3.3.88 \(\int (f+g x^3)^3 \log (c (d+e x^2)^p) \, dx\) [288]

3.3.88.1 Optimal result
3.3.88.2 Mathematica [A] (verified)
3.3.88.3 Rubi [A] (verified)
3.3.88.4 Maple [A] (verified)
3.3.88.5 Fricas [A] (verification not implemented)
3.3.88.6 Sympy [F(-1)]
3.3.88.7 Maxima [F(-2)]
3.3.88.8 Giac [A] (verification not implemented)
3.3.88.9 Mupad [B] (verification not implemented)

3.3.88.1 Optimal result

Integrand size = 22, antiderivative size = 366 \[ \int \left (f+g x^3\right )^3 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-2 f^3 p x+\frac {6 d^3 f g^2 p x}{7 e^3}+\frac {3 d f^2 g p x^2}{4 e}-\frac {d^4 g^3 p x^2}{10 e^4}-\frac {2 d^2 f g^2 p x^3}{7 e^2}-\frac {3}{8} f^2 g p x^4+\frac {d^3 g^3 p x^4}{20 e^3}+\frac {6 d f g^2 p x^5}{35 e}-\frac {d^2 g^3 p x^6}{30 e^2}-\frac {6}{49} f g^2 p x^7+\frac {d g^3 p x^8}{40 e}-\frac {1}{50} g^3 p x^{10}+\frac {2 \sqrt {d} f^3 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}-\frac {6 d^{7/2} f g^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{7 e^{7/2}}-\frac {3 d^2 f^2 g p \log \left (d+e x^2\right )}{4 e^2}+\frac {d^5 g^3 p \log \left (d+e x^2\right )}{10 e^5}+f^3 x \log \left (c \left (d+e x^2\right )^p\right )+\frac {3}{4} f^2 g x^4 \log \left (c \left (d+e x^2\right )^p\right )+\frac {3}{7} f g^2 x^7 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{10} g^3 x^{10} \log \left (c \left (d+e x^2\right )^p\right ) \]

output
-2*f^3*p*x+6/7*d^3*f*g^2*p*x/e^3+3/4*d*f^2*g*p*x^2/e-1/10*d^4*g^3*p*x^2/e^ 
4-2/7*d^2*f*g^2*p*x^3/e^2-3/8*f^2*g*p*x^4+1/20*d^3*g^3*p*x^4/e^3+6/35*d*f* 
g^2*p*x^5/e-1/30*d^2*g^3*p*x^6/e^2-6/49*f*g^2*p*x^7+1/40*d*g^3*p*x^8/e-1/5 
0*g^3*p*x^10-6/7*d^(7/2)*f*g^2*p*arctan(x*e^(1/2)/d^(1/2))/e^(7/2)-3/4*d^2 
*f^2*g*p*ln(e*x^2+d)/e^2+1/10*d^5*g^3*p*ln(e*x^2+d)/e^5+f^3*x*ln(c*(e*x^2+ 
d)^p)+3/4*f^2*g*x^4*ln(c*(e*x^2+d)^p)+3/7*f*g^2*x^7*ln(c*(e*x^2+d)^p)+1/10 
*g^3*x^10*ln(c*(e*x^2+d)^p)+2*f^3*p*arctan(x*e^(1/2)/d^(1/2))*d^(1/2)/e^(1 
/2)
 
3.3.88.2 Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 258, normalized size of antiderivative = 0.70 \[ \int \left (f+g x^3\right )^3 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {-e p x \left (2940 d^4 g^3 x+140 d^2 e^2 g^2 x^2 \left (60 f+7 g x^3\right )-210 d^3 e g^2 \left (120 f+7 g x^3\right )-105 d e^3 g x \left (210 f^2+48 f g x^3+7 g^2 x^6\right )+3 e^4 \left (19600 f^3+3675 f^2 g x^3+1200 f g^2 x^6+196 g^3 x^9\right )\right )-8400 \sqrt {d} e^{3/2} f \left (-7 e^3 f^2+3 d^3 g^2\right ) p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )+1470 d^2 g \left (-15 e^3 f^2+2 d^3 g^2\right ) p \log \left (d+e x^2\right )+210 e^5 x \left (140 f^3+105 f^2 g x^3+60 f g^2 x^6+14 g^3 x^9\right ) \log \left (c \left (d+e x^2\right )^p\right )}{29400 e^5} \]

