Integrand size = 22, antiderivative size = 1865 \[ \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^3\right )^2} \, dx =\text {Too large to display} \] Output:
-(-1)^(1/3)*e*p*ln(e*x^2+d)/(1+(-1)^(1/3))^4/f/(e*f^(2/3)+(-1)^(2/3)*d*g^( 2/3))/g^(1/3)-2/9*p*polylog(2,e^(1/2)*(f^(1/3)+g^(1/3)*x)/(e^(1/2)*f^(1/3) +(-d)^(1/2)*g^(1/3)))/f^(5/3)/g^(1/3)-2/9*p*polylog(2,e^(1/2)*(f^(1/3)+g^( 1/3)*x)/(e^(1/2)*f^(1/3)-(-d)^(1/2)*g^(1/3)))/f^(5/3)/g^(1/3)+2*I*3^(1/2)* p*ln(-(-1)^(1/3)*g^(1/3)*((-d)^(1/2)-e^(1/2)*x)/(e^(1/2)*f^(1/3)-(-1)^(1/3 )*(-d)^(1/2)*g^(1/3)))*ln(-f^(1/3)+(-1)^(1/3)*g^(1/3)*x)/(1+(-1)^(1/3))^5/ f^(5/3)/g^(1/3)+2*I*3^(1/2)*p*ln((-1)^(1/3)*g^(1/3)*((-d)^(1/2)+e^(1/2)*x) /(e^(1/2)*f^(1/3)+(-1)^(1/3)*(-d)^(1/2)*g^(1/3)))*ln(-f^(1/3)+(-1)^(1/3)*g ^(1/3)*x)/(1+(-1)^(1/3))^5/f^(5/3)/g^(1/3)+2*(-1)^(2/3)*d^(1/2)*e^(1/2)*p* arctan(e^(1/2)*x/d^(1/2))/(1+(-1)^(1/3))^4/f^(4/3)/(e*f^(2/3)+(-1)^(2/3)*d *g^(2/3))+2*(-1)^(1/3)*e*p*ln(f^(1/3)-(-1)^(1/3)*g^(1/3)*x)/(1+(-1)^(1/3)) ^4/f/(e*f^(2/3)+(-1)^(2/3)*d*g^(2/3))/g^(1/3)-ln(c*(e*x^2+d)^p)/(1+(-1)^(1 /3))^4/f^(4/3)/g^(1/3)/((-1)^(2/3)*f^(1/3)+g^(1/3)*x)+2*I*3^(1/2)*p*polylo g(2,e^(1/2)*(f^(1/3)-(-1)^(1/3)*g^(1/3)*x)/(e^(1/2)*f^(1/3)+(-1)^(1/3)*(-d )^(1/2)*g^(1/3)))/(1+(-1)^(1/3))^5/f^(5/3)/g^(1/3)+2*I*3^(1/2)*p*polylog(2 ,e^(1/2)*(f^(1/3)-(-1)^(1/3)*g^(1/3)*x)/(e^(1/2)*f^(1/3)-(-1)^(1/3)*(-d)^( 1/2)*g^(1/3)))/(1+(-1)^(1/3))^5/f^(5/3)/g^(1/3)-2*I*3^(1/2)*ln(-f^(1/3)+(- 1)^(1/3)*g^(1/3)*x)*ln(c*(e*x^2+d)^p)/(1+(-1)^(1/3))^5/f^(5/3)/g^(1/3)-2/9 *(-1)^(1/3)*e*p*ln(e*x^2+d)/f/(2*e*f^(2/3)+I*(I-3^(1/2))*d*g^(2/3))/g^(1/3 )+2/9*d^(1/2)*e^(1/2)*p*arctan(e^(1/2)*x/d^(1/2))/f^(4/3)/(e*f^(2/3)+d*...
