\(\int x^2 (f+g x^2) \log (c (d+e x^2)^p) \, dx\) [318]

Optimal result

Integrand size = 23, antiderivative size = 154 \[ \int x^2 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {2 d f p x}{3 e}-\frac {2 d^2 g p x}{5 e^2}-\frac {2}{9} f p x^3+\frac {2 d g p x^3}{15 e}-\frac {2}{25} g p x^5-\frac {2 d^{3/2} f p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 e^{3/2}}+\frac {2 d^{5/2} g p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 e^{5/2}}+\frac {1}{3} f x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{5} g x^5 \log \left (c \left (d+e x^2\right )^p\right ) \]

[Out]

2/3*d*f*p*x/e-2/5*d^2*g*p*x/e^2-2/9*f*p*x^3+2/15*d*g*p*x^3/e-2/25*g*p*x^5-2/3*d^(3/2)*f*p*arctan(x*e^(1/2)/d^(
1/2))/e^(3/2)+2/5*d^(5/2)*g*p*arctan(x*e^(1/2)/d^(1/2))/e^(5/2)+1/3*f*x^3*ln(c*(e*x^2+d)^p)+1/5*g*x^5*ln(c*(e*
x^2+d)^p)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2526, 2505, 308, 211} \[ \int x^2 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-\frac {2 d^{3/2} f p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 e^{3/2}}+\frac {2 d^{5/2} g p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 e^{5/2}}+\frac {1}{3} f x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{5} g x^5 \log \left (c \left (d+e x^2\right )^p\right )-\frac {2 d^2 g p x}{5 e^2}+\frac {2 d f p x}{3 e}+\frac {2 d g p x^3}{15 e}-\frac {2}{9} f p x^3-\frac {2}{25} g p x^5 \]

[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 211

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

Rule 308

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 2505

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 2526

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]

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 0.04 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.77 \[ \int x^2 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {30 d^{3/2} (-5 e f+3 d g) p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )+\sqrt {e} x \left (-2 p \left (45 d^2 g-15 d e \left (5 f+g x^2\right )+e^2 x^2 \left (25 f+9 g x^2\right )\right )+15 e^2 x^2 \left (5 f+3 g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )\right )}{225 e^{5/2}} \]

[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))

Maple [A] (verified)

Time = 1.20 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.77

method	result	size
parts	\(\frac {g \,x^{5} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{5}+\frac {f \,x^{3} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{3}-\frac {2 p e \left (\frac {\frac {3}{5} e^{2} g \,x^{5}-d g \,x^{3} e +\frac {5}{3} f \,x^{3} e^{2}+3 d^{2} g x -5 d e f x}{e^{3}}-\frac {d^{2} \left (3 d g -5 e f \right ) \arctan \left (\frac {x e}{\sqrt {d e}}\right )}{e^{3} \sqrt {d e}}\right )}{15}\)	\(118\)
risch	\(\left (\frac {1}{5} g \,x^{5}+\frac {1}{3} f \,x^{3}\right ) \ln \left (\left (e \,x^{2}+d \right )^{p}\right )-\frac {i \pi g \,x^{5} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{10}+\frac {i \pi g \,x^{5} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2}}{10}+\frac {i \pi f \,x^{3} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2}}{6}-\frac {i \pi f \,x^{3} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{6}+\frac {i \pi f \,x^{3} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{6}-\frac {i \pi g \,x^{5} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}}{10}+\frac {i \pi g \,x^{5} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{10}-\frac {i \pi f \,x^{3} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}}{6}+\frac {\ln \left (c \right ) g \,x^{5}}{5}-\frac {2 g p \,x^{5}}{25}+\frac {\ln \left (c \right ) f \,x^{3}}{3}+\frac {2 d g p \,x^{3}}{15 e}-\frac {2 f p \,x^{3}}{9}+\frac {\sqrt {-d e}\, p \,d^{2} \ln \left (-\sqrt {-d e}\, x +d \right ) g}{5 e^{3}}-\frac {\sqrt {-d e}\, p d \ln \left (-\sqrt {-d e}\, x +d \right ) f}{3 e^{2}}-\frac {\sqrt {-d e}\, p \,d^{2} \ln \left (\sqrt {-d e}\, x +d \right ) g}{5 e^{3}}+\frac {\sqrt {-d e}\, p d \ln \left (\sqrt {-d e}\, x +d \right ) f}{3 e^{2}}-\frac {2 d^{2} g p x}{5 e^{2}}+\frac {2 d f p x}{3 e}\)	\(453\)

