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

Optimal. Leaf size=338 \[ f^3 x \log \left (c \left (d+e x^2\right )^p\right )+f^2 g x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {3}{5} f g^2 x^5 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{7} g^3 x^7 \log \left (c \left (d+e x^2\right )^p\right )-\frac {2 d^{3/2} f^2 g p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{e^{3/2}}+\frac {6 d^{5/2} f g^2 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 e^{5/2}}-\frac {2 d^{7/2} g^3 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{7 e^{7/2}}+\frac {2 d^3 g^3 p x}{7 e^3}-\frac {6 d^2 f g^2 p x}{5 e^2}-\frac {2 d^2 g^3 p x^3}{21 e^2}+\frac {2 \sqrt {d} f^3 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}+\frac {2 d f^2 g p x}{e}+\frac {2 d f g^2 p x^3}{5 e}+\frac {2 d g^3 p x^5}{35 e}-2 f^3 p x-\frac {2}{3} f^2 g p x^3-\frac {6}{25} f g^2 p x^5-\frac {2}{49} g^3 p x^7 \]

[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)

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Rubi [A]  time = 0.26, antiderivative size = 338, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2471, 2448, 321, 205, 2455, 302} \[ f^2 g x^3 \log \left (c \left (d+e x^2\right )^p\right )+f^3 x \log \left (c \left (d+e x^2\right )^p\right )+\frac {3}{5} f g^2 x^5 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{7} g^3 x^7 \log \left (c \left (d+e x^2\right )^p\right )-\frac {2 d^{3/2} f^2 g p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{e^{3/2}}-\frac {6 d^2 f g^2 p x}{5 e^2}+\frac {6 d^{5/2} f g^2 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 e^{5/2}}-\frac {2 d^2 g^3 p x^3}{21 e^2}+\frac {2 d^3 g^3 p x}{7 e^3}-\frac {2 d^{7/2} g^3 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{7 e^{7/2}}+\frac {2 d f^2 g p x}{e}+\frac {2 \sqrt {d} f^3 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}+\frac {2 d f g^2 p x^3}{5 e}+\frac {2 d g^3 p x^5}{35 e}-\frac {2}{3} f^2 g p x^3-2 f^3 p x-\frac {6}{25} f g^2 p x^5-\frac {2}{49} g^3 p x^7 \]

Antiderivative was successfully verified.

[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 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]

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 321

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 2448

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 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 2471

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*} \int \left (f+g x^2\right )^3 \log \left (c \left (d+e x^2\right )^p\right ) \, dx &=\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*}

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Mathematica [A]  time = 0.28, size = 215, normalized size = 0.64 \[ \frac {1}{35} x \left (35 f^3+35 f^2 g x^2+21 f g^2 x^4+5 g^3 x^6\right ) \log \left (c \left (d+e x^2\right )^p\right )-\frac {2 p x \left (-525 d^3 g^3+35 d^2 e g^2 \left (63 f+5 g x^2\right )-105 d e^2 g \left (35 f^2+7 f g x^2+g^2 x^4\right )+e^3 \left (3675 f^3+1225 f^2 g x^2+441 f g^2 x^4+75 g^3 x^6\right )\right )}{3675 e^3}-\frac {2 \sqrt {d} p \left (5 d^3 g^3-21 d^2 e f g^2+35 d e^2 f^2 g-35 e^3 f^3\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{35 e^{7/2}} \]

