3.1.17 \(\int x \log (c (a+b x^3)^p) \, dx\) [17]

Optimal. Leaf size=147 \[ -\frac {3 p x^2}{4}-\frac {\sqrt {3} a^{2/3} p \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{2 b^{2/3}}-\frac {a^{2/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{2 b^{2/3}}+\frac {a^{2/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{4 b^{2/3}}+\frac {1}{2} x^2 \log \left (c \left (a+b x^3\right )^p\right ) \]

[Out]

-3/4*p*x^2-1/2*a^(2/3)*p*ln(a^(1/3)+b^(1/3)*x)/b^(2/3)+1/4*a^(2/3)*p*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)
/b^(2/3)+1/2*x^2*ln(c*(b*x^3+a)^p)-1/2*a^(2/3)*p*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)/a^(1/3)*3^(1/2))*3^(1/2)/b^(
2/3)

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Rubi [A]
time = 0.06, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {2505, 327, 298, 31, 648, 631, 210, 642} \begin {gather*} -\frac {\sqrt {3} a^{2/3} p \text {ArcTan}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{2 b^{2/3}}+\frac {a^{2/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{4 b^{2/3}}-\frac {a^{2/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{2 b^{2/3}}+\frac {1}{2} x^2 \log \left (c \left (a+b x^3\right )^p\right )-\frac {3 p x^2}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Log[c*(a + b*x^3)^p],x]

[Out]

(-3*p*x^2)/4 - (Sqrt[3]*a^(2/3)*p*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(2*b^(2/3)) - (a^(2/3)*p*
Log[a^(1/3) + b^(1/3)*x])/(2*b^(2/3)) + (a^(2/3)*p*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(4*b^(2/3))
+ (x^2*Log[c*(a + b*x^3)^p])/2

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 298

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Dist[-(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 2505

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

Rubi steps

\begin {align*} \int x \log \left (c \left (a+b x^3\right )^p\right ) \, dx &=\frac {1}{2} x^2 \log \left (c \left (a+b x^3\right )^p\right )-\frac {1}{2} (3 b p) \int \frac {x^4}{a+b x^3} \, dx\\ &=-\frac {3 p x^2}{4}+\frac {1}{2} x^2 \log \left (c \left (a+b x^3\right )^p\right )+\frac {1}{2} (3 a p) \int \frac {x}{a+b x^3} \, dx\\ &=-\frac {3 p x^2}{4}+\frac {1}{2} x^2 \log \left (c \left (a+b x^3\right )^p\right )-\frac {\left (a^{2/3} p\right ) \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{2 \sqrt [3]{b}}+\frac {\left (a^{2/3} p\right ) \int \frac {\sqrt [3]{a}+\sqrt [3]{b} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{2 \sqrt [3]{b}}\\ &=-\frac {3 p x^2}{4}-\frac {a^{2/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{2 b^{2/3}}+\frac {1}{2} x^2 \log \left (c \left (a+b x^3\right )^p\right )+\frac {\left (a^{2/3} p\right ) \int \frac {-\sqrt [3]{a} \sqrt [3]{b}+2 b^{2/3} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{4 b^{2/3}}+\frac {(3 a p) \int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{4 \sqrt [3]{b}}\\ &=-\frac {3 p x^2}{4}-\frac {a^{2/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{2 b^{2/3}}+\frac {a^{2/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{4 b^{2/3}}+\frac {1}{2} x^2 \log \left (c \left (a+b x^3\right )^p\right )+\frac {\left (3 a^{2/3} p\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{2 b^{2/3}}\\ &=-\frac {3 p x^2}{4}-\frac {\sqrt {3} a^{2/3} p \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{2 b^{2/3}}-\frac {a^{2/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{2 b^{2/3}}+\frac {a^{2/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{4 b^{2/3}}+\frac {1}{2} x^2 \log \left (c \left (a+b x^3\right )^p\right )\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.00, size = 53, normalized size = 0.36 \begin {gather*} -\frac {3 p x^2}{4}+\frac {3}{4} p x^2 \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};-\frac {b x^3}{a}\right )+\frac {1}{2} x^2 \log \left (c \left (a+b x^3\right )^p\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*Log[c*(a + b*x^3)^p],x]

[Out]

