\(\int (d+e x)^2 \log (c (a+b x^2)^p) \, dx\) [185]

Optimal result

Integrand size = 20, antiderivative size = 141 \[ \int (d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=-\frac {2 \left (3 b d^2-a e^2\right ) p x}{3 b}-d e p x^2-\frac {2}{9} e^2 p x^3+\frac {2 \sqrt {a} \left (3 b d^2-a e^2\right ) p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{3 b^{3/2}}-\frac {d \left (b d^2-3 a e^2\right ) p \log \left (a+b x^2\right )}{3 b e}+\frac {(d+e x)^3 \log \left (c \left (a+b x^2\right )^p\right )}{3 e} \]

[Out]

-2/3*(-a*e^2+3*b*d^2)*p*x/b-d*e*p*x^2-2/9*e^2*p*x^3-1/3*d*(-3*a*e^2+b*d^2)*p*ln(b*x^2+a)/b/e+1/3*(e*x+d)^3*ln(
c*(b*x^2+a)^p)/e+2/3*(-a*e^2+3*b*d^2)*p*arctan(x*b^(1/2)/a^(1/2))*a^(1/2)/b^(3/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 141, 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} \[ \int (d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {2 \sqrt {a} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (3 b d^2-a e^2\right )}{3 b^{3/2}}+\frac {(d+e x)^3 \log \left (c \left (a+b x^2\right )^p\right )}{3 e}-\frac {d p \left (b d^2-3 a e^2\right ) \log \left (a+b x^2\right )}{3 b e}-\frac {2 p x \left (3 b d^2-a e^2\right )}{3 b}-d e p x^2-\frac {2}{9} e^2 p x^3 \]

[In]

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

[Out]

(-2*(3*b*d^2 - a*e^2)*p*x)/(3*b) - d*e*p*x^2 - (2*e^2*p*x^3)/9 + (2*Sqrt[a]*(3*b*d^2 - a*e^2)*p*ArcTan[(Sqrt[b
]*x)/Sqrt[a]])/(3*b^(3/2)) - (d*(b*d^2 - 3*a*e^2)*p*Log[a + b*x^2])/(3*b*e) + ((d + e*x)^3*Log[c*(a + b*x^2)^p
])/(3*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)^3 \log \left (c \left (a+b x^2\right )^p\right )}{3 e}-\frac {(2 b p) \int \frac {x (d+e x)^3}{a+b x^2} \, dx}{3 e} \\ & = \frac {(d+e x)^3 \log \left (c \left (a+b x^2\right )^p\right )}{3 e}-\frac {(2 b p) \int \left (\frac {e \left (3 b d^2-a e^2\right )}{b^2}+\frac {3 d e^2 x}{b}+\frac {e^3 x^2}{b}-\frac {a e \left (3 b d^2-a e^2\right )-b d \left (b d^2-3 a e^2\right ) x}{b^2 \left (a+b x^2\right )}\right ) \, dx}{3 e} \\ & = -\frac {2 \left (3 b d^2-a e^2\right ) p x}{3 b}-d e p x^2-\frac {2}{9} e^2 p x^3+\frac {(d+e x)^3 \log \left (c \left (a+b x^2\right )^p\right )}{3 e}+\frac {(2 p) \int \frac {a e \left (3 b d^2-a e^2\right )-b d \left (b d^2-3 a e^2\right ) x}{a+b x^2} \, dx}{3 b e} \\ & = -\frac {2 \left (3 b d^2-a e^2\right ) p x}{3 b}-d e p x^2-\frac {2}{9} e^2 p x^3+\frac {(d+e x)^3 \log \left (c \left (a+b x^2\right )^p\right )}{3 e}-\frac {\left (2 d \left (b d^2-3 a e^2\right ) p\right ) \int \frac {x}{a+b x^2} \, dx}{3 e}+\frac {\left (2 a \left (3 b d^2-a e^2\right ) p\right ) \int \frac {1}{a+b x^2} \, dx}{3 b} \\ & = -\frac {2 \left (3 b d^2-a e^2\right ) p x}{3 b}-d e p x^2-\frac {2}{9} e^2 p x^3+\frac {2 \sqrt {a} \left (3 b d^2-a e^2\right ) p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{3 b^{3/2}}-\frac {d \left (b d^2-3 a e^2\right ) p \log \left (a+b x^2\right )}{3 b e}+\frac {(d+e x)^3 \log \left (c \left (a+b x^2\right )^p\right )}{3 e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.50 \[ \int (d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {3 \left (-b^{3/2} d^3-3 \sqrt {-a} b d^2 e+3 a \sqrt {b} d e^2+\sqrt {-a} a e^3\right ) p \log \left (\sqrt {-a}-\sqrt {b} x\right )-3 \left (b^{3/2} d^3-3 \sqrt {-a} b d^2 e-3 a \sqrt {b} d e^2+\sqrt {-a} a e^3\right ) p \log \left (\sqrt {-a}+\sqrt {b} x\right )+\sqrt {b} \left (6 a e^3 p x-b e p x \left (18 d^2+9 d e x+2 e^2 x^2\right )+3 b (d+e x)^3 \log \left (c \left (a+b x^2\right )^p\right )\right )}{9 b^{3/2} e} \]

