3.318 \(\int x^2 (f+g x^2) \log (c (d+e x^2)^p) \, dx\)

Optimal. Leaf size=154 \[ \frac{1}{3} f x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac{1}{5} g x^5 \log \left (c \left (d+e x^2\right )^p\right )-\frac{2 d^{3/2} f p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{3 e^{3/2}}-\frac{2 d^2 g p x}{5 e^2}+\frac{2 d^{5/2} g p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{5 e^{5/2}}+\frac{2 d f p x}{3 e}+\frac{2 d g p x^3}{15 e}-\frac{2}{9} f p x^3-\frac{2}{25} g p x^5 \]

[Out]

(2*d*f*p*x)/(3*e) - (2*d^2*g*p*x)/(5*e^2) - (2*f*p*x^3)/9 + (2*d*g*p*x^3)/(15*e) - (2*g*p*x^5)/25 - (2*d^(3/2)
*f*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(3*e^(3/2)) + (2*d^(5/2)*g*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(5*e^(5/2)) + (f*x
^3*Log[c*(d + e*x^2)^p])/3 + (g*x^5*Log[c*(d + e*x^2)^p])/5

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Rubi [A]  time = 0.129955, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2476, 2455, 302, 205} \[ \frac{1}{3} f x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac{1}{5} g x^5 \log \left (c \left (d+e x^2\right )^p\right )-\frac{2 d^{3/2} f p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{3 e^{3/2}}-\frac{2 d^2 g p x}{5 e^2}+\frac{2 d^{5/2} g p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{5 e^{5/2}}+\frac{2 d f p x}{3 e}+\frac{2 d g p x^3}{15 e}-\frac{2}{9} f p x^3-\frac{2}{25} g p x^5 \]

Antiderivative was successfully verified.

[In]

Int[x^2*(f + g*x^2)*Log[c*(d + e*x^2)^p],x]

[Out]

(2*d*f*p*x)/(3*e) - (2*d^2*g*p*x)/(5*e^2) - (2*f*p*x^3)/9 + (2*d*g*p*x^3)/(15*e) - (2*g*p*x^5)/25 - (2*d^(3/2)
*f*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(3*e^(3/2)) + (2*d^(5/2)*g*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(5*e^(5/2)) + (f*x
^3*Log[c*(d + e*x^2)^p])/3 + (g*x^5*Log[c*(d + e*x^2)^p])/5

Rule 2476

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x^n)^p])^q, x^m*(f + g*x^s)^r, x], x] /; FreeQ[{a, b, c,
d, e, f, g, m, n, p, q, r, s}, x] && IGtQ[q, 0] && IntegerQ[m] && IntegerQ[r] && IntegerQ[s]

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^2 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx &=\int \left (f x^2 \log \left (c \left (d+e x^2\right )^p\right )+g x^4 \log \left (c \left (d+e x^2\right )^p\right )\right ) \, dx\\ &=f \int x^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx+g \int x^4 \log \left (c \left (d+e x^2\right )^p\right ) \, dx\\ &=\frac{1}{3} f x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac{1}{5} g x^5 \log \left (c \left (d+e x^2\right )^p\right )-\frac{1}{3} (2 e f p) \int \frac{x^4}{d+e x^2} \, dx-\frac{1}{5} (2 e g p) \int \frac{x^6}{d+e x^2} \, dx\\ &=\frac{1}{3} f x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac{1}{5} g x^5 \log \left (c \left (d+e x^2\right )^p\right )-\frac{1}{3} (2 e f p) \int \left (-\frac{d}{e^2}+\frac{x^2}{e}+\frac{d^2}{e^2 \left (d+e x^2\right )}\right ) \, dx-\frac{1}{5} (2 e g p) \int \left (\frac{d^2}{e^3}-\frac{d x^2}{e^2}+\frac{x^4}{e}-\frac{d^3}{e^3 \left (d+e x^2\right )}\right ) \, dx\\ &=\frac{2 d f p x}{3 e}-\frac{2 d^2 g p x}{5 e^2}-\frac{2}{9} f p x^3+\frac{2 d g p x^3}{15 e}-\frac{2}{25} g p x^5+\frac{1}{3} f x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac{1}{5} g x^5 \log \left (c \left (d+e x^2\right )^p\right )-\frac{\left (2 d^2 f p\right ) \int \frac{1}{d+e x^2} \, dx}{3 e}+\frac{\left (2 d^3 g p\right ) \int \frac{1}{d+e x^2} \, dx}{5 e^2}\\ &=\frac{2 d f p x}{3 e}-\frac{2 d^2 g p x}{5 e^2}-\frac{2}{9} f p x^3+\frac{2 d g p x^3}{15 e}-\frac{2}{25} g p x^5-\frac{2 d^{3/2} f p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{3 e^{3/2}}+\frac{2 d^{5/2} g p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{5 e^{5/2}}+\frac{1}{3} f x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac{1}{5} g x^5 \log \left (c \left (d+e x^2\right )^p\right )\\ \end{align*}

