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

Optimal. Leaf size=231 \[ f^2 x \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{2} f g x^4 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{7} g^2 x^7 \log \left (c \left (d+e x^2\right )^p\right )-\frac {2 d^{7/2} g^2 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{7 e^{7/2}}+\frac {2 d^3 g^2 p x}{7 e^3}-\frac {d^2 f g p \log \left (d+e x^2\right )}{2 e^2}-\frac {2 d^2 g^2 p x^3}{21 e^2}+\frac {2 \sqrt {d} f^2 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}+\frac {d f g p x^2}{2 e}+\frac {2 d g^2 p x^5}{35 e}-2 f^2 p x-\frac {1}{4} f g p x^4-\frac {2}{49} g^2 p x^7 \]

[Out]

-2*f^2*p*x+2/7*d^3*g^2*p*x/e^3+1/2*d*f*g*p*x^2/e-2/21*d^2*g^2*p*x^3/e^2-1/4*f*g*p*x^4+2/35*d*g^2*p*x^5/e-2/49*
g^2*p*x^7-2/7*d^(7/2)*g^2*p*arctan(x*e^(1/2)/d^(1/2))/e^(7/2)-1/2*d^2*f*g*p*ln(e*x^2+d)/e^2+f^2*x*ln(c*(e*x^2+
d)^p)+1/2*f*g*x^4*ln(c*(e*x^2+d)^p)+1/7*g^2*x^7*ln(c*(e*x^2+d)^p)+2*f^2*p*arctan(x*e^(1/2)/d^(1/2))*d^(1/2)/e^
(1/2)

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Rubi [A]  time = 0.18, antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {2471, 2448, 321, 205, 2454, 2395, 43, 2455, 302} \[ f^2 x \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{2} f g x^4 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{7} g^2 x^7 \log \left (c \left (d+e x^2\right )^p\right )-\frac {d^2 f g p \log \left (d+e x^2\right )}{2 e^2}-\frac {2 d^2 g^2 p x^3}{21 e^2}+\frac {2 d^3 g^2 p x}{7 e^3}-\frac {2 d^{7/2} g^2 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{7 e^{7/2}}+\frac {2 \sqrt {d} f^2 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}+\frac {d f g p x^2}{2 e}+\frac {2 d g^2 p x^5}{35 e}-2 f^2 p x-\frac {1}{4} f g p x^4-\frac {2}{49} g^2 p x^7 \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2*f^2*p*x + (2*d^3*g^2*p*x)/(7*e^3) + (d*f*g*p*x^2)/(2*e) - (2*d^2*g^2*p*x^3)/(21*e^2) - (f*g*p*x^4)/4 + (2*d
*g^2*p*x^5)/(35*e) - (2*g^2*p*x^7)/49 + (2*Sqrt[d]*f^2*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/Sqrt[e] - (2*d^(7/2)*g^2
*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(7*e^(7/2)) - (d^2*f*g*p*Log[d + e*x^2])/(2*e^2) + f^2*x*Log[c*(d + e*x^2)^p]
+ (f*g*x^4*Log[c*(d + e*x^2)^p])/2 + (g^2*x^7*Log[c*(d + e*x^2)^p])/7

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

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 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

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

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Mathematica [A]  time = 0.21, size = 178, normalized size = 0.77 \[ \frac {1}{14} x \left (14 f^2+7 f g x^3+2 g^2 x^6\right ) \log \left (c \left (d+e x^2\right )^p\right )-\frac {2 \sqrt {d} p \left (d^3 g^2-7 e^3 f^2\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{7 e^{7/2}}-\frac {d^2 f g p \log \left (d+e x^2\right )}{2 e^2}+\frac {p x \left (840 d^3 g^2-280 d^2 e g^2 x^2+42 d e^2 g x \left (35 f+4 g x^3\right )-15 e^3 \left (392 f^2+49 f g x^3+8 g^2 x^6\right )\right )}{2940 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(p*x*(840*d^3*g^2 - 280*d^2*e*g^2*x^2 + 42*d*e^2*g*x*(35*f + 4*g*x^3) - 15*e^3*(392*f^2 + 49*f*g*x^3 + 8*g^2*x
^6)))/(2940*e^3) - (2*Sqrt[d]*(-7*e^3*f^2 + d^3*g^2)*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(7*e^(7/2)) - (d^2*f*g*p*L
og[d + e*x^2])/(2*e^2) + (x*(14*f^2 + 7*f*g*x^3 + 2*g^2*x^6)*Log[c*(d + e*x^2)^p])/14