(-3*p*x^2)/4 + (3*p*x^2*Hypergeometric2F1[2/3, 1, 5/3, -((b*x^3)/a)])/4 + (x^2*Log[c*(a + b*x^3)^p])/2

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.40, size = 184, normalized size = 1.25

method	result	size
risch	\(\frac {x^{2} \ln \left (\left (x^{3} b +a \right )^{p}\right )}{2}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (i \left (x^{3} b +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right ) \mathrm {csgn}\left (i c \right )}{4}+\frac {i \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right )^{2} x^{2} \pi }{4}+\frac {i \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right )^{2} \mathrm {csgn}\left (i \left (x^{3} b +a \right )^{p}\right ) x^{2} \pi }{4}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right )^{3}}{4}+\frac {\ln \left (c \right ) x^{2}}{2}-\frac {3 p \,x^{2}}{4}+\frac {a p \left (\munderset {\textit {\_R} =\RootOf \left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}}\right )}{2 b}\)	\(184\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*ln((b*x^3+a)^p)-1/4*I*Pi*x^2*csgn(I*(b*x^3+a)^p)*csgn(I*c*(b*x^3+a)^p)*csgn(I*c)+1/4*I*csgn(I*c)*csgn(
I*c*(b*x^3+a)^p)^2*x^2*Pi+1/4*I*csgn(I*c*(b*x^3+a)^p)^2*csgn(I*(b*x^3+a)^p)*x^2*Pi-1/4*I*Pi*x^2*csgn(I*c*(b*x^
3+a)^p)^3+1/2*ln(c)*x^2-3/4*p*x^2+1/2*a*p/b*sum(1/_R*ln(x-_R),_R=RootOf(_Z^3*b+a))

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Maxima [A]
time = 0.49, size = 131, normalized size = 0.89 \begin {gather*} -\frac {1}{4} \, b p {\left (\frac {3 \, x^{2}}{b} - \frac {2 \, \sqrt {3} a \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{b^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {a \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{b^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {2 \, a \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{b^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )} + \frac {1}{2} \, x^{2} \log \left ({\left (b x^{3} + a\right )}^{p} c\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*b*p*(3*x^2/b - 2*sqrt(3)*a*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(b^2*(a/b)^(1/3)) - a*log(
x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(b^2*(a/b)^(1/3)) + 2*a*log(x + (a/b)^(1/3))/(b^2*(a/b)^(1/3))) + 1/2*x^2*l
og((b*x^3 + a)^p*c)

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Fricas [A]
time = 0.36, size = 150, normalized size = 1.02 \begin {gather*} \frac {1}{2} \, p x^{2} \log \left (b x^{3} + a\right ) - \frac {3}{4} \, p x^{2} + \frac {1}{2} \, x^{2} \log \left (c\right ) + \frac {1}{2} \, \sqrt {3} p \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} b x \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} + \sqrt {3} a}{3 \, a}\right ) - \frac {1}{4} \, p \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a x^{2} - b x \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}} - a \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}}\right ) + \frac {1}{2} \, p \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a x + b \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*p*x^2*log(b*x^3 + a) - 3/4*p*x^2 + 1/2*x^2*log(c) + 1/2*sqrt(3)*p*(-a^2/b^2)^(1/3)*arctan(1/3*(2*sqrt(3)*b
*x*(-a^2/b^2)^(1/3) + sqrt(3)*a)/a) - 1/4*p*(-a^2/b^2)^(1/3)*log(a*x^2 - b*x*(-a^2/b^2)^(2/3) - a*(-a^2/b^2)^(
1/3)) + 1/2*p*(-a^2/b^2)^(1/3)*log(a*x + b*(-a^2/b^2)^(2/3))