Mathematica [A]  time = 0.0562203, size = 118, normalized size = 0.77 \[ \frac{\sqrt{e} x \left (15 e^2 x^2 \left (5 f+3 g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )-2 p \left (45 d^2 g-15 d e \left (5 f+g x^2\right )+e^2 x^2 \left (25 f+9 g x^2\right )\right )\right )+30 d^{3/2} p (3 d g-5 e f) \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{225 e^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(f + g*x^2)*Log[c*(d + e*x^2)^p],x]

[Out]

(30*d^(3/2)*(-5*e*f + 3*d*g)*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]] + Sqrt[e]*x*(-2*p*(45*d^2*g - 15*d*e*(5*f + g*x^2)
+ e^2*x^2*(25*f + 9*g*x^2)) + 15*e^2*x^2*(5*f + 3*g*x^2)*Log[c*(d + e*x^2)^p]))/(225*e^(5/2))

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Maple [C]  time = 0.586, size = 453, normalized size = 2.9 \begin{align*} \left ({\frac{g{x}^{5}}{5}}+{\frac{f{x}^{3}}{3}} \right ) \ln \left ( \left ( e{x}^{2}+d \right ) ^{p} \right ) -{\frac{i}{10}}\pi \,g{x}^{5} \left ({\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{3}+{\frac{i}{10}}\pi \,g{x}^{5} \left ({\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) -{\frac{i}{10}}\pi \,g{x}^{5}{\it csgn} \left ( i \left ( e{x}^{2}+d \right ) ^{p} \right ){\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ){\it csgn} \left ( ic \right ) +{\frac{i}{10}}\pi \,g{x}^{5}{\it csgn} \left ( i \left ( e{x}^{2}+d \right ) ^{p} \right ) \left ({\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{2}-{\frac{i}{6}}\pi \,f{x}^{3} \left ({\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{3}+{\frac{i}{6}}\pi \,f{x}^{3} \left ({\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) -{\frac{i}{6}}\pi \,f{x}^{3}{\it csgn} \left ( i \left ( e{x}^{2}+d \right ) ^{p} \right ){\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ){\it csgn} \left ( ic \right ) +{\frac{i}{6}}\pi \,f{x}^{3}{\it csgn} \left ( i \left ( e{x}^{2}+d \right ) ^{p} \right ) \left ({\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{2}+{\frac{\ln \left ( c \right ) g{x}^{5}}{5}}-{\frac{2\,gp{x}^{5}}{25}}+{\frac{\ln \left ( c \right ) f{x}^{3}}{3}}+{\frac{2\,dgp{x}^{3}}{15\,e}}-{\frac{2\,fp{x}^{3}}{9}}-{\frac{{d}^{2}gp}{5\,{e}^{3}}\sqrt{-de}\ln \left ( \sqrt{-de}x+d \right ) }+{\frac{dpf}{3\,{e}^{2}}\sqrt{-de}\ln \left ( \sqrt{-de}x+d \right ) }+{\frac{{d}^{2}gp}{5\,{e}^{3}}\sqrt{-de}\ln \left ( -\sqrt{-de}x+d \right ) }-{\frac{dpf}{3\,{e}^{2}}\sqrt{-de}\ln \left ( -\sqrt{-de}x+d \right ) }-{\frac{2\,{d}^{2}gpx}{5\,{e}^{2}}}+{\frac{2\,dfpx}{3\,e}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(g*x^2+f)*ln(c*(e*x^2+d)^p),x)

[Out]

(1/5*g*x^5+1/3*f*x^3)*ln((e*x^2+d)^p)-1/10*I*Pi*g*x^5*csgn(I*c*(e*x^2+d)^p)^3+1/10*I*Pi*g*x^5*csgn(I*c*(e*x^2+
d)^p)^2*csgn(I*c)-1/10*I*Pi*g*x^5*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)+1/10*I*Pi*g*x^5*csgn(I*(
e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2-1/6*I*Pi*f*x^3*csgn(I*c*(e*x^2+d)^p)^3+1/6*I*Pi*f*x^3*csgn(I*c*(e*x^2+d)^p
)^2*csgn(I*c)-1/6*I*Pi*f*x^3*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)+1/6*I*Pi*f*x^3*csgn(I*(e*x^2+
d)^p)*csgn(I*c*(e*x^2+d)^p)^2+1/5*ln(c)*g*x^5-2/25*g*p*x^5+1/3*ln(c)*f*x^3+2/15*d*g*p*x^3/e-2/9*f*p*x^3-1/5/e^
3*(-d*e)^(1/2)*p*d^2*ln((-d*e)^(1/2)*x+d)*g+1/3/e^2*(-d*e)^(1/2)*p*d*ln((-d*e)^(1/2)*x+d)*f+1/5/e^3*(-d*e)^(1/
2)*p*d^2*ln(-(-d*e)^(1/2)*x+d)*g-1/3/e^2*(-d*e)^(1/2)*p*d*ln(-(-d*e)^(1/2)*x+d)*f-2/5*d^2*g*p*x/e^2+2/3*d*f*p*
x/e

