Integrand size = 20, antiderivative size = 141 \[ \int (d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=-\frac {2 \left (3 b d^2-a e^2\right ) p x}{3 b}-d e p x^2-\frac {2}{9} e^2 p x^3+\frac {2 \sqrt {a} \left (3 b d^2-a e^2\right ) p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{3 b^{3/2}}-\frac {d \left (b d^2-3 a e^2\right ) p \log \left (a+b x^2\right )}{3 b e}+\frac {(d+e x)^3 \log \left (c \left (a+b x^2\right )^p\right )}{3 e} \] Output:
-2/3*(-a*e^2+3*b*d^2)*p*x/b-d*e*p*x^2-2/9*e^2*p*x^3+2/3*a^(1/2)*(-a*e^2+3* b*d^2)*p*arctan(b^(1/2)*x/a^(1/2))/b^(3/2)-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
Time = 0.54 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.50 \[ \int (d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {3 \left (-b^{3/2} d^3-3 \sqrt {-a} b d^2 e+3 a \sqrt {b} d e^2+\sqrt {-a} a e^3\right ) p \log \left (\sqrt {-a}-\sqrt {b} x\right )-3 \left (b^{3/2} d^3-3 \sqrt {-a} b d^2 e-3 a \sqrt {b} d e^2+\sqrt {-a} a e^3\right ) p \log \left (\sqrt {-a}+\sqrt {b} x\right )+\sqrt {b} \left (6 a e^3 p x-b e p x \left (18 d^2+9 d e x+2 e^2 x^2\right )+3 b (d+e x)^3 \log \left (c \left (a+b x^2\right )^p\right )\right )}{9 b^{3/2} e} \] Input:
Integrate[(d + e*x)^2*Log[c*(a + b*x^2)^p],x]
Output:
(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)
Time = 0.71 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.07, 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.
\(\displaystyle \int (d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right ) \, dx\) |
\(\Big \downarrow \) 2913 |
\(\displaystyle \frac {(d+e x)^3 \log \left (c \left (a+b x^2\right )^p\right )}{3 e}-\frac {2 b p \int \frac {x (d+e x)^3}{b x^2+a}dx}{3 e}\) |
\(\Big \downarrow \) 525 |
\(\displaystyle \frac {(d+e x)^3 \log \left (c \left (a+b x^2\right )^p\right )}{3 e}-\frac {2 b p \left (\frac {\int \frac {x \left (b d^3+3 b e^2 x^2 d+e \left (3 b d^2-a e^2\right ) x\right )}{b x^2+a}dx}{b}+\frac {e^3 x^3}{3 b}\right )}{3 e}\) |
\(\Big \downarrow \) 2333 |
\(\displaystyle \frac {(d+e x)^3 \log \left (c \left (a+b x^2\right )^p\right )}{3 e}-\frac {2 b p \left (\frac {\int \left (3 d x e^2+\left (3 d^2-\frac {a e^2}{b}\right ) e-\frac {a e \left (3 b d^2-a e^2\right )-b d \left (b d^2-3 a e^2\right ) x}{b \left (b x^2+a\right )}\right )dx}{b}+\frac {e^3 x^3}{3 b}\right )}{3 e}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(d+e x)^3 \log \left (c \left (a+b x^2\right )^p\right )}{3 e}-\frac {2 b p \left (\frac {-\frac {\sqrt {a} e \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (3 b d^2-a e^2\right )}{b^{3/2}}+\frac {d \left (b d^2-3 a e^2\right ) \log \left (a+b x^2\right )}{2 b}+e x \left (3 d^2-\frac {a e^2}{b}\right )+\frac {3}{2} d e^2 x^2}{b}+\frac {e^3 x^3}{3 b}\right )}{3 e}\) |
Input:
Int[(d + e*x)^2*Log[c*(a + b*x^2)^p],x]
Output:
(-2*b*p*((e^3*x^3)/(3*b) + (e*(3*d^2 - (a*e^2)/b)*x + (3*d*e^2*x^2)/2 - (S qrt[a]*e*(3*b*d^2 - a*e^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/b^(3/2) + (d*(b*d^ 2 - 3*a*e^2)*Log[a + b*x^2])/(2*b))/b))/(3*e) + ((d + e*x)^3*Log[c*(a + b* x^2)^p])/(3*e)
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 ]
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]
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]
Time = 1.39 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.33
method | result | size |
parts | \(\frac {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) e^{2} x^{3}}{3}+\ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) e d \,x^{2}+d^{2} x \ln \left (c \left (b \,x^{2}+a \right )^{p}\right )+\frac {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) d^{3}}{3 e}-\frac {2 p b \left (-\frac {e \left (-\frac {1}{3} x^{3} b \,e^{2}-\frac {3}{2} b d e \,x^{2}+x a \,e^{2}-3 b x \,d^{2}\right )}{b^{2}}+\frac {\frac {\left (-3 a b d \,e^{2}+b^{2} d^{3}\right ) \ln \left (b \,x^{2}+a \right )}{2 b}+\frac {\left (a^{2} e^{3}-3 a b \,d^{2} e \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}}}{b^{2}}\right )}{3 e}\) | \(187\) |
