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

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

[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 = 0.11, 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 = {2513, 815, 649, 211, 266} \begin {gather*} -\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 \text {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 \end {gather*}

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 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*} \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 = 0.54, size = 249, normalized size = 1.40 \begin {gather*} \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} \end {gather*}

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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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.80, size = 1330, normalized size = 7.47

method	result	size
risch	\(\text {Expression too large to display}\)	\(1330\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/b*a*e^3*p*x^2-2*d^3*p*x-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+e^2*
ln(c)*d*x^3+3/2*e*ln(c)*d^2*x^2+1/4*e^3*ln(c)*x^4-1/4*e^3/b^2*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/e/b^2*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)-1/4*e^3/b^2*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/e/b^2*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)-1/2*I*
Pi*d^3*csgn(I*c*(b*x^2+a)^p)^3*x-1/8*I*e^3*Pi*x^4*csgn(I*c*(b*x^2+a)^p)^3-1/8*e^3*p*x^4+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*csg
n(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)^3-1/2*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)-1/2*I*Pi*d^3*c
sgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*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
)*csgn(I*c)+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)+ln(c)*d^3*x-3/2*d^2*e*p*x^2+2/b*a*d*p*e^2*x-2/3*d*e^2*p*x^3+3/2*e/b*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*e/b*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+1/4*(e*x+d)^4/e*ln((b*x^2+a)^p)

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Maxima [A]
time = 0.58, size = 172, normalized size = 0.97 \begin {gather*} \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 x^{4} e^{3} + 16 \, b d x^{3} e^{2} + 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 (x^{4} e^{3} + 4 \, d x^{3} e^{2} + 6 \, d^{2} x^{2} e + 4 \, d^{3} x\right )} \log \left ({\left (b x^{2} + a\right )}^{p} c\right ) \end {gather*}

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*x^4*e^3 + 16*b*d*x^3*e^2 + 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*(x^
4*e^3 + 4*d*x^3*e^2 + 6*d^2*x^2*e + 4*d^3*x)*log((b*x^2 + a)^p*c)

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Fricas [A]
time = 0.40, size = 486, normalized size = 2.73 \begin {gather*} \left [-\frac {36 \, b^{2} d^{2} p x^{2} e + 48 \, b^{2} d^{3} p x - 24 \, {\left (b^{2} d^{3} p - a b d p e^{2}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} + 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) + 3 \, {\left (b^{2} p x^{4} - 2 \, a b p x^{2}\right )} e^{3} + 16 \, {\left (b^{2} d p x^{3} - 3 \, a b d p x\right )} e^{2} - 6 \, {\left (4 \, b^{2} d p x^{3} e^{2} + 4 \, b^{2} d^{3} p x + {\left (b^{2} p x^{4} - a^{2} p\right )} e^{3} + 6 \, {\left (b^{2} d^{2} p x^{2} + a b d^{2} p\right )} e\right )} \log \left (b x^{2} + a\right ) - 6 \, {\left (b^{2} x^{4} e^{3} + 4 \, b^{2} d x^{3} e^{2} + 6 \, b^{2} d^{2} x^{2} e + 4 \, b^{2} d^{3} x\right )} \log \left (c\right )}{24 \, b^{2}}, -\frac {36 \, b^{2} d^{2} p x^{2} e + 48 \, b^{2} d^{3} p x - 48 \, {\left (b^{2} d^{3} p - a b d p e^{2}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) + 3 \, {\left (b^{2} p x^{4} - 2 \, a b p x^{2}\right )} e^{3} + 16 \, {\left (b^{2} d p x^{3} - 3 \, a b d p x\right )} e^{2} - 6 \, {\left (4 \, b^{2} d p x^{3} e^{2} + 4 \, b^{2} d^{3} p x + {\left (b^{2} p x^{4} - a^{2} p\right )} e^{3} + 6 \, {\left (b^{2} d^{2} p x^{2} + a b d^{2} p\right )} e\right )} \log \left (b x^{2} + a\right ) - 6 \, {\left (b^{2} x^{4} e^{3} + 4 \, b^{2} d x^{3} e^{2} + 6 \, b^{2} d^{2} x^{2} e + 4 \, b^{2} d^{3} x\right )} \log \left (c\right )}{24 \, b^{2}}\right ] \end {gather*}

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 527 vs. \(2 (170) = 340\).
time = 19.44, size = 527, normalized size = 2.96 \begin {gather*} \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} \end {gather*}

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(((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))

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Giac [A]
time = 3.86, size = 273, normalized size = 1.53 \begin {gather*} \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 \left (c\right ) + 24 \, b^{2} d x^{3} e^{2} \log \left (c\right ) + 36 \, b^{2} d^{2} x^{2} e \log \left (c\right ) - 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 \left (c\right ) + 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}} \end {gather*}

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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Mupad [B]
time = 0.46, size = 222, normalized size = 1.25 \begin {gather*} \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} \end {gather*}

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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