3.1.14 \(\int x^4 \log (c (a+b x^3)^p) \, dx\) [14]

Optimal. Leaf size=159 \[ \frac {3 a p x^2}{10 b}-\frac {3 p x^5}{25}+\frac {\sqrt {3} a^{5/3} p \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{5 b^{5/3}}+\frac {a^{5/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{5 b^{5/3}}-\frac {a^{5/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{10 b^{5/3}}+\frac {1}{5} x^5 \log \left (c \left (a+b x^3\right )^p\right ) \]

[Out]

3/10*a*p*x^2/b-3/25*p*x^5+1/5*a^(5/3)*p*ln(a^(1/3)+b^(1/3)*x)/b^(5/3)-1/10*a^(5/3)*p*ln(a^(2/3)-a^(1/3)*b^(1/3
)*x+b^(2/3)*x^2)/b^(5/3)+1/5*x^5*ln(c*(b*x^3+a)^p)+1/5*a^(5/3)*p*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)/a^(1/3)*3^(1
/2))*3^(1/2)/b^(5/3)

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Rubi [A]
time = 0.08, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2505, 308, 298, 31, 648, 631, 210, 642} \begin {gather*} \frac {\sqrt {3} a^{5/3} p \text {ArcTan}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{5 b^{5/3}}-\frac {a^{5/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{10 b^{5/3}}+\frac {a^{5/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{5 b^{5/3}}+\frac {1}{5} x^5 \log \left (c \left (a+b x^3\right )^p\right )+\frac {3 a p x^2}{10 b}-\frac {3 p x^5}{25} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4*Log[c*(a + b*x^3)^p],x]

[Out]

(3*a*p*x^2)/(10*b) - (3*p*x^5)/25 + (Sqrt[3]*a^(5/3)*p*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(5*b
^(5/3)) + (a^(5/3)*p*Log[a^(1/3) + b^(1/3)*x])/(5*b^(5/3)) - (a^(5/3)*p*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2
/3)*x^2])/(10*b^(5/3)) + (x^5*Log[c*(a + b*x^3)^p])/5

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 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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^4 \log \left (c \left (a+b x^3\right )^p\right ) \, dx &=\frac {1}{5} x^5 \log \left (c \left (a+b x^3\right )^p\right )-\frac {1}{5} (3 b p) \int \frac {x^7}{a+b x^3} \, dx\\ &=\frac {1}{5} x^5 \log \left (c \left (a+b x^3\right )^p\right )-\frac {1}{5} (3 b p) \int \left (-\frac {a x}{b^2}+\frac {x^4}{b}+\frac {a^2 x}{b^2 \left (a+b x^3\right )}\right ) \, dx\\ &=\frac {3 a p x^2}{10 b}-\frac {3 p x^5}{25}+\frac {1}{5} x^5 \log \left (c \left (a+b x^3\right )^p\right )-\frac {\left (3 a^2 p\right ) \int \frac {x}{a+b x^3} \, dx}{5 b}\\ &=\frac {3 a p x^2}{10 b}-\frac {3 p x^5}{25}+\frac {1}{5} x^5 \log \left (c \left (a+b x^3\right )^p\right )+\frac {\left (a^{5/3} p\right ) \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{5 b^{4/3}}-\frac {\left (a^{5/3} 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}{5 b^{4/3}}\\ &=\frac {3 a p x^2}{10 b}-\frac {3 p x^5}{25}+\frac {a^{5/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{5 b^{5/3}}+\frac {1}{5} x^5 \log \left (c \left (a+b x^3\right )^p\right )-\frac {\left (a^{5/3} 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}{10 b^{5/3}}-\frac {\left (3 a^2 p\right ) \int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{10 b^{4/3}}\\ &=\frac {3 a p x^2}{10 b}-\frac {3 p x^5}{25}+\frac {a^{5/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{5 b^{5/3}}-\frac {a^{5/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{10 b^{5/3}}+\frac {1}{5} x^5 \log \left (c \left (a+b x^3\right )^p\right )-\frac {\left (3 a^{5/3} p\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{5 b^{5/3}}\\ &=\frac {3 a p x^2}{10 b}-\frac {3 p x^5}{25}+\frac {\sqrt {3} a^{5/3} p \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{5 b^{5/3}}+\frac {a^{5/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{5 b^{5/3}}-\frac {a^{5/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{10 b^{5/3}}+\frac {1}{5} x^5 \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 = 69, normalized size = 0.43 \begin {gather*} \frac {3 a p x^2}{10 b}-\frac {3 p x^5}{25}-\frac {3 a p x^2 \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};-\frac {b x^3}{a}\right )}{10 b}+\frac {1}{5} x^5 \log \left (c \left (a+b x^3\right )^p\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4*Log[c*(a + b*x^3)^p],x]

