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

Optimal. Leaf size=141 \[ \frac {2 \sqrt {a} p \left (3 b d^2-a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{3 b^{3/2}}+\frac {(d+e x)^3 \log \left (c \left (a+b x^2\right )^p\right )}{3 e}-\frac {d p \left (b d^2-3 a e^2\right ) \log \left (a+b x^2\right )}{3 b e}-\frac {2 p x \left (3 b d^2-a e^2\right )}{3 b}-d e p x^2-\frac {2}{9} e^2 p x^3 \]

[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)

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Rubi [A]  time = 0.13, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2463, 801, 635, 205, 260} \[ \frac {2 \sqrt {a} p \left (3 b d^2-a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{3 b^{3/2}}+\frac {(d+e x)^3 \log \left (c \left (a+b x^2\right )^p\right )}{3 e}-\frac {d p \left (b d^2-3 a e^2\right ) \log \left (a+b x^2\right )}{3 b e}-\frac {2 p x \left (3 b d^2-a e^2\right )}{3 b}-d e p x^2-\frac {2}{9} e^2 p x^3 \]

Antiderivative was successfully verified.

[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 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 260

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 635

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 801

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 2463

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*} \int (d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right ) \, dx &=\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*}

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Mathematica [A]  time = 0.48, size = 211, normalized size = 1.50 \[ \frac {3 p \left (-3 \sqrt {-a} b d^2 e+3 a \sqrt {b} d e^2+\sqrt {-a} a e^3-b^{3/2} d^3\right ) \log \left (\sqrt {-a}-\sqrt {b} x\right )-3 p \left (-3 \sqrt {-a} b d^2 e-3 a \sqrt {b} d e^2+\sqrt {-a} a e^3+b^{3/2} d^3\right ) \log \left (\sqrt {-a}+\sqrt {b} x\right )+\sqrt {b} \left (3 b (d+e x)^3 \log \left (c \left (a+b x^2\right )^p\right )+6 a e^3 p x-b e p x \left (18 d^2+9 d e x+2 e^2 x^2\right )\right )}{9 b^{3/2} e} \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.45, size = 320, normalized size = 2.27 \[ \left [-\frac {2 \, b e^{2} p x^{3} + 9 \, b d e p x^{2} - 3 \, {\left (3 \, b d^{2} - a e^{2}\right )} p \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} + 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) + 6 \, {\left (3 \, b d^{2} - a e^{2}\right )} p x - 3 \, {\left (b e^{2} p x^{3} + 3 \, b d e p x^{2} + 3 \, b d^{2} p x + 3 \, a d e p\right )} \log \left (b x^{2} + a\right ) - 3 \, {\left (b e^{2} x^{3} + 3 \, b d e x^{2} + 3 \, b d^{2} x\right )} \log \relax (c)}{9 \, b}, -\frac {2 \, b e^{2} p x^{3} + 9 \, b d e p x^{2} - 6 \, {\left (3 \, b d^{2} - a e^{2}\right )} p \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) + 6 \, {\left (3 \, b d^{2} - a e^{2}\right )} p x - 3 \, {\left (b e^{2} p x^{3} + 3 \, b d e p x^{2} + 3 \, b d^{2} p x + 3 \, a d e p\right )} \log \left (b x^{2} + a\right ) - 3 \, {\left (b e^{2} x^{3} + 3 \, b d e x^{2} + 3 \, b d^{2} x\right )} \log \relax (c)}{9 \, b}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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]

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giac [A]  time = 0.20, size = 173, normalized size = 1.23 \[ \frac {2 \, {\left (3 \, a b d^{2} p - a^{2} p e^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{3 \, \sqrt {a b} b} + \frac {3 \, b p x^{3} e^{2} \log \left (b x^{2} + a\right ) + 9 \, b d p x^{2} e \log \left (b x^{2} + a\right ) - 2 \, b p x^{3} e^{2} - 9 \, b d p x^{2} e + 9 \, b d^{2} p x \log \left (b x^{2} + a\right ) + 3 \, b x^{3} e^{2} \log \relax (c) + 9 \, b d x^{2} e \log \relax (c) - 18 \, b d^{2} p x + 9 \, a d p e \log \left (b x^{2} + a\right ) + 9 \, b d^{2} x \log \relax (c) + 6 \, a p x e^{2}}{9 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(3*a*b*d^2*p - a^2*p*e^2)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b) + 1/9*(3*b*p*x^3*e^2*log(b*x^2 + a) + 9*b*d*
p*x^2*e*log(b*x^2 + a) - 2*b*p*x^3*e^2 - 9*b*d*p*x^2*e + 9*b*d^2*p*x*log(b*x^2 + a) + 3*b*x^3*e^2*log(c) + 9*b
*d*x^2*e*log(c) - 18*b*d^2*p*x + 9*a*d*p*e*log(b*x^2 + a) + 9*b*d^2*x*log(c) + 6*a*p*x*e^2)/b