[In]

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

[Out]

1/5*g*x^5*ln(c*(e*x^2+d)^p)+1/3*f*x^3*ln(c*(e*x^2+d)^p)-2/15*p*e*(1/e^3*(3/5*e^2*g*x^5-d*g*x^3*e+5/3*f*x^3*e^2
+3*d^2*g*x-5*d*e*f*x)-d^2*(3*d*g-5*e*f)/e^3/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.95 \[ \int x^2 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\left [-\frac {18 \, e^{2} g p x^{5} + 10 \, {\left (5 \, e^{2} f - 3 \, d e g\right )} p x^{3} + 15 \, {\left (5 \, d e f - 3 \, d^{2} g\right )} p \sqrt {-\frac {d}{e}} \log \left (\frac {e x^{2} + 2 \, e x \sqrt {-\frac {d}{e}} - d}{e x^{2} + d}\right ) - 30 \, {\left (5 \, d e f - 3 \, d^{2} g\right )} p x - 15 \, {\left (3 \, e^{2} g p x^{5} + 5 \, e^{2} f p x^{3}\right )} \log \left (e x^{2} + d\right ) - 15 \, {\left (3 \, e^{2} g x^{5} + 5 \, e^{2} f x^{3}\right )} \log \left (c\right )}{225 \, e^{2}}, -\frac {18 \, e^{2} g p x^{5} + 10 \, {\left (5 \, e^{2} f - 3 \, d e g\right )} p x^{3} + 30 \, {\left (5 \, d e f - 3 \, d^{2} g\right )} p \sqrt {\frac {d}{e}} \arctan \left (\frac {e x \sqrt {\frac {d}{e}}}{d}\right ) - 30 \, {\left (5 \, d e f - 3 \, d^{2} g\right )} p x - 15 \, {\left (3 \, e^{2} g p x^{5} + 5 \, e^{2} f p x^{3}\right )} \log \left (e x^{2} + d\right ) - 15 \, {\left (3 \, e^{2} g x^{5} + 5 \, e^{2} f x^{3}\right )} \log \left (c\right )}{225 \, e^{2}}\right ] \]

[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]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 320 vs. \(2 (158) = 316\).

Time = 31.34 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.08 \[ \int x^2 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\begin {cases} \left (\frac {f x^{3}}{3} + \frac {g x^{5}}{5}\right ) \log {\left (0^{p} c \right )} & \text {for}\: d = 0 \wedge e = 0 \\\left (\frac {f x^{3}}{3} + \frac {g x^{5}}{5}\right ) \log {\left (c d^{p} \right )} & \text {for}\: e = 0 \\- \frac {2 f p x^{3}}{9} + \frac {f x^{3} \log {\left (c \left (e x^{2}\right )^{p} \right )}}{3} - \frac {2 g p x^{5}}{25} + \frac {g x^{5} \log {\left (c \left (e x^{2}\right )^{p} \right )}}{5} & \text {for}\: d = 0 \\\frac {2 d^{3} g p \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{5 e^{3} \sqrt {- \frac {d}{e}}} - \frac {d^{3} g \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{5 e^{3} \sqrt {- \frac {d}{e}}} - \frac {2 d^{2} f p \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{3 e^{2} \sqrt {- \frac {d}{e}}} + \frac {d^{2} f \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{3 e^{2} \sqrt {- \frac {d}{e}}} - \frac {2 d^{2} g p x}{5 e^{2}} + \frac {2 d f p x}{3 e} + \frac {2 d g p x^{3}}{15 e} - \frac {2 f p x^{3}}{9} + \frac {f x^{3} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{3} - \frac {2 g p x^{5}}{25} + \frac {g x^{5} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{5} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise(((f*x**3/3 + g*x**5/5)*log(0**p*c), Eq(d, 0) & Eq(e, 0)), ((f*x**3/3 + g*x**5/5)*log(c*d**p), Eq(e,
0)), (-2*f*p*x**3/9 + f*x**3*log(c*(e*x**2)**p)/3 - 2*g*p*x**5/25 + g*x**5*log(c*(e*x**2)**p)/5, Eq(d, 0)), (2
*d**3*g*p*log(x - sqrt(-d/e))/(5*e**3*sqrt(-d/e)) - d**3*g*log(c*(d + e*x**2)**p)/(5*e**3*sqrt(-d/e)) - 2*d**2
*f*p*log(x - sqrt(-d/e))/(3*e**2*sqrt(-d/e)) + d**2*f*log(c*(d + e*x**2)**p)/(3*e**2*sqrt(-d/e)) - 2*d**2*g*p*
x/(5*e**2) + 2*d*f*p*x/(3*e) + 2*d*g*p*x**3/(15*e) - 2*f*p*x**3/9 + f*x**3*log(c*(d + e*x**2)**p)/3 - 2*g*p*x*
*5/25 + g*x**5*log(c*(d + e*x**2)**p)/5, True))