[Out]

(3*a*p*x^2)/(10*b) - (3*p*x^5)/25 - (3*a*p*x^2*Hypergeometric2F1[2/3, 1, 5/3, -((b*x^3)/a)])/(10*b) + (x^5*Log
[c*(a + b*x^3)^p])/5

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.41, size = 196, normalized size = 1.23

method	result	size
risch	\(\frac {x^{5} \ln \left (\left (x^{3} b +a \right )^{p}\right )}{5}-\frac {i \pi \,x^{5} \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right )^{3}}{10}+\frac {i \pi \,x^{5} \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right )}{10}+\frac {i \pi \,x^{5} \mathrm {csgn}\left (i \left (x^{3} b +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right )^{2}}{10}-\frac {i \pi \,x^{5} \mathrm {csgn}\left (i \left (x^{3} b +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right ) \mathrm {csgn}\left (i c \right )}{10}+\frac {\ln \left (c \right ) x^{5}}{5}-\frac {3 p \,x^{5}}{25}+\frac {3 a p \,x^{2}}{10 b}-\frac {a^{2} p \left (\munderset {\textit {\_R} =\RootOf \left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}}\right )}{5 b^{2}}\)	\(196\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*ln(c*(b*x^3+a)^p),x,method=_RETURNVERBOSE)

[Out]

1/5*x^5*ln((b*x^3+a)^p)-1/10*I*Pi*x^5*csgn(I*c*(b*x^3+a)^p)^3+1/10*I*Pi*x^5*csgn(I*c*(b*x^3+a)^p)^2*csgn(I*c)+
1/10*I*Pi*x^5*csgn(I*(b*x^3+a)^p)*csgn(I*c*(b*x^3+a)^p)^2-1/10*I*Pi*x^5*csgn(I*(b*x^3+a)^p)*csgn(I*c*(b*x^3+a)
^p)*csgn(I*c)+1/5*ln(c)*x^5-3/25*p*x^5+3/10*a*p*x^2/b-1/5/b^2*a^2*p*sum(1/_R*ln(x-_R),_R=RootOf(_Z^3*b+a))

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Maxima [A]
time = 0.50, size = 147, normalized size = 0.92 \begin {gather*} \frac {1}{5} \, x^{5} \log \left ({\left (b x^{3} + a\right )}^{p} c\right ) - \frac {1}{50} \, b p {\left (\frac {10 \, \sqrt {3} a^{2} \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^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {5 \, a^{2} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{b^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {10 \, a^{2} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{b^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {3 \, {\left (2 \, b x^{5} - 5 \, a x^{2}\right )}}{b^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*log(c*(b*x^3+a)^p),x, algorithm="maxima")

[Out]

1/5*x^5*log((b*x^3 + a)^p*c) - 1/50*b*p*(10*sqrt(3)*a^2*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(b
^3*(a/b)^(1/3)) + 5*a^2*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(b^3*(a/b)^(1/3)) - 10*a^2*log(x + (a/b)^(1/3))
/(b^3*(a/b)^(1/3)) + 3*(2*b*x^5 - 5*a*x^2)/b^2)

