3.1.93 \(\int x \log ^3(c (a+b x^2)^p) \, dx\) [93]

Optimal. Leaf size=93 \[ -3 p^3 x^2+\frac {3 p^2 \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b}-\frac {3 p \left (a+b x^2\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )}{2 b}+\frac {\left (a+b x^2\right ) \log ^3\left (c \left (a+b x^2\right )^p\right )}{2 b} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2504, 2436, 2333, 2332} \begin {gather*} \frac {3 p^2 \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b}+\frac {\left (a+b x^2\right ) \log ^3\left (c \left (a+b x^2\right )^p\right )}{2 b}-\frac {3 p \left (a+b x^2\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )}{2 b}-3 p^3 x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Log[c*(a + b*x^2)^p]^3,x]

[Out]

-3*p^3*x^2 + (3*p^2*(a + b*x^2)*Log[c*(a + b*x^2)^p])/b - (3*p*(a + b*x^2)*Log[c*(a + b*x^2)^p]^2)/(2*b) + ((a
+ b*x^2)*Log[c*(a + b*x^2)^p]^3)/(2*b)

Rule 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2333

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2436

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2504

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rubi steps

\begin {align*} \int x \log ^3\left (c \left (a+b x^2\right )^p\right ) \, dx &=\frac {1}{2} \text {Subst}\left (\int \log ^3\left (c (a+b x)^p\right ) \, dx,x,x^2\right )\\ &=\frac {\text {Subst}\left (\int \log ^3\left (c x^p\right ) \, dx,x,a+b x^2\right )}{2 b}\\ &=\frac {\left (a+b x^2\right ) \log ^3\left (c \left (a+b x^2\right )^p\right )}{2 b}-\frac {(3 p) \text {Subst}\left (\int \log ^2\left (c x^p\right ) \, dx,x,a+b x^2\right )}{2 b}\\ &=-\frac {3 p \left (a+b x^2\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )}{2 b}+\frac {\left (a+b x^2\right ) \log ^3\left (c \left (a+b x^2\right )^p\right )}{2 b}+\frac {\left (3 p^2\right ) \text {Subst}\left (\int \log \left (c x^p\right ) \, dx,x,a+b x^2\right )}{b}\\ &=-3 p^3 x^2+\frac {3 p^2 \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b}-\frac {3 p \left (a+b x^2\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )}{2 b}+\frac {\left (a+b x^2\right ) \log ^3\left (c \left (a+b x^2\right )^p\right )}{2 b}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 87, normalized size = 0.94 \begin {gather*} \frac {-6 b p^3 x^2+6 p^2 \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )-3 p \left (a+b x^2\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )+\left (a+b x^2\right ) \log ^3\left (c \left (a+b x^2\right )^p\right )}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*Log[c*(a + b*x^2)^p]^3,x]

[Out]

(-6*b*p^3*x^2 + 6*p^2*(a + b*x^2)*Log[c*(a + b*x^2)^p] - 3*p*(a + b*x^2)*Log[c*(a + b*x^2)^p]^2 + (a + b*x^2)*
Log[c*(a + b*x^2)^p]^3)/(2*b)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 1.45, size = 3925, normalized size = 42.20