Time = 7.28 (sec) , antiderivative size = 2168, normalized size of antiderivative = 1.16 \[ \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^3\right )^2} \, dx=\text {Result too large to show} \] Input:
Integrate[Log[c*(d + e*x^2)^p]/(f + g*x^3)^2,x]
Output:
(x*(-(p*Log[d + e*x^2]) + Log[c*(d + e*x^2)^p]))/(3*f*(f + g*x^3)) + (2*Ar cTan[(-f^(1/3) + 2*g^(1/3)*x)/(Sqrt[3]*f^(1/3))]*(-(p*Log[d + e*x^2]) + Lo g[c*(d + e*x^2)^p]))/(3*Sqrt[3]*f^(5/3)*g^(1/3)) + (2*Log[f^(1/3) + g^(1/3 )*x]*(-(p*Log[d + e*x^2]) + Log[c*(d + e*x^2)^p]))/(9*f^(5/3)*g^(1/3)) - ( (-(p*Log[d + e*x^2]) + Log[c*(d + e*x^2)^p])*Log[f^(2/3) - f^(1/3)*g^(1/3) *x + g^(2/3)*x^2])/(9*f^(5/3)*g^(1/3)) + p*(-1/3*((-1 + (-1)^(1/3))*(-(Log [((-I)*Sqrt[d])/Sqrt[e] + x]/((-1)^(2/3)*f^(1/3) + g^(1/3)*x)) + (Sqrt[e]* (Log[I*Sqrt[d] - Sqrt[e]*x] - Log[-((-1)^(2/3)*f^(1/3)) - g^(1/3)*x]))/((- 1)^(2/3)*Sqrt[e]*f^(1/3) + I*Sqrt[d]*g^(1/3))))/((1 + (-1)^(1/3))^2*f^(4/3 )*g^(1/3)) - ((-1 + (-1)^(1/3))*(-(Log[(I*Sqrt[d])/Sqrt[e] + x]/((-1)^(2/3 )*f^(1/3) + g^(1/3)*x)) + (Sqrt[e]*(Log[I*Sqrt[d] + Sqrt[e]*x] - Log[-((-1 )^(2/3)*f^(1/3)) - g^(1/3)*x]))/((-1)^(2/3)*Sqrt[e]*f^(1/3) - I*Sqrt[d]*g^ (1/3))))/(3*(1 + (-1)^(1/3))^2*f^(4/3)*g^(1/3)) + ((-1)^(1/3)*(-(Log[((-I) *Sqrt[d])/Sqrt[e] + x]/(f^(1/3) + g^(1/3)*x)) + (Sqrt[e]*(Log[I*Sqrt[d] - Sqrt[e]*x] - Log[f^(1/3) + g^(1/3)*x]))/(Sqrt[e]*f^(1/3) + I*Sqrt[d]*g^(1/ 3))))/(3*(1 + (-1)^(1/3))^2*f^(4/3)*g^(1/3)) + ((-1)^(1/3)*(-(Log[(I*Sqrt[ d])/Sqrt[e] + x]/(f^(1/3) + g^(1/3)*x)) + (Sqrt[e]*(Log[I*Sqrt[d] + Sqrt[e ]*x] - Log[f^(1/3) + g^(1/3)*x]))/(Sqrt[e]*f^(1/3) - I*Sqrt[d]*g^(1/3))))/ (3*(1 + (-1)^(1/3))^2*f^(4/3)*g^(1/3)) - (Log[((-I)*Sqrt[d])/Sqrt[e] + x]/ ((-1)^(1/3)*f^(1/3) - g^(1/3)*x) + (Sqrt[e]*(-Log[I*Sqrt[d] - Sqrt[e]*x...