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fricas [A]  time = 0.76, size = 454, normalized size = 1.97 \[ \left [-\frac {120 \, e^{3} g^{2} p x^{7} - 168 \, d e^{2} g^{2} p x^{5} + 735 \, e^{3} f g p x^{4} + 280 \, d^{2} e g^{2} p x^{3} - 1470 \, d e^{2} f g p x^{2} + 420 \, {\left (7 \, e^{3} f^{2} - d^{3} g^{2}\right )} p \sqrt {-\frac {d}{e}} \log \left (\frac {e x^{2} - 2 \, e x \sqrt {-\frac {d}{e}} - d}{e x^{2} + d}\right ) + 840 \, {\left (7 \, e^{3} f^{2} - d^{3} g^{2}\right )} p x - 210 \, {\left (2 \, e^{3} g^{2} p x^{7} + 7 \, e^{3} f g p x^{4} + 14 \, e^{3} f^{2} p x - 7 \, d^{2} e f g p\right )} \log \left (e x^{2} + d\right ) - 210 \, {\left (2 \, e^{3} g^{2} x^{7} + 7 \, e^{3} f g x^{4} + 14 \, e^{3} f^{2} x\right )} \log \relax (c)}{2940 \, e^{3}}, -\frac {120 \, e^{3} g^{2} p x^{7} - 168 \, d e^{2} g^{2} p x^{5} + 735 \, e^{3} f g p x^{4} + 280 \, d^{2} e g^{2} p x^{3} - 1470 \, d e^{2} f g p x^{2} - 840 \, {\left (7 \, e^{3} f^{2} - d^{3} g^{2}\right )} p \sqrt {\frac {d}{e}} \arctan \left (\frac {e x \sqrt {\frac {d}{e}}}{d}\right ) + 840 \, {\left (7 \, e^{3} f^{2} - d^{3} g^{2}\right )} p x - 210 \, {\left (2 \, e^{3} g^{2} p x^{7} + 7 \, e^{3} f g p x^{4} + 14 \, e^{3} f^{2} p x - 7 \, d^{2} e f g p\right )} \log \left (e x^{2} + d\right ) - 210 \, {\left (2 \, e^{3} g^{2} x^{7} + 7 \, e^{3} f g x^{4} + 14 \, e^{3} f^{2} x\right )} \log \relax (c)}{2940 \, e^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2940*(120*e^3*g^2*p*x^7 - 168*d*e^2*g^2*p*x^5 + 735*e^3*f*g*p*x^4 + 280*d^2*e*g^2*p*x^3 - 1470*d*e^2*f*g*p
*x^2 + 420*(7*e^3*f^2 - d^3*g^2)*p*sqrt(-d/e)*log((e*x^2 - 2*e*x*sqrt(-d/e) - d)/(e*x^2 + d)) + 840*(7*e^3*f^2
- d^3*g^2)*p*x - 210*(2*e^3*g^2*p*x^7 + 7*e^3*f*g*p*x^4 + 14*e^3*f^2*p*x - 7*d^2*e*f*g*p)*log(e*x^2 + d) - 21
0*(2*e^3*g^2*x^7 + 7*e^3*f*g*x^4 + 14*e^3*f^2*x)*log(c))/e^3, -1/2940*(120*e^3*g^2*p*x^7 - 168*d*e^2*g^2*p*x^5
+ 735*e^3*f*g*p*x^4 + 280*d^2*e*g^2*p*x^3 - 1470*d*e^2*f*g*p*x^2 - 840*(7*e^3*f^2 - d^3*g^2)*p*sqrt(d/e)*arct
an(e*x*sqrt(d/e)/d) + 840*(7*e^3*f^2 - d^3*g^2)*p*x - 210*(2*e^3*g^2*p*x^7 + 7*e^3*f*g*p*x^4 + 14*e^3*f^2*p*x
- 7*d^2*e*f*g*p)*log(e*x^2 + d) - 210*(2*e^3*g^2*x^7 + 7*e^3*f*g*x^4 + 14*e^3*f^2*x)*log(c))/e^3]