input
Integrate[(f + g*x^3)^3*Log[c*(d + e*x^2)^p],x]
 
output
(-(e*p*x*(2940*d^4*g^3*x + 140*d^2*e^2*g^2*x^2*(60*f + 7*g*x^3) - 210*d^3* 
e*g^2*(120*f + 7*g*x^3) - 105*d*e^3*g*x*(210*f^2 + 48*f*g*x^3 + 7*g^2*x^6) 
 + 3*e^4*(19600*f^3 + 3675*f^2*g*x^3 + 1200*f*g^2*x^6 + 196*g^3*x^9))) - 8 
400*Sqrt[d]*e^(3/2)*f*(-7*e^3*f^2 + 3*d^3*g^2)*p*ArcTan[(Sqrt[e]*x)/Sqrt[d 
]] + 1470*d^2*g*(-15*e^3*f^2 + 2*d^3*g^2)*p*Log[d + e*x^2] + 210*e^5*x*(14 
0*f^3 + 105*f^2*g*x^3 + 60*f*g^2*x^6 + 14*g^3*x^9)*Log[c*(d + e*x^2)^p])/( 
29400*e^5)
 
3.3.88.3 Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 366, 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.

\(\displaystyle \int \left (f+g x^3\right )^3 \log \left (c \left (d+e x^2\right )^p\right ) \, dx\)

\(\Big \downarrow \) 2921

\(\displaystyle \int \left (f^3 \log \left (c \left (d+e x^2\right )^p\right )+3 f^2 g x^3 \log \left (c \left (d+e x^2\right )^p\right )+3 f g^2 x^6 \log \left (c \left (d+e x^2\right )^p\right )+g^3 x^9 \log \left (c \left (d+e x^2\right )^p\right )\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {6 d^{7/2} f g^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{7 e^{7/2}}+\frac {2 \sqrt {d} f^3 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}+f^3 x \log \left (c \left (d+e x^2\right )^p\right )+\frac {3}{4} f^2 g x^4 \log \left (c \left (d+e x^2\right )^p\right )+\frac {3}{7} f g^2 x^7 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{10} g^3 x^{10} \log \left (c \left (d+e x^2\right )^p\right )+\frac {d^5 g^3 p \log \left (d+e x^2\right )}{10 e^5}-\frac {d^4 g^3 p x^2}{10 e^4}+\frac {6 d^3 f g^2 p x}{7 e^3}+\frac {d^3 g^3 p x^4}{20 e^3}-\frac {3 d^2 f^2 g p \log \left (d+e x^2\right )}{4 e^2}-\frac {2 d^2 f g^2 p x^3}{7 e^2}-\frac {d^2 g^3 p x^6}{30 e^2}+\frac {3 d f^2 g p x^2}{4 e}+\frac {6 d f g^2 p x^5}{35 e}+\frac {d g^3 p x^8}{40 e}-2 f^3 p x-\frac {3}{8} f^2 g p x^4-\frac {6}{49} f g^2 p x^7-\frac {1}{50} g^3 p x^{10}\)

input
Int[(f + g*x^3)^3*Log[c*(d + e*x^2)^p],x]
 
output
-2*f^3*p*x + (6*d^3*f*g^2*p*x)/(7*e^3) + (3*d*f^2*g*p*x^2)/(4*e) - (d^4*g^ 
3*p*x^2)/(10*e^4) - (2*d^2*f*g^2*p*x^3)/(7*e^2) - (3*f^2*g*p*x^4)/8 + (d^3 
*g^3*p*x^4)/(20*e^3) + (6*d*f*g^2*p*x^5)/(35*e) - (d^2*g^3*p*x^6)/(30*e^2) 
 - (6*f*g^2*p*x^7)/49 + (d*g^3*p*x^8)/(40*e) - (g^3*p*x^10)/50 + (2*Sqrt[d 
]*f^3*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/Sqrt[e] - (6*d^(7/2)*f*g^2*p*ArcTan[( 
Sqrt[e]*x)/Sqrt[d]])/(7*e^(7/2)) - (3*d^2*f^2*g*p*Log[d + e*x^2])/(4*e^2) 
+ (d^5*g^3*p*Log[d + e*x^2])/(10*e^5) + f^3*x*Log[c*(d + e*x^2)^p] + (3*f^ 
2*g*x^4*Log[c*(d + e*x^2)^p])/4 + (3*f*g^2*x^7*Log[c*(d + e*x^2)^p])/7 + ( 
g^3*x^10*Log[c*(d + e*x^2)^p])/10
 