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Sympy [A]
time = 63.48, size = 178, normalized size = 1.21 \begin {gather*} \begin {cases} \frac {x^{2} \log {\left (0^{p} c \right )}}{2} & \text {for}\: a = 0 \wedge b = 0 \\\frac {x^{2} \log {\left (a^{p} c \right )}}{2} & \text {for}\: b = 0 \\- \frac {3 p x^{2}}{4} + \frac {x^{2} \log {\left (c \left (b x^{3}\right )^{p} \right )}}{2} & \text {for}\: a = 0 \\- \frac {3 p x^{2}}{4} + \frac {3 p \left (- \frac {a}{b}\right )^{\frac {2}{3}} \log {\left (4 x^{2} + 4 x \sqrt [3]{- \frac {a}{b}} + 4 \left (- \frac {a}{b}\right )^{\frac {2}{3}} \right )}}{4} - \frac {\sqrt {3} p \left (- \frac {a}{b}\right )^{\frac {2}{3}} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3 \sqrt [3]{- \frac {a}{b}}} + \frac {\sqrt {3}}{3} \right )}}{2} + \frac {x^{2} \log {\left (c \left (a + b x^{3}\right )^{p} \right )}}{2} - \frac {\left (- \frac {a}{b}\right )^{\frac {2}{3}} \log {\left (c \left (a + b x^{3}\right )^{p} \right )}}{2} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**2*log(0**p*c)/2, Eq(a, 0) & Eq(b, 0)), (x**2*log(a**p*c)/2, Eq(b, 0)), (-3*p*x**2/4 + x**2*log(c
*(b*x**3)**p)/2, Eq(a, 0)), (-3*p*x**2/4 + 3*p*(-a/b)**(2/3)*log(4*x**2 + 4*x*(-a/b)**(1/3) + 4*(-a/b)**(2/3))
/4 - sqrt(3)*p*(-a/b)**(2/3)*atan(2*sqrt(3)*x/(3*(-a/b)**(1/3)) + sqrt(3)/3)/2 + x**2*log(c*(a + b*x**3)**p)/2
- (-a/b)**(2/3)*log(c*(a + b*x**3)**p)/2, True))

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Giac [A]
time = 4.44, size = 150, normalized size = 1.02 \begin {gather*} -\frac {1}{4} \, a b^{2} p {\left (\frac {2 \, \left (-\frac {a}{b}\right )^{\frac {2}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{a b^{2}} + \frac {2 \, \sqrt {3} \left (-a b^{2}\right )^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{a b^{4}} - \frac {\left (-a b^{2}\right )^{\frac {2}{3}} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{a b^{4}}\right )} + \frac {1}{2} \, p x^{2} \log \left (b x^{3} + a\right ) - \frac {1}{4} \, {\left (3 \, p - 2 \, \log \left (c\right )\right )} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*a*b^2*p*(2*(-a/b)^(2/3)*log(abs(x - (-a/b)^(1/3)))/(a*b^2) + 2*sqrt(3)*(-a*b^2)^(2/3)*arctan(1/3*sqrt(3)*
(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/(a*b^4) - (-a*b^2)^(2/3)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/(a*b^4))
+ 1/2*p*x^2*log(b*x^3 + a) - 1/4*(3*p - 2*log(c))*x^2

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Mupad [B]
time = 2.38, size = 121, normalized size = 0.82 \begin {gather*} \frac {x^2\,\ln \left (c\,{\left (b\,x^3+a\right )}^p\right )}{2}-\frac {3\,p\,x^2}{4}-\frac {a^{2/3}\,p\,\ln \left (b^{1/3}\,x+a^{1/3}\right )}{2\,b^{2/3}}-\frac {a^{2/3}\,p\,\ln \left (4\,b^{1/3}\,x-2\,a^{1/3}-\sqrt {3}\,a^{1/3}\,2{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2\,b^{2/3}}+\frac {a^{2/3}\,p\,\ln \left (4\,b^{1/3}\,x-2\,a^{1/3}+\sqrt {3}\,a^{1/3}\,2{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2\,b^{2/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*log(c*(a + b*x^3)^p),x)

[Out]

(x^2*log(c*(a + b*x^3)^p))/2 - (3*p*x^2)/4 - (a^(2/3)*p*log(b^(1/3)*x + a^(1/3)))/(2*b^(2/3)) - (a^(2/3)*p*log
(4*b^(1/3)*x - 3^(1/2)*a^(1/3)*2i - 2*a^(1/3))*((3^(1/2)*1i)/2 - 1/2))/(2*b^(2/3)) + (a^(2/3)*p*log(3^(1/2)*a^
(1/3)*2i + 4*b^(1/3)*x - 2*a^(1/3))*((3^(1/2)*1i)/2 + 1/2))/(2*b^(2/3))

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