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

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2513, 815, 649, 211, 266} \[ \int (d+e x)^3 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=-\frac {p \left (a^2 e^4-6 a b d^2 e^2+b^2 d^4\right ) \log \left (a+b x^2\right )}{4 b^2 e}+\frac {2 \sqrt {a} d p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (b d^2-a e^2\right )}{b^{3/2}}+\frac {(d+e x)^4 \log \left (c \left (a+b x^2\right )^p\right )}{4 e}-\frac {e p x^2 \left (6 b d^2-a e^2\right )}{4 b}-\frac {2 d p x \left (b d^2-a e^2\right )}{b}-\frac {2}{3} d e^2 p x^3-\frac {1}{8} e^3 p x^4 \]

[In]

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

[Out]

(-2*d*(b*d^2 - a*e^2)*p*x)/b - (e*(6*b*d^2 - a*e^2)*p*x^2)/(4*b) - (2*d*e^2*p*x^3)/3 - (e^3*p*x^4)/8 + (2*Sqrt
[a]*d*(b*d^2 - a*e^2)*p*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/b^(3/2) - ((b^2*d^4 - 6*a*b*d^2*e^2 + a^2*e^4)*p*Log[a +
b*x^2])/(4*b^2*e) + ((d + e*x)^4*Log[c*(a + b*x^2)^p])/(4*e)

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 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 649

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[(-a)*c]

Rule 815

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m*((f + g*x)/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 2513

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] - Dist[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]

Rubi steps \begin{align*} \text {integral}& = \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}{a+b x^2} \, dx}{2 e} \\ & = \frac {(d+e x)^4 \log \left (c \left (a+b x^2\right )^p\right )}{4 e}-\frac {(b p) \int \left (\frac {4 d e \left (b d^2-a e^2\right )}{b^2}+\frac {e^2 \left (6 b d^2-a e^2\right ) x}{b^2}+\frac {4 d e^3 x^2}{b}+\frac {e^4 x^3}{b}-\frac {4 a d e \left (b d^2-a e^2\right )-\left (b^2 d^4-6 a b d^2 e^2+a^2 e^4\right ) x}{b^2 \left (a+b x^2\right )}\right ) \, dx}{2 e} \\ & = -\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 {(d+e x)^4 \log \left (c \left (a+b x^2\right )^p\right )}{4 e}+\frac {p \int \frac {4 a d e \left (b d^2-a e^2\right )-\left (b^2 d^4-6 a b d^2 e^2+a^2 e^4\right ) x}{a+b x^2} \, dx}{2 b e} \\ & = -\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 {(d+e x)^4 \log \left (c \left (a+b x^2\right )^p\right )}{4 e}+\frac {\left (2 a d \left (b d^2-a e^2\right ) p\right ) \int \frac {1}{a+b x^2} \, dx}{b}+\frac {\left (\left (-b^2 d^4+6 a b d^2 e^2-a^2 e^4\right ) p\right ) \int \frac {x}{a+b x^2} \, dx}{2 b e} \\ & = -\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 \tan ^{-1}\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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.40 \[ \int (d+e x)^3 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {-6 \left (b^2 d^4+4 \sqrt {-a} b^{3/2} d^3 e-6 a b d^2 e^2+4 (-a)^{3/2} \sqrt {b} d e^3+a^2 e^4\right ) p \log \left (\sqrt {-a}-\sqrt {b} x\right )-6 \left (b^2 d^4-4 \sqrt {-a} b^{3/2} d^3 e-6 a b d^2 e^2+4 \sqrt {-a} a \sqrt {b} d e^3+a^2 e^4\right ) p \log \left (\sqrt {-a}+\sqrt {b} x\right )+b \left (6 a e^3 p x (8 d+e x)-b e p x \left (48 d^3+36 d^2 e x+16 d e^2 x^2+3 e^3 x^3\right )+6 b (d+e x)^4 \log \left (c \left (a+b x^2\right )^p\right )\right )}{24 b^2 e} \]

[In]

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

[Out]

(-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)*Sqrt[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)

Maple [A] (verified)

Time = 1.79 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.37

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

[In]

