Integrand size = 20, antiderivative size = 292 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{(d+e x)^2} \, dx=-\frac {\sqrt {3} \sqrt [3]{a} \sqrt [3]{b} p \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{b^{2/3} d^2+\sqrt [3]{a} \sqrt [3]{b} d e+a^{2/3} e^2}+\frac {\sqrt [3]{a} \sqrt [3]{b} \left (\sqrt [3]{b} d+\sqrt [3]{a} e\right ) p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{b d^3-a e^3}-\frac {3 b d^2 p \log (d+e x)}{e \left (b d^3-a e^3\right )}-\frac {\sqrt [3]{a} \sqrt [3]{b} \left (\sqrt [3]{b} d+\sqrt [3]{a} e\right ) p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{2 \left (b d^3-a e^3\right )}+\frac {b d^2 p \log \left (a+b x^3\right )}{e \left (b d^3-a e^3\right )}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{e (d+e x)} \] Output:
-3^(1/2)*a^(1/3)*b^(1/3)*p*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)*3^(1/2)/a^(1/3 ))/(b^(2/3)*d^2+a^(1/3)*b^(1/3)*d*e+a^(2/3)*e^2)+a^(1/3)*b^(1/3)*(b^(1/3)* d+a^(1/3)*e)*p*ln(a^(1/3)+b^(1/3)*x)/(-a*e^3+b*d^3)-3*b*d^2*p*ln(e*x+d)/e/ (-a*e^3+b*d^3)-a^(1/3)*b^(1/3)*(b^(1/3)*d+a^(1/3)*e)*p*ln(a^(2/3)-a^(1/3)* b^(1/3)*x+b^(2/3)*x^2)/(-2*a*e^3+2*b*d^3)+b*d^2*p*ln(b*x^3+a)/e/(-a*e^3+b* d^3)-ln(c*(b*x^3+a)^p)/e/(e*x+d)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.35 (sec) , antiderivative size = 254, normalized size of antiderivative = 0.87 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{(d+e x)^2} \, dx=-\frac {2 \sqrt {3} \sqrt [3]{a} b^{2/3} d e p (d+e x) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )+3 b e^2 p x^2 (d+e x) \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},-\frac {b x^3}{a}\right )-2 \sqrt [3]{a} b^{2/3} d e p (d+e x) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )+6 b d^2 p (d+e x) \log (d+e x)+\sqrt [3]{a} b^{2/3} d e p (d+e x) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )-2 b d^2 p (d+e x) \log \left (a+b x^3\right )+2 \left (b d^3-a e^3\right ) \log \left (c \left (a+b x^3\right )^p\right )}{2 e \left (b d^3-a e^3\right ) (d+e x)} \] Input:
Integrate[Log[c*(a + b*x^3)^p]/(d + e*x)^2,x]
Output:
-1/2*(2*Sqrt[3]*a^(1/3)*b^(2/3)*d*e*p*(d + e*x)*ArcTan[(1 - (2*b^(1/3)*x)/ a^(1/3))/Sqrt[3]] + 3*b*e^2*p*x^2*(d + e*x)*Hypergeometric2F1[2/3, 1, 5/3, -((b*x^3)/a)] - 2*a^(1/3)*b^(2/3)*d*e*p*(d + e*x)*Log[a^(1/3) + b^(1/3)*x ] + 6*b*d^2*p*(d + e*x)*Log[d + e*x] + a^(1/3)*b^(2/3)*d*e*p*(d + e*x)*Log [a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2] - 2*b*d^2*p*(d + e*x)*Log[a + b*x^3] + 2*(b*d^3 - a*e^3)*Log[c*(a + b*x^3)^p])/(e*(b*d^3 - a*e^3)*(d + e *x))
Time = 1.28 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.01, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2913, 7276, 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.
