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

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

[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)

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Rubi [A]
time = 0.09, antiderivative size = 154, normalized size of antiderivative = 1.00, 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 = {2526, 2505, 308, 211} \begin {gather*} -\frac {2 d^{3/2} f p \text {ArcTan}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 e^{3/2}}+\frac {2 d^{5/2} g p \text {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 \end {gather*}

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 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*} \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*}

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Mathematica [A]
time = 0.04, size = 118, normalized size = 0.77 \begin {gather*} \frac {30 d^{3/2} (-5 e f+3 d g) p \tan ^{-1}\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}} \end {gather*}

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] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.34, size = 453, normalized size = 2.94

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

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,method=_RETURNVERBOSE)

[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*(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)*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+1/10*I*Pi*g*x^5*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2+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*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*c*(e*x^2+d)^p)^3+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*ln(-(-e*d)^(1/2)*x+d)*(-e*d)^(1/2)*d^2*g*p-1/3/e^2*(-e*d)^(1/2)*p*d*ln(-(-e*d)^(1/2)*x+d)*f-1/5/e^3*ln((-e*d
)^(1/2)*x+d)*(-e*d)^(1/2)*d^2*g*p+1/3/e^2*(-e*d)^(1/2)*p*d*ln((-e*d)^(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 [A]
time = 0.52, size = 111, normalized size = 0.72 \begin {gather*} \frac {2}{225} \, {\left (\frac {15 \, {\left (3 \, d^{3} g - 5 \, d^{2} f e\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {7}{2}\right )}}{\sqrt {d}} - {\left (9 \, g x^{5} e^{2} - 5 \, {\left (3 \, d g e - 5 \, f e^{2}\right )} x^{3} + 15 \, {\left (3 \, d^{2} g - 5 \, d f e\right )} x\right )} e^{\left (-3\right )}\right )} p e + \frac {1}{15} \, {\left (3 \, g x^{5} + 5 \, f x^{3}\right )} \log \left ({\left (x^{2} e + d\right )}^{p} c\right ) \end {gather*}

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]

2/225*(15*(3*d^3*g - 5*d^2*f*e)*arctan(x*e^(1/2)/sqrt(d))*e^(-7/2)/sqrt(d) - (9*g*x^5*e^2 - 5*(3*d*g*e - 5*f*e
^2)*x^3 + 15*(3*d^2*g - 5*d*f*e)*x)*e^(-3))*p*e + 1/15*(3*g*x^5 + 5*f*x^3)*log((x^2*e + d)^p*c)

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Fricas [A]
time = 0.39, size = 283, normalized size = 1.84 \begin {gather*} \left [-\frac {1}{225} \, {\left (90 \, d^{2} g p x - 15 \, {\left (3 \, g p x^{5} + 5 \, f p x^{3}\right )} e^{2} \log \left (x^{2} e + d\right ) - 15 \, {\left (3 \, g x^{5} + 5 \, f x^{3}\right )} e^{2} \log \left (c\right ) + 15 \, {\left (3 \, d^{2} g p - 5 \, d f p e\right )} \sqrt {-d e^{\left (-1\right )}} \log \left (\frac {x^{2} e - 2 \, \sqrt {-d e^{\left (-1\right )}} x e - d}{x^{2} e + d}\right ) + 2 \, {\left (9 \, g p x^{5} + 25 \, f p x^{3}\right )} e^{2} - 30 \, {\left (d g p x^{3} + 5 \, d f p x\right )} e\right )} e^{\left (-2\right )}, -\frac {1}{225} \, {\left (90 \, d^{2} g p x - 30 \, {\left (3 \, d^{2} g p - 5 \, d f p e\right )} \sqrt {d} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {1}{2}\right )} - 15 \, {\left (3 \, g p x^{5} + 5 \, f p x^{3}\right )} e^{2} \log \left (x^{2} e + d\right ) - 15 \, {\left (3 \, g x^{5} + 5 \, f x^{3}\right )} e^{2} \log \left (c\right ) + 2 \, {\left (9 \, g p x^{5} + 25 \, f p x^{3}\right )} e^{2} - 30 \, {\left (d g p x^{3} + 5 \, d f p x\right )} e\right )} e^{\left (-2\right )}\right ] \end {gather*}

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*(90*d^2*g*p*x - 15*(3*g*p*x^5 + 5*f*p*x^3)*e^2*log(x^2*e + d) - 15*(3*g*x^5 + 5*f*x^3)*e^2*log(c) + 15
*(3*d^2*g*p - 5*d*f*p*e)*sqrt(-d*e^(-1))*log((x^2*e - 2*sqrt(-d*e^(-1))*x*e - d)/(x^2*e + d)) + 2*(9*g*p*x^5 +
25*f*p*x^3)*e^2 - 30*(d*g*p*x^3 + 5*d*f*p*x)*e)*e^(-2), -1/225*(90*d^2*g*p*x - 30*(3*d^2*g*p - 5*d*f*p*e)*sqr
t(d)*arctan(x*e^(1/2)/sqrt(d))*e^(-1/2) - 15*(3*g*p*x^5 + 5*f*p*x^3)*e^2*log(x^2*e + d) - 15*(3*g*x^5 + 5*f*x^
3)*e^2*log(c) + 2*(9*g*p*x^5 + 25*f*p*x^3)*e^2 - 30*(d*g*p*x^3 + 5*d*f*p*x)*e)*e^(-2)]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 320 vs. \(2 (158) = 316\).
time = 34.48, size = 320, normalized size = 2.08 \begin {gather*} \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} \end {gather*}

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]

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))

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Giac [A]
time = 5.43, size = 138, normalized size = 0.90 \begin {gather*} \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 {gather*}

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)

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Mupad [B]
time = 0.32, size = 126, normalized size = 0.82 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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