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maple [C]  time = 0.60, size = 965, normalized size = 6.84 \[ \frac {\left (e x +d \right )^{3} \ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{3 e}+d e \,x^{2} \ln \relax (c )-\frac {d^{3} 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 )}{3 e}-\frac {d^{3} 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 )}{3 e}+\frac {e^{2} x^{3} \ln \relax (c )}{3}+d^{2} x \ln \relax (c )-\frac {2 e^{2} p \,x^{3}}{9}-2 d^{2} p x -d e p \,x^{2}-\frac {i \pi d e \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}{2}+\frac {i \pi \,e^{2} x^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{2}}{6}+\frac {i \pi \,e^{2} x^{3} \mathrm {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{2}}{6}-\frac {i \pi d e \,x^{2} \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{3}}{2}+\frac {i \pi \,d^{2} x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{2}}{2}+\frac {i \pi \,d^{2} x \,\mathrm {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{2}}{2}+\frac {2 a \,e^{2} p x}{3 b}-\frac {i \pi \,e^{2} x^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}{6}+\frac {i \pi d e \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{2}}{2}+\frac {i \pi d e \,x^{2} \mathrm {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{2}}{2}-\frac {i \pi \,d^{2} x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}{2}-\frac {i \pi \,e^{2} x^{3} \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{3}}{6}-\frac {i \pi \,d^{2} x \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{3}}{2}+\frac {\sqrt {-a^{3} b \,e^{6}+6 a^{2} b^{2} d^{2} e^{4}-9 a \,b^{3} d^{4} e^{2}}\, 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 )}{3 b^{2} e}-\frac {\sqrt {-a^{3} b \,e^{6}+6 a^{2} b^{2} d^{2} e^{4}-9 a \,b^{3} d^{4} e^{2}}\, 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 )}{3 b^{2} e}+\frac {a d 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 )}{b}+\frac {a d 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 )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(e*x+d)^3/e*ln((b*x^2+a)^p)+d*e*x^2*ln(c)-1/3/e*p*ln(-a^2*e^3+3*a*b*d^2*e-(-a^3*b*e^6+6*a^2*b^2*d^2*e^4-9*
a*b^3*d^4*e^2)^(1/2)*x)*d^3-1/3/e*p*ln(-a^2*e^3+3*a*b*d^2*e+(-a^3*b*e^6+6*a^2*b^2*d^2*e^4-9*a*b^3*d^4*e^2)^(1/
2)*x)*d^3+1/3*e^2*x^3*ln(c)+d^2*x*ln(c)-2/9*e^2*p*x^3-2*d^2*p*x-d*e*p*x^2-1/2*I*e*Pi*d*x^2*csgn(I*(b*x^2+a)^p)
*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)+2/3/b*a*p*e^2*x+1/6*I*e^2*Pi*x^3*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2+
1/6*I*e^2*Pi*x^3*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)-1/2*I*e*Pi*d*x^2*csgn(I*c*(b*x^2+a)^p)^3+1/3/b^2/e*p*ln(-a^
2*e^3+3*a*b*d^2*e-(-a^3*b*e^6+6*a^2*b^2*d^2*e^4-9*a*b^3*d^4*e^2)^(1/2)*x)*(-a^3*b*e^6+6*a^2*b^2*d^2*e^4-9*a*b^
3*d^4*e^2)^(1/2)-1/3/b^2/e*p*ln(-a^2*e^3+3*a*b*d^2*e+(-a^3*b*e^6+6*a^2*b^2*d^2*e^4-9*a*b^3*d^4*e^2)^(1/2)*x)*(
-a^3*b*e^6+6*a^2*b^2*d^2*e^4-9*a*b^3*d^4*e^2)^(1/2)-1/6*I*e^2*Pi*x^3*csgn(I*c*(b*x^2+a)^p)^3-1/2*I*Pi*d^2*csgn
(I*c*(b*x^2+a)^p)^3*x-1/6*I*e^2*Pi*x^3*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)+1/2*I*e*Pi*d*x^2*cs
gn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2+1/2*I*e*Pi*d*x^2*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)-1/2*I*Pi*d^2*csgn
(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)*x+1/b*e*p*ln(-a^2*e^3+3*a*b*d^2*e-(-a^3*b*e^6+6*a^2*b^2*d^2*e^
4-9*a*b^3*d^4*e^2)^(1/2)*x)*a*d+1/b*e*p*ln(-a^2*e^3+3*a*b*d^2*e+(-a^3*b*e^6+6*a^2*b^2*d^2*e^4-9*a*b^3*d^4*e^2)
^(1/2)*x)*a*d+1/2*I*Pi*d^2*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)*x+1/2*I*Pi*d^2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^
2+a)^p)^2*x