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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^2*(g*x^2+f)*log(c*(e*x^2+d)^p),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.37542, size = 693, normalized size = 4.5 \begin{align*} \left [-\frac{18 \, e^{2} g p x^{5} + 10 \,{\left (5 \, e^{2} f - 3 \, d e g\right )} p x^{3} + 15 \,{\left (5 \, d e f - 3 \, d^{2} g\right )} p \sqrt{-\frac{d}{e}} \log \left (\frac{e x^{2} + 2 \, e x \sqrt{-\frac{d}{e}} - d}{e x^{2} + d}\right ) - 30 \,{\left (5 \, d e f - 3 \, d^{2} g\right )} p x - 15 \,{\left (3 \, e^{2} g p x^{5} + 5 \, e^{2} f p x^{3}\right )} \log \left (e x^{2} + d\right ) - 15 \,{\left (3 \, e^{2} g x^{5} + 5 \, e^{2} f x^{3}\right )} \log \left (c\right )}{225 \, e^{2}}, -\frac{18 \, e^{2} g p x^{5} + 10 \,{\left (5 \, e^{2} f - 3 \, d e g\right )} p x^{3} + 30 \,{\left (5 \, d e f - 3 \, d^{2} g\right )} p \sqrt{\frac{d}{e}} \arctan \left (\frac{e x \sqrt{\frac{d}{e}}}{d}\right ) - 30 \,{\left (5 \, d e f - 3 \, d^{2} g\right )} p x - 15 \,{\left (3 \, e^{2} g p x^{5} + 5 \, e^{2} f p x^{3}\right )} \log \left (e x^{2} + d\right ) - 15 \,{\left (3 \, e^{2} g x^{5} + 5 \, e^{2} f x^{3}\right )} \log \left (c\right )}{225 \, e^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(g*x^2+f)*log(c*(e*x^2+d)^p),x, algorithm="fricas")

[Out]

[-1/225*(18*e^2*g*p*x^5 + 10*(5*e^2*f - 3*d*e*g)*p*x^3 + 15*(5*d*e*f - 3*d^2*g)*p*sqrt(-d/e)*log((e*x^2 + 2*e*
x*sqrt(-d/e) - d)/(e*x^2 + d)) - 30*(5*d*e*f - 3*d^2*g)*p*x - 15*(3*e^2*g*p*x^5 + 5*e^2*f*p*x^3)*log(e*x^2 + d
) - 15*(3*e^2*g*x^5 + 5*e^2*f*x^3)*log(c))/e^2, -1/225*(18*e^2*g*p*x^5 + 10*(5*e^2*f - 3*d*e*g)*p*x^3 + 30*(5*
d*e*f - 3*d^2*g)*p*sqrt(d/e)*arctan(e*x*sqrt(d/e)/d) - 30*(5*d*e*f - 3*d^2*g)*p*x - 15*(3*e^2*g*p*x^5 + 5*e^2*
f*p*x^3)*log(e*x^2 + d) - 15*(3*e^2*g*x^5 + 5*e^2*f*x^3)*log(c))/e^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**2*(g*x**2+f)*ln(c*(e*x**2+d)**p),x)

[Out]

Timed out

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Giac [A]  time = 1.48279, size = 186, normalized size = 1.21 \begin{align*} \frac{2 \,{\left (3 \, d^{3} g p - 5 \, d^{2} f p e\right )} \arctan \left (\frac{x e^{\frac{1}{2}}}{\sqrt{d}}\right ) e^{\left (-\frac{5}{2}\right )}}{15 \, \sqrt{d}} + \frac{1}{225} \,{\left (45 \, g p x^{5} e^{2} \log \left (x^{2} e + d\right ) - 18 \, g p x^{5} e^{2} + 45 \, g x^{5} e^{2} \log \left (c\right ) + 30 \, d g p x^{3} e + 75 \, f p x^{3} e^{2} \log \left (x^{2} e + d\right ) - 50 \, f p x^{3} e^{2} + 75 \, f x^{3} e^{2} \log \left (c\right ) - 90 \, d^{2} g p x + 150 \, d f p x e\right )} e^{\left (-2\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(g*x^2+f)*log(c*(e*x^2+d)^p),x, algorithm="giac")

[Out]

2/15*(3*d^3*g*p - 5*d^2*f*p*e)*arctan(x*e^(1/2)/sqrt(d))*e^(-5/2)/sqrt(d) + 1/225*(45*g*p*x^5*e^2*log(x^2*e +
d) - 18*g*p*x^5*e^2 + 45*g*x^5*e^2*log(c) + 30*d*g*p*x^3*e + 75*f*p*x^3*e^2*log(x^2*e + d) - 50*f*p*x^3*e^2 +
75*f*x^3*e^2*log(c) - 90*d^2*g*p*x + 150*d*f*p*x*e)*e^(-2)