method	result	size
risch	\(\text {Expression too large to display}\)	\(3925\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*I*Pi^3*x^2*csgn(I*c*(b*x^2+a)^p)^9-3/2/b*ln(c)*a*p^2*ln(b*x^2+a)^2+3/2/b*ln(c)^2*ln(b*x^2+a)*a*p-3/b*ln(c
)*ln(b*x^2+a)*a*p^2-3/8*ln(c)*Pi^2*x^2*csgn(I*(b*x^2+a)^p)^2*csgn(I*c*(b*x^2+a)^p)^4+3/4*ln(c)*Pi^2*x^2*csgn(I
*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^5+3/4*ln(c)*Pi^2*x^2*csgn(I*c*(b*x^2+a)^p)^5*csgn(I*c)-3/8*ln(c)*Pi^2*x^2*
csgn(I*c*(b*x^2+a)^p)^4*csgn(I*c)^2+3/8*Pi^2*p*x^2*csgn(I*(b*x^2+a)^p)^2*csgn(I*c*(b*x^2+a)^p)^4-3/4*Pi^2*p*x^
2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^5-3/4*Pi^2*p*x^2*csgn(I*c*(b*x^2+a)^p)^5*csgn(I*c)+3/8*Pi^2*p*x^2*
csgn(I*c*(b*x^2+a)^p)^4*csgn(I*c)^2-1/16*I*Pi^3*x^2*csgn(I*(b*x^2+a)^p)^3*csgn(I*c*(b*x^2+a)^p)^6+3/16*I*Pi^3*
x^2*csgn(I*(b*x^2+a)^p)^2*csgn(I*c*(b*x^2+a)^p)^7-3/2*I*ln(c)*Pi*p*x^2*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)-3/2*I
*Pi*p^2*x^2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)-3/16*I*Pi^3*x^2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(
b*x^2+a)^p)^8-3/16*I*Pi^3*x^2*csgn(I*c*(b*x^2+a)^p)^8*csgn(I*c)+3/16*I*Pi^3*x^2*csgn(I*c*(b*x^2+a)^p)^7*csgn(I
*c)^2-1/16*I*Pi^3*x^2*csgn(I*c*(b*x^2+a)^p)^6*csgn(I*c)^3-3/4*I*ln(c)^2*Pi*x^2*csgn(I*c*(b*x^2+a)^p)^3-3/2*I*P
i*p^2*x^2*csgn(I*c*(b*x^2+a)^p)^3+3/4/b*Pi^2*ln(b*x^2+a)*a*p*csgn(I*c*(b*x^2+a)^p)^5*csgn(I*c)-3/8/b*Pi^2*ln(b
*x^2+a)*a*p*csgn(I*c*(b*x^2+a)^p)^4*csgn(I*c)^2-3/8/b*Pi^2*ln(b*x^2+a)*a*p*csgn(I*(b*x^2+a)^p)^2*csgn(I*c*(b*x
^2+a)^p)^4+3/4/b*Pi^2*ln(b*x^2+a)*a*p*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^5+3/4*I/b*Pi*a*p^2*csgn(I*c*(b
*x^2+a)^p)^3*ln(b*x^2+a)^2+3/2*I/b*Pi*ln(b*x^2+a)*a*p^2*csgn(I*c*(b*x^2+a)^p)^3-3/4*I*ln(c)^2*Pi*x^2*csgn(I*(b
*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)-3/2*I*ln(c)*Pi*p*x^2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2-3/
2*I/b*ln(c)*Pi*ln(b*x^2+a)*a*p*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)+3*a*p^3/b*ln(b*x^2+a)-3/2*l
n(c)^2*p*x^2+3*ln(c)*p^2*x^2+1/2*x^2*ln((b*x^2+a)^p)^3-3*p^3*x^2+3/4/b*Pi^2*ln(b*x^2+a)*a*p*csgn(I*(b*x^2+a)^p
)^2*csgn(I*c*(b*x^2+a)^p)^3*csgn(I*c)-3/8/b*Pi^2*ln(b*x^2+a)*a*p*csgn(I*(b*x^2+a)^p)^2*csgn(I*c*(b*x^2+a)^p)^2
*csgn(I*c)^2-3/2/b*Pi^2*ln(b*x^2+a)*a*p*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^4*csgn(I*c)+3/4/b*Pi^2*ln(b*
x^2+a)*a*p*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^3*csgn(I*c)^2-3/4*I/b*Pi*a*p^2*csgn(I*(b*x^2+a)^p)*csgn(I
*c*(b*x^2+a)^p)^2*ln(b*x^2+a)^2-3/4*I/b*Pi*a*p^2*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)*ln(b*x^2+a)^2-3/2*I/b*ln(c)
*Pi*ln(b*x^2+a)*a*p*csgn(I*c*(b*x^2+a)^p)^3-3/2*I/b*Pi*ln(b*x^2+a)*a*p^2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a
)^p)^2-3/2*I/b*Pi*ln(b*x^2+a)*a*p^2*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)+3/2*I*ln(c)*Pi*p*x^2*csgn(I*(b*x^2+a)^p)
*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)-3/8*ln(c)*Pi^2*x^2*csgn(I*c*(b*x^2+a)^p)^6+3/8*Pi^2*p*x^2*csgn(I*c*(b*x^2+a)^
p)^6+1/2/b*a*p^3*ln(b*x^2+a)^3+3/2/b*a*p^3*ln(b*x^2+a)^2+3/4*I/b*Pi*a*p^2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+
a)^p)*csgn(I*c)*ln(b*x^2+a)^2+3/2*I/b*ln(c)*Pi*ln(b*x^2+a)*a*p*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2+3/2
*I/b*ln(c)*Pi*ln(b*x^2+a)*a*p*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)+3/2*I/b*Pi*ln(b*x^2+a)*a*p^2*csgn(I*(b*x^2+a)^
p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)+3/4*(I*Pi*b*x^2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2-I*Pi*b*x^2*csgn
(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)-I*Pi*b*x^2*csgn(I*c*(b*x^2+a)^p)^3+I*Pi*b*x^2*csgn(I*c*(b*x^2+
a)^p)^2*csgn(I*c)+2*ln(c)*b*x^2-2*x^2*p*b+2*p*a*ln(b*x^2+a))/b*ln((b*x^2+a)^p)^2+3/8*(8*x^2*b*p^2-8*ln(c)*b*p*
x^2+8*ln(c)*ln(b*x^2+a)*a*p-Pi^2*b*x^2*csgn(I*c*(b*x^2+a)^p)^4*csgn(I*c)^2+2*Pi^2*b*x^2*csgn(I*c*(b*x^2+a)^p)^
5*csgn(I*c)+4*ln(c)^2*b*x^2-4*a*p^2*ln(b*x^2+a)^2-Pi^2*b*x^2*csgn(I*c*(b*x^2+a)^p)^6+4*I*Pi*b*p*x^2*csgn(I*(b*
x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)-4*I*Pi*b*p*x^2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2-4*I*Pi*b*
p*x^2*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)+4*I*Pi*ln(b*x^2+a)*a*p*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2-8*l
n(b*x^2+a)*a*p^2-4*I*Pi*ln(b*x^2+a)*a*p*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)-Pi^2*b*x^2*csgn(I*
(b*x^2+a)^p)^2*csgn(I*c*(b*x^2+a)^p)^4+2*Pi^2*b*x^2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^5-4*I*ln(c)*Pi*b
*x^2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)+4*I*Pi*ln(b*x^2+a)*a*p*csgn(I*c*(b*x^2+a)^p)^2*csgn(I
*c)+4*I*ln(c)*Pi*b*x^2*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)+4*I*ln(c)*Pi*b*x^2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^
2+a)^p)^2+2*Pi^2*b*x^2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^3*csgn(I*c)^2+2*Pi^2*b*x^2*csgn(I*(b*x^2+a)^p
)^2*csgn(I*c*(b*x^2+a)^p)^3*csgn(I*c)-4*I*Pi*ln(b*x^2+a)*a*p*csgn(I*c*(b*x^2+a)^p)^3-4*I*ln(c)*Pi*b*x^2*csgn(I
*c*(b*x^2+a)^p)^3+4*I*Pi*b*p*x^2*csgn(I*c*(b*x^2+a)^p)^3-Pi^2*b*x^2*csgn(I*(b*x^2+a)^p)^2*csgn(I*c*(b*x^2+a)^p
)^2*csgn(I*c)^2-4*Pi^2*b*x^2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^4*csgn(I*c))/b*ln((b*x^2+a)^p)+1/2*ln(c
)^3*x^2+3/4*ln(c)*Pi^2*x^2*csgn(I*(b*x^2+a)^p)^2*csgn(I*c*(b*x^2+a)^p)^3*csgn(I*c)-3/8*ln(c)*Pi^2*x^2*csgn(I*(
b*x^2+a)^p)^2*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)^2-3/2*ln(c)*Pi^2*x^2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)
^4*csgn(I*c)+3/4*ln(c)*Pi^2*x^2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^3*csgn(I*c)^2-3/4*Pi^2*p*x^2*csgn(I*
(b*x^2+a)^p)^2*csgn(I*c*(b*x^2+a)^p)^3*csgn(I*c...