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Fricas [A]
time = 0.40, size = 161, normalized size = 1.01 \begin {gather*} \frac {10 \, b p x^{5} \log \left (b x^{3} + a\right ) - 6 \, b p x^{5} + 10 \, b x^{5} \log \left (c\right ) + 15 \, a p x^{2} - 10 \, \sqrt {3} a 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 ) - 5 \, a 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 ) + 10 \, a 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 )}{50 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*log(c*(b*x^3+a)^p),x, algorithm="fricas")

[Out]

1/50*(10*b*p*x^5*log(b*x^3 + a) - 6*b*p*x^5 + 10*b*x^5*log(c) + 15*a*p*x^2 - 10*sqrt(3)*a*p*(a^2/b^2)^(1/3)*ar
ctan(1/3*(2*sqrt(3)*b*x*(a^2/b^2)^(1/3) - sqrt(3)*a)/a) - 5*a*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)) + 10*a*p*(a^2/b^2)^(1/3)*log(a*x + b*(a^2/b^2)^(2/3)))/b

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*ln(c*(b*x**3+a)**p),x)

[Out]

Timed out

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Giac [A]
time = 2.86, size = 162, normalized size = 1.02 \begin {gather*} \frac {1}{10} \, a^{2} b^{4} 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^{5}} + \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^{7}} - \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^{7}}\right )} + \frac {1}{5} \, p x^{5} \log \left (b x^{3} + a\right ) - \frac {1}{25} \, {\left (3 \, p - 5 \, \log \left (c\right )\right )} x^{5} + \frac {3 \, a p x^{2}}{10 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*log(c*(b*x^3+a)^p),x, algorithm="giac")

[Out]

1/10*a^2*b^4*p*(2*(-a/b)^(2/3)*log(abs(x - (-a/b)^(1/3)))/(a*b^5) + 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^7) - (-a*b^2)^(2/3)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/(a*b^7)
) + 1/5*p*x^5*log(b*x^3 + a) - 1/25*(3*p - 5*log(c))*x^5 + 3/10*a*p*x^2/b

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Mupad [B]
time = 2.50, size = 157, normalized size = 0.99 \begin {gather*} \frac {x^5\,\ln \left (c\,{\left (b\,x^3+a\right )}^p\right )}{5}-\frac {3\,p\,x^5}{25}+\frac {a^{5/3}\,p\,\ln \left (b^{1/3}\,x+a^{1/3}\right )}{5\,b^{5/3}}+\frac {3\,a\,p\,x^2}{10\,b}+\frac {a^{5/3}\,p\,\ln \left (\frac {9\,a^4\,p^2\,x}{25\,b}+\frac {9\,a^{13/3}\,p^2\,{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2}{25\,b^{4/3}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{5\,b^{5/3}}-\frac {a^{5/3}\,p\,\ln \left (\frac {9\,a^4\,p^2\,x}{25\,b}+\frac {9\,a^{13/3}\,p^2\,{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2}{25\,b^{4/3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{5\,b^{5/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*log(c*(a + b*x^3)^p),x)

[Out]

(x^5*log(c*(a + b*x^3)^p))/5 - (3*p*x^5)/25 + (a^(5/3)*p*log(b^(1/3)*x + a^(1/3)))/(5*b^(5/3)) + (3*a*p*x^2)/(
10*b) + (a^(5/3)*p*log((9*a^4*p^2*x)/(25*b) + (9*a^(13/3)*p^2*((3^(1/2)*1i)/2 - 1/2)^2)/(25*b^(4/3)))*((3^(1/2
)*1i)/2 - 1/2))/(5*b^(5/3)) - (a^(5/3)*p*log((9*a^4*p^2*x)/(25*b) + (9*a^(13/3)*p^2*((3^(1/2)*1i)/2 + 1/2)^2)/
(25*b^(4/3)))*((3^(1/2)*1i)/2 + 1/2))/(5*b^(5/3))

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