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

Optimal. Leaf size=178 \[ -\frac {p \left (a^2 e^4-6 a b d^2 e^2+b^2 d^4\right ) \log \left (a+b x^2\right )}{4 b^2 e}+\frac {2 \sqrt {a} d p \left (b d^2-a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{3/2}}+\frac {(d+e x)^4 \log \left (c \left (a+b x^2\right )^p\right )}{4 e}-\frac {e p x^2 \left (6 b d^2-a e^2\right )}{4 b}-\frac {2 d p x \left (b d^2-a e^2\right )}{b}-\frac {2}{3} d e^2 p x^3-\frac {1}{8} e^3 p x^4 \]

[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)

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Rubi [A]  time = 0.16, antiderivative size = 178, 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} \[ -\frac {p \left (a^2 e^4-6 a b d^2 e^2+b^2 d^4\right ) \log \left (a+b x^2\right )}{4 b^2 e}+\frac {2 \sqrt {a} d p \left (b d^2-a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{3/2}}+\frac {(d+e x)^4 \log \left (c \left (a+b x^2\right )^p\right )}{4 e}-\frac {e p x^2 \left (6 b d^2-a e^2\right )}{4 b}-\frac {2 d p x \left (b d^2-a e^2\right )}{b}-\frac {2}{3} d e^2 p x^3-\frac {1}{8} e^3 p x^4 \]

Antiderivative was successfully verified.

[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 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)^3 \log \left (c \left (a+b x^2\right )^p\right ) \, dx &=\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*}

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Mathematica [A]  time = 0.78, size = 249, normalized size = 1.40 \[ \frac {-6 p \left (a^2 e^4+4 \sqrt {-a} b^{3/2} d^3 e-6 a b d^2 e^2+4 (-a)^{3/2} \sqrt {b} d e^3+b^2 d^4\right ) \log \left (\sqrt {-a}-\sqrt {b} x\right )-6 p \left (a^2 e^4-4 \sqrt {-a} b^{3/2} d^3 e-6 a b d^2 e^2+4 \sqrt {-a} a \sqrt {b} d e^3+b^2 d^4\right ) \log \left (\sqrt {-a}+\sqrt {b} x\right )+b \left (6 b (d+e x)^4 \log \left (c \left (a+b x^2\right )^p\right )+6 a e^3 p x (8 d+e x)-b e p x \left (48 d^3+36 d^2 e x+16 d e^2 x^2+3 e^3 x^3\right )\right )}{24 b^2 e} \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.47, size = 498, normalized size = 2.80 \[ \left [-\frac {3 \, b^{2} e^{3} p x^{4} + 16 \, b^{2} d e^{2} p x^{3} + 6 \, {\left (6 \, b^{2} d^{2} e - a b e^{3}\right )} p x^{2} - 24 \, {\left (b^{2} d^{3} - a b d e^{2}\right )} p \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} + 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) + 48 \, {\left (b^{2} d^{3} - a b d e^{2}\right )} p x - 6 \, {\left (b^{2} e^{3} p x^{4} + 4 \, b^{2} d e^{2} p x^{3} + 6 \, b^{2} d^{2} e p x^{2} + 4 \, b^{2} d^{3} p x + {\left (6 \, a b d^{2} e - a^{2} e^{3}\right )} p\right )} \log \left (b x^{2} + a\right ) - 6 \, {\left (b^{2} e^{3} x^{4} + 4 \, b^{2} d e^{2} x^{3} + 6 \, b^{2} d^{2} e x^{2} + 4 \, b^{2} d^{3} x\right )} \log \relax (c)}{24 \, b^{2}}, -\frac {3 \, b^{2} e^{3} p x^{4} + 16 \, b^{2} d e^{2} p x^{3} + 6 \, {\left (6 \, b^{2} d^{2} e - a b e^{3}\right )} p x^{2} - 48 \, {\left (b^{2} d^{3} - a b d e^{2}\right )} p \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) + 48 \, {\left (b^{2} d^{3} - a b d e^{2}\right )} p x - 6 \, {\left (b^{2} e^{3} p x^{4} + 4 \, b^{2} d e^{2} p x^{3} + 6 \, b^{2} d^{2} e p x^{2} + 4 \, b^{2} d^{3} p x + {\left (6 \, a b d^{2} e - a^{2} e^{3}\right )} p\right )} \log \left (b x^{2} + a\right ) - 6 \, {\left (b^{2} e^{3} x^{4} + 4 \, b^{2} d e^{2} x^{3} + 6 \, b^{2} d^{2} e x^{2} + 4 \, b^{2} d^{3} x\right )} \log \relax (c)}{24 \, b^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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]

