\(\int (d+e x)^3 \log (c (a+b x^2)^p) \, dx\) [184]

Optimal result

Integrand size = 20, antiderivative size = 178 \[ \int (d+e x)^3 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=-\frac {2 d \left (b d^2-a e^2\right ) p x}{b}-\frac {e \left (6 b d^2-a e^2\right ) p x^2}{4 b}-\frac {2}{3} d e^2 p x^3-\frac {1}{8} e^3 p x^4+\frac {2 \sqrt {a} d \left (b d^2-a e^2\right ) p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{3/2}}-\frac {\left (b^2 d^4-6 a b d^2 e^2+a^2 e^4\right ) p \log \left (a+b x^2\right )}{4 b^2 e}+\frac {(d+e x)^4 \log \left (c \left (a+b x^2\right )^p\right )}{4 e} \] Output:

-2*d*(-a*e^2+b*d^2)*p*x/b-1/4*e*(-a*e^2+6*b*d^2)*p*x^2/b-2/3*d*e^2*p*x^3-1 
/8*e^3*p*x^4+2*a^(1/2)*d*(-a*e^2+b*d^2)*p*arctan(b^(1/2)*x/a^(1/2))/b^(3/2 
)-1/4*(a^2*e^4-6*a*b*d^2*e^2+b^2*d^4)*p*ln(b*x^2+a)/b^2/e+1/4*(e*x+d)^4*ln 
(c*(b*x^2+a)^p)/e
 

Mathematica [A] (verified)

Time = 0.84 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.40 \[ \int (d+e x)^3 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {-6 \left (b^2 d^4+4 \sqrt {-a} b^{3/2} d^3 e-6 a b d^2 e^2+4 (-a)^{3/2} \sqrt {b} d e^3+a^2 e^4\right ) p \log \left (\sqrt {-a}-\sqrt {b} x\right )-6 \left (b^2 d^4-4 \sqrt {-a} b^{3/2} d^3 e-6 a b d^2 e^2+4 \sqrt {-a} a \sqrt {b} d e^3+a^2 e^4\right ) p \log \left (\sqrt {-a}+\sqrt {b} x\right )+b \left (6 a e^3 p x (8 d+e x)-b e p x \left (48 d^3+36 d^2 e x+16 d e^2 x^2+3 e^3 x^3\right )+6 b (d+e x)^4 \log \left (c \left (a+b x^2\right )^p\right )\right )}{24 b^2 e} \] Input:

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

Output:

(-6*(b^2*d^4 + 4*Sqrt[-a]*b^(3/2)*d^3*e - 6*a*b*d^2*e^2 + 4*(-a)^(3/2)*Sqr 
t[b]*d*e^3 + a^2*e^4)*p*Log[Sqrt[-a] - Sqrt[b]*x] - 6*(b^2*d^4 - 4*Sqrt[-a 
]*b^(3/2)*d^3*e - 6*a*b*d^2*e^2 + 4*Sqrt[-a]*a*Sqrt[b]*d*e^3 + a^2*e^4)*p* 
Log[Sqrt[-a] + Sqrt[b]*x] + b*(6*a*e^3*p*x*(8*d + e*x) - b*e*p*x*(48*d^3 + 
 36*d^2*e*x + 16*d*e^2*x^2 + 3*e^3*x^3) + 6*b*(d + e*x)^4*Log[c*(a + b*x^2 
)^p]))/(24*b^2*e)
 

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2913, 525, 2333, 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.

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

\(\Big \downarrow \) 2913

\(\displaystyle \frac {(d+e x)^4 \log \left (c \left (a+b x^2\right )^p\right )}{4 e}-\frac {b p \int \frac {x (d+e x)^4}{b x^2+a}dx}{2 e}\)

\(\Big \downarrow \) 525

\(\displaystyle \frac {(d+e x)^4 \log \left (c \left (a+b x^2\right )^p\right )}{4 e}-\frac {b p \left (\frac {\int \frac {x \left (b d^4+4 b e x d^3+4 b e^3 x^3 d+e^2 \left (6 b d^2-a e^2\right ) x^2\right )}{b x^2+a}dx}{b}+\frac {e^4 x^4}{4 b}\right )}{2 e}\)