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

[Out]

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

none

Time = 0.37 (sec) , antiderivative size = 498, normalized size of antiderivative = 2.80 \[ \int (d+e x)^3 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\left [-\frac {3 \, b^{2} e^{3} p x^{4} + 16 \, b^{2} d e^{2} p x^{3} + 6 \, {\left (6 \, b^{2} d^{2} e - a b e^{3}\right )} p x^{2} - 24 \, {\left (b^{2} d^{3} - a b d e^{2}\right )} p \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} + 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) + 48 \, {\left (b^{2} d^{3} - a b d e^{2}\right )} p x - 6 \, {\left (b^{2} e^{3} p x^{4} + 4 \, b^{2} d e^{2} p x^{3} + 6 \, b^{2} d^{2} e p x^{2} + 4 \, b^{2} d^{3} p x + {\left (6 \, a b d^{2} e - a^{2} e^{3}\right )} p\right )} \log \left (b x^{2} + a\right ) - 6 \, {\left (b^{2} e^{3} x^{4} + 4 \, b^{2} d e^{2} x^{3} + 6 \, b^{2} d^{2} e x^{2} + 4 \, b^{2} d^{3} x\right )} \log \left (c\right )}{24 \, b^{2}}, -\frac {3 \, b^{2} e^{3} p x^{4} + 16 \, b^{2} d e^{2} p x^{3} + 6 \, {\left (6 \, b^{2} d^{2} e - a b e^{3}\right )} p x^{2} - 48 \, {\left (b^{2} d^{3} - a b d e^{2}\right )} p \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) + 48 \, {\left (b^{2} d^{3} - a b d e^{2}\right )} p x - 6 \, {\left (b^{2} e^{3} p x^{4} + 4 \, b^{2} d e^{2} p x^{3} + 6 \, b^{2} d^{2} e p x^{2} + 4 \, b^{2} d^{3} p x + {\left (6 \, a b d^{2} e - a^{2} e^{3}\right )} p\right )} \log \left (b x^{2} + a\right ) - 6 \, {\left (b^{2} e^{3} x^{4} + 4 \, b^{2} d e^{2} x^{3} + 6 \, b^{2} d^{2} e x^{2} + 4 \, b^{2} d^{3} x\right )} \log \left (c\right )}{24 \, b^{2}}\right ] \]

[In]

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

[Out]

[-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*sq
rt(-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)*arctan(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 = 17.80 (sec) , antiderivative size = 527, normalized size of antiderivative = 2.96 \[ \int (d+e x)^3 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\begin {cases} \left (d^{3} x + \frac {3 d^{2} e x^{2}}{2} + d e^{2} x^{3} + \frac {e^{3} x^{4}}{4}\right ) \log {\left (0^{p} c \right )} & \text {for}\: a = 0 \wedge b = 0 \\\left (d^{3} x + \frac {3 d^{2} e x^{2}}{2} + d e^{2} x^{3} + \frac {e^{3} x^{4}}{4}\right ) \log {\left (a^{p} c \right )} & \text {for}\: b = 0 \\- 2 d^{3} p x + d^{3} x \log {\left (c \left (b x^{2}\right )^{p} \right )} - \frac {3 d^{2} e p x^{2}}{2} + \frac {3 d^{2} e x^{2} \log {\left (c \left (b x^{2}\right )^{p} \right )}}{2} - \frac {2 d e^{2} p x^{3}}{3} + d e^{2} x^{3} \log {\left (c \left (b x^{2}\right )^{p} \right )} - \frac {e^{3} p x^{4}}{8} + \frac {e^{3} x^{4} \log {\left (c \left (b x^{2}\right )^{p} \right )}}{4} & \text {for}\: a = 0 \\- \frac {2 a^{2} d e^{2} p \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{b^{2} \sqrt {- \frac {a}{b}}} + \frac {a^{2} d e^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{b^{2} \sqrt {- \frac {a}{b}}} - \frac {a^{2} e^{3} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{4 b^{2}} + \frac {2 a d^{3} p \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{b \sqrt {- \frac {a}{b}}} - \frac {a d^{3} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{b \sqrt {- \frac {a}{b}}} + \frac {3 a d^{2} e \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{2 b} + \frac {2 a d e^{2} p x}{b} + \frac {a e^{3} p x^{2}}{4 b} - 2 d^{3} p x + d^{3} x \log {\left (c \left (a + b x^{2}\right )^{p} \right )} - \frac {3 d^{2} e p x^{2}}{2} + \frac {3 d^{2} e x^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{2} - \frac {2 d e^{2} p x^{3}}{3} + d e^{2} x^{3} \log {\left (c \left (a + b x^{2}\right )^{p} \right )} - \frac {e^{3} p x^{4}}{8} + \frac {e^{3} x^{4} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{4} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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*sqr
t(-a/b)) + a**2*d*e**2*log(c*(a + b*x**2)**p)/(b**2*sqrt(-a/b)) - a**2*e**3*log(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)