3.3.88.3.1 Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2921
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]))
 
3.3.88.4 Maple [A] (verified)

Time = 7.72 (sec) , antiderivative size = 309, normalized size of antiderivative = 0.84

method result size
parts \(\frac {g^{3} x^{10} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{10}+\frac {3 f \,g^{2} x^{7} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{7}+\frac {3 f^{2} g \,x^{4} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{4}+f^{3} x \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )-\frac {p e \left (\frac {\frac {7}{5} e^{4} g^{3} x^{10}-\frac {7}{4} d \,e^{3} g^{3} x^{8}+\frac {60}{7} e^{4} f \,g^{2} x^{7}+\frac {7}{3} d^{2} e^{2} g^{3} x^{6}-12 d \,e^{3} f \,g^{2} x^{5}-\frac {7}{2} d^{3} e \,g^{3} x^{4}+\frac {105}{4} e^{4} f^{2} g \,x^{4}+20 d^{2} e^{2} f \,g^{2} x^{3}+7 d^{4} g^{3} x^{2}-\frac {105}{2} d \,f^{2} g \,x^{2} e^{3}-60 x \,d^{3} f \,g^{2} e +140 x \,e^{4} f^{3}}{e^{5}}-\frac {d \left (\frac {\left (14 d^{4} g^{3}-105 d \,e^{3} f^{2} g \right ) \ln \left (e \,x^{2}+d \right )}{2 e}+\frac {\left (-60 d^{3} f \,g^{2} e +140 e^{4} f^{3}\right ) \arctan \left (\frac {x e}{\sqrt {d e}}\right )}{\sqrt {d e}}\right )}{e^{5}}\right )}{70}\) \(309\)
risch \(\text {Expression too large to display}\) \(1311\)

input
int((g*x^3+f)^3*ln(c*(e*x^2+d)^p),x,method=_RETURNVERBOSE)
 
output
1/10*g^3*x^10*ln(c*(e*x^2+d)^p)+3/7*f*g^2*x^7*ln(c*(e*x^2+d)^p)+3/4*f^2*g* 
x^4*ln(c*(e*x^2+d)^p)+f^3*x*ln(c*(e*x^2+d)^p)-1/70*p*e*(1/e^5*(7/5*e^4*g^3 
*x^10-7/4*d*e^3*g^3*x^8+60/7*e^4*f*g^2*x^7+7/3*d^2*e^2*g^3*x^6-12*d*e^3*f* 
g^2*x^5-7/2*d^3*e*g^3*x^4+105/4*e^4*f^2*g*x^4+20*d^2*e^2*f*g^2*x^3+7*d^4*g 
^3*x^2-105/2*d*f^2*g*x^2*e^3-60*x*d^3*f*g^2*e+140*x*e^4*f^3)-d/e^5*(1/2*(1 
4*d^4*g^3-105*d*e^3*f^2*g)/e*ln(e*x^2+d)+(-60*d^3*e*f*g^2+140*e^4*f^3)/(d* 
e)^(1/2)*arctan(x*e/(d*e)^(1/2))))
 
