3.36 \(\int x^4 \log (c (a+\frac{b}{x^2})^p) \, dx\)

Optimal. Leaf size=72 \[ -\frac{2 b^2 p x}{5 a^2}+\frac{2 b^{5/2} p \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{5 a^{5/2}}+\frac{1}{5} x^5 \log \left (c \left (a+\frac{b}{x^2}\right )^p\right )+\frac{2 b p x^3}{15 a} \]

[Out]

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

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Rubi [A]  time = 0.0375931, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2455, 263, 302, 205} \[ -\frac{2 b^2 p x}{5 a^2}+\frac{2 b^{5/2} p \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{5 a^{5/2}}+\frac{1}{5} x^5 \log \left (c \left (a+\frac{b}{x^2}\right )^p\right )+\frac{2 b p x^3}{15 a} \]

Antiderivative was successfully verified.

[In]

Int[x^4*Log[c*(a + b/x^2)^p],x]

[Out]

(-2*b^2*p*x)/(5*a^2) + (2*b*p*x^3)/(15*a) + (2*b^(5/2)*p*ArcTan[(Sqrt[a]*x)/Sqrt[b]])/(5*a^(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 263

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

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

Mathematica [C]  time = 0.0060882, size = 49, normalized size = 0.68 \[ \frac{1}{5} x^5 \log \left (c \left (a+\frac{b}{x^2}\right )^p\right )+\frac{2 b p x^3 \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};-\frac{b}{a x^2}\right )}{15 a} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*Log[c*(a + b/x^2)^p],x]

[Out]

(2*b*p*x^3*Hypergeometric2F1[-3/2, 1, -1/2, -(b/(a*x^2))])/(15*a) + (x^5*Log[c*(a + b/x^2)^p])/5

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Maple [F]  time = 0.333, size = 0, normalized size = 0. \begin{align*} \int{x}^{4}\ln \left ( c \left ( a+{\frac{b}{{x}^{2}}} \right ) ^{p} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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*(a+b/x^2)^p),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.32446, size = 401, normalized size = 5.57 \begin{align*} \left [\frac{3 \, a^{2} p x^{5} \log \left (\frac{a x^{2} + b}{x^{2}}\right ) + 3 \, a^{2} x^{5} \log \left (c\right ) + 2 \, a b p x^{3} + 3 \, b^{2} p \sqrt{-\frac{b}{a}} \log \left (\frac{a x^{2} + 2 \, a x \sqrt{-\frac{b}{a}} - b}{a x^{2} + b}\right ) - 6 \, b^{2} p x}{15 \, a^{2}}, \frac{3 \, a^{2} p x^{5} \log \left (\frac{a x^{2} + b}{x^{2}}\right ) + 3 \, a^{2} x^{5} \log \left (c\right ) + 2 \, a b p x^{3} + 6 \, b^{2} p \sqrt{\frac{b}{a}} \arctan \left (\frac{a x \sqrt{\frac{b}{a}}}{b}\right ) - 6 \, b^{2} p x}{15 \, a^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.19816, size = 101, normalized size = 1.4 \begin{align*} \frac{1}{5} \, p x^{5} \log \left (a x^{2} + b\right ) - \frac{1}{5} \, p x^{5} \log \left (x^{2}\right ) + \frac{1}{5} \, x^{5} \log \left (c\right ) + \frac{2 \, b p x^{3}}{15 \, a} + \frac{2 \, b^{3} p \arctan \left (\frac{a x}{\sqrt{a b}}\right )}{5 \, \sqrt{a b} a^{2}} - \frac{2 \, b^{2} p x}{5 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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