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maxima [A]  time = 0.99, size = 131, normalized size = 0.93 \[ \frac {1}{9} \, {\left (\frac {9 \, a d e \log \left (b x^{2} + a\right )}{b^{2}} + \frac {6 \, {\left (3 \, a b d^{2} - a^{2} e^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} - \frac {2 \, b e^{2} x^{3} + 9 \, b d e x^{2} + 6 \, {\left (3 \, b d^{2} - a e^{2}\right )} x}{b^{2}}\right )} b p + \frac {1}{3} \, {\left (e^{2} x^{3} + 3 \, d e x^{2} + 3 \, d^{2} x\right )} \log \left ({\left (b x^{2} + a\right )}^{p} c\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 3.28, size = 263, normalized size = 1.87 \[ \frac {e^2\,x^3\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}{3}-2\,d^2\,p\,x-\frac {2\,e^2\,p\,x^3}{9}+d^2\,x\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )+d\,e\,x^2\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )-d\,e\,p\,x^2+\frac {2\,a\,e^2\,p\,x}{3\,b}-\frac {2\,\sqrt {a}\,d^2\,p\,\mathrm {atan}\left (\frac {3\,\sqrt {a}\,b^{3/2}\,d^2\,p\,x}{a^2\,e^2\,p-3\,a\,b\,d^2\,p}-\frac {a^{3/2}\,\sqrt {b}\,e^2\,p\,x}{a^2\,e^2\,p-3\,a\,b\,d^2\,p}\right )}{\sqrt {b}}+\frac {2\,a^{3/2}\,e^2\,p\,\mathrm {atan}\left (\frac {3\,\sqrt {a}\,b^{3/2}\,d^2\,p\,x}{a^2\,e^2\,p-3\,a\,b\,d^2\,p}-\frac {a^{3/2}\,\sqrt {b}\,e^2\,p\,x}{a^2\,e^2\,p-3\,a\,b\,d^2\,p}\right )}{3\,b^{3/2}}+\frac {a\,d\,e\,p\,\ln \left (b\,x^2+a\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 24.02, size = 309, normalized size = 2.19 \[ \begin {cases} - \frac {i a^{\frac {3}{2}} e^{2} p \log {\left (a + b x^{2} \right )}}{3 b^{2} \sqrt {\frac {1}{b}}} + \frac {2 i a^{\frac {3}{2}} e^{2} p \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{3 b^{2} \sqrt {\frac {1}{b}}} + \frac {i \sqrt {a} d^{2} p \log {\left (a + b x^{2} \right )}}{b \sqrt {\frac {1}{b}}} - \frac {2 i \sqrt {a} d^{2} p \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{b \sqrt {\frac {1}{b}}} + \frac {a d e p \log {\left (a + b x^{2} \right )}}{b} + \frac {2 a e^{2} p x}{3 b} + d^{2} p x \log {\left (a + b x^{2} \right )} - 2 d^{2} p x + d^{2} x \log {\relax (c )} + d e p x^{2} \log {\left (a + b x^{2} \right )} - d e p x^{2} + d e x^{2} \log {\relax (c )} + \frac {e^{2} p x^{3} \log {\left (a + b x^{2} \right )}}{3} - \frac {2 e^{2} p x^{3}}{9} + \frac {e^{2} x^{3} \log {\relax (c )}}{3} & \text {for}\: b \neq 0 \\\left (d^{2} x + d e x^{2} + \frac {e^{2} x^{3}}{3}\right ) \log {\left (a^{p} c \right )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-I*a**(3/2)*e**2*p*log(a + b*x**2)/(3*b**2*sqrt(1/b)) + 2*I*a**(3/2)*e**2*p*log(-I*sqrt(a)*sqrt(1/b
) + x)/(3*b**2*sqrt(1/b)) + I*sqrt(a)*d**2*p*log(a + b*x**2)/(b*sqrt(1/b)) - 2*I*sqrt(a)*d**2*p*log(-I*sqrt(a)
*sqrt(1/b) + x)/(b*sqrt(1/b)) + a*d*e*p*log(a + b*x**2)/b + 2*a*e**2*p*x/(3*b) + d**2*p*x*log(a + b*x**2) - 2*
d**2*p*x + d**2*x*log(c) + d*e*p*x**2*log(a + b*x**2) - d*e*p*x**2 + d*e*x**2*log(c) + e**2*p*x**3*log(a + b*x
**2)/3 - 2*e**2*p*x**3/9 + e**2*x**3*log(c)/3, Ne(b, 0)), ((d**2*x + d*e*x**2 + e**2*x**3/3)*log(a**p*c), True
))

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