[In]

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

[Out]

(3*(-(b^(3/2)*d^3) - 3*Sqrt[-a]*b*d^2*e + 3*a*Sqrt[b]*d*e^2 + Sqrt[-a]*a*e^3)*p*Log[Sqrt[-a] - Sqrt[b]*x] - 3*
(b^(3/2)*d^3 - 3*Sqrt[-a]*b*d^2*e - 3*a*Sqrt[b]*d*e^2 + Sqrt[-a]*a*e^3)*p*Log[Sqrt[-a] + Sqrt[b]*x] + Sqrt[b]*
(6*a*e^3*p*x - b*e*p*x*(18*d^2 + 9*d*e*x + 2*e^2*x^2) + 3*b*(d + e*x)^3*Log[c*(a + b*x^2)^p]))/(9*b^(3/2)*e)

Maple [A] (verified)

Time = 1.22 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.33

method	result	size
parts	\(\frac {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) e^{2} x^{3}}{3}+\ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) e d \,x^{2}+d^{2} \ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) x +\frac {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) d^{3}}{3 e}-\frac {2 p b \left (-\frac {e \left (-\frac {1}{3} x^{3} b \,e^{2}-\frac {3}{2} b d e \,x^{2}+x a \,e^{2}-3 b \,d^{2} x \right )}{b^{2}}+\frac {\frac {\left (-3 a b d \,e^{2}+b^{2} d^{3}\right ) \ln \left (b \,x^{2}+a \right )}{2 b}+\frac {\left (a^{2} e^{3}-3 a b \,d^{2} e \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}}}{b^{2}}\right )}{3 e}\)	\(187\)
risch	\(-\frac {i e^{2} \pi \,x^{3} {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{3}}{6}-\frac {i x \pi \,d^{2} {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{3}}{2}-\frac {i e \pi d \,x^{2} \operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}+\frac {2 x a p \,e^{2}}{3 b}+\frac {p \ln \left (-a^{2} e^{3}+3 a b \,d^{2} e -\sqrt {-a^{3} b \,e^{6}+6 a^{2} b^{2} d^{2} e^{4}-9 a \,b^{3} d^{4} e^{2}}\, x \right ) \sqrt {-a^{3} b \,e^{6}+6 a^{2} b^{2} d^{2} e^{4}-9 a \,b^{3} d^{4} e^{2}}}{3 e \,b^{2}}-\frac {p \ln \left (-a^{2} e^{3}+3 a b \,d^{2} e +\sqrt {-a^{3} b \,e^{6}+6 a^{2} b^{2} d^{2} e^{4}-9 a \,b^{3} d^{4} e^{2}}\, x \right ) \sqrt {-a^{3} b \,e^{6}+6 a^{2} b^{2} d^{2} e^{4}-9 a \,b^{3} d^{4} e^{2}}}{3 e \,b^{2}}+\frac {\left (e x +d \right )^{3} \ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{3 e}-\frac {p \ln \left (-a^{2} e^{3}+3 a b \,d^{2} e -\sqrt {-a^{3} b \,e^{6}+6 a^{2} b^{2} d^{2} e^{4}-9 a \,b^{3} d^{4} e^{2}}\, x \right ) d^{3}}{3 e}-\frac {p \ln \left (-a^{2} e^{3}+3 a b \,d^{2} e +\sqrt {-a^{3} b \,e^{6}+6 a^{2} b^{2} d^{2} e^{4}-9 a \,b^{3} d^{4} e^{2}}\, x \right ) d^{3}}{3 e}+e \ln \left (c \right ) d \,x^{2}+\frac {i e^{2} \pi \,x^{3} {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{6}+\frac {i x \pi \,d^{2} {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\frac {i x \pi \,d^{2} \operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2}}{2}-\frac {i e \pi d \,x^{2} {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{3}}{2}+\frac {i e^{2} \pi \,x^{3} \operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2}}{6}+\frac {\ln \left (c \right ) e^{2} x^{3}}{3}+\frac {e p \ln \left (-a^{2} e^{3}+3 a b \,d^{2} e -\sqrt {-a^{3} b \,e^{6}+6 a^{2} b^{2} d^{2} e^{4}-9 a \,b^{3} d^{4} e^{2}}\, x \right ) a d}{b}+\frac {e p \ln \left (-a^{2} e^{3}+3 a b \,d^{2} e +\sqrt {-a^{3} b \,e^{6}+6 a^{2} b^{2} d^{2} e^{4}-9 a \,b^{3} d^{4} e^{2}}\, x \right ) a d}{b}+\ln \left (c \right ) d^{2} x -\frac {i x \pi \,d^{2} \operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i e^{2} \pi \,x^{3} \operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{6}+\frac {i e \pi d \,x^{2} {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\frac {i e \pi d \,x^{2} \operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2}}{2}-d e p \,x^{2}-\frac {2 e^{2} p \,x^{3}}{9}-2 d^{2} p x\)	\(965\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.27 \[ \int (d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\left [-\frac {2 \, b e^{2} p x^{3} + 9 \, b d e p x^{2} - 3 \, {\left (3 \, b d^{2} - a e^{2}\right )} p \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} + 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) + 6 \, {\left (3 \, b d^{2} - a e^{2}\right )} p x - 3 \, {\left (b e^{2} p x^{3} + 3 \, b d e p x^{2} + 3 \, b d^{2} p x + 3 \, a d e p\right )} \log \left (b x^{2} + a\right ) - 3 \, {\left (b e^{2} x^{3} + 3 \, b d e x^{2} + 3 \, b d^{2} x\right )} \log \left (c\right )}{9 \, b}, -\frac {2 \, b e^{2} p x^{3} + 9 \, b d e p x^{2} - 6 \, {\left (3 \, b d^{2} - a e^{2}\right )} p \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) + 6 \, {\left (3 \, b d^{2} - a e^{2}\right )} p x - 3 \, {\left (b e^{2} p x^{3} + 3 \, b d e p x^{2} + 3 \, b d^{2} p x + 3 \, a d e p\right )} \log \left (b x^{2} + a\right ) - 3 \, {\left (b e^{2} x^{3} + 3 \, b d e x^{2} + 3 \, b d^{2} x\right )} \log \left (c\right )}{9 \, b}\right ] \]

[In]

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

[Out]

[-1/9*(2*b*e^2*p*x^3 + 9*b*d*e*p*x^2 - 3*(3*b*d^2 - a*e^2)*p*sqrt(-a/b)*log((b*x^2 + 2*b*x*sqrt(-a/b) - a)/(b*
x^2 + a)) + 6*(3*b*d^2 - a*e^2)*p*x - 3*(b*e^2*p*x^3 + 3*b*d*e*p*x^2 + 3*b*d^2*p*x + 3*a*d*e*p)*log(b*x^2 + a)
- 3*(b*e^2*x^3 + 3*b*d*e*x^2 + 3*b*d^2*x)*log(c))/b, -1/9*(2*b*e^2*p*x^3 + 9*b*d*e*p*x^2 - 6*(3*b*d^2 - a*e^2
)*p*sqrt(a/b)*arctan(b*x*sqrt(a/b)/a) + 6*(3*b*d^2 - a*e^2)*p*x - 3*(b*e^2*p*x^3 + 3*b*d*e*p*x^2 + 3*b*d^2*p*x
+ 3*a*d*e*p)*log(b*x^2 + a) - 3*(b*e^2*x^3 + 3*b*d*e*x^2 + 3*b*d^2*x)*log(c))/b]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 374 vs. \(2 (131) = 262\).