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.92, size = 596, normalized size = 1.76 \[ \left [-\frac {150 \, e^{3} g^{3} p x^{7} + 42 \, {\left (21 \, e^{3} f g^{2} - 5 \, d e^{2} g^{3}\right )} p x^{5} + 70 \, {\left (35 \, e^{3} f^{2} g - 21 \, d e^{2} f g^{2} + 5 \, d^{2} e g^{3}\right )} p x^{3} + 105 \, {\left (35 \, e^{3} f^{3} - 35 \, d e^{2} f^{2} g + 21 \, d^{2} e f g^{2} - 5 \, d^{3} g^{3}\right )} p \sqrt {-\frac {d}{e}} \log \left (\frac {e x^{2} - 2 \, e x \sqrt {-\frac {d}{e}} - d}{e x^{2} + d}\right ) + 210 \, {\left (35 \, e^{3} f^{3} - 35 \, d e^{2} f^{2} g + 21 \, d^{2} e f g^{2} - 5 \, d^{3} g^{3}\right )} p x - 105 \, {\left (5 \, e^{3} g^{3} p x^{7} + 21 \, e^{3} f g^{2} p x^{5} + 35 \, e^{3} f^{2} g p x^{3} + 35 \, e^{3} f^{3} p x\right )} \log \left (e x^{2} + d\right ) - 105 \, {\left (5 \, e^{3} g^{3} x^{7} + 21 \, e^{3} f g^{2} x^{5} + 35 \, e^{3} f^{2} g x^{3} + 35 \, e^{3} f^{3} x\right )} \log \relax (c)}{3675 \, e^{3}}, -\frac {150 \, e^{3} g^{3} p x^{7} + 42 \, {\left (21 \, e^{3} f g^{2} - 5 \, d e^{2} g^{3}\right )} p x^{5} + 70 \, {\left (35 \, e^{3} f^{2} g - 21 \, d e^{2} f g^{2} + 5 \, d^{2} e g^{3}\right )} p x^{3} - 210 \, {\left (35 \, e^{3} f^{3} - 35 \, d e^{2} f^{2} g + 21 \, d^{2} e f g^{2} - 5 \, d^{3} g^{3}\right )} p \sqrt {\frac {d}{e}} \arctan \left (\frac {e x \sqrt {\frac {d}{e}}}{d}\right ) + 210 \, {\left (35 \, e^{3} f^{3} - 35 \, d e^{2} f^{2} g + 21 \, d^{2} e f g^{2} - 5 \, d^{3} g^{3}\right )} p x - 105 \, {\left (5 \, e^{3} g^{3} p x^{7} + 21 \, e^{3} f g^{2} p x^{5} + 35 \, e^{3} f^{2} g p x^{3} + 35 \, e^{3} f^{3} p x\right )} \log \left (e x^{2} + d\right ) - 105 \, {\left (5 \, e^{3} g^{3} x^{7} + 21 \, e^{3} f g^{2} x^{5} + 35 \, e^{3} f^{2} g x^{3} + 35 \, e^{3} f^{3} x\right )} \log \relax (c)}{3675 \, e^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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]

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giac [A]  time = 0.21, size = 309, normalized size = 0.91 \[ -\frac {2 \, {\left (5 \, d^{4} g^{3} p - 21 \, d^{3} f g^{2} p e + 35 \, d^{2} f^{2} g p e^{2} - 35 \, d f^{3} p e^{3}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {7}{2}\right )}}{35 \, \sqrt {d}} + \frac {1}{3675} \, {\left (525 \, g^{3} p x^{7} e^{3} \log \left (x^{2} e + d\right ) - 150 \, g^{3} p x^{7} e^{3} + 525 \, g^{3} x^{7} e^{3} \log \relax (c) + 210 \, d g^{3} p x^{5} e^{2} + 2205 \, f g^{2} p x^{5} e^{3} \log \left (x^{2} e + d\right ) - 882 \, f g^{2} p x^{5} e^{3} - 350 \, d^{2} g^{3} p x^{3} e + 2205 \, f g^{2} x^{5} e^{3} \log \relax (c) + 1470 \, d f g^{2} p x^{3} e^{2} + 3675 \, f^{2} g p x^{3} e^{3} \log \left (x^{2} e + d\right ) + 1050 \, d^{3} g^{3} p x - 2450 \, f^{2} g p x^{3} e^{3} - 4410 \, d^{2} f g^{2} p x e + 3675 \, f^{2} g x^{3} e^{3} \log \relax (c) + 7350 \, d f^{2} g p x e^{2} + 3675 \, f^{3} p x e^{3} \log \left (x^{2} e + d\right ) - 7350 \, f^{3} p x e^{3} + 3675 \, f^{3} x e^{3} \log \relax (c)\right )} e^{\left (-3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/35*(5*d^4*g^3*p - 21*d^3*f*g^2*p*e + 35*d^2*f^2*g*p*e^2 - 35*d*f^3*p*e^3)*arctan(x*e^(1/2)/sqrt(d))*e^(-7/2
)/sqrt(d) + 1/3675*(525*g^3*p*x^7*e^3*log(x^2*e + d) - 150*g^3*p*x^7*e^3 + 525*g^3*x^7*e^3*log(c) + 210*d*g^3*
p*x^5*e^2 + 2205*f*g^2*p*x^5*e^3*log(x^2*e + d) - 882*f*g^2*p*x^5*e^3 - 350*d^2*g^3*p*x^3*e + 2205*f*g^2*x^5*e
^3*log(c) + 1470*d*f*g^2*p*x^3*e^2 + 3675*f^2*g*p*x^3*e^3*log(x^2*e + d) + 1050*d^3*g^3*p*x - 2450*f^2*g*p*x^3
*e^3 - 4410*d^2*f*g^2*p*x*e + 3675*f^2*g*x^3*e^3*log(c) + 7350*d*f^2*g*p*x*e^2 + 3675*f^3*p*x*e^3*log(x^2*e +
d) - 7350*f^3*p*x*e^3 + 3675*f^3*x*e^3*log(c))*e^(-3)