Maxima [F(-2)]

Exception generated. \[ \int x^2 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.79 \[ \int x^2 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-\frac {1}{25} \, {\left (2 \, g p - 5 \, g \log \left (c\right )\right )} x^{5} - \frac {{\left (10 \, e f p - 6 \, d g p - 15 \, e f \log \left (c\right )\right )} x^{3}}{45 \, e} + \frac {1}{15} \, {\left (3 \, g p x^{5} + 5 \, f p x^{3}\right )} \log \left (e x^{2} + d\right ) + \frac {2 \, {\left (5 \, d e f p - 3 \, d^{2} g p\right )} x}{15 \, e^{2}} - \frac {2 \, {\left (5 \, d^{2} e f p - 3 \, d^{3} g p\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{15 \, \sqrt {d e} e^{2}} \]

[In]

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

[Out]

-1/25*(2*g*p - 5*g*log(c))*x^5 - 1/45*(10*e*f*p - 6*d*g*p - 15*e*f*log(c))*x^3/e + 1/15*(3*g*p*x^5 + 5*f*p*x^3
)*log(e*x^2 + d) + 2/15*(5*d*e*f*p - 3*d^2*g*p)*x/e^2 - 2/15*(5*d^2*e*f*p - 3*d^3*g*p)*arctan(e*x/sqrt(d*e))/(
sqrt(d*e)*e^2)

Mupad [B] (verification not implemented)

Time = 1.65 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.82 \[ \int x^2 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )\,\left (\frac {g\,x^5}{5}+\frac {f\,x^3}{3}\right )-x^3\,\left (\frac {2\,f\,p}{9}-\frac {2\,d\,g\,p}{15\,e}\right )-\frac {2\,g\,p\,x^5}{25}+\frac {d\,x\,\left (\frac {2\,f\,p}{3}-\frac {2\,d\,g\,p}{5\,e}\right )}{e}+\frac {2\,d^{3/2}\,p\,\mathrm {atan}\left (\frac {d^{3/2}\,\sqrt {e}\,p\,x\,\left (3\,d\,g-5\,e\,f\right )}{3\,d^3\,g\,p-5\,d^2\,e\,f\,p}\right )\,\left (3\,d\,g-5\,e\,f\right )}{15\,e^{5/2}} \]

[In]

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

[Out]

log(c*(d + e*x^2)^p)*((f*x^3)/3 + (g*x^5)/5) - x^3*((2*f*p)/9 - (2*d*g*p)/(15*e)) - (2*g*p*x^5)/25 + (d*x*((2*
f*p)/3 - (2*d*g*p)/(5*e)))/e + (2*d^(3/2)*p*atan((d^(3/2)*e^(1/2)*p*x*(3*d*g - 5*e*f))/(3*d^3*g*p - 5*d^2*e*f*
p))*(3*d*g - 5*e*f))/(15*e^(5/2))