrt(-c)*c^2)
Time = 15.21 (sec) , antiderivative size = 597, normalized size of antiderivative = 3.78 \[ \int \frac {1}{x^4 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {\ln \left (\frac {\left (\sqrt {d\,x^3+c}-\sqrt {c}\right )\,{\left (\sqrt {d\,x^3+c}+\sqrt {c}\right )}^3}{x^6}\right )\,\left (3\,a\,d+2\,b\,c\right )}{6\,a^2\,c^{5/2}}-\frac {\sqrt {d\,x^3+c}}{3\,a\,c^2\,x^3}-\frac {\frac {c\,\left (\frac {c\,\left (\frac {c\,\left (\frac {3\,a^2\,d^4+24\,a\,b\,c\,d^3+15\,b^2\,c^2\,d^2}{8\,a^3\,c^5}+\frac {c\,\left (\frac {c\,\left (\frac {3\,b^2\,d^4}{8\,a^3\,c^5}+\frac {b^2\,d^4\,\left (5\,a\,d-3\,b\,c\right )}{8\,a^3\,c^4\,\left (b\,c^2-a\,c\,d\right )}-\frac {b\,d^4\,\left (a\,d+2\,b\,c\right )\,\left (5\,a\,d-3\,b\,c\right )}{4\,a^3\,c^5\,\left (b\,c^2-a\,c\,d\right )}\right )}{d}-\frac {3\,b\,d^3\,\left (a\,d+2\,b\,c\right )}{4\,a^3\,c^5}+\frac {d\,\left (5\,a\,d-3\,b\,c\right )\,\left (3\,a^2\,d^4+24\,a\,b\,c\,d^3+15\,b^2\,c^2\,d^2\right )}{24\,a^3\,c^5\,\left (b\,c^2-a\,c\,d\right )}\right )}{d}-\frac {d^2\,\left (5\,a\,d-3\,b\,c\right )\,\left (6\,a^2\,d^2+14\,a\,b\,c\,d+3\,b^2\,c^2\right )}{12\,a^3\,c^4\,\left (b\,c^2-a\,c\,d\right )}\right )}{d}-\frac {d\,\left (6\,a^2\,d^2+14\,a\,b\,c\,d+3\,b^2\,c^2\right )}{4\,a^3\,c^4}+\frac {d^2\,\left (5\,a\,d-3\,b\,c\right )\,\left (13\,a\,d+18\,b\,c\right )}{24\,a^2\,c^3\,\left (b\,c^2-a\,c\,d\right )}\right )}{d}+\frac {d\,\left (13\,a\,d+18\,b\,c\right )}{8\,a^2\,c^3}-\frac {d\,\left (3\,a\,d+2\,b\,c\right )\,\left (5\,a\,d-3\,b\,c\right )}{6\,a^2\,c^2\,\left (b\,c^2-a\,c\,d\right )}\right )}{d}-\frac {3\,a\,d+2\,b\,c}{2\,a^2\,c^2}}{\sqrt {d\,x^3+c}}+\frac {b^{5/2}\,\ln \left (\frac {a\,d-2\,b\,c-b\,d\,x^3+\sqrt {b}\,\sqrt {d\,x^3+c}\,\sqrt {a\,d-b\,c}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,1{}\mathrm {i}}{3\,a^2\,{\left (a\,d-b\,c\right )}^{3/2}} \]
(log((((c + d*x^3)^(1/2) - c^(1/2))*((c + d*x^3)^(1/2) + c^(1/2))^3)/x^6)* (3*a*d + 2*b*c))/(6*a^2*c^(5/2)) - (c + d*x^3)^(1/2)/(3*a*c^2*x^3) - ((c*( (c*((c*((3*a^2*d^4 + 15*b^2*c^2*d^2 + 24*a*b*c*d^3)/(8*a^3*c^5) + (c*((c*( (3*b^2*d^4)/(8*a^3*c^5) + (b^2*d^4*(5*a*d - 3*b*c))/(8*a^3*c^4*(b*c^2 - a* c*d)) - (b*d^4*(a*d + 2*b*c)*(5*a*d - 3*b*c))/(4*a^3*c^5*(b*c^2 - a*c*d))) )/d - (3*b*d^3*(a*d + 2*b*c))/(4*a^3*c^5) + (d*(5*a*d - 3*b*c)*(3*a^2*d^4 + 15*b^2*c^2*d^2 + 24*a*b*c*d^3))/(24*a^3*c^5*(b*c^2 - a*c*d))))/d - (d^2* (5*a*d - 3*b*c)*(6*a^2*d^2 + 3*b^2*c^2 + 14*a*b*c*d))/(12*a^3*c^4*(b*c^2 - a*c*d))))/d - (d*(6*a^2*d^2 + 3*b^2*c^2 + 14*a*b*c*d))/(4*a^3*c^4) + (d^2 *(5*a*d - 3*b*c)*(13*a*d + 18*b*c))/(24*a^2*c^3*(b*c^2 - a*c*d))))/d + (d* (13*a*d + 18*b*c))/(8*a^2*c^3) - (d*(3*a*d + 2*b*c)*(5*a*d - 3*b*c))/(6*a^ 2*c^2*(b*c^2 - a*c*d))))/d - (3*a*d + 2*b*c)/(2*a^2*c^2))/(c + d*x^3)^(1/2 ) + (b^(5/2)*log((a*d - 2*b*c + b^(1/2)*(c + d*x^3)^(1/2)*(a*d - b*c)^(1/2 )*2i - b*d*x^3)/(a + b*x^3))*1i)/(3*a^2*(a*d - b*c)^(3/2))