\(\int (f+g x^2)^3 \log (c (d+e x^2)^p) \, dx\) [268]

Optimal result

Integrand size = 22, antiderivative size = 338 \[ \int \left (f+g x^2\right )^3 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-2 f^3 p x+\frac {2 d f^2 g p x}{e}-\frac {6 d^2 f g^2 p x}{5 e^2}+\frac {2 d^3 g^3 p x}{7 e^3}-\frac {2}{3} f^2 g p x^3+\frac {2 d f g^2 p x^3}{5 e}-\frac {2 d^2 g^3 p x^3}{21 e^2}-\frac {6}{25} f g^2 p x^5+\frac {2 d g^3 p x^5}{35 e}-\frac {2}{49} g^3 p x^7+\frac {2 \sqrt {d} f^3 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}-\frac {2 d^{3/2} f^2 g p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{e^{3/2}}+\frac {6 d^{5/2} f g^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 e^{5/2}}-\frac {2 d^{7/2} g^3 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{7 e^{7/2}}+f^3 x \log \left (c \left (d+e x^2\right )^p\right )+f^2 g x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {3}{5} f g^2 x^5 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{7} g^3 x^7 \log \left (c \left (d+e x^2\right )^p\right ) \]

[Out]

-2*f^3*p*x+2*d*f^2*g*p*x/e-6/5*d^2*f*g^2*p*x/e^2+2/7*d^3*g^3*p*x/e^3-2/3*f^2*g*p*x^3+2/5*d*f*g^2*p*x^3/e-2/21*
d^2*g^3*p*x^3/e^2-6/25*f*g^2*p*x^5+2/35*d*g^3*p*x^5/e-2/49*g^3*p*x^7-2*d^(3/2)*f^2*g*p*arctan(x*e^(1/2)/d^(1/2
))/e^(3/2)+6/5*d^(5/2)*f*g^2*p*arctan(x*e^(1/2)/d^(1/2))/e^(5/2)-2/7*d^(7/2)*g^3*p*arctan(x*e^(1/2)/d^(1/2))/e
^(7/2)+f^3*x*ln(c*(e*x^2+d)^p)+f^2*g*x^3*ln(c*(e*x^2+d)^p)+3/5*f*g^2*x^5*ln(c*(e*x^2+d)^p)+1/7*g^3*x^7*ln(c*(e
*x^2+d)^p)+2*f^3*p*arctan(x*e^(1/2)/d^(1/2))*d^(1/2)/e^(1/2)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2521, 2498, 327, 211, 2505, 308} \[ \int \left (f+g x^2\right )^3 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-\frac {2 d^{3/2} f^2 g p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{e^{3/2}}+\frac {6 d^{5/2} f g^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 e^{5/2}}-\frac {2 d^{7/2} g^3 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{7 e^{7/2}}+\frac {2 \sqrt {d} f^3 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}+f^3 x \log \left (c \left (d+e x^2\right )^p\right )+f^2 g x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {3}{5} f g^2 x^5 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{7} g^3 x^7 \log \left (c \left (d+e x^2\right )^p\right )+\frac {2 d^3 g^3 p x}{7 e^3}-\frac {6 d^2 f g^2 p x}{5 e^2}-\frac {2 d^2 g^3 p x^3}{21 e^2}+\frac {2 d f^2 g p x}{e}+\frac {2 d f g^2 p x^3}{5 e}+\frac {2 d g^3 p x^5}{35 e}-2 f^3 p x-\frac {2}{3} f^2 g p x^3-\frac {6}{25} f g^2 p x^5-\frac {2}{49} g^3 p x^7 \]

[In]

Int[(f + g*x^2)^3*Log[c*(d + e*x^2)^p],x]

[Out]

