\(\int (f+g x^2)^3 \log (c (d+e x^2)^p) \, dx\) [268]

Optimal result

Integrand size = 22, antiderivative size = 338 \[ \int \left (f+g x^2\right )^3 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-2 f^3 p x+\frac {2 d f^2 g p x}{e}-\frac {6 d^2 f g^2 p x}{5 e^2}+\frac {2 d^3 g^3 p x}{7 e^3}-\frac {2}{3} f^2 g p x^3+\frac {2 d f g^2 p x^3}{5 e}-\frac {2 d^2 g^3 p x^3}{21 e^2}-\frac {6}{25} f g^2 p x^5+\frac {2 d g^3 p x^5}{35 e}-\frac {2}{49} g^3 p x^7+\frac {2 \sqrt {d} f^3 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}-\frac {2 d^{3/2} f^2 g p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{e^{3/2}}+\frac {6 d^{5/2} f g^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 e^{5/2}}-\frac {2 d^{7/2} g^3 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{7 e^{7/2}}+f^3 x \log \left (c \left (d+e x^2\right )^p\right )+f^2 g x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {3}{5} f g^2 x^5 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{7} g^3 x^7 \log \left (c \left (d+e x^2\right )^p\right ) \] Output:

-2*f^3*p*x+2*d*f^2*g*p*x/e-6/5*d^2*f*g^2*p*x/e^2+2/7*d^3*g^3*p*x/e^3-2/3*f 
^2*g*p*x^3+2/5*d*f*g^2*p*x^3/e-2/21*d^2*g^3*p*x^3/e^2-6/25*f*g^2*p*x^5+2/3 
5*d*g^3*p*x^5/e-2/49*g^3*p*x^7+2*d^(1/2)*f^3*p*arctan(e^(1/2)*x/d^(1/2))/e 
^(1/2)-2*d^(3/2)*f^2*g*p*arctan(e^(1/2)*x/d^(1/2))/e^(3/2)+6/5*d^(5/2)*f*g 
^2*p*arctan(e^(1/2)*x/d^(1/2))/e^(5/2)-2/7*d^(7/2)*g^3*p*arctan(e^(1/2)*x/ 
d^(1/2))/e^(7/2)+f^3*x*ln(c*(e*x^2+d)^p)+f^2*g*x^3*ln(c*(e*x^2+d)^p)+3/5*f 
*g^2*x^5*ln(c*(e*x^2+d)^p)+1/7*g^3*x^7*ln(c*(e*x^2+d)^p)
 

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.64 \[ \int \left (f+g x^2\right )^3 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-\frac {2 p x \left (-525 d^3 g^3+35 d^2 e g^2 \left (63 f+5 g x^2\right )-105 d e^2 g \left (35 f^2+7 f g x^2+g^2 x^4\right )+e^3 \left (3675 f^3+1225 f^2 g x^2+441 f g^2 x^4+75 g^3 x^6\right )\right )}{3675 e^3}-\frac {2 \sqrt {d} \left (-35 e^3 f^3+35 d e^2 f^2 g-21 d^2 e f g^2+5 d^3 g^3\right ) p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{35 e^{7/2}}+\frac {1}{35} x \left (35 f^3+35 f^2 g x^2+21 f g^2 x^4+5 g^3 x^6\right ) \log \left (c \left (d+e x^2\right )^p\right ) \] Input:

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

Output:

(-2*p*x*(-525*d^3*g^3 + 35*d^2*e*g^2*(63*f + 5*g*x^2) - 105*d*e^2*g*(35*f^ 
2 + 7*f*g*x^2 + g^2*x^4) + e^3*(3675*f^3 + 1225*f^2*g*x^2 + 441*f*g^2*x^4 
+ 75*g^3*x^6)))/(3675*e^3) - (2*Sqrt[d]*(-35*e^3*f^3 + 35*d*e^2*f^2*g - 21 
*d^2*e*f*g^2 + 5*d^3*g^3)*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(35*e^(7/2)) + (x 
*(35*f^3 + 35*f^2*g*x^2 + 21*f*g^2*x^4 + 5*g^3*x^6)*Log[c*(d + e*x^2)^p])/ 
35
 

Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 338, 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.