3.3.88.5 Fricas [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 708, normalized size of antiderivative = 1.93 \[ \int \left (f+g x^3\right )^3 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\left [-\frac {588 \, e^{5} g^{3} p x^{10} - 735 \, d e^{4} g^{3} p x^{8} + 3600 \, e^{5} f g^{2} p x^{7} + 980 \, d^{2} e^{3} g^{3} p x^{6} - 5040 \, d e^{4} f g^{2} p x^{5} + 8400 \, d^{2} e^{3} f g^{2} p x^{3} + 735 \, {\left (15 \, e^{5} f^{2} g - 2 \, d^{3} e^{2} g^{3}\right )} p x^{4} - 1470 \, {\left (15 \, d e^{4} f^{2} g - 2 \, d^{4} e g^{3}\right )} p x^{2} + 4200 \, {\left (7 \, e^{5} f^{3} - 3 \, d^{3} e^{2} f g^{2}\right )} p \sqrt {-\frac {d}{e}} \log \left (\frac {e x^{2} - 2 \, e x \sqrt {-\frac {d}{e}} - d}{e x^{2} + d}\right ) + 8400 \, {\left (7 \, e^{5} f^{3} - 3 \, d^{3} e^{2} f g^{2}\right )} p x - 210 \, {\left (14 \, e^{5} g^{3} p x^{10} + 60 \, e^{5} f g^{2} p x^{7} + 105 \, e^{5} f^{2} g p x^{4} + 140 \, e^{5} f^{3} p x - 7 \, {\left (15 \, d^{2} e^{3} f^{2} g - 2 \, d^{5} g^{3}\right )} p\right )} \log \left (e x^{2} + d\right ) - 210 \, {\left (14 \, e^{5} g^{3} x^{10} + 60 \, e^{5} f g^{2} x^{7} + 105 \, e^{5} f^{2} g x^{4} + 140 \, e^{5} f^{3} x\right )} \log \left (c\right )}{29400 \, e^{5}}, -\frac {588 \, e^{5} g^{3} p x^{10} - 735 \, d e^{4} g^{3} p x^{8} + 3600 \, e^{5} f g^{2} p x^{7} + 980 \, d^{2} e^{3} g^{3} p x^{6} - 5040 \, d e^{4} f g^{2} p x^{5} + 8400 \, d^{2} e^{3} f g^{2} p x^{3} + 735 \, {\left (15 \, e^{5} f^{2} g - 2 \, d^{3} e^{2} g^{3}\right )} p x^{4} - 1470 \, {\left (15 \, d e^{4} f^{2} g - 2 \, d^{4} e g^{3}\right )} p x^{2} - 8400 \, {\left (7 \, e^{5} f^{3} - 3 \, d^{3} e^{2} f g^{2}\right )} p \sqrt {\frac {d}{e}} \arctan \left (\frac {e x \sqrt {\frac {d}{e}}}{d}\right ) + 8400 \, {\left (7 \, e^{5} f^{3} - 3 \, d^{3} e^{2} f g^{2}\right )} p x - 210 \, {\left (14 \, e^{5} g^{3} p x^{10} + 60 \, e^{5} f g^{2} p x^{7} + 105 \, e^{5} f^{2} g p x^{4} + 140 \, e^{5} f^{3} p x - 7 \, {\left (15 \, d^{2} e^{3} f^{2} g - 2 \, d^{5} g^{3}\right )} p\right )} \log \left (e x^{2} + d\right ) - 210 \, {\left (14 \, e^{5} g^{3} x^{10} + 60 \, e^{5} f g^{2} x^{7} + 105 \, e^{5} f^{2} g x^{4} + 140 \, e^{5} f^{3} x\right )} \log \left (c\right )}{29400 \, e^{5}}\right ] \]

input
integrate((g*x^3+f)^3*log(c*(e*x^2+d)^p),x, algorithm="fricas")
 