Time = 5.07 (sec) , antiderivative size = 1867, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2921, 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.
| \(\displaystyle \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 2921 |
| \(\displaystyle \int \left (\frac {2 \log \left (c \left (d+e x^2\right )^p\right )}{9 f^{5/3} \left (\sqrt [3]{f}+\sqrt [3]{g} x\right )}-\frac {2 (-1)^{5/6} \sqrt {3} \log \left (c \left (d+e x^2\right )^p\right )}{\left (1+\sqrt [3]{-1}\right )^5 f^{5/3} \left (\sqrt [3]{-1} \sqrt [3]{g} x-\sqrt [3]{f}\right )}+\frac {2 (-1)^{2/3} \log \left (c \left (d+e x^2\right )^p\right )}{\left (1+\sqrt [3]{-1}\right )^4 f^{5/3} \left (\sqrt [3]{f}+(-1)^{2/3} \sqrt [3]{g} x\right )}+\frac {\log \left (c \left (d+e x^2\right )^p\right )}{9 f^{4/3} \left (\sqrt [3]{f}+\sqrt [3]{g} x\right )^2}+\frac {(-1)^{2/3} \log \left (c \left (d+e x^2\right )^p\right )}{\left (1+\sqrt [3]{-1}\right )^4 f^{4/3} \left (\sqrt [3]{-1} \sqrt [3]{g} x-\sqrt [3]{f}\right )^2}+\frac {\log \left (c \left (d+e x^2\right )^p\right )}{\left (\sqrt [3]{-1}-1\right )^2 \left (1+\sqrt [3]{-1}\right )^4 f^{4/3} \left (\sqrt [3]{f}+(-1)^{2/3} \sqrt [3]{g} x\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 \sqrt {d} \sqrt {e} p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{9 f^{4/3} \left (g^{2/3} d+e f^{2/3}\right )}+\frac {2 (-1)^{2/3} \sqrt {d} \sqrt {e} p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\left (1+\sqrt [3]{-1}\right )^4 f^{4/3} \left ((-1)^{2/3} g^{2/3} d+e f^{2/3}\right )}+\frac {4 \sqrt {d} \sqrt {e} p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{9 f^{4/3} \left (2 e f^{2/3}-\left (1+i \sqrt {3}\right ) d g^{2/3}\right )}+\frac {2 \sqrt [3]{-1} e p \log \left (-\sqrt [3]{g} x-(-1)^{2/3} \sqrt [3]{f}\right )}{\left (1+\sqrt [3]{-1}\right )^4 f \left ((-1)^{2/3} g^{2/3} d+e f^{2/3}\right ) \sqrt [3]{g}}-\frac {2 e p \log \left (\sqrt [3]{g} x+\sqrt [3]{f}\right )}{9 f \left (g^{2/3} d+e f^{2/3}\right ) \sqrt [3]{g}}-\frac {2 p \log \left (\frac {\sqrt [3]{g} \left (\sqrt {-d}-\sqrt {e} x\right )}{\sqrt [3]{g} \sqrt {-d}+\sqrt {e} \sqrt [3]{f}}\right ) \log \left (\sqrt [3]{g} x+\sqrt [3]{f}\right )}{9 f^{5/3} \sqrt [3]{g}}-\frac {2 p \log \left (-\frac {\sqrt [3]{g} \left (\sqrt {e} x+\sqrt {-d}\right )}{\sqrt {e} \sqrt [3]{f}-\sqrt {-d} \sqrt [3]{g}}\right ) \log \left (\sqrt [3]{g} x+\sqrt [3]{f}\right )}{9 f^{5/3} \sqrt [3]{g}}+\frac {2 i \sqrt {3} p \log \left (-\frac {\sqrt [3]{-1} \sqrt [3]{g} \left (\sqrt {-d}-\sqrt {e} x\right )}{\sqrt {e} \sqrt [3]{f}-\sqrt [3]{-1} \sqrt {-d} \sqrt [3]{g}}\right ) \log \left (\sqrt [3]{-1} \sqrt [3]{g} x-\sqrt [3]{f}\right )}{\left (1+\sqrt [3]{-1}\right )^5 f^{5/3} \sqrt [3]{g}}+\frac {2 i \sqrt {3} p \log \left (\frac {\sqrt [3]{-1} \sqrt [3]{g} \left (\sqrt {e} x+\sqrt {-d}\right )}{\sqrt [3]{-1} \sqrt [3]{g} \sqrt {-d}+\sqrt {e} \sqrt [3]{f}}\right ) \log \left (\sqrt [3]{-1} \sqrt [3]{g} x-\sqrt [3]{f}\right )}{\left (1+\sqrt [3]{-1}\right )^5 f^{5/3} \sqrt [3]{g}}+\frac {4 \sqrt [3]{-1} e p \log \left ((-1)^{2/3} \sqrt [3]{g} x+\sqrt [3]{f}\right )}{9 f \left (i \left (i-\sqrt {3}\right ) g^{2/3} d+2 e f^{2/3}\right ) \sqrt [3]{g}}-\frac {2 p \log \left (\frac {(-1)^{2/3} \sqrt [3]{g} \left (\sqrt {-d}-\sqrt {e} x\right )}{(-1)^{2/3} \sqrt [3]{g} \sqrt {-d}+\sqrt {e} \sqrt [3]{f}}\right ) \log \left ((-1)^{2/3} \sqrt [3]{g} x+\sqrt [3]{f}\right )}{\left (1+\sqrt [3]{-1}\right )^4 f^{5/3} \sqrt [3]{g}}-\frac {2 p \log \left (-\frac {(-1)^{2/3} \sqrt [3]{g} \left (\sqrt {e} x+\sqrt {-d}\right )}{\sqrt {e} \sqrt [3]{f}-(-1)^{2/3} \sqrt {-d} \sqrt [3]{g}}\right ) \log \left ((-1)^{2/3} \sqrt [3]{g} x+\sqrt [3]{f}\right )}{\left (1+\sqrt [3]{-1}\right )^4 f^{5/3} \sqrt [3]{g}}+\frac {e p \log \left (e x^2+d\right )}{9 f \left (g^{2/3} d+e f^{2/3}\right ) \sqrt [3]{g}}-\frac {\sqrt [3]{-1} e p \log \left (e x^2+d\right )}{\left (1+\sqrt [3]{-1}\right )^4 f \left ((-1)^{2/3} g^{2/3} d+e f^{2/3}\right ) \sqrt [3]{g}}-\frac {2 \sqrt [3]{-1} e p \log \left (e x^2+d\right )}{9 f \left (i \left (i-\sqrt {3}\right ) g^{2/3} d+2 e f^{2/3}\right ) \sqrt [3]{g}}+\frac {2 \log \left (\sqrt [3]{g} x+\sqrt [3]{f}\right ) \log \left (c \left (e x^2+d\right )^p\right )}{9 f^{5/3} \sqrt [3]{g}}-\frac {2 i \sqrt {3} \log \left (\sqrt [3]{-1} \sqrt [3]{g} x-\sqrt [3]{f}\right ) \log \left (c \left (e x^2+d\right )^p\right )}{\left (1+\sqrt [3]{-1}\right )^5 f^{5/3} \sqrt [3]{g}}+\frac {2 \log \left ((-1)^{2/3} \sqrt [3]{g} x+\sqrt [3]{f}\right ) \log \left (c \left (e x^2+d\right )^p\right )}{\left (1+\sqrt [3]{-1}\right )^4 f^{5/3} \sqrt [3]{g}}-\frac {\log \left (c \left (e x^2+d\right )^p\right )}{9 f^{4/3} \sqrt [3]{g} \left (\sqrt [3]{g} x+\sqrt [3]{f}\right )}-\frac {\log \left (c \left (e x^2+d\right )^p\right )}{\left (1+\sqrt [3]{-1}\right )^4 f^{4/3} \sqrt [3]{g} \left (\sqrt [3]{g} x+(-1)^{2/3} \sqrt [3]{f}\right )}+\frac {\sqrt [3]{-1} \log \left (c \left (e x^2+d\right )^p\right )}{9 f^{4/3} \sqrt [3]{g} \left ((-1)^{2/3} \sqrt [3]{g} x+\sqrt [3]{f}\right )}-\frac {2 p \operatorname {PolyLog}\left (2,\frac {\sqrt {e} \left (\sqrt [3]{g} x+\sqrt [3]{f}\right )}{\sqrt {e} \sqrt [3]{f}-\sqrt {-d} \sqrt [3]{g}}\right )}{9 f^{5/3} \sqrt [3]{g}}-\frac {2 p \operatorname {PolyLog}\left (2,\frac {\sqrt {e} \left (\sqrt [3]{g} x+\sqrt [3]{f}\right )}{\sqrt [3]{g} \sqrt {-d}+\sqrt {e} \sqrt [3]{f}}\right )}{9 f^{5/3} \sqrt [3]{g}}+\frac {2 i \sqrt {3} p \operatorname {PolyLog}\left (2,\frac {\sqrt {e} \left (\sqrt [3]{f}-\sqrt [3]{-1} \sqrt [3]{g} x\right )}{\sqrt {e} \sqrt [3]{f}-\sqrt [3]{-1} \sqrt {-d} \sqrt [3]{g}}\right )}{\left (1+\sqrt [3]{-1}\right )^5 f^{5/3} \sqrt [3]{g}}+\frac {2 i \sqrt {3} p \operatorname {PolyLog}\left (2,\frac {\sqrt {e} \left (\sqrt [3]{f}-\sqrt [3]{-1} \sqrt [3]{g} x\right )}{\sqrt [3]{-1} \sqrt [3]{g} \sqrt {-d}+\sqrt {e} \sqrt [3]{f}}\right )}{\left (1+\sqrt [3]{-1}\right )^5 f^{5/3} \sqrt [3]{g}}-\frac {2 p \operatorname {PolyLog}\left (2,\frac {\sqrt {e} \left ((-1)^{2/3} \sqrt [3]{g} x+\sqrt [3]{f}\right )}{\sqrt {e} \sqrt [3]{f}-(-1)^{2/3} \sqrt {-d} \sqrt [3]{g}}\right )}{\left (1+\sqrt [3]{-1}\right )^4 f^{5/3} \sqrt [3]{g}}-\frac {2 p \operatorname {PolyLog}\left (2,\frac {\sqrt {e} \left ((-1)^{2/3} \sqrt [3]{g} x+\sqrt [3]{f}\right )}{(-1)^{2/3} \sqrt [3]{g} \sqrt {-d}+\sqrt {e} \sqrt [3]{f}}\right )}{\left (1+\sqrt [3]{-1}\right )^4 f^{5/3} \sqrt [3]{g}}\) |
Input:
Int[Log[c*(d + e*x^2)^p]/(f + g*x^3)^2,x]
Output:
(2*Sqrt[d]*Sqrt[e]*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(9*f^(4/3)*(e*f^(2/3) + d*g^(2/3))) + (2*(-1)^(2/3)*Sqrt[d]*Sqrt[e]*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]]) /((1 + (-1)^(1/3))^4*f^(4/3)*(e*f^(2/3) + (-1)^(2/3)*d*g^(2/3))) + (4*Sqrt [d]*Sqrt[e]*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(9*f^(4/3)*(2*e*f^(2/3) - (1 + I*Sqrt[3])*d*g^(2/3))) + (2*(-1)^(1/3)*e*p*Log[-((-1)^(2/3)*f^(1/3)) - g^( 1/3)*x])/((1 + (-1)^(1/3))^4*f*(e*f^(2/3) + (-1)^(2/3)*d*g^(2/3))*g^(1/3)) - (2*e*p*Log[f^(1/3) + g^(1/3)*x])/(9*f*(e*f^(2/3) + d*g^(2/3))*g^(1/3)) - (2*p*Log[(g^(1/3)*(Sqrt[-d] - Sqrt[e]*x))/(Sqrt[e]*f^(1/3) + Sqrt[-d]*g^ (1/3))]*Log[f^(1/3) + g^(1/3)*x])/(9*f^(5/3)*g^(1/3)) - (2*p*Log[-((g^(1/3 )*(Sqrt[-d] + Sqrt[e]*x))/(Sqrt[e]*f^(1/3) - Sqrt[-d]*g^(1/3)))]*Log[f^(1/ 3) + g^(1/3)*x])/(9*f^(5/3)*g^(1/3)) + ((2*I)*Sqrt[3]*p*Log[-(((-1)^(1/3)* g^(1/3)*(Sqrt[-d] - Sqrt[e]*x))/(Sqrt[e]*f^(1/3) - (-1)^(1/3)*Sqrt[-d]*g^( 1/3)))]*Log[-f^(1/3) + (-1)^(1/3)*g^(1/3)*x])/((1 + (-1)^(1/3))^5*f^(5/3)* g^(1/3)) + ((2*I)*Sqrt[3]*p*Log[((-1)^(1/3)*g^(1/3)*(Sqrt[-d] + Sqrt[e]*x) )/(Sqrt[e]*f^(1/3) + (-1)^(1/3)*Sqrt[-d]*g^(1/3))]*Log[-f^(1/3) + (-1)^(1/ 3)*g^(1/3)*x])/((1 + (-1)^(1/3))^5*f^(5/3)*g^(1/3)) + (4*(-1)^(1/3)*e*p*Lo g[f^(1/3) + (-1)^(2/3)*g^(1/3)*x])/(9*f*(2*e*f^(2/3) + I*(I - Sqrt[3])*d*g ^(2/3))*g^(1/3)) - (2*p*Log[((-1)^(2/3)*g^(1/3)*(Sqrt[-d] - Sqrt[e]*x))/(S qrt[e]*f^(1/3) + (-1)^(2/3)*Sqrt[-d]*g^(1/3))]*Log[f^(1/3) + (-1)^(2/3)*g^ (1/3)*x])/((1 + (-1)^(1/3))^4*f^(5/3)*g^(1/3)) - (2*p*Log[-(((-1)^(2/3)...
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*((f_) + (g_.)*(x_)^(s_))^(r_.), x_Symbol] :> With[{t = ExpandIntegrand[(a + b*Log[ c*(d + e*x^n)^p])^q, (f + g*x^s)^r, x]}, Int[t, x] /; SumQ[t]] /; FreeQ[{a, b, c, d, e, f, g, n, p, q, r, s}, x] && IntegerQ[n] && IGtQ[q, 0] && Integ erQ[r] && IntegerQ[s] && (EqQ[q, 1] || (GtQ[r, 0] && GtQ[s, 1]) || (LtQ[s, 0] && LtQ[r, 0]))
\[\int \frac {\ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{\left (g \,x^{3}+f \right )^{2}}d x\]
Input:
int(ln(c*(e*x^2+d)^p)/(g*x^3+f)^2,x)
Output:
int(ln(c*(e*x^2+d)^p)/(g*x^3+f)^2,x)
\[ \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^3\right )^2} \, dx=\int { \frac {\log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{{\left (g x^{3} + f\right )}^{2}} \,d x } \] Input:
integrate(log(c*(e*x^2+d)^p)/(g*x^3+f)^2,x, algorithm="fricas")
Output:
integral(log((e*x^2 + d)^p*c)/(g^2*x^6 + 2*f*g*x^3 + f^2), x)
Timed out. \[ \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^3\right )^2} \, dx=\text {Timed out} \] Input:
integrate(ln(c*(e*x**2+d)**p)/(g*x**3+f)**2,x)
Output:
Timed out
Exception generated. \[ \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^3\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(log(c*(e*x^2+d)^p)/(g*x^3+f)^2,x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