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

\(\Big \downarrow \) 2921

\(\displaystyle \int \left (f^3 \log \left (c \left (d+e x^2\right )^p\right )+3 f^2 g x^2 \log \left (c \left (d+e x^2\right )^p\right )+3 f g^2 x^4 \log \left (c \left (d+e x^2\right )^p\right )+g^3 x^6 \log \left (c \left (d+e x^2\right )^p\right )\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 d^{3/2} f^2 g p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{e^{3/2}}+\frac {6 d^{5/2} f g^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 e^{5/2}}-\frac {2 d^{7/2} g^3 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{7 e^{7/2}}+\frac {2 \sqrt {d} f^3 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}+f^3 x \log \left (c \left (d+e x^2\right )^p\right )+f^2 g x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {3}{5} f g^2 x^5 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{7} g^3 x^7 \log \left (c \left (d+e x^2\right )^p\right )+\frac {2 d^3 g^3 p x}{7 e^3}-\frac {6 d^2 f g^2 p x}{5 e^2}-\frac {2 d^2 g^3 p x^3}{21 e^2}+\frac {2 d f^2 g p x}{e}+\frac {2 d f g^2 p x^3}{5 e}+\frac {2 d g^3 p x^5}{35 e}-2 f^3 p x-\frac {2}{3} f^2 g p x^3-\frac {6}{25} f g^2 p x^5-\frac {2}{49} g^3 p x^7\)

Input:

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

Output:

-2*f^3*p*x + (2*d*f^2*g*p*x)/e - (6*d^2*f*g^2*p*x)/(5*e^2) + (2*d^3*g^3*p* 
x)/(7*e^3) - (2*f^2*g*p*x^3)/3 + (2*d*f*g^2*p*x^3)/(5*e) - (2*d^2*g^3*p*x^ 
3)/(21*e^2) - (6*f*g^2*p*x^5)/25 + (2*d*g^3*p*x^5)/(35*e) - (2*g^3*p*x^7)/ 
49 + (2*Sqrt[d]*f^3*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/Sqrt[e] - (2*d^(3/2)*f^ 
2*g*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/e^(3/2) + (6*d^(5/2)*f*g^2*p*ArcTan[(Sq 
rt[e]*x)/Sqrt[d]])/(5*e^(5/2)) - (2*d^(7/2)*g^3*p*ArcTan[(Sqrt[e]*x)/Sqrt[ 
d]])/(7*e^(7/2)) + f^3*x*Log[c*(d + e*x^2)^p] + f^2*g*x^3*Log[c*(d + e*x^2 
)^p] + (3*f*g^2*x^5*Log[c*(d + e*x^2)^p])/5 + (g^3*x^7*Log[c*(d + e*x^2)^p 
])/7
 

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]))
 

Maple [A] (verified)

Time = 3.90 (sec) , antiderivative size = 258, normalized size of antiderivative = 0.76

method	result	size
parts	\(\frac {g^{3} x^{7} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{7}+\frac {3 f \,g^{2} x^{5} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{5}+f^{2} g \,x^{3} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )+f^{3} x \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )-\frac {2 e p \left (-\frac {-\frac {5}{7} e^{3} g^{3} x^{7}+d \,e^{2} g^{3} x^{5}-\frac {21}{5} e^{3} f \,g^{2} x^{5}-\frac {5}{3} d^{2} e \,g^{3} x^{3}+7 d \,e^{2} f \,g^{2} x^{3}-\frac {35}{3} e^{3} f^{2} g \,x^{3}+5 x \,d^{3} g^{3}-21 x e f \,g^{2} d^{2}+35 x d \,e^{2} f^{2} g -35 x \,e^{3} f^{3}}{e^{4}}+\frac {d \left (5 d^{3} g^{3}-21 e f \,g^{2} d^{2}+35 d \,e^{2} f^{2} g -35 e^{3} f^{3}\right ) \arctan \left (\frac {x e}{\sqrt {d e}}\right )}{e^{4} \sqrt {d e}}\right )}{35}\)	\(258\)
risch	\(\text {Expression too large to display}\)	\(995\)

Input:

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

Output:

1/7*g^3*x^7*ln(c*(e*x^2+d)^p)+3/5*f*g^2*x^5*ln(c*(e*x^2+d)^p)+f^2*g*x^3*ln 
(c*(e*x^2+d)^p)+f^3*x*ln(c*(e*x^2+d)^p)-2/35*e*p*(-1/e^4*(-5/7*e^3*g^3*x^7 
+d*e^2*g^3*x^5-21/5*e^3*f*g^2*x^5-5/3*d^2*e*g^3*x^3+7*d*e^2*f*g^2*x^3-35/3 
*e^3*f^2*g*x^3+5*x*d^3*g^3-21*x*e*f*g^2*d^2+35*x*d*e^2*f^2*g-35*x*e^3*f^3) 
+d*(5*d^3*g^3-21*d^2*e*f*g^2+35*d*e^2*f^2*g-35*e^3*f^3)/e^4/(d*e)^(1/2)*ar 
ctan(x*e/(d*e)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 596, normalized size of antiderivative = 1.76 \[ \int \left (f+g x^2\right )^3 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\left [-\frac {150 \, e^{3} g^{3} p x^{7} + 42 \, {\left (21 \, e^{3} f g^{2} - 5 \, d e^{2} g^{3}\right )} p x^{5} + 70 \, {\left (35 \, e^{3} f^{2} g - 21 \, d e^{2} f g^{2} + 5 \, d^{2} e g^{3}\right )} p x^{3} + 105 \, {\left (35 \, e^{3} f^{3} - 35 \, d e^{2} f^{2} g + 21 \, d^{2} e f g^{2} - 5 \, d^{3} g^{3}\right )} p \sqrt {-\frac {d}{e}} \log \left (\frac {e x^{2} - 2 \, e x \sqrt {-\frac {d}{e}} - d}{e x^{2} + d}\right ) + 210 \, {\left (35 \, e^{3} f^{3} - 35 \, d e^{2} f^{2} g + 21 \, d^{2} e f g^{2} - 5 \, d^{3} g^{3}\right )} p x - 105 \, {\left (5 \, e^{3} g^{3} p x^{7} + 21 \, e^{3} f g^{2} p x^{5} + 35 \, e^{3} f^{2} g p x^{3} + 35 \, e^{3} f^{3} p x\right )} \log \left (e x^{2} + d\right ) - 105 \, {\left (5 \, e^{3} g^{3} x^{7} + 21 \, e^{3} f g^{2} x^{5} + 35 \, e^{3} f^{2} g x^{3} + 35 \, e^{3} f^{3} x\right )} \log \left (c\right )}{3675 \, e^{3}}, -\frac {150 \, e^{3} g^{3} p x^{7} + 42 \, {\left (21 \, e^{3} f g^{2} - 5 \, d e^{2} g^{3}\right )} p x^{5} + 70 \, {\left (35 \, e^{3} f^{2} g - 21 \, d e^{2} f g^{2} + 5 \, d^{2} e g^{3}\right )} p x^{3} - 210 \, {\left (35 \, e^{3} f^{3} - 35 \, d e^{2} f^{2} g + 21 \, d^{2} e f g^{2} - 5 \, d^{3} g^{3}\right )} p \sqrt {\frac {d}{e}} \arctan \left (\frac {e x \sqrt {\frac {d}{e}}}{d}\right ) + 210 \, {\left (35 \, e^{3} f^{3} - 35 \, d e^{2} f^{2} g + 21 \, d^{2} e f g^{2} - 5 \, d^{3} g^{3}\right )} p x - 105 \, {\left (5 \, e^{3} g^{3} p x^{7} + 21 \, e^{3} f g^{2} p x^{5} + 35 \, e^{3} f^{2} g p x^{3} + 35 \, e^{3} f^{3} p x\right )} \log \left (e x^{2} + d\right ) - 105 \, {\left (5 \, e^{3} g^{3} x^{7} + 21 \, e^{3} f g^{2} x^{5} + 35 \, e^{3} f^{2} g x^{3} + 35 \, e^{3} f^{3} x\right )} \log \left (c\right )}{3675 \, e^{3}}\right ] \] Input:

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

Output:

[-1/3675*(150*e^3*g^3*p*x^7 + 42*(21*e^3*f*g^2 - 5*d*e^2*g^3)*p*x^5 + 70*( 
35*e^3*f^2*g - 21*d*e^2*f*g^2 + 5*d^2*e*g^3)*p*x^3 + 105*(35*e^3*f^3 - 35* 
d*e^2*f^2*g + 21*d^2*e*f*g^2 - 5*d^3*g^3)*p*sqrt(-d/e)*log((e*x^2 - 2*e*x* 
sqrt(-d/e) - d)/(e*x^2 + d)) + 210*(35*e^3*f^3 - 35*d*e^2*f^2*g + 21*d^2*e 
*f*g^2 - 5*d^3*g^3)*p*x - 105*(5*e^3*g^3*p*x^7 + 21*e^3*f*g^2*p*x^5 + 35*e 
^3*f^2*g*p*x^3 + 35*e^3*f^3*p*x)*log(e*x^2 + d) - 105*(5*e^3*g^3*x^7 + 21* 
e^3*f*g^2*x^5 + 35*e^3*f^2*g*x^3 + 35*e^3*f^3*x)*log(c))/e^3, -1/3675*(150 
*e^3*g^3*p*x^7 + 42*(21*e^3*f*g^2 - 5*d*e^2*g^3)*p*x^5 + 70*(35*e^3*f^2*g 
- 21*d*e^2*f*g^2 + 5*d^2*e*g^3)*p*x^3 - 210*(35*e^3*f^3 - 35*d*e^2*f^2*g + 
 21*d^2*e*f*g^2 - 5*d^3*g^3)*p*sqrt(d/e)*arctan(e*x*sqrt(d/e)/d) + 210*(35 
*e^3*f^3 - 35*d*e^2*f^2*g + 21*d^2*e*f*g^2 - 5*d^3*g^3)*p*x - 105*(5*e^3*g 
^3*p*x^7 + 21*e^3*f*g^2*p*x^5 + 35*e^3*f^2*g*p*x^3 + 35*e^3*f^3*p*x)*log(e 
*x^2 + d) - 105*(5*e^3*g^3*x^7 + 21*e^3*f*g^2*x^5 + 35*e^3*f^2*g*x^3 + 35* 
e^3*f^3*x)*log(c))/e^3]
 

Sympy [A] (verification not implemented)

Time = 125.04 (sec) , antiderivative size = 697, normalized size of antiderivative = 2.06 \[ \int \left (f+g x^2\right )^3 \log \left (c \left (d+e x^2\right )^p\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

Piecewise(((f**3*x + f**2*g*x**3 + 3*f*g**2*x**5/5 + g**3*x**7/7)*log(0**p 
*c), Eq(d, 0) & Eq(e, 0)), ((f**3*x + f**2*g*x**3 + 3*f*g**2*x**5/5 + g**3 
*x**7/7)*log(c*d**p), Eq(e, 0)), (-2*f**3*p*x + f**3*x*log(c*(e*x**2)**p) 
- 2*f**2*g*p*x**3/3 + f**2*g*x**3*log(c*(e*x**2)**p) - 6*f*g**2*p*x**5/25 
+ 3*f*g**2*x**5*log(c*(e*x**2)**p)/5 - 2*g**3*p*x**7/49 + g**3*x**7*log(c* 
(e*x**2)**p)/7, Eq(d, 0)), (-2*d**4*g**3*p*log(x - sqrt(-d/e))/(7*e**4*sqr 
t(-d/e)) + d**4*g**3*log(c*(d + e*x**2)**p)/(7*e**4*sqrt(-d/e)) + 6*d**3*f 
*g**2*p*log(x - sqrt(-d/e))/(5*e**3*sqrt(-d/e)) - 3*d**3*f*g**2*log(c*(d + 
 e*x**2)**p)/(5*e**3*sqrt(-d/e)) + 2*d**3*g**3*p*x/(7*e**3) - 2*d**2*f**2* 
g*p*log(x - sqrt(-d/e))/(e**2*sqrt(-d/e)) + d**2*f**2*g*log(c*(d + e*x**2) 
**p)/(e**2*sqrt(-d/e)) - 6*d**2*f*g**2*p*x/(5*e**2) - 2*d**2*g**3*p*x**3/( 
21*e**2) + 2*d*f**3*p*log(x - sqrt(-d/e))/(e*sqrt(-d/e)) - d*f**3*log(c*(d 
 + e*x**2)**p)/(e*sqrt(-d/e)) + 2*d*f**2*g*p*x/e + 2*d*f*g**2*p*x**3/(5*e) 
 + 2*d*g**3*p*x**5/(35*e) - 2*f**3*p*x + f**3*x*log(c*(d + e*x**2)**p) - 2 
*f**2*g*p*x**3/3 + f**2*g*x**3*log(c*(d + e*x**2)**p) - 6*f*g**2*p*x**5/25 
 + 3*f*g**2*x**5*log(c*(d + e*x**2)**p)/5 - 2*g**3*p*x**7/49 + g**3*x**7*l 
og(c*(d + e*x**2)**p)/7, True))
 