none

Time = 0.30 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.99 \[ \int (d+e x)^3 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {1}{24} \, b p {\left (\frac {48 \, {\left (a b d^{3} - a^{2} d e^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} - \frac {3 \, b e^{3} x^{4} + 16 \, b d e^{2} x^{3} + 6 \, {\left (6 \, b d^{2} e - a e^{3}\right )} x^{2} + 48 \, {\left (b d^{3} - a d e^{2}\right )} x}{b^{2}} + \frac {6 \, {\left (6 \, a b d^{2} e - a^{2} e^{3}\right )} \log \left (b x^{2} + a\right )}{b^{3}}\right )} + \frac {1}{4} \, {\left (e^{3} x^{4} + 4 \, d e^{2} x^{3} + 6 \, d^{2} e x^{2} + 4 \, d^{3} x\right )} \log \left ({\left (b x^{2} + a\right )}^{p} c\right ) \]

[In]

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

[Out]

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)

none

Time = 0.33 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.22 \[ \int (d+e x)^3 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=-\frac {1}{8} \, {\left (e^{3} p - 2 \, e^{3} \log \left (c\right )\right )} x^{4} - \frac {1}{3} \, {\left (2 \, d e^{2} p - 3 \, d e^{2} \log \left (c\right )\right )} x^{3} - \frac {{\left (6 \, b d^{2} e p - a e^{3} p - 6 \, b d^{2} e \log \left (c\right )\right )} x^{2}}{4 \, b} + \frac {1}{4} \, {\left (e^{3} p x^{4} + 4 \, d e^{2} p x^{3} + 6 \, d^{2} e p x^{2} + 4 \, d^{3} p x\right )} \log \left (b x^{2} + a\right ) - \frac {{\left (2 \, b d^{3} p - 2 \, a d e^{2} p - b d^{3} \log \left (c\right )\right )} x}{b} + \frac {2 \, {\left (a b d^{3} p - a^{2} d e^{2} p\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b} + \frac {{\left (6 \, a b d^{2} e p - a^{2} e^{3} p\right )} \log \left (b x^{2} + a\right )}{4 \, b^{2}} \]

[In]

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

[Out]

-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 = 1.44 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.25 \[ \int (d+e x)^3 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {e^3\,x^4\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}{4}-2\,d^3\,p\,x-\frac {e^3\,p\,x^4}{8}+d^3\,x\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )+\frac {3\,d^2\,e\,x^2\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}{2}+d\,e^2\,x^3\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )-\frac {3\,d^2\,e\,p\,x^2}{2}-\frac {2\,d\,e^2\,p\,x^3}{3}+\frac {a\,e^3\,p\,x^2}{4\,b}+\frac {2\,\sqrt {a}\,d^3\,p\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{\sqrt {b}}-\frac {a^2\,e^3\,p\,\ln \left (b\,x^2+a\right )}{4\,b^2}-\frac {2\,a^{3/2}\,d\,e^2\,p\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{b^{3/2}}+\frac {2\,a\,d\,e^2\,p\,x}{b}+\frac {3\,a\,d^2\,e\,p\,\ln \left (b\,x^2+a\right )}{2\,b} \]

[In]

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

[Out]

(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)