\(\Big \downarrow \) 2333

\(\displaystyle \frac {(d+e x)^4 \log \left (c \left (a+b x^2\right )^p\right )}{4 e}-\frac {b p \left (\frac {\int \left (4 d x^2 e^3+\frac {\left (6 b d^2-a e^2\right ) x e^2}{b}+4 d \left (d^2-\frac {a e^2}{b}\right ) e-\frac {4 a d e \left (b d^2-a e^2\right )-\left (b^2 d^4-6 a b e^2 d^2+a^2 e^4\right ) x}{b \left (b x^2+a\right )}\right )dx}{b}+\frac {e^4 x^4}{4 b}\right )}{2 e}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {(d+e x)^4 \log \left (c \left (a+b x^2\right )^p\right )}{4 e}-\frac {b p \left (\frac {\frac {\left (a^2 e^4-6 a b d^2 e^2+b^2 d^4\right ) \log \left (a+b x^2\right )}{2 b^2}-\frac {4 \sqrt {a} d e \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (b d^2-a e^2\right )}{b^{3/2}}+\frac {e^2 x^2 \left (6 b d^2-a e^2\right )}{2 b}+4 d e x \left (d^2-\frac {a e^2}{b}\right )+\frac {4}{3} d e^3 x^3}{b}+\frac {e^4 x^4}{4 b}\right )}{2 e}\)

Input:

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

Output:

-1/2*(b*p*((e^4*x^4)/(4*b) + (4*d*e*(d^2 - (a*e^2)/b)*x + (e^2*(6*b*d^2 - 
a*e^2)*x^2)/(2*b) + (4*d*e^3*x^3)/3 - (4*Sqrt[a]*d*e*(b*d^2 - a*e^2)*ArcTa 
n[(Sqrt[b]*x)/Sqrt[a]])/b^(3/2) + ((b^2*d^4 - 6*a*b*d^2*e^2 + a^2*e^4)*Log 
[a + b*x^2])/(2*b^2))/b))/e + ((d + e*x)^4*Log[c*(a + b*x^2)^p])/(4*e)
 

Defintions of rubi rules used

rule 525

Int[((x_)^(m_.)*((c_) + (d_.)*(x_))^(n_))/((a_) + (b_.)*(x_)^2), x_Symbol] 
:> Simp[d^n*(x^(m + n - 1)/(b*(m + n - 1))), x] + Simp[1/b   Int[x^m*(Expan 
dToSum[b*(c + d*x)^n - b*d^n*x^n - a*d^n*x^(n - 2), x]/(a + b*x^2)), x], x] 
 /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 1] && IGtQ[m, -2] && NeQ[m + n - 1, 0 
]
 

rule 2009

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

rule 2333

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ 
ExpandIntegrand[(c*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] 
&& PolyQ[Pq, x] && IGtQ[p, -2]
 

rule 2913

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]
 

Maple [A] (verified)

Time = 2.02 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.37

method	result	size
parts	\(\frac {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) e^{3} x^{4}}{4}+\ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) e^{2} d \,x^{3}+\frac {3 \ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) e \,d^{2} x^{2}}{2}+d^{3} x \ln \left (c \left (b \,x^{2}+a \right )^{p}\right )+\frac {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) d^{4}}{4 e}-\frac {p b \left (-\frac {e \left (-\frac {1}{4} x^{4} b \,e^{3}-\frac {4}{3} x^{3} b d \,e^{2}+\frac {1}{2} a \,e^{3} x^{2}-3 e \,d^{2} b \,x^{2}+4 x \,e^{2} d a -4 b x \,d^{3}\right )}{b^{2}}+\frac {\frac {\left (a^{2} e^{4}-6 a b \,d^{2} e^{2}+b^{2} d^{4}\right ) \ln \left (b \,x^{2}+a \right )}{2 b}+\frac {\left (4 a^{2} d \,e^{3}-4 a b \,d^{3} e \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}}}{b^{2}}\right )}{2 e}\)	\(244\)
risch	\(\text {Expression too large to display}\)	\(1330\)

Input:

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

Output:

1/4*ln(c*(b*x^2+a)^p)*e^3*x^4+ln(c*(b*x^2+a)^p)*e^2*d*x^3+3/2*ln(c*(b*x^2+ 
a)^p)*e*d^2*x^2+d^3*x*ln(c*(b*x^2+a)^p)+1/4*ln(c*(b*x^2+a)^p)/e*d^4-1/2*p* 
b/e*(-e/b^2*(-1/4*x^4*b*e^3-4/3*x^3*b*d*e^2+1/2*a*e^3*x^2-3*e*d^2*b*x^2+4* 
x*e^2*d*a-4*b*x*d^3)+1/b^2*(1/2*(a^2*e^4-6*a*b*d^2*e^2+b^2*d^4)/b*ln(b*x^2 
+a)+(4*a^2*d*e^3-4*a*b*d^3*e)/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 498, normalized size of antiderivative = 2.80 \[ \int (d+e x)^3 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\left [-\frac {3 \, b^{2} e^{3} p x^{4} + 16 \, b^{2} d e^{2} p x^{3} + 6 \, {\left (6 \, b^{2} d^{2} e - a b e^{3}\right )} p x^{2} + 24 \, {\left (b^{2} d^{3} - a b d e^{2}\right )} p \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} - 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) + 48 \, {\left (b^{2} d^{3} - a b d e^{2}\right )} p x - 6 \, {\left (b^{2} e^{3} p x^{4} + 4 \, b^{2} d e^{2} p x^{3} + 6 \, b^{2} d^{2} e p x^{2} + 4 \, b^{2} d^{3} p x + {\left (6 \, a b d^{2} e - a^{2} e^{3}\right )} p\right )} \log \left (b x^{2} + a\right ) - 6 \, {\left (b^{2} e^{3} x^{4} + 4 \, b^{2} d e^{2} x^{3} + 6 \, b^{2} d^{2} e x^{2} + 4 \, b^{2} d^{3} x\right )} \log \left (c\right )}{24 \, b^{2}}, -\frac {3 \, b^{2} e^{3} p x^{4} + 16 \, b^{2} d e^{2} p x^{3} + 6 \, {\left (6 \, b^{2} d^{2} e - a b e^{3}\right )} p x^{2} - 48 \, {\left (b^{2} d^{3} - a b d e^{2}\right )} p \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) + 48 \, {\left (b^{2} d^{3} - a b d e^{2}\right )} p x - 6 \, {\left (b^{2} e^{3} p x^{4} + 4 \, b^{2} d e^{2} p x^{3} + 6 \, b^{2} d^{2} e p x^{2} + 4 \, b^{2} d^{3} p x + {\left (6 \, a b d^{2} e - a^{2} e^{3}\right )} p\right )} \log \left (b x^{2} + a\right ) - 6 \, {\left (b^{2} e^{3} x^{4} + 4 \, b^{2} d e^{2} x^{3} + 6 \, b^{2} d^{2} e x^{2} + 4 \, b^{2} d^{3} x\right )} \log \left (c\right )}{24 \, b^{2}}\right ] \] Input:

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

Output:

[-1/24*(3*b^2*e^3*p*x^4 + 16*b^2*d*e^2*p*x^3 + 6*(6*b^2*d^2*e - a*b*e^3)*p 
*x^2 + 24*(b^2*d^3 - a*b*d*e^2)*p*sqrt(-a/b)*log((b*x^2 - 2*b*x*sqrt(-a/b) 
 - a)/(b*x^2 + a)) + 48*(b^2*d^3 - a*b*d*e^2)*p*x - 6*(b^2*e^3*p*x^4 + 4*b 
^2*d*e^2*p*x^3 + 6*b^2*d^2*e*p*x^2 + 4*b^2*d^3*p*x + (6*a*b*d^2*e - a^2*e^ 
3)*p)*log(b*x^2 + a) - 6*(b^2*e^3*x^4 + 4*b^2*d*e^2*x^3 + 6*b^2*d^2*e*x^2 
+ 4*b^2*d^3*x)*log(c))/b^2, -1/24*(3*b^2*e^3*p*x^4 + 16*b^2*d*e^2*p*x^3 + 
6*(6*b^2*d^2*e - a*b*e^3)*p*x^2 - 48*(b^2*d^3 - a*b*d*e^2)*p*sqrt(a/b)*arc 
tan(b*x*sqrt(a/b)/a) + 48*(b^2*d^3 - a*b*d*e^2)*p*x - 6*(b^2*e^3*p*x^4 + 4 
*b^2*d*e^2*p*x^3 + 6*b^2*d^2*e*p*x^2 + 4*b^2*d^3*p*x + (6*a*b*d^2*e - a^2* 
e^3)*p)*log(b*x^2 + a) - 6*(b^2*e^3*x^4 + 4*b^2*d*e^2*x^3 + 6*b^2*d^2*e*x^ 
2 + 4*b^2*d^3*x)*log(c))/b^2]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 527 vs. \(2 (170) = 340\).