| \(\displaystyle \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{(d+e x)^2} \, dx\) |
| \(\Big \downarrow \) 2913 |
| \(\displaystyle \frac {3 b p \int \frac {x^2}{(d+e x) \left (b x^3+a\right )}dx}{e}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{e (d+e x)}\) |
| \(\Big \downarrow \) 7276 |
| \(\displaystyle \frac {3 b p \int \left (\frac {-a x e^2+a d e+b d^2 x^2}{\left (b d^3-a e^3\right ) \left (b x^3+a\right )}-\frac {d^2 e}{\left (b d^3-a e^3\right ) (d+e x)}\right )dx}{e}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{e (d+e x)}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {3 b p \left (-\frac {\sqrt [3]{a} e \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{2/3} \left (a^{2/3} e^2+\sqrt [3]{a} \sqrt [3]{b} d e+b^{2/3} d^2\right )}-\frac {\sqrt [3]{a} e \left (\sqrt [3]{a} e+\sqrt [3]{b} d\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{2/3} \left (b d^3-a e^3\right )}+\frac {\sqrt [3]{a} e \left (\sqrt [3]{a} e+\sqrt [3]{b} d\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{2/3} \left (b d^3-a e^3\right )}+\frac {d^2 \log \left (a+b x^3\right )}{3 \left (b d^3-a e^3\right )}-\frac {d^2 \log (d+e x)}{b d^3-a e^3}\right )}{e}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{e (d+e x)}\) |
Input:
Int[Log[c*(a + b*x^3)^p]/(d + e*x)^2,x]
Output:
(3*b*p*(-((a^(1/3)*e*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(S qrt[3]*b^(2/3)*(b^(2/3)*d^2 + a^(1/3)*b^(1/3)*d*e + a^(2/3)*e^2))) + (a^(1 /3)*e*(b^(1/3)*d + a^(1/3)*e)*Log[a^(1/3) + b^(1/3)*x])/(3*b^(2/3)*(b*d^3 - a*e^3)) - (d^2*Log[d + e*x])/(b*d^3 - a*e^3) - (a^(1/3)*e*(b^(1/3)*d + a ^(1/3)*e)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(6*b^(2/3)*(b*d^ 3 - a*e^3)) + (d^2*Log[a + b*x^3])/(3*(b*d^3 - a*e^3))))/e - Log[c*(a + b* x^3)^p]/(e*(d + e*x))
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.) + (g_. )*(x_))^(r_.), x_Symbol] :> Simp[(f + g*x)^(r + 1)*((a + b*Log[c*(d + e*x^n )^p])/(g*(r + 1))), x] - Simp[b*e*n*(p/(g*(r + 1))) Int[x^(n - 1)*((f + g *x)^(r + 1)/(d + e*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, r}, x ] && (IGtQ[r, 0] || RationalQ[n]) && NeQ[r, -1]
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 2.21 (sec) , antiderivative size = 277, normalized size of antiderivative = 0.95
| method | result | size |
| parts | \(-\frac {\ln \left (c \left (b \,x^{3}+a \right )^{p}\right )}{e \left (e x +d \right )}+\frac {3 p b \left (\frac {d^{2} \ln \left (e x +d \right )}{a \,e^{3}-b \,d^{3}}+\frac {-d e a \left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )+a \,e^{2} \left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )-\frac {d^{2} \ln \left (b \,x^{3}+a \right )}{3}}{a \,e^{3}-b \,d^{3}}\right )}{e}\) | \(277\) |
| risch | \(\text {Expression too large to display}\) | \(1068\) |
Input:
int(ln(c*(b*x^3+a)^p)/(e*x+d)^2,x,method=_RETURNVERBOSE)
Output:
-ln(c*(b*x^3+a)^p)/e/(e*x+d)+3*p*b/e*(d^2/(a*e^3-b*d^3)*ln(e*x+d)+(-d*e*a* (1/3/b/(1/b*a)^(2/3)*ln(x+(1/b*a)^(1/3))-1/6/b/(1/b*a)^(2/3)*ln(x^2-(1/b*a )^(1/3)*x+(1/b*a)^(2/3))+1/3/b/(1/b*a)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2 /(1/b*a)^(1/3)*x-1)))+a*e^2*(-1/3/b/(1/b*a)^(1/3)*ln(x+(1/b*a)^(1/3))+1/6/ b/(1/b*a)^(1/3)*ln(x^2-(1/b*a)^(1/3)*x+(1/b*a)^(2/3))+1/3*3^(1/2)/b/(1/b*a )^(1/3)*arctan(1/3*3^(1/2)*(2/(1/b*a)^(1/3)*x-1)))-1/3*d^2*ln(b*x^3+a))/(a *e^3-b*d^3))
Result contains complex when optimal does not.
