3.23 \(\int \frac{\log (c (a+b x^3)^p)}{x^5} \, dx\)

Optimal. Leaf size=151 \[ -\frac{b^{4/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{8 a^{4/3}}+\frac{b^{4/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{4 a^{4/3}}+\frac{\sqrt{3} b^{4/3} p \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{4 a^{4/3}}-\frac{\log \left (c \left (a+b x^3\right )^p\right )}{4 x^4}-\frac{3 b p}{4 a x} \]

[Out]

(-3*b*p)/(4*a*x) + (Sqrt[3]*b^(4/3)*p*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(4*a^(4/3)) + (b^(4/3
)*p*Log[a^(1/3) + b^(1/3)*x])/(4*a^(4/3)) - (b^(4/3)*p*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(8*a^(4
/3)) - Log[c*(a + b*x^3)^p]/(4*x^4)

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Rubi [A]  time = 0.0945938, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {2455, 325, 292, 31, 634, 617, 204, 628} \[ -\frac{b^{4/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{8 a^{4/3}}+\frac{b^{4/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{4 a^{4/3}}+\frac{\sqrt{3} b^{4/3} p \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{4 a^{4/3}}-\frac{\log \left (c \left (a+b x^3\right )^p\right )}{4 x^4}-\frac{3 b p}{4 a x} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b*x^3)^p]/x^5,x]

[Out]

(-3*b*p)/(4*a*x) + (Sqrt[3]*b^(4/3)*p*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(4*a^(4/3)) + (b^(4/3
)*p*Log[a^(1/3) + b^(1/3)*x])/(4*a^(4/3)) - (b^(4/3)*p*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(8*a^(4
/3)) - Log[c*(a + b*x^3)^p]/(4*x^4)

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 325

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

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 \frac{\log \left (c \left (a+b x^3\right )^p\right )}{x^5} \, dx &=-\frac{\log \left (c \left (a+b x^3\right )^p\right )}{4 x^4}+\frac{1}{4} (3 b p) \int \frac{1}{x^2 \left (a+b x^3\right )} \, dx\\ &=-\frac{3 b p}{4 a x}-\frac{\log \left (c \left (a+b x^3\right )^p\right )}{4 x^4}-\frac{\left (3 b^2 p\right ) \int \frac{x}{a+b x^3} \, dx}{4 a}\\ &=-\frac{3 b p}{4 a x}-\frac{\log \left (c \left (a+b x^3\right )^p\right )}{4 x^4}+\frac{\left (b^{5/3} p\right ) \int \frac{1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{4 a^{4/3}}-\frac{\left (b^{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}{4 a^{4/3}}\\ &=-\frac{3 b p}{4 a x}+\frac{b^{4/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{4 a^{4/3}}-\frac{\log \left (c \left (a+b x^3\right )^p\right )}{4 x^4}-\frac{\left (b^{4/3} 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}{8 a^{4/3}}-\frac{\left (3 b^{5/3} p\right ) \int \frac{1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{8 a}\\ &=-\frac{3 b p}{4 a x}+\frac{b^{4/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{4 a^{4/3}}-\frac{b^{4/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{8 a^{4/3}}-\frac{\log \left (c \left (a+b x^3\right )^p\right )}{4 x^4}-\frac{\left (3 b^{4/3} p\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{4 a^{4/3}}\\ &=-\frac{3 b p}{4 a x}+\frac{\sqrt{3} b^{4/3} p \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{4 a^{4/3}}+\frac{b^{4/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{4 a^{4/3}}-\frac{b^{4/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{8 a^{4/3}}-\frac{\log \left (c \left (a+b x^3\right )^p\right )}{4 x^4}\\ \end{align*}

Mathematica [C]  time = 0.0028287, size = 49, normalized size = 0.32 \[ -\frac{\log \left (c \left (a+b x^3\right )^p\right )}{4 x^4}-\frac{3 b p \, _2F_1\left (-\frac{1}{3},1;\frac{2}{3};-\frac{b x^3}{a}\right )}{4 a x} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(a + b*x^3)^p]/x^5,x]

[Out]