-2*f^3*p*x + (2*d*f^2*g*p*x)/e - (6*d^2*f*g^2*p*x)/(5*e^2) + (2*d^3*g^3*p*x)/(7*e^3) - (2*f^2*g*p*x^3)/3 + (2*
d*f*g^2*p*x^3)/(5*e) - (2*d^2*g^3*p*x^3)/(21*e^2) - (6*f*g^2*p*x^5)/25 + (2*d*g^3*p*x^5)/(35*e) - (2*g^3*p*x^7
)/49 + (2*Sqrt[d]*f^3*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/Sqrt[e] - (2*d^(3/2)*f^2*g*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])
/e^(3/2) + (6*d^(5/2)*f*g^2*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(5*e^(5/2)) - (2*d^(7/2)*g^3*p*ArcTan[(Sqrt[e]*x)/S
qrt[d]])/(7*e^(7/2)) + f^3*x*Log[c*(d + e*x^2)^p] + f^2*g*x^3*Log[c*(d + e*x^2)^p] + (3*f*g^2*x^5*Log[c*(d + e
*x^2)^p])/5 + (g^3*x^7*Log[c*(d + e*x^2)^p])/7

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 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2498

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)], x_Symbol] :> Simp[x*Log[c*(d + e*x^n)^p], x] - Dist[e*n*p, Int[
x^n/(d + e*x^n), x], x] /; FreeQ[{c, d, e, n, p}, x]

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 2521

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*((f_) + (g_.)*(x_)^(s_))^(r_.), x_Symbol]
:> With[{t = ExpandIntegrand[(a + b*Log[c*(d + e*x^n)^p])^q, (f + g*x^s)^r, x]}, Int[t, x] /; SumQ[t]] /; Free
Q[{a, b, c, d, e, f, g, n, p, q, r, s}, x] && IntegerQ[n] && IGtQ[q, 0] && IntegerQ[r] && IntegerQ[s] && (EqQ[
q, 1] || (GtQ[r, 0] && GtQ[s, 1]) || (LtQ[s, 0] && LtQ[r, 0]))