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Maxima [A]
time = 0.29, size = 164, normalized size = 1.76 \begin {gather*} -\frac {3}{2} \, b p {\left (\frac {x^{2}}{b} - \frac {a \log \left (b x^{2} + a\right )}{b^{2}}\right )} \log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2} + \frac {1}{2} \, x^{2} \log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{3} + \frac {1}{2} \, b p {\left (\frac {{\left (a \log \left (b x^{2} + a\right )^{3} - 6 \, b x^{2} + 3 \, a \log \left (b x^{2} + a\right )^{2} + 6 \, a \log \left (b x^{2} + a\right )\right )} p^{2}}{b^{2}} + \frac {3 \, {\left (2 \, b x^{2} - a \log \left (b x^{2} + a\right )^{2} - 2 \, a \log \left (b x^{2} + a\right )\right )} p \log \left ({\left (b x^{2} + a\right )}^{p} c\right )}{b^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*b*p*(x^2/b - a*log(b*x^2 + a)/b^2)*log((b*x^2 + a)^p*c)^2 + 1/2*x^2*log((b*x^2 + a)^p*c)^3 + 1/2*b*p*((a*
log(b*x^2 + a)^3 - 6*b*x^2 + 3*a*log(b*x^2 + a)^2 + 6*a*log(b*x^2 + a))*p^2/b^2 + 3*(2*b*x^2 - a*log(b*x^2 + a
)^2 - 2*a*log(b*x^2 + a))*p*log((b*x^2 + a)^p*c)/b^2)

