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

Optimal result

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

[Out]

-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
/50*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)

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 366, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {2521, 2498, 327, 211, 2504, 2442, 45, 2505, 308} \[ \int \left (f+g x^3\right )^3 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-\frac {6 d^{7/2} f g^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{7 e^{7/2}}+\frac {2 \sqrt {d} f^3 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}+f^3 x \log \left (c \left (d+e x^2\right )^p\right )+\frac {3}{4} f^2 g x^4 \log \left (c \left (d+e x^2\right )^p\right )+\frac {3}{7} f g^2 x^7 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{10} g^3 x^{10} \log \left (c \left (d+e x^2\right )^p\right )+\frac {d^5 g^3 p \log \left (d+e x^2\right )}{10 e^5}-\frac {d^4 g^3 p x^2}{10 e^4}+\frac {6 d^3 f g^2 p x}{7 e^3}+\frac {d^3 g^3 p x^4}{20 e^3}-\frac {3 d^2 f^2 g p \log \left (d+e x^2\right )}{4 e^2}-\frac {2 d^2 f g^2 p x^3}{7 e^2}-\frac {d^2 g^3 p x^6}{30 e^2}+\frac {3 d f^2 g p x^2}{4 e}+\frac {6 d f g^2 p x^5}{35 e}+\frac {d g^3 p x^8}{40 e}-2 f^3 p x-\frac {3}{8} f^2 g p x^4-\frac {6}{49} f g^2 p x^7-\frac {1}{50} g^3 p x^{10} \]

[In]

Int[(f + g*x^3)^3*Log[c*(d + e*x^2)^p],x]

[Out]

-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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2442

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[(f + g*
x)^(q + 1)*((a + b*Log[c*(d + e*x)^n])/(g*(q + 1))), x] - Dist[b*e*(n/(g*(q + 1))), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 2498

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)], x_Symbol] :> Simp[x*Log[c*(d + e*x^n)^p], x] - Dist[e*n*p, Int[
x^n/(d + e*x^n), x], x] /; FreeQ[{c, d, e, n, p}, x]

Rule 2504

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rule 2505

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.)*(x_))^(m_.), x_Symbol] :> Simp[(f*x)^(m +
1)*((a + b*Log[c*(d + e*x^n)^p])/(f*(m + 1))), x] - Dist[b*e*n*(p/(f*(m + 1))), Int[x^(n - 1)*((f*x)^(m + 1)/
(d + e*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && NeQ[m, -1]

Rule 2521

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]] /; Free
Q[{a, b, c, d, e, f, g, n, p, q, r, s}, x] && IntegerQ[n] && IGtQ[q, 0] && IntegerQ[r] && IntegerQ[s] && (EqQ[
q, 1] || (GtQ[r, 0] && GtQ[s, 1]) || (LtQ[s, 0] && LtQ[r, 0]))

Rubi steps \begin{align*} \text {integral}& = \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 \\ & = f^3 \int \log \left (c \left (d+e x^2\right )^p\right ) \, dx+\left (3 f^2 g\right ) \int x^3 \log \left (c \left (d+e x^2\right )^p\right ) \, dx+\left (3 f g^2\right ) \int x^6 \log \left (c \left (d+e x^2\right )^p\right ) \, dx+g^3 \int x^9 \log \left (c \left (d+e x^2\right )^p\right ) \, dx \\ & = f^3 x \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}{2} \left (3 f^2 g\right ) \text {Subst}\left (\int x \log \left (c (d+e x)^p\right ) \, dx,x,x^2\right )+\frac {1}{2} g^3 \text {Subst}\left (\int x^4 \log \left (c (d+e x)^p\right ) \, dx,x,x^2\right )-\left (2 e f^3 p\right ) \int \frac {x^2}{d+e x^2} \, dx-\frac {1}{7} \left (6 e f g^2 p\right ) \int \frac {x^8}{d+e x^2} \, dx \\ & = -2 f^3 p x+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 )+\left (2 d f^3 p\right ) \int \frac {1}{d+e x^2} \, dx-\frac {1}{4} \left (3 e f^2 g p\right ) \text {Subst}\left (\int \frac {x^2}{d+e x} \, dx,x,x^2\right )-\frac {1}{7} \left (6 e f g^2 p\right ) \int \left (-\frac {d^3}{e^4}+\frac {d^2 x^2}{e^3}-\frac {d x^4}{e^2}+\frac {x^6}{e}+\frac {d^4}{e^4 \left (d+e x^2\right )}\right ) \, dx-\frac {1}{10} \left (e g^3 p\right ) \text {Subst}\left (\int \frac {x^5}{d+e x} \, dx,x,x^2\right ) \\ & = -2 f^3 p x+\frac {6 d^3 f g^2 p x}{7 e^3}-\frac {2 d^2 f g^2 p x^3}{7 e^2}+\frac {6 d f g^2 p x^5}{35 e}-\frac {6}{49} f g^2 p x^7+\frac {2 \sqrt {d} f^3 p \tan ^{-1}\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 {1}{4} \left (3 e f^2 g p\right ) \text {Subst}\left (\int \left (-\frac {d}{e^2}+\frac {x}{e}+\frac {d^2}{e^2 (d+e x)}\right ) \, dx,x,x^2\right )-\frac {\left (6 d^4 f g^2 p\right ) \int \frac {1}{d+e x^2} \, dx}{7 e^3}-\frac {1}{10} \left (e g^3 p\right ) \text {Subst}\left (\int \left (\frac {d^4}{e^5}-\frac {d^3 x}{e^4}+\frac {d^2 x^2}{e^3}-\frac {d x^3}{e^2}+\frac {x^4}{e}-\frac {d^5}{e^5 (d+e x)}\right ) \, dx,x,x^2\right ) \\ & = -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 \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}-\frac {6 d^{7/2} f g^2 p \tan ^{-1}\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 ) \\ \end{align*}

Mathematica [A] (verified)

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

[In]

Integrate[(f + g*x^3)^3*Log[c*(d + e*x^2)^p],x]

[Out]

(-(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))) -
8400*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*(140*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)

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	\(\text {Expression too large to display}\)	\(1311\)

[In]

int((g*x^3+f)^3*ln(c*(e*x^2+d)^p),x,method=_RETURNVERBOSE)

[Out]

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*(14*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))))

Fricas [A] (verification not implemented)

none

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 ] \]

[In]

integrate((g*x^3+f)^3*log(c*(e*x^2+d)^p),x, algorithm="fricas")

[Out]

[-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 - 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]

Sympy [F(-1)]

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

[In]

integrate((g*x**3+f)**3*ln(c*(e*x**2+d)**p),x)

[Out]

Timed out

Maxima [F(-2)]

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

[In]

integrate((g*x^3+f)^3*log(c*(e*x^2+d)^p),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [A] (verification not implemented)

none

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

[In]

integrate((g*x^3+f)^3*log(c*(e*x^2+d)^p),x, algorithm="giac")

[Out]

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*
log(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

Mupad [B] (verification not implemented)

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

[In]

int(log(c*(d + e*x^2)^p)*(f + g*x^3)^3,x)

[Out]

(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*log(c*(d + e*x^2)^p) + (3*f^2*g*x^4*l
og(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)