Rubi steps \begin{align*} \text {integral}& = \int \left (f^3 \log \left (c \left (d+e x^2\right )^p\right )+3 f^2 g x^2 \log \left (c \left (d+e x^2\right )^p\right )+3 f g^2 x^4 \log \left (c \left (d+e x^2\right )^p\right )+g^3 x^6 \log \left (c \left (d+e x^2\right )^p\right )\right ) \, dx \\ & = f^3 \int \log \left (c \left (d+e x^2\right )^p\right ) \, dx+\left (3 f^2 g\right ) \int x^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx+\left (3 f g^2\right ) \int x^4 \log \left (c \left (d+e x^2\right )^p\right ) \, dx+g^3 \int x^6 \log \left (c \left (d+e x^2\right )^p\right ) \, dx \\ & = f^3 x \log \left (c \left (d+e x^2\right )^p\right )+f^2 g x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {3}{5} f g^2 x^5 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{7} g^3 x^7 \log \left (c \left (d+e x^2\right )^p\right )-\left (2 e f^3 p\right ) \int \frac {x^2}{d+e x^2} \, dx-\left (2 e f^2 g p\right ) \int \frac {x^4}{d+e x^2} \, dx-\frac {1}{5} \left (6 e f g^2 p\right ) \int \frac {x^6}{d+e x^2} \, dx-\frac {1}{7} \left (2 e g^3 p\right ) \int \frac {x^8}{d+e x^2} \, dx \\ & = -2 f^3 p x+f^3 x \log \left (c \left (d+e x^2\right )^p\right )+f^2 g x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {3}{5} f g^2 x^5 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{7} g^3 x^7 \log \left (c \left (d+e x^2\right )^p\right )+\left (2 d f^3 p\right ) \int \frac {1}{d+e x^2} \, dx-\left (2 e f^2 g p\right ) \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} \left (6 e f g^2 p\right ) \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 {1}{7} \left (2 e g^3 p\right ) \int \left (-\frac {d^3}{e^4}+\frac {d^2 x^2}{e^3}-\frac {d x^4}{e^2}+\frac {x^6}{e}+\frac {d^4}{e^4 \left (d+e x^2\right )}\right ) \, dx \\ & = -2 f^3 p x+\frac {2 d f^2 g p x}{e}-\frac {6 d^2 f g^2 p x}{5 e^2}+\frac {2 d^3 g^3 p x}{7 e^3}-\frac {2}{3} f^2 g p x^3+\frac {2 d f g^2 p x^3}{5 e}-\frac {2 d^2 g^3 p x^3}{21 e^2}-\frac {6}{25} f g^2 p x^5+\frac {2 d g^3 p x^5}{35 e}-\frac {2}{49} g^3 p x^7+\frac {2 \sqrt {d} f^3 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}+f^3 x \log \left (c \left (d+e x^2\right )^p\right )+f^2 g x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {3}{5} f g^2 x^5 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{7} g^3 x^7 \log \left (c \left (d+e x^2\right )^p\right )-\frac {\left (2 d^2 f^2 g p\right ) \int \frac {1}{d+e x^2} \, dx}{e}+\frac {\left (6 d^3 f g^2 p\right ) \int \frac {1}{d+e x^2} \, dx}{5 e^2}-\frac {\left (2 d^4 g^3 p\right ) \int \frac {1}{d+e x^2} \, dx}{7 e^3} \\ & = -2 f^3 p x+\frac {2 d f^2 g p x}{e}-\frac {6 d^2 f g^2 p x}{5 e^2}+\frac {2 d^3 g^3 p x}{7 e^3}-\frac {2}{3} f^2 g p x^3+\frac {2 d f g^2 p x^3}{5 e}-\frac {2 d^2 g^3 p x^3}{21 e^2}-\frac {6}{25} f g^2 p x^5+\frac {2 d g^3 p x^5}{35 e}-\frac {2}{49} g^3 p x^7+\frac {2 \sqrt {d} f^3 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}-\frac {2 d^{3/2} f^2 g p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{e^{3/2}}+\frac {6 d^{5/2} f g^2 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 e^{5/2}}-\frac {2 d^{7/2} g^3 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{7 e^{7/2}}+f^3 x \log \left (c \left (d+e x^2\right )^p\right )+f^2 g x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {3}{5} f g^2 x^5 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{7} g^3 x^7 \log \left (c \left (d+e x^2\right )^p\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.64 \[ \int \left (f+g x^2\right )^3 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-\frac {2 p x \left (-525 d^3 g^3+35 d^2 e g^2 \left (63 f+5 g x^2\right )-105 d e^2 g \left (35 f^2+7 f g x^2+g^2 x^4\right )+e^3 \left (3675 f^3+1225 f^2 g x^2+441 f g^2 x^4+75 g^3 x^6\right )\right )}{3675 e^3}-\frac {2 \sqrt {d} \left (-35 e^3 f^3+35 d e^2 f^2 g-21 d^2 e f g^2+5 d^3 g^3\right ) p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{35 e^{7/2}}+\frac {1}{35} x \left (35 f^3+35 f^2 g x^2+21 f g^2 x^4+5 g^3 x^6\right ) \log \left (c \left (d+e x^2\right )^p\right ) \]

[In]

Integrate[(f + g*x^2)^3*Log[c*(d + e*x^2)^p],x]

[Out]

(-2*p*x*(-525*d^3*g^3 + 35*d^2*e*g^2*(63*f + 5*g*x^2) - 105*d*e^2*g*(35*f^2 + 7*f*g*x^2 + g^2*x^4) + e^3*(3675
*f^3 + 1225*f^2*g*x^2 + 441*f*g^2*x^4 + 75*g^3*x^6)))/(3675*e^3) - (2*Sqrt[d]*(-35*e^3*f^3 + 35*d*e^2*f^2*g -
21*d^2*e*f*g^2 + 5*d^3*g^3)*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(35*e^(7/2)) + (x*(35*f^3 + 35*f^2*g*x^2 + 21*f*g^2
*x^4 + 5*g^3*x^6)*Log[c*(d + e*x^2)^p])/35