output
[-1/29400*(588*e^5*g^3*p*x^10 - 735*d*e^4*g^3*p*x^8 + 3600*e^5*f*g^2*p*x^7 
 + 980*d^2*e^3*g^3*p*x^6 - 5040*d*e^4*f*g^2*p*x^5 + 8400*d^2*e^3*f*g^2*p*x 
^3 + 735*(15*e^5*f^2*g - 2*d^3*e^2*g^3)*p*x^4 - 1470*(15*d*e^4*f^2*g - 2*d 
^4*e*g^3)*p*x^2 + 4200*(7*e^5*f^3 - 3*d^3*e^2*f*g^2)*p*sqrt(-d/e)*log((e*x 
^2 - 2*e*x*sqrt(-d/e) - d)/(e*x^2 + d)) + 8400*(7*e^5*f^3 - 3*d^3*e^2*f*g^ 
2)*p*x - 210*(14*e^5*g^3*p*x^10 + 60*e^5*f*g^2*p*x^7 + 105*e^5*f^2*g*p*x^4 
 + 140*e^5*f^3*p*x - 7*(15*d^2*e^3*f^2*g - 2*d^5*g^3)*p)*log(e*x^2 + d) - 
210*(14*e^5*g^3*x^10 + 60*e^5*f*g^2*x^7 + 105*e^5*f^2*g*x^4 + 140*e^5*f^3* 
x)*log(c))/e^5, -1/29400*(588*e^5*g^3*p*x^10 - 735*d*e^4*g^3*p*x^8 + 3600* 
e^5*f*g^2*p*x^7 + 980*d^2*e^3*g^3*p*x^6 - 5040*d*e^4*f*g^2*p*x^5 + 8400*d^ 
2*e^3*f*g^2*p*x^3 + 735*(15*e^5*f^2*g - 2*d^3*e^2*g^3)*p*x^4 - 1470*(15*d* 
e^4*f^2*g - 2*d^4*e*g^3)*p*x^2 - 8400*(7*e^5*f^3 - 3*d^3*e^2*f*g^2)*p*sqrt 
(d/e)*arctan(e*x*sqrt(d/e)/d) + 8400*(7*e^5*f^3 - 3*d^3*e^2*f*g^2)*p*x - 2 
10*(14*e^5*g^3*p*x^10 + 60*e^5*f*g^2*p*x^7 + 105*e^5*f^2*g*p*x^4 + 140*e^5 
*f^3*p*x - 7*(15*d^2*e^3*f^2*g - 2*d^5*g^3)*p)*log(e*x^2 + d) - 210*(14*e^ 
5*g^3*x^10 + 60*e^5*f*g^2*x^7 + 105*e^5*f^2*g*x^4 + 140*e^5*f^3*x)*log(c)) 
/e^5]
 
3.3.88.6 Sympy [F(-1)]

Timed out. \[ \int \left (f+g x^3\right )^3 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\text {Timed out} \]

input
integrate((g*x**3+f)**3*ln(c*(e*x**2+d)**p),x)
 
output
Timed out
 
3.3.88.7 Maxima [F(-2)]

Exception generated. \[ \int \left (f+g x^3\right )^3 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\text {Exception raised: ValueError} \]

input
integrate((g*x^3+f)^3*log(c*(e*x^2+d)^p),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
 
3.3.88.8 Giac [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 325, normalized size of antiderivative = 0.89 \[ \int \left (f+g x^3\right )^3 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {d g^{3} p x^{8}}{40 \, e} - \frac {1}{50} \, {\left (g^{3} p - 5 \, g^{3} \log \left (c\right )\right )} x^{10} - \frac {d^{2} g^{3} p x^{6}}{30 \, e^{2}} + \frac {6 \, d f g^{2} p x^{5}}{35 \, e} - \frac {3}{49} \, {\left (2 \, f g^{2} p - 7 \, f g^{2} \log \left (c\right )\right )} x^{7} - \frac {2 \, d^{2} f g^{2} p x^{3}}{7 \, e^{2}} - \frac {{\left (15 \, e^{3} f^{2} g p - 2 \, d^{3} g^{3} p - 30 \, e^{3} f^{2} g \log \left (c\right )\right )} x^{4}}{40 \, e^{3}} + \frac {1}{140} \, {\left (14 \, g^{3} p x^{10} + 60 \, f g^{2} p x^{7} + 105 \, f^{2} g p x^{4} + 140 \, f^{3} p x\right )} \log \left (e x^{2} + d\right ) - \frac {{\left (14 \, e^{3} f^{3} p - 6 \, d^{3} f g^{2} p - 7 \, e^{3} f^{3} \log \left (c\right )\right )} x}{7 \, e^{3}} + \frac {{\left (15 \, d e^{3} f^{2} g p - 2 \, d^{4} g^{3} p\right )} x^{2}}{20 \, e^{4}} + \frac {2 \, {\left (7 \, d e^{3} f^{3} p - 3 \, d^{4} f g^{2} p\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{7 \, \sqrt {d e} e^{3}} - \frac {{\left (15 \, d^{2} e^{3} f^{2} g p - 2 \, d^{5} g^{3} p\right )} \log \left (e x^{2} + d\right )}{20 \, e^{5}} \]

input
integrate((g*x^3+f)^3*log(c*(e*x^2+d)^p),x, algorithm="giac")
 