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giac [A]  time = 0.21, size = 273, normalized size = 1.53 \[ \frac {2 \, {\left (a b d^{3} p - a^{2} d p e^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b} + \frac {6 \, b^{2} p x^{4} e^{3} \log \left (b x^{2} + a\right ) + 24 \, b^{2} d p x^{3} e^{2} \log \left (b x^{2} + a\right ) + 36 \, b^{2} d^{2} p x^{2} e \log \left (b x^{2} + a\right ) - 3 \, b^{2} p x^{4} e^{3} - 16 \, b^{2} d p x^{3} e^{2} - 36 \, b^{2} d^{2} p x^{2} e + 24 \, b^{2} d^{3} p x \log \left (b x^{2} + a\right ) + 6 \, b^{2} x^{4} e^{3} \log \relax (c) + 24 \, b^{2} d x^{3} e^{2} \log \relax (c) + 36 \, b^{2} d^{2} x^{2} e \log \relax (c) - 48 \, b^{2} d^{3} p x + 36 \, a b d^{2} p e \log \left (b x^{2} + a\right ) + 24 \, b^{2} d^{3} x \log \relax (c) + 6 \, a b p x^{2} e^{3} + 48 \, a b d p x e^{2} - 6 \, a^{2} p e^{3} \log \left (b x^{2} + a\right )}{24 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(a*b*d^3*p - a^2*d*p*e^2)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b) + 1/24*(6*b^2*p*x^4*e^3*log(b*x^2 + a) + 24*b^
2*d*p*x^3*e^2*log(b*x^2 + a) + 36*b^2*d^2*p*x^2*e*log(b*x^2 + a) - 3*b^2*p*x^4*e^3 - 16*b^2*d*p*x^3*e^2 - 36*b
^2*d^2*p*x^2*e + 24*b^2*d^3*p*x*log(b*x^2 + a) + 6*b^2*x^4*e^3*log(c) + 24*b^2*d*x^3*e^2*log(c) + 36*b^2*d^2*x
^2*e*log(c) - 48*b^2*d^3*p*x + 36*a*b*d^2*p*e*log(b*x^2 + a) + 24*b^2*d^3*x*log(c) + 6*a*b*p*x^2*e^3 + 48*a*b*
d*p*x*e^2 - 6*a^2*p*e^3*log(b*x^2 + a))/b^2