Maple [A] (verified)

Time = 3.39 (sec) , antiderivative size = 258, normalized size of antiderivative = 0.76

method	result	size
parts	\(\frac {g^{3} x^{7} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{7}+\frac {3 f \,g^{2} x^{5} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{5}+f^{2} g \,x^{3} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )+f^{3} x \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )-\frac {2 p e \left (-\frac {-\frac {5}{7} e^{3} g^{3} x^{7}+d \,e^{2} g^{3} x^{5}-\frac {21}{5} e^{3} f \,g^{2} x^{5}-\frac {5}{3} d^{2} e \,g^{3} x^{3}+7 d \,e^{2} f \,g^{2} x^{3}-\frac {35}{3} e^{3} f^{2} g \,x^{3}+5 d^{3} x \,g^{3}-21 d^{2} e x f \,g^{2}+35 d \,e^{2} x \,f^{2} g -35 x \,e^{3} f^{3}}{e^{4}}+\frac {d \left (5 d^{3} g^{3}-21 d^{2} e f \,g^{2}+35 d \,e^{2} f^{2} g -35 e^{3} f^{3}\right ) \arctan \left (\frac {x e}{\sqrt {d e}}\right )}{e^{4} \sqrt {d e}}\right )}{35}\)	\(258\)
risch	\(\text {Expression too large to display}\)	\(995\)

[In]

int((g*x^2+f)^3*ln(c*(e*x^2+d)^p),x,method=_RETURNVERBOSE)

[Out]

1/7*g^3*x^7*ln(c*(e*x^2+d)^p)+3/5*f*g^2*x^5*ln(c*(e*x^2+d)^p)+f^2*g*x^3*ln(c*(e*x^2+d)^p)+f^3*x*ln(c*(e*x^2+d)
^p)-2/35*p*e*(-1/e^4*(-5/7*e^3*g^3*x^7+d*e^2*g^3*x^5-21/5*e^3*f*g^2*x^5-5/3*d^2*e*g^3*x^3+7*d*e^2*f*g^2*x^3-35
/3*e^3*f^2*g*x^3+5*d^3*x*g^3-21*d^2*e*x*f*g^2+35*d*e^2*x*f^2*g-35*x*e^3*f^3)+d*(5*d^3*g^3-21*d^2*e*f*g^2+35*d*
e^2*f^2*g-35*e^3*f^3)/e^4/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 596, normalized size of antiderivative = 1.76 \[ \int \left (f+g x^2\right )^3 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\left [-\frac {150 \, e^{3} g^{3} p x^{7} + 42 \, {\left (21 \, e^{3} f g^{2} - 5 \, d e^{2} g^{3}\right )} p x^{5} + 70 \, {\left (35 \, e^{3} f^{2} g - 21 \, d e^{2} f g^{2} + 5 \, d^{2} e g^{3}\right )} p x^{3} + 105 \, {\left (35 \, e^{3} f^{3} - 35 \, d e^{2} f^{2} g + 21 \, d^{2} e f g^{2} - 5 \, d^{3} g^{3}\right )} p \sqrt {-\frac {d}{e}} \log \left (\frac {e x^{2} - 2 \, e x \sqrt {-\frac {d}{e}} - d}{e x^{2} + d}\right ) + 210 \, {\left (35 \, e^{3} f^{3} - 35 \, d e^{2} f^{2} g + 21 \, d^{2} e f g^{2} - 5 \, d^{3} g^{3}\right )} p x - 105 \, {\left (5 \, e^{3} g^{3} p x^{7} + 21 \, e^{3} f g^{2} p x^{5} + 35 \, e^{3} f^{2} g p x^{3} + 35 \, e^{3} f^{3} p x\right )} \log \left (e x^{2} + d\right ) - 105 \, {\left (5 \, e^{3} g^{3} x^{7} + 21 \, e^{3} f g^{2} x^{5} + 35 \, e^{3} f^{2} g x^{3} + 35 \, e^{3} f^{3} x\right )} \log \left (c\right )}{3675 \, e^{3}}, -\frac {150 \, e^{3} g^{3} p x^{7} + 42 \, {\left (21 \, e^{3} f g^{2} - 5 \, d e^{2} g^{3}\right )} p x^{5} + 70 \, {\left (35 \, e^{3} f^{2} g - 21 \, d e^{2} f g^{2} + 5 \, d^{2} e g^{3}\right )} p x^{3} - 210 \, {\left (35 \, e^{3} f^{3} - 35 \, d e^{2} f^{2} g + 21 \, d^{2} e f g^{2} - 5 \, d^{3} g^{3}\right )} p \sqrt {\frac {d}{e}} \arctan \left (\frac {e x \sqrt {\frac {d}{e}}}{d}\right ) + 210 \, {\left (35 \, e^{3} f^{3} - 35 \, d e^{2} f^{2} g + 21 \, d^{2} e f g^{2} - 5 \, d^{3} g^{3}\right )} p x - 105 \, {\left (5 \, e^{3} g^{3} p x^{7} + 21 \, e^{3} f g^{2} p x^{5} + 35 \, e^{3} f^{2} g p x^{3} + 35 \, e^{3} f^{3} p x\right )} \log \left (e x^{2} + d\right ) - 105 \, {\left (5 \, e^{3} g^{3} x^{7} + 21 \, e^{3} f g^{2} x^{5} + 35 \, e^{3} f^{2} g x^{3} + 35 \, e^{3} f^{3} x\right )} \log \left (c\right )}{3675 \, e^{3}}\right ] \]

