Integrand size = 29, antiderivative size = 93 \[ \int \frac {1}{x \left (a c+b c x^3+d \sqrt {a+b x^3}\right )} \, dx=\frac {2 d \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{3 \sqrt {a} \left (a c^2-d^2\right )}+\frac {c \log (x)}{a c^2-d^2}-\frac {2 c \log \left (d+c \sqrt {a+b x^3}\right )}{3 \left (a c^2-d^2\right )} \]
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*arctan h((b*x^3+a)^(1/2)/a^(1/2))/(a*c^2-d^2)/a^(1/2)
Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.74 \[ \int \frac {1}{x \left (a c+b c x^3+d \sqrt {a+b x^3}\right )} \, dx=\frac {\frac {2 d \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{\sqrt {a}}+c \log \left (b x^3\right )-2 c \log \left (d+c \sqrt {a+b x^3}\right )}{3 a c^2-3 d^2} \]
((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)
Time = 0.44 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.97, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2586, 7267, 25, 479, 452, 219, 240}
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.
\(\displaystyle \int \frac {1}{x \left (d \sqrt {a+b x^3}+a c+b c x^3\right )} \, dx\) |
\(\Big \downarrow \) 2586 |
\(\displaystyle \frac {1}{3} \int \frac {1}{x^3 \left (b c x^3+a c+d \sqrt {b x^3+a}\right )}dx^3\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle \frac {2}{3} \int -\frac {1}{\left (a-x^6\right ) \left (\sqrt {b x^3+a} c+d\right )}d\sqrt {b x^3+a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {2}{3} \int \frac {1}{\left (a-x^6\right ) \left (\sqrt {b x^3+a} c+d\right )}d\sqrt {b x^3+a}\) |
\(\Big \downarrow \) 479 |
\(\displaystyle \frac {2}{3} \left (\frac {\int \frac {d-c \sqrt {b x^3+a}}{a-x^6}d\sqrt {b x^3+a}}{a c^2-d^2}-\frac {c \log \left (c \sqrt {a+b x^3}+d\right )}{a c^2-d^2}\right )\) |
\(\Big \downarrow \) 452 |
\(\displaystyle \frac {2}{3} \left (\frac {d \int \frac {1}{a-x^6}d\sqrt {b x^3+a}-c \int \frac {\sqrt {b x^3+a}}{a-x^6}d\sqrt {b x^3+a}}{a c^2-d^2}-\frac {c \log \left (c \sqrt {a+b x^3}+d\right )}{a c^2-d^2}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {2}{3} \left (\frac {\frac {d \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{\sqrt {a}}-c \int \frac {\sqrt {b x^3+a}}{a-x^6}d\sqrt {b x^3+a}}{a c^2-d^2}-\frac {c \log \left (c \sqrt {a+b x^3}+d\right )}{a c^2-d^2}\right )\) |
\(\Big \downarrow \) 240 |
\(\displaystyle \frac {2}{3} \left (\frac {\frac {d \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {1}{2} c \log \left (a-x^6\right )}{a c^2-d^2}-\frac {c \log \left (c \sqrt {a+b x^3}+d\right )}{a c^2-d^2}\right )\) |
(2*(((d*ArcTanh[Sqrt[a + b*x^3]/Sqrt[a]])/Sqrt[a] + (c*Log[a - x^6])/2)/(a *c^2 - d^2) - (c*Log[d + c*Sqrt[a + b*x^3]])/(a*c^2 - d^2)))/3
3.6.55.3.1 Defintions of rubi rules used
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x ^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
Int[((c_) + (d_.)*(x_))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[c Int[1/ (a + b*x^2), x], x] + Simp[d Int[x/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c^2 + a*d^2, 0]
Int[1/(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)), x_Symbol] :> Simp[d*(Log [RemoveContent[c + d*x, x]]/(b*c^2 + a*d^2)), x] + Simp[b/(b*c^2 + a*d^2) Int[(c - d*x)/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d}, x]
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. \(222\) vs. \(2(81)=162\).