Time = 8.66 (sec) , antiderivative size = 374, normalized size of antiderivative = 2.65 \[ \int (d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\begin {cases} \left (d^{2} x + d e x^{2} + \frac {e^{2} x^{3}}{3}\right ) \log {\left (0^{p} c \right )} & \text {for}\: a = 0 \wedge b = 0 \\\left (d^{2} x + d e x^{2} + \frac {e^{2} x^{3}}{3}\right ) \log {\left (a^{p} c \right )} & \text {for}\: b = 0 \\- 2 d^{2} p x + d^{2} x \log {\left (c \left (b x^{2}\right )^{p} \right )} - d e p x^{2} + d e x^{2} \log {\left (c \left (b x^{2}\right )^{p} \right )} - \frac {2 e^{2} p x^{3}}{9} + \frac {e^{2} x^{3} \log {\left (c \left (b x^{2}\right )^{p} \right )}}{3} & \text {for}\: a = 0 \\- \frac {2 a^{2} e^{2} p \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{3 b^{2} \sqrt {- \frac {a}{b}}} + \frac {a^{2} e^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{3 b^{2} \sqrt {- \frac {a}{b}}} + \frac {2 a d^{2} p \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{b \sqrt {- \frac {a}{b}}} - \frac {a d^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{b \sqrt {- \frac {a}{b}}} + \frac {a d e \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{b} + \frac {2 a e^{2} p x}{3 b} - 2 d^{2} p x + d^{2} x \log {\left (c \left (a + b x^{2}\right )^{p} \right )} - d e p x^{2} + d e x^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )} - \frac {2 e^{2} p x^{3}}{9} + \frac {e^{2} x^{3} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{3} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise(((d**2*x + d*e*x**2 + e**2*x**3/3)*log(0**p*c), Eq(a, 0) & Eq(b, 0)), ((d**2*x + d*e*x**2 + e**2*x**
3/3)*log(a**p*c), Eq(b, 0)), (-2*d**2*p*x + d**2*x*log(c*(b*x**2)**p) - d*e*p*x**2 + d*e*x**2*log(c*(b*x**2)**
p) - 2*e**2*p*x**3/9 + e**2*x**3*log(c*(b*x**2)**p)/3, Eq(a, 0)), (-2*a**2*e**2*p*log(x - sqrt(-a/b))/(3*b**2*
sqrt(-a/b)) + a**2*e**2*log(c*(a + b*x**2)**p)/(3*b**2*sqrt(-a/b)) + 2*a*d**2*p*log(x - sqrt(-a/b))/(b*sqrt(-a
/b)) - a*d**2*log(c*(a + b*x**2)**p)/(b*sqrt(-a/b)) + a*d*e*log(c*(a + b*x**2)**p)/b + 2*a*e**2*p*x/(3*b) - 2*
d**2*p*x + d**2*x*log(c*(a + b*x**2)**p) - d*e*p*x**2 + d*e*x**2*log(c*(a + b*x**2)**p) - 2*e**2*p*x**3/9 + e*
*2*x**3*log(c*(a + b*x**2)**p)/3, True))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.93 \[ \int (d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {1}{9} \, {\left (\frac {9 \, a d e \log \left (b x^{2} + a\right )}{b^{2}} + \frac {6 \, {\left (3 \, a b d^{2} - a^{2} e^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} - \frac {2 \, b e^{2} x^{3} + 9 \, b d e x^{2} + 6 \, {\left (3 \, b d^{2} - a e^{2}\right )} x}{b^{2}}\right )} b p + \frac {1}{3} \, {\left (e^{2} x^{3} + 3 \, d e x^{2} + 3 \, d^{2} x\right )} \log \left ({\left (b x^{2} + a\right )}^{p} c\right ) \]

[In]

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

[Out]

