3.14 \(\int x^4 \log (c (a+b x^3)^p) \, dx\)

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

[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

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Rubi [A]  time = 0.129039, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {2455, 302, 292, 31, 634, 617, 204, 628} \[ -\frac{a^{5/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{10 b^{5/3}}+\frac{a^{5/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{5 b^{5/3}}+\frac{\sqrt{3} a^{5/3} p \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{5 b^{5/3}}+\frac{1}{5} x^5 \log \left (c \left (a+b x^3\right )^p\right )+\frac{3 a p x^2}{10 b}-\frac{3 p x^5}{25} \]

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 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]

Rule 302

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 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 31

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

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 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 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 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]

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 ) \operatorname{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*}

Mathematica [C]  time = 0.0032979, size = 69, normalized size = 0.43 \[ \frac{1}{5} x^5 \log \left (c \left (a+b x^3\right )^p\right )-\frac{3 a p x^2 \, _2F_1\left (\frac{2}{3},1;\frac{5}{3};-\frac{b x^3}{a}\right )}{10 b}+\frac{3 a p x^2}{10 b}-\frac{3 p x^5}{25} \]

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]  time = 0.452, size = 196, normalized size = 1.2 \begin{align*}{\frac{{x}^{5}\ln \left ( \left ( b{x}^{3}+a \right ) ^{p} \right ) }{5}}-{\frac{i}{10}}\pi \,{x}^{5}{\it csgn} \left ( i \left ( b{x}^{3}+a \right ) ^{p} \right ){\it csgn} \left ( ic \left ( b{x}^{3}+a \right ) ^{p} \right ){\it csgn} \left ( ic \right ) +{\frac{i}{10}}\pi \,{x}^{5} \left ({\it csgn} \left ( ic \left ( b{x}^{3}+a \right ) ^{p} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) +{\frac{i}{10}}\pi \,{x}^{5}{\it csgn} \left ( i \left ( b{x}^{3}+a \right ) ^{p} \right ) \left ({\it csgn} \left ( ic \left ( b{x}^{3}+a \right ) ^{p} \right ) \right ) ^{2}-{\frac{i}{10}}\pi \,{x}^{5} \left ({\it csgn} \left ( ic \left ( b{x}^{3}+a \right ) ^{p} \right ) \right ) ^{3}+{\frac{\ln \left ( c \right ){x}^{5}}{5}}-{\frac{3\,p{x}^{5}}{25}}+{\frac{3\,ap{x}^{2}}{10\,b}}-{\frac{{a}^{2}p}{5\,{b}^{2}}\sum _{{\it \_R}={\it RootOf} \left ( b{{\it \_Z}}^{3}+a \right ) }{\frac{\ln \left ( x-{\it \_R} \right ) }{{\it \_R}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*x^5*ln((b*x^3+a)^p)-1/10*I*Pi*x^5*csgn(I*(b*x^3+a)^p)*csgn(I*c*(b*x^3+a)^p)*csgn(I*c)+1/10*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*c*(
b*x^3+a)^p)^3+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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 2.14814, size = 406, normalized size = 2.55 \begin{align*} \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{align*}

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)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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 = 1.27326, size = 219, normalized size = 1.38 \begin{align*} \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{align*}

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