[In]

integrate((g*x^2+f)^3*log(c*(e*x^2+d)^p),x, algorithm="fricas")

[Out]

[-1/3675*(150*e^3*g^3*p*x^7 + 42*(21*e^3*f*g^2 - 5*d*e^2*g^3)*p*x^5 + 70*(35*e^3*f^2*g - 21*d*e^2*f*g^2 + 5*d^
2*e*g^3)*p*x^3 + 105*(35*e^3*f^3 - 35*d*e^2*f^2*g + 21*d^2*e*f*g^2 - 5*d^3*g^3)*p*sqrt(-d/e)*log((e*x^2 - 2*e*
x*sqrt(-d/e) - d)/(e*x^2 + d)) + 210*(35*e^3*f^3 - 35*d*e^2*f^2*g + 21*d^2*e*f*g^2 - 5*d^3*g^3)*p*x - 105*(5*e
^3*g^3*p*x^7 + 21*e^3*f*g^2*p*x^5 + 35*e^3*f^2*g*p*x^3 + 35*e^3*f^3*p*x)*log(e*x^2 + d) - 105*(5*e^3*g^3*x^7 +
21*e^3*f*g^2*x^5 + 35*e^3*f^2*g*x^3 + 35*e^3*f^3*x)*log(c))/e^3, -1/3675*(150*e^3*g^3*p*x^7 + 42*(21*e^3*f*g^
2 - 5*d*e^2*g^3)*p*x^5 + 70*(35*e^3*f^2*g - 21*d*e^2*f*g^2 + 5*d^2*e*g^3)*p*x^3 - 210*(35*e^3*f^3 - 35*d*e^2*f
^2*g + 21*d^2*e*f*g^2 - 5*d^3*g^3)*p*sqrt(d/e)*arctan(e*x*sqrt(d/e)/d) + 210*(35*e^3*f^3 - 35*d*e^2*f^2*g + 21
*d^2*e*f*g^2 - 5*d^3*g^3)*p*x - 105*(5*e^3*g^3*p*x^7 + 21*e^3*f*g^2*p*x^5 + 35*e^3*f^2*g*p*x^3 + 35*e^3*f^3*p*
x)*log(e*x^2 + d) - 105*(5*e^3*g^3*x^7 + 21*e^3*f*g^2*x^5 + 35*e^3*f^2*g*x^3 + 35*e^3*f^3*x)*log(c))/e^3]