Time = 1.11 (sec) , antiderivative size = 223, normalized size of antiderivative = 2.40
method | result | size |
default | \(\frac {c \ln \left (x \right )}{a \,c^{2}-d^{2}}-\frac {a \,c^{3} \ln \left (b \,c^{2} x^{3}+a \,c^{2}-d^{2}\right )}{3 \left (a \,c^{2}-d^{2}\right ) d^{2}}+\frac {c \ln \left (b \,c^{2} x^{3}+a \,c^{2}-d^{2}\right )}{3 d^{2}}-d \left (\frac {\frac {2 \sqrt {b \,x^{3}+a}}{3}-\frac {2 \sqrt {a}\, \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{3}}{a \left (a \,c^{2}-d^{2}\right )}+\frac {2 \sqrt {b \,x^{3}+a}}{3 a \,d^{2}}-\frac {c \left (d \ln \left (c \sqrt {b \,x^{3}+a}-d \right )-d \ln \left (d +c \sqrt {b \,x^{3}+a}\right )+2 c \sqrt {b \,x^{3}+a}\right )}{3 \left (a \,c^{2}-d^{2}\right ) d^{2}}\right )\) | \(223\) |
elliptic | \(\text {Expression too large to display}\) | \(1768\) |
c*ln(x)/(a*c^2-d^2)-1/3*a*c^3/(a*c^2-d^2)/d^2*ln(b*c^2*x^3+a*c^2-d^2)+1/3* c/d^2*ln(b*c^2*x^3+a*c^2-d^2)-d*(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)))+2/3/a/d^2*(b*x^3+a)^(1/2)-1/3*c/ (a*c^2-d^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 = 0.30 (sec) , antiderivative size = 232, normalized size of antiderivative = 2.49 \[ \int \frac {1}{x \left (a c+b c x^3+d \sqrt {a+b x^3}\right )} \, dx=\left [-\frac {a c \log \left (b c^{2} x^{3} + a c^{2} - d^{2}\right ) + a c \log \left (\sqrt {b x^{3} + a} c + d\right ) - a c \log \left (\sqrt {b x^{3} + a} c - d\right ) - 3 \, a c \log \left (x\right ) - \sqrt {a} d \log \left (\frac {b x^{3} + 2 \, \sqrt {b x^{3} + a} \sqrt {a} + 2 \, a}{x^{3}}\right )}{3 \, {\left (a^{2} c^{2} - a d^{2}\right )}}, -\frac {a c \log \left (b c^{2} x^{3} + a c^{2} - d^{2}\right ) + a c \log \left (\sqrt {b x^{3} + a} c + d\right ) - a c \log \left (\sqrt {b x^{3} + a} c - d\right ) - 3 \, a c \log \left (x\right ) + 2 \, \sqrt {-a} d \arctan \left (\frac {\sqrt {b x^{3} + a} \sqrt {-a}}{a}\right )}{3 \, {\left (a^{2} c^{2} - a d^{2}\right )}}\right ] \]
[-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*sqrt(-a)*d*arctan(sqrt(b*x^3 + a)*sqrt(-a) /a))/(a^2*c^2 - a*d^2)]
Time = 2.97 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.61 \[ \int \frac {1}{x \left (a c+b c x^3+d \sqrt {a+b x^3}\right )} \, dx=\begin {cases} \frac {2 \left (- \frac {b c^{2} \left (\begin {cases} \frac {\sqrt {a + b x^{3}}}{d} & \text {for}\: c = 0 \\\frac {\log {\left (c \sqrt {a + b x^{3}} + d \right )}}{c} & \text {otherwise} \end {cases}\right )}{3 \left (a c^{2} - d^{2}\right )} - \frac {b \left (- \frac {c \log {\left (- b x^{3} \right )}}{2} + \frac {d \operatorname {atan}{\left (\frac {\sqrt {a + b x^{3}}}{\sqrt {- a}} \right )}}{\sqrt {- a}}\right )}{3 \left (a c^{2} - d^{2}\right )}\right )}{b} & \text {for}\: b \neq 0 \\\begin {cases} \frac {x^{3} \log {\left (x^{3} \right )}}{3 \sqrt {a} d x^{3} + 3 a c x^{3}} & \text {for}\: 3 \sqrt {a} d + 3 a c \neq 0 \\\tilde {\infty } x^{3} & \text {otherwise} \end {cases} & \text {otherwise} \end {cases} \]
Piecewise((2*(-b*c**2*Piecewise((sqrt(a + b*x**3)/d, Eq(c, 0)), (log(c*sqr t(a + b*x**3) + d)/c, True))/(3*(a*c**2 - d**2)) - b*(-c*log(-b*x**3)/2 + d*atan(sqrt(a + b*x**3)/sqrt(-a))/sqrt(-a))/(3*(a*c**2 - d**2)))/b, Ne(b, 0)), (Piecewise((x**3*log(x**3)/(3*sqrt(a)*d*x**3 + 3*a*c*x**3), Ne(3*sqrt (a)*d + 3*a*c, 0)), (zoo*x**3, True)), True))
\[ \int \frac {1}{x \left (a c+b c x^3+d \sqrt {a+b x^3}\right )} \, dx=\int { \frac {1}{{\left (b c x^{3} + a c + \sqrt {b x^{3} + a} d\right )} x} \,d x } \]
Time = 0.38 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.01 \[ \int \frac {1}{x \left (a c+b c x^3+d \sqrt {a+b x^3}\right )} \, dx=-\frac {2 \, c^{2} \log \left ({\left | \sqrt {b x^{3} + a} c + d \right |}\right )}{3 \, {\left (a c^{3} - c d^{2}\right )}} + \frac {c \log \left (b x^{3}\right )}{3 \, {\left (a c^{2} - d^{2}\right )}} - \frac {2 \, d \arctan \left (\frac {\sqrt {b x^{3} + a}}{\sqrt {-a}}\right )}{3 \, {\left (a c^{2} - d^{2}\right )} \sqrt {-a}} \]
-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)*sq rt(-a))
Time = 17.76 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.68 \[ \int \frac {1}{x \left (a c+b c x^3+d \sqrt {a+b x^3}\right )} \, dx=\frac {c\,\ln \left (x\right )}{a\,c^2-d^2}+\frac {c\,\ln \left (\frac {d-c\,\sqrt {b\,x^3+a}}{d+c\,\sqrt {b\,x^3+a}}\right )}{3\,\left (a\,c^2-d^2\right )}-\frac {c\,\ln \left (b\,c^2\,x^3+a\,c^2-d^2\right )}{3\,a\,c^2-3\,d^2}+\frac {d\,\ln \left (\frac {\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )\,{\left (\sqrt {b\,x^3+a}+\sqrt {a}\right )}^3}{x^6}\right )}{3\,\sqrt {a}\,\left (a\,c^2-d^2\right )} \]