Time = 17.51 (sec) , antiderivative size = 527, normalized size of antiderivative = 2.96 \[ \int (d+e x)^3 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\begin {cases} \left (d^{3} x + \frac {3 d^{2} e x^{2}}{2} + d e^{2} x^{3} + \frac {e^{3} x^{4}}{4}\right ) \log {\left (0^{p} c \right )} & \text {for}\: a = 0 \wedge b = 0 \\\left (d^{3} x + \frac {3 d^{2} e x^{2}}{2} + d e^{2} x^{3} + \frac {e^{3} x^{4}}{4}\right ) \log {\left (a^{p} c \right )} & \text {for}\: b = 0 \\- 2 d^{3} p x + d^{3} x \log {\left (c \left (b x^{2}\right )^{p} \right )} - \frac {3 d^{2} e p x^{2}}{2} + \frac {3 d^{2} e x^{2} \log {\left (c \left (b x^{2}\right )^{p} \right )}}{2} - \frac {2 d e^{2} p x^{3}}{3} + d e^{2} x^{3} \log {\left (c \left (b x^{2}\right )^{p} \right )} - \frac {e^{3} p x^{4}}{8} + \frac {e^{3} x^{4} \log {\left (c \left (b x^{2}\right )^{p} \right )}}{4} & \text {for}\: a = 0 \\- \frac {2 a^{2} d e^{2} p \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{b^{2} \sqrt {- \frac {a}{b}}} + \frac {a^{2} d e^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{b^{2} \sqrt {- \frac {a}{b}}} - \frac {a^{2} e^{3} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{4 b^{2}} + \frac {2 a d^{3} p \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{b \sqrt {- \frac {a}{b}}} - \frac {a d^{3} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{b \sqrt {- \frac {a}{b}}} + \frac {3 a d^{2} e \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{2 b} + \frac {2 a d e^{2} p x}{b} + \frac {a e^{3} p x^{2}}{4 b} - 2 d^{3} p x + d^{3} x \log {\left (c \left (a + b x^{2}\right )^{p} \right )} - \frac {3 d^{2} e p x^{2}}{2} + \frac {3 d^{2} e x^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{2} - \frac {2 d e^{2} p x^{3}}{3} + d e^{2} x^{3} \log {\left (c \left (a + b x^{2}\right )^{p} \right )} - \frac {e^{3} p x^{4}}{8} + \frac {e^{3} x^{4} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{4} & \text {otherwise} \end {cases} \] Input:

integrate((e*x+d)**3*ln(c*(b*x**2+a)**p),x)
 

Output:

Piecewise(((d**3*x + 3*d**2*e*x**2/2 + d*e**2*x**3 + e**3*x**4/4)*log(0**p 
*c), Eq(a, 0) & Eq(b, 0)), ((d**3*x + 3*d**2*e*x**2/2 + d*e**2*x**3 + e**3 
*x**4/4)*log(a**p*c), Eq(b, 0)), (-2*d**3*p*x + d**3*x*log(c*(b*x**2)**p) 
- 3*d**2*e*p*x**2/2 + 3*d**2*e*x**2*log(c*(b*x**2)**p)/2 - 2*d*e**2*p*x**3 
/3 + d*e**2*x**3*log(c*(b*x**2)**p) - e**3*p*x**4/8 + e**3*x**4*log(c*(b*x 
**2)**p)/4, Eq(a, 0)), (-2*a**2*d*e**2*p*log(x - sqrt(-a/b))/(b**2*sqrt(-a 
/b)) + a**2*d*e**2*log(c*(a + b*x**2)**p)/(b**2*sqrt(-a/b)) - a**2*e**3*lo 
g(c*(a + b*x**2)**p)/(4*b**2) + 2*a*d**3*p*log(x - sqrt(-a/b))/(b*sqrt(-a/ 
b)) - a*d**3*log(c*(a + b*x**2)**p)/(b*sqrt(-a/b)) + 3*a*d**2*e*log(c*(a + 
 b*x**2)**p)/(2*b) + 2*a*d*e**2*p*x/b + a*e**3*p*x**2/(4*b) - 2*d**3*p*x + 
 d**3*x*log(c*(a + b*x**2)**p) - 3*d**2*e*p*x**2/2 + 3*d**2*e*x**2*log(c*( 
a + b*x**2)**p)/2 - 2*d*e**2*p*x**3/3 + d*e**2*x**3*log(c*(a + b*x**2)**p) 
 - e**3*p*x**4/8 + e**3*x**4*log(c*(a + b*x**2)**p)/4, True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.99 \[ \int (d+e x)^3 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {1}{24} \, b p {\left (\frac {48 \, {\left (a b d^{3} - a^{2} d e^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} - \frac {3 \, b e^{3} x^{4} + 16 \, b d e^{2} x^{3} + 6 \, {\left (6 \, b d^{2} e - a e^{3}\right )} x^{2} + 48 \, {\left (b d^{3} - a d e^{2}\right )} x}{b^{2}} + \frac {6 \, {\left (6 \, a b d^{2} e - a^{2} e^{3}\right )} \log \left (b x^{2} + a\right )}{b^{3}}\right )} + \frac {1}{4} \, {\left (e^{3} x^{4} + 4 \, d e^{2} x^{3} + 6 \, d^{2} e x^{2} + 4 \, d^{3} x\right )} \log \left ({\left (b x^{2} + a\right )}^{p} c\right ) \] Input:

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

Output:

1/24*b*p*(48*(a*b*d^3 - a^2*d*e^2)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^2) - 
 (3*b*e^3*x^4 + 16*b*d*e^2*x^3 + 6*(6*b*d^2*e - a*e^3)*x^2 + 48*(b*d^3 - a 
*d*e^2)*x)/b^2 + 6*(6*a*b*d^2*e - a^2*e^3)*log(b*x^2 + a)/b^3) + 1/4*(e^3* 
x^4 + 4*d*e^2*x^3 + 6*d^2*e*x^2 + 4*d^3*x)*log((b*x^2 + a)^p*c)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.22 \[ \int (d+e x)^3 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=-\frac {1}{8} \, {\left (e^{3} p - 2 \, e^{3} \log \left (c\right )\right )} x^{4} - \frac {1}{3} \, {\left (2 \, d e^{2} p - 3 \, d e^{2} \log \left (c\right )\right )} x^{3} - \frac {{\left (6 \, b d^{2} e p - a e^{3} p - 6 \, b d^{2} e \log \left (c\right )\right )} x^{2}}{4 \, b} + \frac {1}{4} \, {\left (e^{3} p x^{4} + 4 \, d e^{2} p x^{3} + 6 \, d^{2} e p x^{2} + 4 \, d^{3} p x\right )} \log \left (b x^{2} + a\right ) - \frac {{\left (2 \, b d^{3} p - 2 \, a d e^{2} p - b d^{3} \log \left (c\right )\right )} x}{b} + \frac {2 \, {\left (a b d^{3} p - a^{2} d e^{2} p\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b} + \frac {{\left (6 \, a b d^{2} e p - a^{2} e^{3} p\right )} \log \left (b x^{2} + a\right )}{4 \, b^{2}} \] Input:

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

Output:

-1/8*(e^3*p - 2*e^3*log(c))*x^4 - 1/3*(2*d*e^2*p - 3*d*e^2*log(c))*x^3 - 1 
/4*(6*b*d^2*e*p - a*e^3*p - 6*b*d^2*e*log(c))*x^2/b + 1/4*(e^3*p*x^4 + 4*d 
*e^2*p*x^3 + 6*d^2*e*p*x^2 + 4*d^3*p*x)*log(b*x^2 + a) - (2*b*d^3*p - 2*a* 
d*e^2*p - b*d^3*log(c))*x/b + 2*(a*b*d^3*p - a^2*d*e^2*p)*arctan(b*x/sqrt( 
a*b))/(sqrt(a*b)*b) + 1/4*(6*a*b*d^2*e*p - a^2*e^3*p)*log(b*x^2 + a)/b^2
 

