3.1 \(\int x^4 \log (c (a+b x^2)^p) \, dx\)

Optimal. Leaf size=80 \[ -\frac{2 a^2 p x}{5 b^2}+\frac{2 a^{5/2} p \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{5 b^{5/2}}+\frac{1}{5} x^5 \log \left (c \left (a+b x^2\right )^p\right )+\frac{2 a p x^3}{15 b}-\frac{2 p x^5}{25} \]

[Out]

(-2*a^2*p*x)/(5*b^2) + (2*a*p*x^3)/(15*b) - (2*p*x^5)/25 + (2*a^(5/2)*p*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(5*b^(5/2
)) + (x^5*Log[c*(a + b*x^2)^p])/5

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Rubi [A]  time = 0.0465048, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {2455, 302, 205} \[ -\frac{2 a^2 p x}{5 b^2}+\frac{2 a^{5/2} p \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{5 b^{5/2}}+\frac{1}{5} x^5 \log \left (c \left (a+b x^2\right )^p\right )+\frac{2 a p x^3}{15 b}-\frac{2 p x^5}{25} \]

Antiderivative was successfully verified.

[In]

Int[x^4*Log[c*(a + b*x^2)^p],x]

[Out]

(-2*a^2*p*x)/(5*b^2) + (2*a*p*x^3)/(15*b) - (2*p*x^5)/25 + (2*a^(5/2)*p*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(5*b^(5/2
)) + (x^5*Log[c*(a + b*x^2)^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 205

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

Rubi steps

\begin{align*} \int x^4 \log \left (c \left (a+b x^2\right )^p\right ) \, dx &=\frac{1}{5} x^5 \log \left (c \left (a+b x^2\right )^p\right )-\frac{1}{5} (2 b p) \int \frac{x^6}{a+b x^2} \, dx\\ &=\frac{1}{5} x^5 \log \left (c \left (a+b x^2\right )^p\right )-\frac{1}{5} (2 b p) \int \left (\frac{a^2}{b^3}-\frac{a x^2}{b^2}+\frac{x^4}{b}-\frac{a^3}{b^3 \left (a+b x^2\right )}\right ) \, dx\\ &=-\frac{2 a^2 p x}{5 b^2}+\frac{2 a p x^3}{15 b}-\frac{2 p x^5}{25}+\frac{1}{5} x^5 \log \left (c \left (a+b x^2\right )^p\right )+\frac{\left (2 a^3 p\right ) \int \frac{1}{a+b x^2} \, dx}{5 b^2}\\ &=-\frac{2 a^2 p x}{5 b^2}+\frac{2 a p x^3}{15 b}-\frac{2 p x^5}{25}+\frac{2 a^{5/2} p \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{5 b^{5/2}}+\frac{1}{5} x^5 \log \left (c \left (a+b x^2\right )^p\right )\\ \end{align*}

Mathematica [A]  time = 0.0444517, size = 74, normalized size = 0.92 \[ \frac{1}{75} \left (-\frac{30 a^2 p x}{b^2}+\frac{30 a^{5/2} p \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{b^{5/2}}+15 x^5 \log \left (c \left (a+b x^2\right )^p\right )+\frac{10 a p x^3}{b}-6 p x^5\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*Log[c*(a + b*x^2)^p],x]

[Out]

((-30*a^2*p*x)/b^2 + (10*a*p*x^3)/b - 6*p*x^5 + (30*a^(5/2)*p*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/b^(5/2) + 15*x^5*Lo
g[c*(a + b*x^2)^p])/75

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Maple [C]  time = 0.499, size = 229, normalized size = 2.9 \begin{align*}{\frac{{x}^{5}\ln \left ( \left ( b{x}^{2}+a \right ) ^{p} \right ) }{5}}-{\frac{i}{10}}\pi \,{x}^{5} \left ({\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ) \right ) ^{3}+{\frac{i}{10}}\pi \,{x}^{5} \left ({\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) +{\frac{i}{10}}\pi \,{x}^{5}{\it csgn} \left ( i \left ( b{x}^{2}+a \right ) ^{p} \right ) \left ({\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ) \right ) ^{2}-{\frac{i}{10}}\pi \,{x}^{5}{\it csgn} \left ( i \left ( b{x}^{2}+a \right ) ^{p} \right ){\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ){\it csgn} \left ( ic \right ) +{\frac{\ln \left ( c \right ){x}^{5}}{5}}-{\frac{2\,p{x}^{5}}{25}}+{\frac{2\,ap{x}^{3}}{15\,b}}+{\frac{{a}^{2}p}{5\,{b}^{3}}\sqrt{-ab}\ln \left ( -\sqrt{-ab}x+a \right ) }-{\frac{{a}^{2}p}{5\,{b}^{3}}\sqrt{-ab}\ln \left ( \sqrt{-ab}x+a \right ) }-{\frac{2\,{a}^{2}px}{5\,{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*ln(c*(b*x^2+a)^p),x)