Sympy [A] (verification not implemented)

Time = 124.72 (sec) , antiderivative size = 697, normalized size of antiderivative = 2.06 \[ \int \left (f+g x^2\right )^3 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\begin {cases} \left (f^{3} x + f^{2} g x^{3} + \frac {3 f g^{2} x^{5}}{5} + \frac {g^{3} x^{7}}{7}\right ) \log {\left (0^{p} c \right )} & \text {for}\: d = 0 \wedge e = 0 \\\left (f^{3} x + f^{2} g x^{3} + \frac {3 f g^{2} x^{5}}{5} + \frac {g^{3} x^{7}}{7}\right ) \log {\left (c d^{p} \right )} & \text {for}\: e = 0 \\- 2 f^{3} p x + f^{3} x \log {\left (c \left (e x^{2}\right )^{p} \right )} - \frac {2 f^{2} g p x^{3}}{3} + f^{2} g x^{3} \log {\left (c \left (e x^{2}\right )^{p} \right )} - \frac {6 f g^{2} p x^{5}}{25} + \frac {3 f g^{2} x^{5} \log {\left (c \left (e x^{2}\right )^{p} \right )}}{5} - \frac {2 g^{3} p x^{7}}{49} + \frac {g^{3} x^{7} \log {\left (c \left (e x^{2}\right )^{p} \right )}}{7} & \text {for}\: d = 0 \\- \frac {2 d^{4} g^{3} p \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{7 e^{4} \sqrt {- \frac {d}{e}}} + \frac {d^{4} g^{3} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{7 e^{4} \sqrt {- \frac {d}{e}}} + \frac {6 d^{3} f g^{2} p \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{5 e^{3} \sqrt {- \frac {d}{e}}} - \frac {3 d^{3} f g^{2} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{5 e^{3} \sqrt {- \frac {d}{e}}} + \frac {2 d^{3} g^{3} p x}{7 e^{3}} - \frac {2 d^{2} f^{2} g p \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{e^{2} \sqrt {- \frac {d}{e}}} + \frac {d^{2} f^{2} g \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{e^{2} \sqrt {- \frac {d}{e}}} - \frac {6 d^{2} f g^{2} p x}{5 e^{2}} - \frac {2 d^{2} g^{3} p x^{3}}{21 e^{2}} + \frac {2 d f^{3} p \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{e \sqrt {- \frac {d}{e}}} - \frac {d f^{3} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{e \sqrt {- \frac {d}{e}}} + \frac {2 d f^{2} g p x}{e} + \frac {2 d f g^{2} p x^{3}}{5 e} + \frac {2 d g^{3} p x^{5}}{35 e} - 2 f^{3} p x + f^{3} x \log {\left (c \left (d + e x^{2}\right )^{p} \right )} - \frac {2 f^{2} g p x^{3}}{3} + f^{2} g x^{3} \log {\left (c \left (d + e x^{2}\right )^{p} \right )} - \frac {6 f g^{2} p x^{5}}{25} + \frac {3 f g^{2} x^{5} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{5} - \frac {2 g^{3} p x^{7}}{49} + \frac {g^{3} x^{7} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{7} & \text {otherwise} \end {cases} \]

[In]

integrate((g*x**2+f)**3*ln(c*(e*x**2+d)**p),x)

[Out]