Mupad [B] (verification not implemented)

Time = 26.02 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.25 \[ \int (d+e x)^3 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {e^3\,x^4\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}{4}-2\,d^3\,p\,x-\frac {e^3\,p\,x^4}{8}+d^3\,x\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )+\frac {3\,d^2\,e\,x^2\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}{2}+d\,e^2\,x^3\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )-\frac {3\,d^2\,e\,p\,x^2}{2}-\frac {2\,d\,e^2\,p\,x^3}{3}+\frac {a\,e^3\,p\,x^2}{4\,b}+\frac {2\,\sqrt {a}\,d^3\,p\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{\sqrt {b}}-\frac {a^2\,e^3\,p\,\ln \left (b\,x^2+a\right )}{4\,b^2}-\frac {2\,a^{3/2}\,d\,e^2\,p\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{b^{3/2}}+\frac {2\,a\,d\,e^2\,p\,x}{b}+\frac {3\,a\,d^2\,e\,p\,\ln \left (b\,x^2+a\right )}{2\,b} \] Input:

int(log(c*(a + b*x^2)^p)*(d + e*x)^3,x)
 

Output:

(e^3*x^4*log(c*(a + b*x^2)^p))/4 - 2*d^3*p*x - (e^3*p*x^4)/8 + d^3*x*log(c 
*(a + b*x^2)^p) + (3*d^2*e*x^2*log(c*(a + b*x^2)^p))/2 + d*e^2*x^3*log(c*( 
a + b*x^2)^p) - (3*d^2*e*p*x^2)/2 - (2*d*e^2*p*x^3)/3 + (a*e^3*p*x^2)/(4*b 
) + (2*a^(1/2)*d^3*p*atan((b^(1/2)*x)/a^(1/2)))/b^(1/2) - (a^2*e^3*p*log(a 
 + b*x^2))/(4*b^2) - (2*a^(3/2)*d*e^2*p*atan((b^(1/2)*x)/a^(1/2)))/b^(3/2) 
 + (2*a*d*e^2*p*x)/b + (3*a*d^2*e*p*log(a + b*x^2))/(2*b)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.43 \[ \int (d+e x)^3 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {-48 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a d \,e^{2} p +48 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b \,d^{3} p -6 \,\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) a^{2} e^{3}+36 \,\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) a b \,d^{2} e +24 \,\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) b^{2} d^{3} x +36 \,\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) b^{2} d^{2} e \,x^{2}+24 \,\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) b^{2} d \,e^{2} x^{3}+6 \,\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) b^{2} e^{3} x^{4}+48 a b d \,e^{2} p x +6 a b \,e^{3} p \,x^{2}-48 b^{2} d^{3} p x -36 b^{2} d^{2} e p \,x^{2}-16 b^{2} d \,e^{2} p \,x^{3}-3 b^{2} e^{3} p \,x^{4}}{24 b^{2}} \] Input:

int((e*x+d)^3*log(c*(b*x^2+a)^p),x)
 

Output:

( - 48*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*d*e**2*p + 48*sqrt( 
b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*b*d**3*p - 6*log((a + b*x**2)**p* 
c)*a**2*e**3 + 36*log((a + b*x**2)**p*c)*a*b*d**2*e + 24*log((a + b*x**2)* 
*p*c)*b**2*d**3*x + 36*log((a + b*x**2)**p*c)*b**2*d**2*e*x**2 + 24*log((a 
 + b*x**2)**p*c)*b**2*d*e**2*x**3 + 6*log((a + b*x**2)**p*c)*b**2*e**3*x** 
4 + 48*a*b*d*e**2*p*x + 6*a*b*e**3*p*x**2 - 48*b**2*d**3*p*x - 36*b**2*d** 
2*e*p*x**2 - 16*b**2*d*e**2*p*x**3 - 3*b**2*e**3*p*x**4)/(24*b**2)