[Out]

1/5*x^5*ln((b*x^2+a)^p)-1/10*I*Pi*x^5*csgn(I*c*(b*x^2+a)^p)^3+1/10*I*Pi*x^5*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)+
1/10*I*Pi*x^5*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2-1/10*I*Pi*x^5*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)
^p)*csgn(I*c)+1/5*ln(c)*x^5-2/25*p*x^5+2/15*a*p*x^3/b+1/5/b^3*(-a*b)^(1/2)*a^2*p*ln(-(-a*b)^(1/2)*x+a)-1/5/b^3
*(-a*b)^(1/2)*a^2*p*ln((-a*b)^(1/2)*x+a)-2/5*a^2*p*x/b^2

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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^2+a)^p),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.10261, size = 436, normalized size = 5.45 \begin{align*} \left [\frac{15 \, b^{2} p x^{5} \log \left (b x^{2} + a\right ) - 6 \, b^{2} p x^{5} + 15 \, b^{2} x^{5} \log \left (c\right ) + 10 \, a b p x^{3} + 15 \, a^{2} p \sqrt{-\frac{a}{b}} \log \left (\frac{b x^{2} + 2 \, b x \sqrt{-\frac{a}{b}} - a}{b x^{2} + a}\right ) - 30 \, a^{2} p x}{75 \, b^{2}}, \frac{15 \, b^{2} p x^{5} \log \left (b x^{2} + a\right ) - 6 \, b^{2} p x^{5} + 15 \, b^{2} x^{5} \log \left (c\right ) + 10 \, a b p x^{3} + 30 \, a^{2} p \sqrt{\frac{a}{b}} \arctan \left (\frac{b x \sqrt{\frac{a}{b}}}{a}\right ) - 30 \, a^{2} p x}{75 \, b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*log(c*(b*x^2+a)^p),x, algorithm="fricas")

[Out]

[1/75*(15*b^2*p*x^5*log(b*x^2 + a) - 6*b^2*p*x^5 + 15*b^2*x^5*log(c) + 10*a*b*p*x^3 + 15*a^2*p*sqrt(-a/b)*log(
(b*x^2 + 2*b*x*sqrt(-a/b) - a)/(b*x^2 + a)) - 30*a^2*p*x)/b^2, 1/75*(15*b^2*p*x^5*log(b*x^2 + a) - 6*b^2*p*x^5
+ 15*b^2*x^5*log(c) + 10*a*b*p*x^3 + 30*a^2*p*sqrt(a/b)*arctan(b*x*sqrt(a/b)/a) - 30*a^2*p*x)/b^2]

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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**2+a)**p),x)

[Out]

Timed out

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Giac [A]  time = 1.21173, size = 96, normalized size = 1.2 \begin{align*} \frac{1}{5} \, p x^{5} \log \left (b x^{2} + a\right ) - \frac{1}{25} \,{\left (2 \, p - 5 \, \log \left (c\right )\right )} x^{5} + \frac{2 \, a p x^{3}}{15 \, b} + \frac{2 \, a^{3} p \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{5 \, \sqrt{a b} b^{2}} - \frac{2 \, a^{2} p x}{5 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*log(c*(b*x^2+a)^p),x, algorithm="giac")

[Out]

1/5*p*x^5*log(b*x^2 + a) - 1/25*(2*p - 5*log(c))*x^5 + 2/15*a*p*x^3/b + 2/5*a^3*p*arctan(b*x/sqrt(a*b))/(sqrt(
a*b)*b^2) - 2/5*a^2*p*x/b^2