output
1/40*d*g^3*p*x^8/e - 1/50*(g^3*p - 5*g^3*log(c))*x^10 - 1/30*d^2*g^3*p*x^6 
/e^2 + 6/35*d*f*g^2*p*x^5/e - 3/49*(2*f*g^2*p - 7*f*g^2*log(c))*x^7 - 2/7* 
d^2*f*g^2*p*x^3/e^2 - 1/40*(15*e^3*f^2*g*p - 2*d^3*g^3*p - 30*e^3*f^2*g*lo 
g(c))*x^4/e^3 + 1/140*(14*g^3*p*x^10 + 60*f*g^2*p*x^7 + 105*f^2*g*p*x^4 + 
140*f^3*p*x)*log(e*x^2 + d) - 1/7*(14*e^3*f^3*p - 6*d^3*f*g^2*p - 7*e^3*f^ 
3*log(c))*x/e^3 + 1/20*(15*d*e^3*f^2*g*p - 2*d^4*g^3*p)*x^2/e^4 + 2/7*(7*d 
*e^3*f^3*p - 3*d^4*f*g^2*p)*arctan(e*x/sqrt(d*e))/(sqrt(d*e)*e^3) - 1/20*( 
15*d^2*e^3*f^2*g*p - 2*d^5*g^3*p)*log(e*x^2 + d)/e^5
 
3.3.88.9 Mupad [B] (verification not implemented)

Time = 4.73 (sec) , antiderivative size = 316, normalized size of antiderivative = 0.86 \[ \int \left (f+g x^3\right )^3 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {g^3\,x^{10}\,\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )}{10}-2\,f^3\,p\,x-\frac {g^3\,p\,x^{10}}{50}+f^3\,x\,\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )+\frac {3\,f^2\,g\,x^4\,\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )}{4}+\frac {3\,f\,g^2\,x^7\,\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )}{7}-\frac {3\,f^2\,g\,p\,x^4}{8}-\frac {6\,f\,g^2\,p\,x^7}{49}+\frac {d\,g^3\,p\,x^8}{40\,e}+\frac {2\,\sqrt {d}\,f^3\,p\,\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )}{\sqrt {e}}+\frac {d^5\,g^3\,p\,\ln \left (e\,x^2+d\right )}{10\,e^5}-\frac {d^2\,g^3\,p\,x^6}{30\,e^2}+\frac {d^3\,g^3\,p\,x^4}{20\,e^3}-\frac {d^4\,g^3\,p\,x^2}{10\,e^4}-\frac {6\,d^{7/2}\,f\,g^2\,p\,\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )}{7\,e^{7/2}}-\frac {3\,d^2\,f^2\,g\,p\,\ln \left (e\,x^2+d\right )}{4\,e^2}-\frac {2\,d^2\,f\,g^2\,p\,x^3}{7\,e^2}+\frac {3\,d\,f^2\,g\,p\,x^2}{4\,e}+\frac {6\,d\,f\,g^2\,p\,x^5}{35\,e}+\frac {6\,d^3\,f\,g^2\,p\,x}{7\,e^3} \]

input
int(log(c*(d + e*x^2)^p)*(f + g*x^3)^3,x)
 
output
(g^3*x^10*log(c*(d + e*x^2)^p))/10 - 2*f^3*p*x - (g^3*p*x^10)/50 + f^3*x*l 
og(c*(d + e*x^2)^p) + (3*f^2*g*x^4*log(c*(d + e*x^2)^p))/4 + (3*f*g^2*x^7* 
log(c*(d + e*x^2)^p))/7 - (3*f^2*g*p*x^4)/8 - (6*f*g^2*p*x^7)/49 + (d*g^3* 
p*x^8)/(40*e) + (2*d^(1/2)*f^3*p*atan((e^(1/2)*x)/d^(1/2)))/e^(1/2) + (d^5 
*g^3*p*log(d + e*x^2))/(10*e^5) - (d^2*g^3*p*x^6)/(30*e^2) + (d^3*g^3*p*x^ 
4)/(20*e^3) - (d^4*g^3*p*x^2)/(10*e^4) - (6*d^(7/2)*f*g^2*p*atan((e^(1/2)* 
x)/d^(1/2)))/(7*e^(7/2)) - (3*d^2*f^2*g*p*log(d + e*x^2))/(4*e^2) - (2*d^2 
*f*g^2*p*x^3)/(7*e^2) + (3*d*f^2*g*p*x^2)/(4*e) + (6*d*f*g^2*p*x^5)/(35*e) 
 + (6*d^3*f*g^2*p*x)/(7*e^3)