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maple [C]  time = 0.52, size = 995, normalized size = 2.94 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*ln(c)*g^3*x^7+ln(c)*f^3*x-2*f^3*p*x-2/49*g^3*p*x^7+ln(c)*f^2*g*x^3+3/5*ln(c)*f*g^2*x^5+(1/7*g^3*x^7+3/5*f*
g^2*x^5+f^2*g*x^3+f^3*x)*ln((e*x^2+d)^p)+1/e*(-d*e)^(1/2)*p*ln((-d*e)^(1/2)*x-d)*f^3-1/e*(-d*e)^(1/2)*p*ln(-(-
d*e)^(1/2)*x-d)*f^3-1/14*I*Pi*g^3*x^7*csgn(I*c*(e*x^2+d)^p)^3-1/2*I*Pi*f^3*csgn(I*c*(e*x^2+d)^p)^3*x-2/3*f^2*g
*p*x^3-6/25*f*g^2*p*x^5-3/5/e^3*(-d*e)^(1/2)*p*ln(-(-d*e)^(1/2)*x-d)*f*g^2*d^2-1/e^2*(-d*e)^(1/2)*p*ln((-d*e)^
(1/2)*x-d)*f^2*g*d+3/5/e^3*(-d*e)^(1/2)*p*ln((-d*e)^(1/2)*x-d)*f*g^2*d^2+1/e^2*(-d*e)^(1/2)*p*ln(-(-d*e)^(1/2)
*x-d)*f^2*g*d-1/2*I*Pi*f^3*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)*x+1/2*I*Pi*f^2*g*x^3*csgn(I*(e*
x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2+1/2*I*Pi*f^2*g*x^3*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)-1/14*I*Pi*g^3*x^7*csgn(
I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)+3/10*I*Pi*f*g^2*x^5*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2
+3/10*I*Pi*f*g^2*x^5*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)+2/7*d^3*g^3*p*x/e^3-2/21*d^2*g^3*p*x^3/e^2+2/35*d*g^3*p
*x^5/e-3/10*I*Pi*f*g^2*x^5*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)-1/2*I*Pi*f^2*g*x^3*csgn(I*(e*x^
2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)+1/7/e^4*(-d*e)^(1/2)*p*ln(-(-d*e)^(1/2)*x-d)*g^3*d^3-1/7/e^4*(-d*e)^(1
/2)*p*ln((-d*e)^(1/2)*x-d)*g^3*d^3+1/2*I*Pi*f^3*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)*x+1/14*I*Pi*g^3*x^7*csgn(I*(
e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2+1/2*I*Pi*f^3*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2*x+2*d*f^2*g*p*x/e
-6/5*d^2*f*g^2*p*x/e^2+2/5*d*f*g^2*p*x^3/e+1/14*I*Pi*g^3*x^7*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)-3/10*I*Pi*f*g^2
*x^5*csgn(I*c*(e*x^2+d)^p)^3-1/2*I*Pi*f^2*g*x^3*csgn(I*c*(e*x^2+d)^p)^3