Piecewise(((f**3*x + f**2*g*x**3 + 3*f*g**2*x**5/5 + g**3*x**7/7)*log(0**p*c), Eq(d, 0) & Eq(e, 0)), ((f**3*x
+ f**2*g*x**3 + 3*f*g**2*x**5/5 + g**3*x**7/7)*log(c*d**p), Eq(e, 0)), (-2*f**3*p*x + f**3*x*log(c*(e*x**2)**p
) - 2*f**2*g*p*x**3/3 + f**2*g*x**3*log(c*(e*x**2)**p) - 6*f*g**2*p*x**5/25 + 3*f*g**2*x**5*log(c*(e*x**2)**p)
/5 - 2*g**3*p*x**7/49 + g**3*x**7*log(c*(e*x**2)**p)/7, Eq(d, 0)), (-2*d**4*g**3*p*log(x - sqrt(-d/e))/(7*e**4
*sqrt(-d/e)) + d**4*g**3*log(c*(d + e*x**2)**p)/(7*e**4*sqrt(-d/e)) + 6*d**3*f*g**2*p*log(x - sqrt(-d/e))/(5*e
**3*sqrt(-d/e)) - 3*d**3*f*g**2*log(c*(d + e*x**2)**p)/(5*e**3*sqrt(-d/e)) + 2*d**3*g**3*p*x/(7*e**3) - 2*d**2
*f**2*g*p*log(x - sqrt(-d/e))/(e**2*sqrt(-d/e)) + d**2*f**2*g*log(c*(d + e*x**2)**p)/(e**2*sqrt(-d/e)) - 6*d**
2*f*g**2*p*x/(5*e**2) - 2*d**2*g**3*p*x**3/(21*e**2) + 2*d*f**3*p*log(x - sqrt(-d/e))/(e*sqrt(-d/e)) - d*f**3*
log(c*(d + e*x**2)**p)/(e*sqrt(-d/e)) + 2*d*f**2*g*p*x/e + 2*d*f*g**2*p*x**3/(5*e) + 2*d*g**3*p*x**5/(35*e) -
2*f**3*p*x + f**3*x*log(c*(d + e*x**2)**p) - 2*f**2*g*p*x**3/3 + f**2*g*x**3*log(c*(d + e*x**2)**p) - 6*f*g**2
*p*x**5/25 + 3*f*g**2*x**5*log(c*(d + e*x**2)**p)/5 - 2*g**3*p*x**7/49 + g**3*x**7*log(c*(d + e*x**2)**p)/7, T
rue))

Maxima [F(-2)]

Exception generated. \[ \int \left (f+g x^2\right )^3 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((g*x^2+f)^3*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 = 0.32 (sec) , antiderivative size = 268, normalized size of antiderivative = 0.79 \[ \int \left (f+g x^2\right )^3 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-\frac {1}{49} \, {\left (2 \, g^{3} p - 7 \, g^{3} \log \left (c\right )\right )} x^{7} - \frac {{\left (42 \, e f g^{2} p - 10 \, d g^{3} p - 105 \, e f g^{2} \log \left (c\right )\right )} x^{5}}{175 \, e} - \frac {{\left (70 \, e^{2} f^{2} g p - 42 \, d e f g^{2} p + 10 \, d^{2} g^{3} p - 105 \, e^{2} f^{2} g \log \left (c\right )\right )} x^{3}}{105 \, e^{2}} + \frac {1}{35} \, {\left (5 \, g^{3} p x^{7} + 21 \, f g^{2} p x^{5} + 35 \, f^{2} g p x^{3} + 35 \, f^{3} p x\right )} \log \left (e x^{2} + d\right ) - \frac {{\left (70 \, e^{3} f^{3} p - 70 \, d e^{2} f^{2} g p + 42 \, d^{2} e f g^{2} p - 10 \, d^{3} g^{3} p - 35 \, e^{3} f^{3} \log \left (c\right )\right )} x}{35 \, e^{3}} + \frac {2 \, {\left (35 \, d e^{3} f^{3} p - 35 \, d^{2} e^{2} f^{2} g p + 21 \, d^{3} e f g^{2} p - 5 \, d^{4} g^{3} p\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{35 \, \sqrt {d e} e^{3}} \]

[In]

integrate((g*x^2+f)^3*log(c*(e*x^2+d)^p),x, algorithm="giac")

[Out]