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

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Maple [C]  time = 0.323, size = 215, normalized size = 1.4 \begin{align*} -{\frac{\ln \left ( \left ( b{x}^{3}+a \right ) ^{p} \right ) }{4\,{x}^{4}}}-{\frac{i\pi \,a{\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}-i\pi \,a{\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 ) -i\pi \,a \left ({\it csgn} \left ( ic \left ( b{x}^{3}+a \right ) ^{p} \right ) \right ) ^{3}+i\pi \,a \left ({\it csgn} \left ( ic \left ( b{x}^{3}+a \right ) ^{p} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) -2\,\sum _{{\it \_R}={\it RootOf} \left ({a}^{4}{{\it \_Z}}^{3}-{b}^{4}{p}^{3} \right ) }{\it \_R}\,\ln \left ( \left ( -4\,{a}^{4}{{\it \_R}}^{3}+3\,{b}^{4}{p}^{3} \right ) x-{a}^{3}p{{\it \_R}}^{2}b \right ) a{x}^{4}+6\,bp{x}^{3}+2\,\ln \left ( c \right ) a}{8\,a{x}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(c*(b*x^3+a)^p)/x^5,x)

[Out]

-1/4/x^4*ln((b*x^3+a)^p)-1/8*(I*Pi*a*csgn(I*(b*x^3+a)^p)*csgn(I*c*(b*x^3+a)^p)^2-I*Pi*a*csgn(I*(b*x^3+a)^p)*cs
gn(I*c*(b*x^3+a)^p)*csgn(I*c)-I*Pi*a*csgn(I*c*(b*x^3+a)^p)^3+I*Pi*a*csgn(I*c*(b*x^3+a)^p)^2*csgn(I*c)-2*sum(_R
*ln((-4*_R^3*a^4+3*b^4*p^3)*x-a^3*p*_R^2*b),_R=RootOf(_Z^3*a^4-b^4*p^3))*a*x^4+6*b*p*x^3+2*ln(c)*a)/a/x^4

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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(log(c*(b*x^3+a)^p)/x^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.45381, size = 348, normalized size = 2.3 \begin{align*} -\frac{2 \, \sqrt{3} b p x^{4} \left (\frac{b}{a}\right )^{\frac{1}{3}} \arctan \left (\frac{2}{3} \, \sqrt{3} x \left (\frac{b}{a}\right )^{\frac{1}{3}} - \frac{1}{3} \, \sqrt{3}\right ) + b p x^{4} \left (\frac{b}{a}\right )^{\frac{1}{3}} \log \left (b x^{2} - a x \left (\frac{b}{a}\right )^{\frac{2}{3}} + a \left (\frac{b}{a}\right )^{\frac{1}{3}}\right ) - 2 \, b p x^{4} \left (\frac{b}{a}\right )^{\frac{1}{3}} \log \left (b x + a \left (\frac{b}{a}\right )^{\frac{2}{3}}\right ) + 6 \, b p x^{3} + 2 \, a p \log \left (b x^{3} + a\right ) + 2 \, a \log \left (c\right )}{8 \, a x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^3+a)^p)/x^5,x, algorithm="fricas")

[Out]

-1/8*(2*sqrt(3)*b*p*x^4*(b/a)^(1/3)*arctan(2/3*sqrt(3)*x*(b/a)^(1/3) - 1/3*sqrt(3)) + b*p*x^4*(b/a)^(1/3)*log(
b*x^2 - a*x*(b/a)^(2/3) + a*(b/a)^(1/3)) - 2*b*p*x^4*(b/a)^(1/3)*log(b*x + a*(b/a)^(2/3)) + 6*b*p*x^3 + 2*a*p*
log(b*x^3 + a) + 2*a*log(c))/(a*x^4)

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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(ln(c*(b*x**3+a)**p)/x**5,x)

[Out]

Timed out

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Giac [A]  time = 1.2632, size = 207, normalized size = 1.37 \begin{align*} \frac{1}{8} \, 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^{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^{2} b^{2}} - \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^{2} b^{2}}\right )} - \frac{p \log \left (b x^{3} + a\right )}{4 \, x^{4}} - \frac{3 \, b p x^{3} + a \log \left (c\right )}{4 \, a x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^3+a)^p)/x^5,x, algorithm="giac")

[Out]

1/8*b^2*p*(2*(-a/b)^(2/3)*log(abs(x - (-a/b)^(1/3)))/a^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^2*b^2) - (-a*b^2)^(2/3)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/(a^2*b^2)) - 1
/4*p*log(b*x^3 + a)/x^4 - 1/4*(3*b*p*x^3 + a*log(c))/(a*x^4)