Time = 1.32 (sec) , antiderivative size = 7010, normalized size of antiderivative = 24.01 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{(d+e x)^2} \, dx=\text {Too large to display} \] Input:
integrate(log(c*(b*x^3+a)^p)/(e*x+d)^2,x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{(d+e x)^2} \, dx=\text {Timed out} \] Input:
integrate(ln(c*(b*x**3+a)**p)/(e*x+d)**2,x)
Output:
Timed out
Time = 0.11 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.07 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{(d+e x)^2} \, dx=-\frac {{\left (\frac {6 \, d^{2} \log \left (e x + d\right )}{b d^{3} - a e^{3}} + \frac {2 \, \sqrt {3} {\left (a e^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}} - a d e \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{{\left (b^{2} d^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}} - a b e^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )} \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {{\left (2 \, b d^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}} - a e^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}} - a d e\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{b^{2} d^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}} - a b e^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {2 \, {\left (b d^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}} + a e^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}} + a d e\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{b^{2} d^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}} - a b e^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )} b p}{2 \, e} - \frac {\log \left ({\left (b x^{3} + a\right )}^{p} c\right )}{{\left (e x + d\right )} e} \] Input:
integrate(log(c*(b*x^3+a)^p)/(e*x+d)^2,x, algorithm="maxima")
Output:
-1/2*(6*d^2*log(e*x + d)/(b*d^3 - a*e^3) + 2*sqrt(3)*(a*e^2*(a/b)^(2/3) - a*d*e*(a/b)^(1/3))*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/((b ^2*d^3*(a/b)^(2/3) - a*b*e^3*(a/b)^(2/3))*(a/b)^(1/3)) - (2*b*d^2*(a/b)^(2 /3) - a*e^2*(a/b)^(1/3) - a*d*e)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(b ^2*d^3*(a/b)^(2/3) - a*b*e^3*(a/b)^(2/3)) - 2*(b*d^2*(a/b)^(2/3) + a*e^2*( a/b)^(1/3) + a*d*e)*log(x + (a/b)^(1/3))/(b^2*d^3*(a/b)^(2/3) - a*b*e^3*(a /b)^(2/3)))*b*p/e - log((b*x^3 + a)^p*c)/((e*x + d)*e)
Time = 0.18 (sec) , antiderivative size = 363, normalized size of antiderivative = 1.24 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{(d+e x)^2} \, dx=-\frac {3 \, b d^{2} p \log \left (e x + d\right )}{b d^{3} e - a e^{4}} + \frac {b d^{2} p \log \left ({\left | b x^{3} + a \right |}\right )}{b d^{3} e - a e^{4}} + \frac {\sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}} b p \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{b^{2} d^{2} - \left (-a b^{2}\right )^{\frac {1}{3}} b d e + \left (-a b^{2}\right )^{\frac {2}{3}} e^{2}} + \frac {{\left (a b^{3} d^{3} e^{3} p \left (-\frac {a}{b}\right )^{\frac {1}{3}} - a^{2} b^{2} e^{6} p \left (-\frac {a}{b}\right )^{\frac {1}{3}} - a b^{3} d^{4} e^{2} p + a^{2} b^{2} d e^{5} p\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{a b^{3} d^{6} e^{2} - 2 \, a^{2} b^{2} d^{3} e^{5} + a^{3} b e^{8}} - \frac {p \log \left (b x^{3} + a\right )}{e^{2} x + d e} + \frac {{\left (\left (-a b^{2}\right )^{\frac {1}{3}} b d p - \left (-a b^{2}\right )^{\frac {2}{3}} e p\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{2 \, {\left (b^{2} d^{3} - a b e^{3}\right )}} - \frac {\log \left (c\right )}{e^{2} x + d e} \] Input:
integrate(log(c*(b*x^3+a)^p)/(e*x+d)^2,x, algorithm="giac")
Output:
-3*b*d^2*p*log(e*x + d)/(b*d^3*e - a*e^4) + b*d^2*p*log(abs(b*x^3 + a))/(b *d^3*e - a*e^4) + sqrt(3)*(-a*b^2)^(1/3)*b*p*arctan(1/3*sqrt(3)*(2*x + (-a /b)^(1/3))/(-a/b)^(1/3))/(b^2*d^2 - (-a*b^2)^(1/3)*b*d*e + (-a*b^2)^(2/3)* e^2) + (a*b^3*d^3*e^3*p*(-a/b)^(1/3) - a^2*b^2*e^6*p*(-a/b)^(1/3) - a*b^3* d^4*e^2*p + a^2*b^2*d*e^5*p)*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/(a*b^ 3*d^6*e^2 - 2*a^2*b^2*d^3*e^5 + a^3*b*e^8) - p*log(b*x^3 + a)/(e^2*x + d*e ) + 1/2*((-a*b^2)^(1/3)*b*d*p - (-a*b^2)^(2/3)*e*p)*log(x^2 + x*(-a/b)^(1/ 3) + (-a/b)^(2/3))/(b^2*d^3 - a*b*e^3) - log(c)/(e^2*x + d*e)
Time = 25.83 (sec) , antiderivative size = 736, normalized size of antiderivative = 2.52 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{(d+e x)^2} \, dx=\left (\sum _{k=1}^3\ln \left (-\frac {27\,a\,b^4\,d\,p^3+27\,a\,b^4\,e\,p^3\,x+{\mathrm {root}\left (b\,d^3\,e^3\,z^3-a\,e^6\,z^3-3\,b\,d^2\,e^2\,p\,z^2+3\,b\,d\,e\,p^2\,z-b\,p^3,z,k\right )}^3\,a\,b^4\,d^4\,e^3\,9+{\mathrm {root}\left (b\,d^3\,e^3\,z^3-a\,e^6\,z^3-3\,b\,d^2\,e^2\,p\,z^2+3\,b\,d\,e\,p^2\,z-b\,p^3,z,k\right )}^3\,a^2\,b^3\,d\,e^6\,45-{\mathrm {root}\left (b\,d^3\,e^3\,z^3-a\,e^6\,z^3-3\,b\,d^2\,e^2\,p\,z^2+3\,b\,d\,e\,p^2\,z-b\,p^3,z,k\right )}^2\,a^2\,b^3\,e^5\,p\,9+{\mathrm {root}\left (b\,d^3\,e^3\,z^3-a\,e^6\,z^3-3\,b\,d^2\,e^2\,p\,z^2+3\,b\,d\,e\,p^2\,z-b\,p^3,z,k\right )}^3\,a^2\,b^3\,e^7\,x\,36+{\mathrm {root}\left (b\,d^3\,e^3\,z^3-a\,e^6\,z^3-3\,b\,d^2\,e^2\,p\,z^2+3\,b\,d\,e\,p^2\,z-b\,p^3,z,k\right )}^2\,a\,b^4\,d^3\,e^2\,p\,9+{\mathrm {root}\left (b\,d^3\,e^3\,z^3-a\,e^6\,z^3-3\,b\,d^2\,e^2\,p\,z^2+3\,b\,d\,e\,p^2\,z-b\,p^3,z,k\right )}^3\,a\,b^4\,d^3\,e^4\,x\,18-\mathrm {root}\left (b\,d^3\,e^3\,z^3-a\,e^6\,z^3-3\,b\,d^2\,e^2\,p\,z^2+3\,b\,d\,e\,p^2\,z-b\,p^3,z,k\right )\,a\,b^4\,d^2\,e\,p^2\,45-\mathrm {root}\left (b\,d^3\,e^3\,z^3-a\,e^6\,z^3-3\,b\,d^2\,e^2\,p\,z^2+3\,b\,d\,e\,p^2\,z-b\,p^3,z,k\right )\,a\,b^4\,d\,e^2\,p^2\,x\,72+{\mathrm {root}\left (b\,d^3\,e^3\,z^3-a\,e^6\,z^3-3\,b\,d^2\,e^2\,p\,z^2+3\,b\,d\,e\,p^2\,z-b\,p^3,z,k\right )}^2\,a\,b^4\,d^2\,e^3\,p\,x\,27}{e^2}\right )\,\mathrm {root}\left (b\,d^3\,e^3\,z^3-a\,e^6\,z^3-3\,b\,d^2\,e^2\,p\,z^2+3\,b\,d\,e\,p^2\,z-b\,p^3,z,k\right )\right )-\frac {\ln \left (c\,{\left (b\,x^3+a\right )}^p\right )}{x\,e^2+d\,e}+\frac {3\,b\,d^2\,p\,\ln \left (d+e\,x\right )}{a\,e^4-b\,d^3\,e} \] Input:
int(log(c*(a + b*x^3)^p)/(d + e*x)^2,x)
Output:
symsum(log(-(27*a*b^4*d*p^3 + 27*a*b^4*e*p^3*x + 9*root(b*d^3*e^3*z^3 - a* e^6*z^3 - 3*b*d^2*e^2*p*z^2 + 3*b*d*e*p^2*z - b*p^3, z, k)^3*a*b^4*d^4*e^3 + 45*root(b*d^3*e^3*z^3 - a*e^6*z^3 - 3*b*d^2*e^2*p*z^2 + 3*b*d*e*p^2*z - b*p^3, z, k)^3*a^2*b^3*d*e^6 - 9*root(b*d^3*e^3*z^3 - a*e^6*z^3 - 3*b*d^2 *e^2*p*z^2 + 3*b*d*e*p^2*z - b*p^3, z, k)^2*a^2*b^3*e^5*p + 36*root(b*d^3* e^3*z^3 - a*e^6*z^3 - 3*b*d^2*e^2*p*z^2 + 3*b*d*e*p^2*z - b*p^3, z, k)^3*a ^2*b^3*e^7*x + 9*root(b*d^3*e^3*z^3 - a*e^6*z^3 - 3*b*d^2*e^2*p*z^2 + 3*b* d*e*p^2*z - b*p^3, z, k)^2*a*b^4*d^3*e^2*p + 18*root(b*d^3*e^3*z^3 - a*e^6 *z^3 - 3*b*d^2*e^2*p*z^2 + 3*b*d*e*p^2*z - b*p^3, z, k)^3*a*b^4*d^3*e^4*x - 45*root(b*d^3*e^3*z^3 - a*e^6*z^3 - 3*b*d^2*e^2*p*z^2 + 3*b*d*e*p^2*z - b*p^3, z, k)*a*b^4*d^2*e*p^2 - 72*root(b*d^3*e^3*z^3 - a*e^6*z^3 - 3*b*d^2 *e^2*p*z^2 + 3*b*d*e*p^2*z - b*p^3, z, k)*a*b^4*d*e^2*p^2*x + 27*root(b*d^ 3*e^3*z^3 - a*e^6*z^3 - 3*b*d^2*e^2*p*z^2 + 3*b*d*e*p^2*z - b*p^3, z, k)^2 *a*b^4*d^2*e^3*p*x)/e^2)*root(b*d^3*e^3*z^3 - a*e^6*z^3 - 3*b*d^2*e^2*p*z^ 2 + 3*b*d*e*p^2*z - b*p^3, z, k), k, 1, 3) - log(c*(a + b*x^3)^p)/(d*e + e ^2*x) + (3*b*d^2*p*log(d + e*x))/(a*e^4 - b*d^3*e)
Time = 0.24 (sec) , antiderivative size = 6006, normalized size of antiderivative = 20.57 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{(d+e x)^2} \, dx =\text {Too large to display} \] Input:
int(log(c*(b*x^3+a)^p)/(e*x+d)^2,x)
Output:
( - 206*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)) )*a**6*b*d**4*e**18*p - 206*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x )/(a**(1/3)*sqrt(3)))*a**6*b*d**3*e**19*p*x + 4216*a**(2/3)*sqrt(3)*atan(( a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**5*b**2*d**7*e**15*p + 4216 *a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**5* b**2*d**6*e**16*p*x + 4420*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x) /(a**(1/3)*sqrt(3)))*a**4*b**3*d**10*e**12*p + 4420*a**(2/3)*sqrt(3)*atan( (a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**4*b**3*d**9*e**13*p*x - 4 420*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a* *3*b**4*d**13*e**9*p - 4420*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x )/(a**(1/3)*sqrt(3)))*a**3*b**4*d**12*e**10*p*x - 4216*a**(2/3)*sqrt(3)*at an((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**2*b**5*d**16*e**6*p - 4216*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a **2*b**5*d**15*e**7*p*x + 206*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3) *x)/(a**(1/3)*sqrt(3)))*a*b**6*d**19*e**3*p + 206*a**(2/3)*sqrt(3)*atan((a **(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a*b**6*d**18*e**4*p*x - 2*b**( 1/3)*a**(1/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a **7*d**2*e**20*p - 2*b**(1/3)*a**(1/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3) *x)/(a**(1/3)*sqrt(3)))*a**7*d*e**21*p*x + 850*b**(1/3)*a**(1/3)*sqrt(3)*a tan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**6*b*d**5*e**17*p +...