-1/49*(2*g^3*p - 7*g^3*log(c))*x^7 - 1/175*(42*e*f*g^2*p - 10*d*g^3*p - 105*e*f*g^2*log(c))*x^5/e - 1/105*(70*
e^2*f^2*g*p - 42*d*e*f*g^2*p + 10*d^2*g^3*p - 105*e^2*f^2*g*log(c))*x^3/e^2 + 1/35*(5*g^3*p*x^7 + 21*f*g^2*p*x
^5 + 35*f^2*g*p*x^3 + 35*f^3*p*x)*log(e*x^2 + d) - 1/35*(70*e^3*f^3*p - 70*d*e^2*f^2*g*p + 42*d^2*e*f*g^2*p -
10*d^3*g^3*p - 35*e^3*f^3*log(c))*x/e^3 + 2/35*(35*d*e^3*f^3*p - 35*d^2*e^2*f^2*g*p + 21*d^3*e*f*g^2*p - 5*d^4
*g^3*p)*arctan(e*x/sqrt(d*e))/(sqrt(d*e)*e^3)

Mupad [B] (verification not implemented)

Time = 1.44 (sec) , antiderivative size = 298, normalized size of antiderivative = 0.88 \[ \int \left (f+g x^2\right )^3 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=x^3\,\left (\frac {d\,\left (\frac {6\,f\,g^2\,p}{5}-\frac {2\,d\,g^3\,p}{7\,e}\right )}{3\,e}-\frac {2\,f^2\,g\,p}{3}\right )-x\,\left (2\,f^3\,p+\frac {d\,\left (\frac {d\,\left (\frac {6\,f\,g^2\,p}{5}-\frac {2\,d\,g^3\,p}{7\,e}\right )}{e}-2\,f^2\,g\,p\right )}{e}\right )-x^5\,\left (\frac {6\,f\,g^2\,p}{25}-\frac {2\,d\,g^3\,p}{35\,e}\right )+\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )\,\left (f^3\,x+f^2\,g\,x^3+\frac {3\,f\,g^2\,x^5}{5}+\frac {g^3\,x^7}{7}\right )-\frac {2\,g^3\,p\,x^7}{49}-\frac {2\,\sqrt {d}\,p\,\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e}\,p\,x\,\left (5\,d^3\,g^3-21\,d^2\,e\,f\,g^2+35\,d\,e^2\,f^2\,g-35\,e^3\,f^3\right )}{5\,p\,d^4\,g^3-21\,p\,d^3\,e\,f\,g^2+35\,p\,d^2\,e^2\,f^2\,g-35\,p\,d\,e^3\,f^3}\right )\,\left (5\,d^3\,g^3-21\,d^2\,e\,f\,g^2+35\,d\,e^2\,f^2\,g-35\,e^3\,f^3\right )}{35\,e^{7/2}} \]

[In]

int(log(c*(d + e*x^2)^p)*(f + g*x^2)^3,x)

[Out]

x^3*((d*((6*f*g^2*p)/5 - (2*d*g^3*p)/(7*e)))/(3*e) - (2*f^2*g*p)/3) - x*(2*f^3*p + (d*((d*((6*f*g^2*p)/5 - (2*
d*g^3*p)/(7*e)))/e - 2*f^2*g*p))/e) - x^5*((6*f*g^2*p)/25 - (2*d*g^3*p)/(35*e)) + log(c*(d + e*x^2)^p)*(f^3*x
+ (g^3*x^7)/7 + f^2*g*x^3 + (3*f*g^2*x^5)/5) - (2*g^3*p*x^7)/49 - (2*d^(1/2)*p*atan((d^(1/2)*e^(1/2)*p*x*(5*d^
3*g^3 - 35*e^3*f^3 + 35*d*e^2*f^2*g - 21*d^2*e*f*g^2))/(5*d^4*g^3*p - 35*d*e^3*f^3*p - 21*d^3*e*f*g^2*p + 35*d
^2*e^2*f^2*g*p))*(5*d^3*g^3 - 35*e^3*f^3 + 35*d*e^2*f^2*g - 21*d^2*e*f*g^2))/(35*e^(7/2))