1/9*(9*a*d*e*log(b*x^2 + a)/b^2 + 6*(3*a*b*d^2 - a^2*e^2)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^2) - (2*b*e^2*x^3
+ 9*b*d*e*x^2 + 6*(3*b*d^2 - a*e^2)*x)/b^2)*b*p + 1/3*(e^2*x^3 + 3*d*e*x^2 + 3*d^2*x)*log((b*x^2 + a)^p*c)

Giac [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.08 \[ \int (d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=-\frac {1}{9} \, {\left (2 \, e^{2} p - 3 \, e^{2} \log \left (c\right )\right )} x^{3} + \frac {a d e p \log \left (b x^{2} + a\right )}{b} - {\left (d e p - d e \log \left (c\right )\right )} x^{2} + \frac {1}{3} \, {\left (e^{2} p x^{3} + 3 \, d e p x^{2} + 3 \, d^{2} p x\right )} \log \left (b x^{2} + a\right ) - \frac {{\left (6 \, b d^{2} p - 2 \, a e^{2} p - 3 \, b d^{2} \log \left (c\right )\right )} x}{3 \, b} + \frac {2 \, {\left (3 \, a b d^{2} p - a^{2} e^{2} p\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{3 \, \sqrt {a b} b} \]

[In]

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

[Out]

-1/9*(2*e^2*p - 3*e^2*log(c))*x^3 + a*d*e*p*log(b*x^2 + a)/b - (d*e*p - d*e*log(c))*x^2 + 1/3*(e^2*p*x^3 + 3*d
*e*p*x^2 + 3*d^2*p*x)*log(b*x^2 + a) - 1/3*(6*b*d^2*p - 2*a*e^2*p - 3*b*d^2*log(c))*x/b + 2/3*(3*a*b*d^2*p - a
^2*e^2*p)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b)

Mupad [B] (verification not implemented)

Time = 4.45 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.87 \[ \int (d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {e^2\,x^3\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}{3}-2\,d^2\,p\,x-\frac {2\,e^2\,p\,x^3}{9}+d^2\,x\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )+d\,e\,x^2\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )-d\,e\,p\,x^2+\frac {2\,a\,e^2\,p\,x}{3\,b}-\frac {2\,\sqrt {a}\,d^2\,p\,\mathrm {atan}\left (\frac {3\,\sqrt {a}\,b^{3/2}\,d^2\,p\,x}{a^2\,e^2\,p-3\,a\,b\,d^2\,p}-\frac {a^{3/2}\,\sqrt {b}\,e^2\,p\,x}{a^2\,e^2\,p-3\,a\,b\,d^2\,p}\right )}{\sqrt {b}}+\frac {2\,a^{3/2}\,e^2\,p\,\mathrm {atan}\left (\frac {3\,\sqrt {a}\,b^{3/2}\,d^2\,p\,x}{a^2\,e^2\,p-3\,a\,b\,d^2\,p}-\frac {a^{3/2}\,\sqrt {b}\,e^2\,p\,x}{a^2\,e^2\,p-3\,a\,b\,d^2\,p}\right )}{3\,b^{3/2}}+\frac {a\,d\,e\,p\,\ln \left (b\,x^2+a\right )}{b} \]

[In]

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

[Out]

(e^2*x^3*log(c*(a + b*x^2)^p))/3 - 2*d^2*p*x - (2*e^2*p*x^3)/9 + d^2*x*log(c*(a + b*x^2)^p) + d*e*x^2*log(c*(a
+ b*x^2)^p) - d*e*p*x^2 + (2*a*e^2*p*x)/(3*b) - (2*a^(1/2)*d^2*p*atan((3*a^(1/2)*b^(3/2)*d^2*p*x)/(a^2*e^2*p
- 3*a*b*d^2*p) - (a^(3/2)*b^(1/2)*e^2*p*x)/(a^2*e^2*p - 3*a*b*d^2*p)))/b^(1/2) + (2*a^(3/2)*e^2*p*atan((3*a^(1
/2)*b^(3/2)*d^2*p*x)/(a^2*e^2*p - 3*a*b*d^2*p) - (a^(3/2)*b^(1/2)*e^2*p*x)/(a^2*e^2*p - 3*a*b*d^2*p)))/(3*b^(3
/2)) + (a*d*e*p*log(a + b*x^2))/b