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maxima [A]  time = 1.01, size = 227, normalized size = 0.67 \[ \frac {2}{3675} \, e p {\left (\frac {105 \, {\left (35 \, d e^{3} f^{3} - 35 \, d^{2} e^{2} f^{2} g + 21 \, d^{3} e f g^{2} - 5 \, d^{4} g^{3}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{\sqrt {d e} e^{4}} - \frac {75 \, e^{3} g^{3} x^{7} + 21 \, {\left (21 \, e^{3} f g^{2} - 5 \, d e^{2} g^{3}\right )} x^{5} + 35 \, {\left (35 \, e^{3} f^{2} g - 21 \, d e^{2} f g^{2} + 5 \, d^{2} e g^{3}\right )} x^{3} + 105 \, {\left (35 \, e^{3} f^{3} - 35 \, d e^{2} f^{2} g + 21 \, d^{2} e f g^{2} - 5 \, d^{3} g^{3}\right )} x}{e^{4}}\right )} + \frac {1}{35} \, {\left (5 \, g^{3} x^{7} + 21 \, f g^{2} x^{5} + 35 \, f^{2} g x^{3} + 35 \, f^{3} x\right )} \log \left ({\left (e x^{2} + d\right )}^{p} c\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3675*e*p*(105*(35*d*e^3*f^3 - 35*d^2*e^2*f^2*g + 21*d^3*e*f*g^2 - 5*d^4*g^3)*arctan(e*x/sqrt(d*e))/(sqrt(d*e
)*e^4) - (75*e^3*g^3*x^7 + 21*(21*e^3*f*g^2 - 5*d*e^2*g^3)*x^5 + 35*(35*e^3*f^2*g - 21*d*e^2*f*g^2 + 5*d^2*e*g
^3)*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)*x)/e^4) + 1/35*(5*g^3*x^7 + 21*f*g^2*
x^5 + 35*f^2*g*x^3 + 35*f^3*x)*log((e*x^2 + d)^p*c)

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mupad [B]  time = 0.38, size = 298, normalized size = 0.88 \[ x^3\,\left (\frac {d\,\left (\frac {6\,f\,g^2\,p}{5}-\frac {2\,d\,g^3\,p}{7\,e}\right )}{3\,e}-\frac {2\,f^2\,g\,p}{3}\right )-x\,\left (2\,f^3\,p+\frac {d\,\left (\frac {d\,\left (\frac {6\,f\,g^2\,p}{5}-\frac {2\,d\,g^3\,p}{7\,e}\right )}{e}-2\,f^2\,g\,p\right )}{e}\right )-x^5\,\left (\frac {6\,f\,g^2\,p}{25}-\frac {2\,d\,g^3\,p}{35\,e}\right )+\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )\,\left (f^3\,x+f^2\,g\,x^3+\frac {3\,f\,g^2\,x^5}{5}+\frac {g^3\,x^7}{7}\right )-\frac {2\,g^3\,p\,x^7}{49}-\frac {2\,\sqrt {d}\,p\,\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e}\,p\,x\,\left (5\,d^3\,g^3-21\,d^2\,e\,f\,g^2+35\,d\,e^2\,f^2\,g-35\,e^3\,f^3\right )}{5\,p\,d^4\,g^3-21\,p\,d^3\,e\,f\,g^2+35\,p\,d^2\,e^2\,f^2\,g-35\,p\,d\,e^3\,f^3}\right )\,\left (5\,d^3\,g^3-21\,d^2\,e\,f\,g^2+35\,d\,e^2\,f^2\,g-35\,e^3\,f^3\right )}{35\,e^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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