\[ \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^3\right )^2} \, dx=\int { \frac {\log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{{\left (g x^{3} + f\right )}^{2}} \,d x } \] Input:
integrate(log(c*(e*x^2+d)^p)/(g*x^3+f)^2,x, algorithm="giac")
Output:
integrate(log((e*x^2 + d)^p*c)/(g*x^3 + f)^2, x)
Timed out. \[ \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^3\right )^2} \, dx=\int \frac {\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )}{{\left (g\,x^3+f\right )}^2} \,d x \] Input:
int(log(c*(d + e*x^2)^p)/(f + g*x^3)^2,x)
Output:
int(log(c*(d + e*x^2)^p)/(f + g*x^3)^2, x)
\[ \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^3\right )^2} \, dx =\text {Too large to display} \] Input:
int(log(c*(e*x^2+d)^p)/(g*x^3+f)^2,x)
Output:
(2*g**(1/3)*f**(2/3)*sqrt(3)*atan((f**(1/3) - 2*g**(1/3)*x)/(f**(1/3)*sqrt (3)))*e**3*f**2*p + 2*g**(1/3)*f**(2/3)*sqrt(3)*atan((f**(1/3) - 2*g**(1/3 )*x)/(f**(1/3)*sqrt(3)))*e**3*f*g*p*x**3 + 2*sqrt(3)*atan((f**(1/3) - 2*g* *(1/3)*x)/(f**(1/3)*sqrt(3)))*d*e**2*f**2*g*p + 2*sqrt(3)*atan((f**(1/3) - 2*g**(1/3)*x)/(f**(1/3)*sqrt(3)))*d*e**2*f*g**2*p*x**3 + 6*g**(2/3)*f**(1 /3)*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*e**2*f**2*p + 6*g**(2/3) *f**(1/3)*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*e**2*f*g*p*x**3 + g**(1/3)*f**(2/3)*log(f**(2/3) - g**(1/3)*f**(1/3)*x + g**(2/3)*x**2)*e**3 *f**2*p + g**(1/3)*f**(2/3)*log(f**(2/3) - g**(1/3)*f**(1/3)*x + g**(2/3)* x**2)*e**3*f*g*p*x**3 - 2*g**(1/3)*f**(2/3)*log(f**(1/3) + g**(1/3)*x)*e** 3*f**2*p - 2*g**(1/3)*f**(2/3)*log(f**(1/3) + g**(1/3)*x)*e**3*f*g*p*x**3 + 6*g**(2/3)*f**(1/3)*int((log((d + e*x**2)**p*c)*x**3)/(f**2 + 2*f*g*x**3 + g**2*x**6),x)*d**3*f*g**3 + 6*g**(2/3)*f**(1/3)*int((log((d + e*x**2)** p*c)*x**3)/(f**2 + 2*f*g*x**3 + g**2*x**6),x)*d**3*g**4*x**3 + 6*g**(2/3)* f**(1/3)*int((log((d + e*x**2)**p*c)*x**3)/(f**2 + 2*f*g*x**3 + g**2*x**6) ,x)*e**3*f**3*g + 6*g**(2/3)*f**(1/3)*int((log((d + e*x**2)**p*c)*x**3)/(f **2 + 2*f*g*x**3 + g**2*x**6),x)*e**3*f**2*g**2*x**3 - 2*g**(2/3)*f**(1/3) *log(f**(2/3) - g**(1/3)*f**(1/3)*x + g**(2/3)*x**2)*d**2*e*f*g*p - 2*g**( 2/3)*f**(1/3)*log(f**(2/3) - g**(1/3)*f**(1/3)*x + g**(2/3)*x**2)*d**2*e*g **2*p*x**3 - 2*g**(2/3)*f**(1/3)*log(f**(1/3) + g**(1/3)*x)*d**2*e*f*g*...