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maple [C]  time = 0.72, size = 1330, normalized size = 7.47 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*e^2*x^3*ln(c)+3/2*d^2*e*x^2*ln(c)-1/4/e*p*ln(-a^2*d*e^3+a*b*d^3*e+(-a^3*b*d^2*e^6+2*a^2*b^2*d^4*e^4-a*b^3*d^
6*e^2)^(1/2)*x)*d^4-1/4/e*p*ln(-a^2*d*e^3+a*b*d^3*e-(-a^3*b*d^2*e^6+2*a^2*b^2*d^4*e^4-a*b^3*d^6*e^2)^(1/2)*x)*
d^4+1/4*(e*x+d)^4/e*ln((b*x^2+a)^p)+1/4*e^3*x^4*ln(c)+d^3*x*ln(c)-1/8*e^3*p*x^4-3/2*d^2*e*p*x^2-1/4/b^2*e^3*p*
ln(-a^2*d*e^3+a*b*d^3*e+(-a^3*b*d^2*e^6+2*a^2*b^2*d^4*e^4-a*b^3*d^6*e^2)^(1/2)*x)*a^2-1/4/b^2*e^3*p*ln(-a^2*d*
e^3+a*b*d^3*e-(-a^3*b*d^2*e^6+2*a^2*b^2*d^4*e^4-a*b^3*d^6*e^2)^(1/2)*x)*a^2-1/b^2/e*p*ln(-a^2*d*e^3+a*b*d^3*e+
(-a^3*b*d^2*e^6+2*a^2*b^2*d^4*e^4-a*b^3*d^6*e^2)^(1/2)*x)*(-a^3*b*d^2*e^6+2*a^2*b^2*d^4*e^4-a*b^3*d^6*e^2)^(1/
2)+1/b^2/e*p*ln(-a^2*d*e^3+a*b*d^3*e-(-a^3*b*d^2*e^6+2*a^2*b^2*d^4*e^4-a*b^3*d^6*e^2)^(1/2)*x)*(-a^3*b*d^2*e^6
+2*a^2*b^2*d^4*e^4-a*b^3*d^6*e^2)^(1/2)-1/8*I*e^3*Pi*x^4*csgn(I*c*(b*x^2+a)^p)^3-1/2*I*Pi*d^3*csgn(I*c*(b*x^2+
a)^p)^3*x+2/b*a*d*p*e^2*x-2*d^3*p*x-2/3*d*e^2*p*x^3+3/2/b*e*p*ln(-a^2*d*e^3+a*b*d^3*e+(-a^3*b*d^2*e^6+2*a^2*b^
2*d^4*e^4-a*b^3*d^6*e^2)^(1/2)*x)*a*d^2+3/2/b*e*p*ln(-a^2*d*e^3+a*b*d^3*e-(-a^3*b*d^2*e^6+2*a^2*b^2*d^4*e^4-a*
b^3*d^6*e^2)^(1/2)*x)*a*d^2+1/2*I*Pi*d^3*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2*x+1/2*I*Pi*d^3*csgn(I*c*(
b*x^2+a)^p)^2*csgn(I*c)*x+1/8*I*e^3*Pi*x^4*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2+1/8*I*e^3*Pi*x^4*csgn(I
*c*(b*x^2+a)^p)^2*csgn(I*c)-1/2*I*e^2*Pi*d*x^3*csgn(I*c*(b*x^2+a)^p)^3-3/4*I*e*Pi*d^2*x^2*csgn(I*c*(b*x^2+a)^p
)^3+1/4/b*a*e^3*p*x^2+1/2*I*e^2*Pi*d*x^3*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2+1/2*I*e^2*Pi*d*x^3*csgn(I
*c*(b*x^2+a)^p)^2*csgn(I*c)+3/4*I*e*Pi*d^2*x^2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2+3/4*I*e*Pi*d^2*x^2*
csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)-1/8*I*e^3*Pi*x^4*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)-1/2*I*P
i*d^3*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)*x-1/2*I*e^2*Pi*d*x^3*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b
*x^2+a)^p)*csgn(I*c)-3/4*I*e*Pi*d^2*x^2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)

