3.17 \(\int x \log (c (a+b x^3)^p) \, dx\)

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

[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 = 0.08, 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 = {2455, 321, 292, 31, 634, 617, 204, 628} \[ \frac {a^{2/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{4 b^{2/3}}-\frac {a^{2/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{2 b^{2/3}}-\frac {\sqrt {3} a^{2/3} p \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{2 b^{2/3}}+\frac {1}{2} x^2 \log \left (c \left (a+b x^3\right )^p\right )-\frac {3 p x^2}{4} \]

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 204

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

Rule 292

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 321

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 617

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 628

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 634

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 2455

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 ) \operatorname {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]  time = 0.00, size = 53, normalized size = 0.36 \[ \frac {1}{2} x^2 \log \left (c \left (a+b x^3\right )^p\right )+\frac {3}{4} p x^2 \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};-\frac {b x^3}{a}\right )-\frac {3 p x^2}{4} \]

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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fricas [A]  time = 0.48, size = 150, normalized size = 1.02 \[ \frac {1}{2} \, p x^{2} \log \left (b x^{3} + a\right ) - \frac {3}{4} \, p x^{2} + \frac {1}{2} \, x^{2} \log \relax (c) + \frac {1}{2} \, \sqrt {3} p \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} b x \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} + \sqrt {3} a}{3 \, a}\right ) - \frac {1}{4} \, p \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a x^{2} - b x \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}} - a \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}}\right ) + \frac {1}{2} \, p \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a x + b \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}}\right ) \]

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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giac [A]  time = 0.20, size = 150, normalized size = 1.02 \[ -\frac {1}{4} \, a b^{2} p {\left (\frac {2 \, \left (-\frac {a}{b}\right )^{\frac {2}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{a b^{2}} + \frac {2 \, \sqrt {3} \left (-a b^{2}\right )^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{a b^{4}} - \frac {\left (-a b^{2}\right )^{\frac {2}{3}} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{a b^{4}}\right )} + \frac {1}{2} \, p x^{2} \log \left (b x^{3} + a\right ) - \frac {1}{4} \, {\left (3 \, p - 2 \, \log \relax (c)\right )} x^{2} \]

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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maple [C]  time = 0.45, size = 184, normalized size = 1.25 \[ -\frac {i \pi \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}{4}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )^{2}}{4}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )^{2}}{4}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )^{3}}{4}-\frac {3 p \,x^{2}}{4}+\frac {x^{2} \ln \relax (c )}{2}+\frac {x^{2} \ln \left (\left (b \,x^{3}+a \right )^{p}\right )}{2}+\frac {a p \ln \left (-\RootOf \left (b \,\textit {\_Z}^{3}+a \right )+x \right )}{2 b \RootOf \left (b \,\textit {\_Z}^{3}+a \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.33, size = 131, normalized size = 0.89 \[ -\frac {1}{4} \, b p {\left (\frac {3 \, x^{2}}{b} - \frac {2 \, \sqrt {3} a \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{b^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {a \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{b^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {2 \, a \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{b^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )} + \frac {1}{2} \, x^{2} \log \left ({\left (b x^{3} + a\right )}^{p} c\right ) \]

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

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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sympy [A]  time = 142.41, size = 260, normalized size = 1.77 \[ \begin {cases} \frac {x^{2} \log {\left (0^{p} c \right )}}{2} & \text {for}\: a = 0 \wedge b = 0 \\\frac {x^{2} \log {\left (a^{p} c \right )}}{2} & \text {for}\: b = 0 \\\frac {p x^{2} \log {\relax (b )}}{2} + \frac {3 p x^{2} \log {\relax (x )}}{2} - \frac {3 p x^{2}}{4} + \frac {x^{2} \log {\relax (c )}}{2} & \text {for}\: a = 0 \\- \frac {\left (-1\right )^{\frac {2}{3}} a^{\frac {2}{3}} p \left (\frac {1}{b}\right )^{\frac {2}{3}} \log {\left (a + b x^{3} \right )}}{2} + \frac {3 \left (-1\right )^{\frac {2}{3}} a^{\frac {2}{3}} p \left (\frac {1}{b}\right )^{\frac {2}{3}} \log {\left (4 \left (-1\right )^{\frac {2}{3}} a^{\frac {2}{3}} \left (\frac {1}{b}\right )^{\frac {2}{3}} + 4 \sqrt [3]{-1} \sqrt [3]{a} x \sqrt [3]{\frac {1}{b}} + 4 x^{2} \right )}}{4} - \frac {\left (-1\right )^{\frac {2}{3}} \sqrt {3} a^{\frac {2}{3}} p \left (\frac {1}{b}\right )^{\frac {2}{3}} \operatorname {atan}{\left (\frac {\sqrt {3}}{3} - \frac {2 \left (-1\right )^{\frac {2}{3}} \sqrt {3} x}{3 \sqrt [3]{a} \sqrt [3]{\frac {1}{b}}} \right )}}{2} + \frac {p x^{2} \log {\left (a + b x^{3} \right )}}{2} - \frac {3 p x^{2}}{4} + \frac {x^{2} \log {\relax (c )}}{2} & \text {otherwise} \end {cases} \]

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

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