risch | \(-d e p \,x^{2}-\frac {i x \pi \,d^{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 c \left (b \,x^{2}+a \right )^{p}\right )}^{3}}{6}+\frac {2 x a p \,e^{2}}{3 b}+\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 e \pi d \,x^{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 {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}}+\ln \left (c \right ) d e \,x^{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 ) 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}-\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 {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 {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}+\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 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 {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 {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 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 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}+x \ln \left (c \right ) d^{2}+\frac {\left (e x +d \right )^{3} \ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{3 e}-\frac {2 e^{2} p \,x^{3}}{9}-2 d^{2} p x\) | \(965\) |
Input:
int((e*x+d)^2*ln(c*(b*x^2+a)^p),x,method=_RETURNVERBOSE)
Output:
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*x*ln(c*(b*x^2+ a)^p)+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*x*d^2)+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))))
Time = 0.12 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.27 \[ \int (d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\left [-\frac {2 \, b e^{2} p x^{3} + 9 \, b d e p x^{2} + 3 \, {\left (3 \, b d^{2} - a e^{2}\right )} p \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} - 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) + 6 \, {\left (3 \, b d^{2} - a e^{2}\right )} p x - 3 \, {\left (b e^{2} p x^{3} + 3 \, b d e p x^{2} + 3 \, b d^{2} p x + 3 \, a d e p\right )} \log \left (b x^{2} + a\right ) - 3 \, {\left (b e^{2} x^{3} + 3 \, b d e x^{2} + 3 \, b d^{2} x\right )} \log \left (c\right )}{9 \, b}, -\frac {2 \, b e^{2} p x^{3} + 9 \, b d e p x^{2} - 6 \, {\left (3 \, b d^{2} - a e^{2}\right )} p \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) + 6 \, {\left (3 \, b d^{2} - a e^{2}\right )} p x - 3 \, {\left (b e^{2} p x^{3} + 3 \, b d e p x^{2} + 3 \, b d^{2} p x + 3 \, a d e p\right )} \log \left (b x^{2} + a\right ) - 3 \, {\left (b e^{2} x^{3} + 3 \, b d e x^{2} + 3 \, b d^{2} x\right )} \log \left (c\right )}{9 \, b}\right ] \] Input:
integrate((e*x+d)^2*log(c*(b*x^2+a)^p),x, algorithm="fricas")
Output:
[-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)*lo g((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]
Leaf count of result is larger than twice the leaf count of optimal. 374 vs. \(2 (131) = 262\).
Time = 8.72 (sec) , antiderivative size = 374, normalized size of antiderivative = 2.65 \[ \int (d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\begin {cases} \left (d^{2} x + d e x^{2} + \frac {e^{2} x^{3}}{3}\right ) \log {\left (0^{p} c \right )} & \text {for}\: a = 0 \wedge b = 0 \\\left (d^{2} x + d e x^{2} + \frac {e^{2} x^{3}}{3}\right ) \log {\left (a^{p} c \right )} & \text {for}\: b = 0 \\- 2 d^{2} p x + d^{2} x \log {\left (c \left (b x^{2}\right )^{p} \right )} - d e p x^{2} + d e x^{2} \log {\left (c \left (b x^{2}\right )^{p} \right )} - \frac {2 e^{2} p x^{3}}{9} + \frac {e^{2} x^{3} \log {\left (c \left (b x^{2}\right )^{p} \right )}}{3} & \text {for}\: a = 0 \\- \frac {2 a^{2} e^{2} p \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{3 b^{2} \sqrt {- \frac {a}{b}}} + \frac {a^{2} e^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{3 b^{2} \sqrt {- \frac {a}{b}}} + \frac {2 a d^{2} p \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{b \sqrt {- \frac {a}{b}}} - \frac {a d^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{b \sqrt {- \frac {a}{b}}} + \frac {a d e \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{b} + \frac {2 a e^{2} p x}{3 b} - 2 d^{2} p x + d^{2} x \log {\left (c \left (a + b x^{2}\right )^{p} \right )} - d e p x^{2} + d e x^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )} - \frac {2 e^{2} p x^{3}}{9} + \frac {e^{2} x^{3} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{3} & \text {otherwise} \end {cases} \] Input:
integrate((e*x+d)**2*ln(c*(b*x**2+a)**p),x)
Output:
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))
Time = 0.11 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.93 \[ \int (d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {1}{9} \, {\left (\frac {9 \, a d e \log \left (b x^{2} + a\right )}{b^{2}} + \frac {6 \, {\left (3 \, a b d^{2} - a^{2} e^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} - \frac {2 \, b e^{2} x^{3} + 9 \, b d e x^{2} + 6 \, {\left (3 \, b d^{2} - a e^{2}\right )} x}{b^{2}}\right )} b p + \frac {1}{3} \, {\left (e^{2} x^{3} + 3 \, d e x^{2} + 3 \, d^{2} x\right )} \log \left ({\left (b x^{2} + a\right )}^{p} c\right ) \] Input:
integrate((e*x+d)^2*log(c*(b*x^2+a)^p),x, algorithm="maxima")
Output:
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)
Time = 0.13 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.08 \[ \int (d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=-\frac {1}{9} \, {\left (2 \, e^{2} p - 3 \, e^{2} \log \left (c\right )\right )} x^{3} + \frac {a d e p \log \left (b x^{2} + a\right )}{b} - {\left (d e p - d e \log \left (c\right )\right )} x^{2} + \frac {1}{3} \, {\left (e^{2} p x^{3} + 3 \, d e p x^{2} + 3 \, d^{2} p x\right )} \log \left (b x^{2} + a\right ) - \frac {{\left (6 \, b d^{2} p - 2 \, a e^{2} p - 3 \, b d^{2} \log \left (c\right )\right )} x}{3 \, b} + \frac {2 \, {\left (3 \, a b d^{2} p - a^{2} e^{2} p\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{3 \, \sqrt {a b} b} \] Input:
integrate((e*x+d)^2*log(c*(b*x^2+a)^p),x, algorithm="giac")
Output:
-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)
Time = 28.78 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.87 \[ \int (d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {e^2\,x^3\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}{3}-2\,d^2\,p\,x-\frac {2\,e^2\,p\,x^3}{9}+d^2\,x\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )+d\,e\,x^2\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )-d\,e\,p\,x^2+\frac {2\,a\,e^2\,p\,x}{3\,b}-\frac {2\,\sqrt {a}\,d^2\,p\,\mathrm {atan}\left (\frac {3\,\sqrt {a}\,b^{3/2}\,d^2\,p\,x}{a^2\,e^2\,p-3\,a\,b\,d^2\,p}-\frac {a^{3/2}\,\sqrt {b}\,e^2\,p\,x}{a^2\,e^2\,p-3\,a\,b\,d^2\,p}\right )}{\sqrt {b}}+\frac {2\,a^{3/2}\,e^2\,p\,\mathrm {atan}\left (\frac {3\,\sqrt {a}\,b^{3/2}\,d^2\,p\,x}{a^2\,e^2\,p-3\,a\,b\,d^2\,p}-\frac {a^{3/2}\,\sqrt {b}\,e^2\,p\,x}{a^2\,e^2\,p-3\,a\,b\,d^2\,p}\right )}{3\,b^{3/2}}+\frac {a\,d\,e\,p\,\ln \left (b\,x^2+a\right )}{b} \] Input:
int(log(c*(a + b*x^2)^p)*(d + e*x)^2,x)
Output:
(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
Time = 0.18 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.26 \[ \int (d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {-6 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,e^{2} p +18 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b \,d^{2} p +9 \,\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) a b d e +9 \,\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) b^{2} d^{2} x +9 \,\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) b^{2} d e \,x^{2}+3 \,\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) b^{2} e^{2} x^{3}+6 a b \,e^{2} p x -18 b^{2} d^{2} p x -9 b^{2} d e p \,x^{2}-2 b^{2} e^{2} p \,x^{3}}{9 b^{2}} \] Input:
int((e*x+d)^2*log(c*(b*x^2+a)^p),x)
Output:
( - 6*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*e**2*p + 18*sqrt(b)* sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*b*d**2*p + 9*log((a + b*x**2)**p*c)* a*b*d*e + 9*log((a + b*x**2)**p*c)*b**2*d**2*x + 9*log((a + b*x**2)**p*c)* b**2*d*e*x**2 + 3*log((a + b*x**2)**p*c)*b**2*e**2*x**3 + 6*a*b*e**2*p*x - 18*b**2*d**2*p*x - 9*b**2*d*e*p*x**2 - 2*b**2*e**2*p*x**3)/(9*b**2)