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maxima [A]  time = 0.99, size = 177, normalized size = 0.99 \[ \frac {1}{24} \, b p {\left (\frac {48 \, {\left (a b d^{3} - a^{2} d e^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} - \frac {3 \, b e^{3} x^{4} + 16 \, b d e^{2} x^{3} + 6 \, {\left (6 \, b d^{2} e - a e^{3}\right )} x^{2} + 48 \, {\left (b d^{3} - a d e^{2}\right )} x}{b^{2}} + \frac {6 \, {\left (6 \, a b d^{2} e - a^{2} e^{3}\right )} \log \left (b x^{2} + a\right )}{b^{3}}\right )} + \frac {1}{4} \, {\left (e^{3} x^{4} + 4 \, d e^{2} x^{3} + 6 \, d^{2} e x^{2} + 4 \, d^{3} x\right )} \log \left ({\left (b x^{2} + a\right )}^{p} c\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.46, size = 222, normalized size = 1.25 \[ \frac {e^3\,x^4\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}{4}-2\,d^3\,p\,x-\frac {e^3\,p\,x^4}{8}+d^3\,x\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )+\frac {3\,d^2\,e\,x^2\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}{2}+d\,e^2\,x^3\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )-\frac {3\,d^2\,e\,p\,x^2}{2}-\frac {2\,d\,e^2\,p\,x^3}{3}+\frac {a\,e^3\,p\,x^2}{4\,b}+\frac {2\,\sqrt {a}\,d^3\,p\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{\sqrt {b}}-\frac {a^2\,e^3\,p\,\ln \left (b\,x^2+a\right )}{4\,b^2}-\frac {2\,a^{3/2}\,d\,e^2\,p\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{b^{3/2}}+\frac {2\,a\,d\,e^2\,p\,x}{b}+\frac {3\,a\,d^2\,e\,p\,\ln \left (b\,x^2+a\right )}{2\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 47.87, size = 422, normalized size = 2.37 \[ \begin {cases} - \frac {i a^{\frac {3}{2}} d e^{2} p \log {\left (a + b x^{2} \right )}}{b^{2} \sqrt {\frac {1}{b}}} + \frac {2 i a^{\frac {3}{2}} d e^{2} p \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{b^{2} \sqrt {\frac {1}{b}}} + \frac {i \sqrt {a} d^{3} p \log {\left (a + b x^{2} \right )}}{b \sqrt {\frac {1}{b}}} - \frac {2 i \sqrt {a} d^{3} p \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{b \sqrt {\frac {1}{b}}} - \frac {a^{2} e^{3} p \log {\left (a + b x^{2} \right )}}{4 b^{2}} + \frac {3 a d^{2} e p \log {\left (a + b x^{2} \right )}}{2 b} + \frac {2 a d e^{2} p x}{b} + \frac {a e^{3} p x^{2}}{4 b} + d^{3} p x \log {\left (a + b x^{2} \right )} - 2 d^{3} p x + d^{3} x \log {\relax (c )} + \frac {3 d^{2} e p x^{2} \log {\left (a + b x^{2} \right )}}{2} - \frac {3 d^{2} e p x^{2}}{2} + \frac {3 d^{2} e x^{2} \log {\relax (c )}}{2} + d e^{2} p x^{3} \log {\left (a + b x^{2} \right )} - \frac {2 d e^{2} p x^{3}}{3} + d e^{2} x^{3} \log {\relax (c )} + \frac {e^{3} p x^{4} \log {\left (a + b x^{2} \right )}}{4} - \frac {e^{3} p x^{4}}{8} + \frac {e^{3} x^{4} \log {\relax (c )}}{4} & \text {for}\: b \neq 0 \\\left (d^{3} x + \frac {3 d^{2} e x^{2}}{2} + d e^{2} x^{3} + \frac {e^{3} x^{4}}{4}\right ) \log {\left (a^{p} c \right )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-I*a**(3/2)*d*e**2*p*log(a + b*x**2)/(b**2*sqrt(1/b)) + 2*I*a**(3/2)*d*e**2*p*log(-I*sqrt(a)*sqrt(1
/b) + x)/(b**2*sqrt(1/b)) + I*sqrt(a)*d**3*p*log(a + b*x**2)/(b*sqrt(1/b)) - 2*I*sqrt(a)*d**3*p*log(-I*sqrt(a)
*sqrt(1/b) + x)/(b*sqrt(1/b)) - a**2*e**3*p*log(a + b*x**2)/(4*b**2) + 3*a*d**2*e*p*log(a + b*x**2)/(2*b) + 2*
a*d*e**2*p*x/b + a*e**3*p*x**2/(4*b) + d**3*p*x*log(a + b*x**2) - 2*d**3*p*x + d**3*x*log(c) + 3*d**2*e*p*x**2
*log(a + b*x**2)/2 - 3*d**2*e*p*x**2/2 + 3*d**2*e*x**2*log(c)/2 + d*e**2*p*x**3*log(a + b*x**2) - 2*d*e**2*p*x
**3/3 + d*e**2*x**3*log(c) + e**3*p*x**4*log(a + b*x**2)/4 - e**3*p*x**4/8 + e**3*x**4*log(c)/4, Ne(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), True))

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