Maxima [F(-2)]

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

integrate((g*x^2+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
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 268, normalized size of antiderivative = 0.79 \[ \int \left (f+g x^2\right )^3 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-\frac {1}{49} \, {\left (2 \, g^{3} p - 7 \, g^{3} \log \left (c\right )\right )} x^{7} - \frac {{\left (42 \, e f g^{2} p - 10 \, d g^{3} p - 105 \, e f g^{2} \log \left (c\right )\right )} x^{5}}{175 \, e} - \frac {{\left (70 \, e^{2} f^{2} g p - 42 \, d e f g^{2} p + 10 \, d^{2} g^{3} p - 105 \, e^{2} f^{2} g \log \left (c\right )\right )} x^{3}}{105 \, e^{2}} + \frac {1}{35} \, {\left (5 \, g^{3} p x^{7} + 21 \, f g^{2} p x^{5} + 35 \, f^{2} g p x^{3} + 35 \, f^{3} p x\right )} \log \left (e x^{2} + d\right ) - \frac {{\left (70 \, e^{3} f^{3} p - 70 \, d e^{2} f^{2} g p + 42 \, d^{2} e f g^{2} p - 10 \, d^{3} g^{3} p - 35 \, e^{3} f^{3} \log \left (c\right )\right )} x}{35 \, e^{3}} + \frac {2 \, {\left (35 \, d e^{3} f^{3} p - 35 \, d^{2} e^{2} f^{2} g p + 21 \, d^{3} e f g^{2} p - 5 \, d^{4} g^{3} p\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{35 \, \sqrt {d e} e^{3}} \] Input:

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

Output:

-1/49*(2*g^3*p - 7*g^3*log(c))*x^7 - 1/175*(42*e*f*g^2*p - 10*d*g^3*p - 10 
5*e*f*g^2*log(c))*x^5/e - 1/105*(70*e^2*f^2*g*p - 42*d*e*f*g^2*p + 10*d^2* 
g^3*p - 105*e^2*f^2*g*log(c))*x^3/e^2 + 1/35*(5*g^3*p*x^7 + 21*f*g^2*p*x^5 
 + 35*f^2*g*p*x^3 + 35*f^3*p*x)*log(e*x^2 + d) - 1/35*(70*e^3*f^3*p - 70*d 
*e^2*f^2*g*p + 42*d^2*e*f*g^2*p - 10*d^3*g^3*p - 35*e^3*f^3*log(c))*x/e^3 
+ 2/35*(35*d*e^3*f^3*p - 35*d^2*e^2*f^2*g*p + 21*d^3*e*f*g^2*p - 5*d^4*g^3 
*p)*arctan(e*x/sqrt(d*e))/(sqrt(d*e)*e^3)
 

Mupad [B] (verification not implemented)

Time = 26.43 (sec) , antiderivative size = 298, normalized size of antiderivative = 0.88 \[ \int \left (f+g x^2\right )^3 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=x^3\,\left (\frac {d\,\left (\frac {6\,f\,g^2\,p}{5}-\frac {2\,d\,g^3\,p}{7\,e}\right )}{3\,e}-\frac {2\,f^2\,g\,p}{3}\right )-x\,\left (2\,f^3\,p+\frac {d\,\left (\frac {d\,\left (\frac {6\,f\,g^2\,p}{5}-\frac {2\,d\,g^3\,p}{7\,e}\right )}{e}-2\,f^2\,g\,p\right )}{e}\right )-x^5\,\left (\frac {6\,f\,g^2\,p}{25}-\frac {2\,d\,g^3\,p}{35\,e}\right )+\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )\,\left (f^3\,x+f^2\,g\,x^3+\frac {3\,f\,g^2\,x^5}{5}+\frac {g^3\,x^7}{7}\right )-\frac {2\,g^3\,p\,x^7}{49}-\frac {2\,\sqrt {d}\,p\,\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e}\,p\,x\,\left (5\,d^3\,g^3-21\,d^2\,e\,f\,g^2+35\,d\,e^2\,f^2\,g-35\,e^3\,f^3\right )}{5\,p\,d^4\,g^3-21\,p\,d^3\,e\,f\,g^2+35\,p\,d^2\,e^2\,f^2\,g-35\,p\,d\,e^3\,f^3}\right )\,\left (5\,d^3\,g^3-21\,d^2\,e\,f\,g^2+35\,d\,e^2\,f^2\,g-35\,e^3\,f^3\right )}{35\,e^{7/2}} \] Input:

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

Output:

x^3*((d*((6*f*g^2*p)/5 - (2*d*g^3*p)/(7*e)))/(3*e) - (2*f^2*g*p)/3) - x*(2 
*f^3*p + (d*((d*((6*f*g^2*p)/5 - (2*d*g^3*p)/(7*e)))/e - 2*f^2*g*p))/e) - 
x^5*((6*f*g^2*p)/25 - (2*d*g^3*p)/(35*e)) + log(c*(d + e*x^2)^p)*(f^3*x + 
(g^3*x^7)/7 + f^2*g*x^3 + (3*f*g^2*x^5)/5) - (2*g^3*p*x^7)/49 - (2*d^(1/2) 
*p*atan((d^(1/2)*e^(1/2)*p*x*(5*d^3*g^3 - 35*e^3*f^3 + 35*d*e^2*f^2*g - 21 
*d^2*e*f*g^2))/(5*d^4*g^3*p - 35*d*e^3*f^3*p - 21*d^3*e*f*g^2*p + 35*d^2*e 
^2*f^2*g*p))*(5*d^3*g^3 - 35*e^3*f^3 + 35*d*e^2*f^2*g - 21*d^2*e*f*g^2))/( 
35*e^(7/2))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 329, normalized size of antiderivative = 0.97 \[ \int \left (f+g x^2\right )^3 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {-1050 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) d^{3} g^{3} p +4410 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) d^{2} e f \,g^{2} p -7350 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) d \,e^{2} f^{2} g p +7350 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) e^{3} f^{3} p +3675 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) e^{4} f^{3} x +3675 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) e^{4} f^{2} g \,x^{3}+2205 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) e^{4} f \,g^{2} x^{5}+525 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) e^{4} g^{3} x^{7}+1050 d^{3} e \,g^{3} p x -4410 d^{2} e^{2} f \,g^{2} p x -350 d^{2} e^{2} g^{3} p \,x^{3}+7350 d \,e^{3} f^{2} g p x +1470 d \,e^{3} f \,g^{2} p \,x^{3}+210 d \,e^{3} g^{3} p \,x^{5}-7350 e^{4} f^{3} p x -2450 e^{4} f^{2} g p \,x^{3}-882 e^{4} f \,g^{2} p \,x^{5}-150 e^{4} g^{3} p \,x^{7}}{3675 e^{4}} \] Input:

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

Output:

( - 1050*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*d**3*g**3*p + 4410* 
sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*d**2*e*f*g**2*p - 7350*sqrt( 
e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*d*e**2*f**2*g*p + 7350*sqrt(e)*sq 
rt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*e**3*f**3*p + 3675*log((d + e*x**2)**p 
*c)*e**4*f**3*x + 3675*log((d + e*x**2)**p*c)*e**4*f**2*g*x**3 + 2205*log( 
(d + e*x**2)**p*c)*e**4*f*g**2*x**5 + 525*log((d + e*x**2)**p*c)*e**4*g**3 
*x**7 + 1050*d**3*e*g**3*p*x - 4410*d**2*e**2*f*g**2*p*x - 350*d**2*e**2*g 
**3*p*x**3 + 7350*d*e**3*f**2*g*p*x + 1470*d*e**3*f*g**2*p*x**3 + 210*d*e* 
*3*g**3*p*x**5 - 7350*e**4*f**3*p*x - 2450*e**4*f**2*g*p*x**3 - 882*e**4*f 
*g**2*p*x**5 - 150*e**4*g**3*p*x**7)/(3675*e**4)