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Fricas [A]
time = 0.45, size = 176, normalized size = 1.89 \begin {gather*} -\frac {6 \, b p^{3} x^{2} - 6 \, b p^{2} x^{2} \log \left (c\right ) + 3 \, b p x^{2} \log \left (c\right )^{2} - b x^{2} \log \left (c\right )^{3} - {\left (b p^{3} x^{2} + a p^{3}\right )} \log \left (b x^{2} + a\right )^{3} + 3 \, {\left (b p^{3} x^{2} + a p^{3} - {\left (b p^{2} x^{2} + a p^{2}\right )} \log \left (c\right )\right )} \log \left (b x^{2} + a\right )^{2} - 3 \, {\left (2 \, b p^{3} x^{2} + 2 \, a p^{3} + {\left (b p x^{2} + a p\right )} \log \left (c\right )^{2} - 2 \, {\left (b p^{2} x^{2} + a p^{2}\right )} \log \left (c\right )\right )} \log \left (b x^{2} + a\right )}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(6*b*p^3*x^2 - 6*b*p^2*x^2*log(c) + 3*b*p*x^2*log(c)^2 - b*x^2*log(c)^3 - (b*p^3*x^2 + a*p^3)*log(b*x^2 +
a)^3 + 3*(b*p^3*x^2 + a*p^3 - (b*p^2*x^2 + a*p^2)*log(c))*log(b*x^2 + a)^2 - 3*(2*b*p^3*x^2 + 2*a*p^3 + (b*p*
x^2 + a*p)*log(c)^2 - 2*(b*p^2*x^2 + a*p^2)*log(c))*log(b*x^2 + a))/b

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Sympy [A]
time = 0.95, size = 143, normalized size = 1.54 \begin {gather*} \begin {cases} \frac {3 a p^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{b} - \frac {3 a p \log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{2}}{2 b} + \frac {a \log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{3}}{2 b} - 3 p^{3} x^{2} + 3 p^{2} x^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )} - \frac {3 p x^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{2}}{2} + \frac {x^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{3}}{2} & \text {for}\: b \neq 0 \\\frac {x^{2} \log {\left (a^{p} c \right )}^{3}}{2} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((3*a*p**2*log(c*(a + b*x**2)**p)/b - 3*a*p*log(c*(a + b*x**2)**p)**2/(2*b) + a*log(c*(a + b*x**2)**p
)**3/(2*b) - 3*p**3*x**2 + 3*p**2*x**2*log(c*(a + b*x**2)**p) - 3*p*x**2*log(c*(a + b*x**2)**p)**2/2 + x**2*lo
g(c*(a + b*x**2)**p)**3/2, Ne(b, 0)), (x**2*log(a**p*c)**3/2, True))

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Giac [A]
time = 4.67, size = 169, normalized size = 1.82 \begin {gather*} \frac {{\left ({\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right )^{3} - 6 \, b x^{2} - 3 \, {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right )^{2} + 6 \, {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right ) - 6 \, a\right )} p^{3} + 3 \, {\left (2 \, b x^{2} + {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right )^{2} - 2 \, {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right ) + 2 \, a\right )} p^{2} \log \left (c\right ) - 3 \, {\left (b x^{2} - {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right ) + a\right )} p \log \left (c\right )^{2} + {\left (b x^{2} + a\right )} \log \left (c\right )^{3}}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(((b*x^2 + a)*log(b*x^2 + a)^3 - 6*b*x^2 - 3*(b*x^2 + a)*log(b*x^2 + a)^2 + 6*(b*x^2 + a)*log(b*x^2 + a) -
6*a)*p^3 + 3*(2*b*x^2 + (b*x^2 + a)*log(b*x^2 + a)^2 - 2*(b*x^2 + a)*log(b*x^2 + a) + 2*a)*p^2*log(c) - 3*(b*
x^2 - (b*x^2 + a)*log(b*x^2 + a) + a)*p*log(c)^2 + (b*x^2 + a)*log(c)^3)/b

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Mupad [B]
time = 0.24, size = 103, normalized size = 1.11 \begin {gather*} {\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}^3\,\left (\frac {a}{2\,b}+\frac {x^2}{2}\right )-{\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}^2\,\left (\frac {3\,p\,x^2}{2}+\frac {3\,a\,p}{2\,b}\right )-3\,p^3\,x^2+3\,p^2\,x^2\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )+\frac {3\,a\,p^3\,\ln \left (b\,x^2+a\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*log(c*(a + b*x^2)^p)^3,x)

[Out]

log(c*(a + b*x^2)^p)^3*(a/(2*b) + x^2/2) - log(c*(a + b*x^2)^p)^2*((3*p*x^2)/2 + (3*a*p)/(2*b)) - 3*p^3*x^2 +
3*p^2*x^2*log(c*(a + b*x^2)^p) + (3*a*p^3*log(a + b*x^2))/b

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