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giac [A]  time = 0.25, size = 225, normalized size = 0.97 \[ -\frac {1}{2} \, d^{2} f g p e^{\left (-2\right )} \log \left (x^{2} e + d\right ) - \frac {2 \, {\left (d^{4} g^{2} p - 7 \, d f^{2} p e^{3}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {7}{2}\right )}}{7 \, \sqrt {d}} + \frac {1}{2940} \, {\left (420 \, g^{2} p x^{7} e^{3} \log \left (x^{2} e + d\right ) - 120 \, g^{2} p x^{7} e^{3} + 420 \, g^{2} x^{7} e^{3} \log \relax (c) + 168 \, d g^{2} p x^{5} e^{2} - 280 \, d^{2} g^{2} p x^{3} e + 1470 \, f g p x^{4} e^{3} \log \left (x^{2} e + d\right ) - 735 \, f g p x^{4} e^{3} + 1470 \, f g x^{4} e^{3} \log \relax (c) + 840 \, d^{3} g^{2} p x + 1470 \, d f g p x^{2} e^{2} + 2940 \, f^{2} p x e^{3} \log \left (x^{2} e + d\right ) - 5880 \, f^{2} p x e^{3} + 2940 \, f^{2} x e^{3} \log \relax (c)\right )} e^{\left (-3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*d^2*f*g*p*e^(-2)*log(x^2*e + d) - 2/7*(d^4*g^2*p - 7*d*f^2*p*e^3)*arctan(x*e^(1/2)/sqrt(d))*e^(-7/2)/sqrt
(d) + 1/2940*(420*g^2*p*x^7*e^3*log(x^2*e + d) - 120*g^2*p*x^7*e^3 + 420*g^2*x^7*e^3*log(c) + 168*d*g^2*p*x^5*
e^2 - 280*d^2*g^2*p*x^3*e + 1470*f*g*p*x^4*e^3*log(x^2*e + d) - 735*f*g*p*x^4*e^3 + 1470*f*g*x^4*e^3*log(c) +
840*d^3*g^2*p*x + 1470*d*f*g*p*x^2*e^2 + 2940*f^2*p*x*e^3*log(x^2*e + d) - 5880*f^2*p*x*e^3 + 2940*f^2*x*e^3*l
og(c))*e^(-3)

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maple [C]  time = 0.54, size = 869, normalized size = 3.76 \[ -\frac {i \pi f g \,x^{4} \mathrm {csgn}\left (i c \right ) \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 )}{4}+\frac {g^{2} x^{7} \ln \relax (c )}{7}+f^{2} x \ln \relax (c )-\frac {d^{2} f g p \ln \left (-d^{4} g^{2}+7 d \,e^{3} f^{2}-\sqrt {-d^{7} e \,g^{4}+14 d^{4} e^{4} f^{2} g^{2}-49 d \,e^{7} f^{4}}\, x \right )}{2 e^{2}}-\frac {d^{2} f g p \ln \left (-d^{4} g^{2}+7 d \,e^{3} f^{2}+\sqrt {-d^{7} e \,g^{4}+14 d^{4} e^{4} f^{2} g^{2}-49 d \,e^{7} f^{4}}\, x \right )}{2 e^{2}}-\frac {2 g^{2} p \,x^{7}}{49}-2 f^{2} p x +\left (\frac {1}{7} g^{2} x^{7}+\frac {1}{2} f g \,x^{4}+f^{2} x \right ) \ln \left (\left (e \,x^{2}+d \right )^{p}\right )+\frac {i \pi \,g^{2} x^{7} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}}{14}+\frac {i \pi \,g^{2} x^{7} \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}}{14}+\frac {\sqrt {-d^{7} e \,g^{4}+14 d^{4} e^{4} f^{2} g^{2}-49 d \,e^{7} f^{4}}\, p \ln \left (-d^{4} g^{2}+7 d \,e^{3} f^{2}-\sqrt {-d^{7} e \,g^{4}+14 d^{4} e^{4} f^{2} g^{2}-49 d \,e^{7} f^{4}}\, x \right )}{7 e^{4}}-\frac {\sqrt {-d^{7} e \,g^{4}+14 d^{4} e^{4} f^{2} g^{2}-49 d \,e^{7} f^{4}}\, p \ln \left (-d^{4} g^{2}+7 d \,e^{3} f^{2}+\sqrt {-d^{7} e \,g^{4}+14 d^{4} e^{4} f^{2} g^{2}-49 d \,e^{7} f^{4}}\, x \right )}{7 e^{4}}+\frac {f g \,x^{4} \ln \relax (c )}{2}+\frac {i \pi \,f^{2} x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}}{2}+\frac {i \pi \,f^{2} x \,\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}}{2}-\frac {f g p \,x^{4}}{4}+\frac {2 d^{3} g^{2} p x}{7 e^{3}}-\frac {2 d^{2} g^{2} p \,x^{3}}{21 e^{2}}+\frac {2 d \,g^{2} p \,x^{5}}{35 e}-\frac {i \pi \,f^{2} x \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}}{2}-\frac {i \pi \,g^{2} x^{7} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}}{14}-\frac {i \pi \,g^{2} x^{7} \mathrm {csgn}\left (i c \right ) \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 )}{14}+\frac {i \pi f g \,x^{4} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}}{4}+\frac {i \pi f g \,x^{4} \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}}{4}-\frac {i \pi \,f^{2} x \,\mathrm {csgn}\left (i c \right ) \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}-\frac {i \pi f g \,x^{4} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}}{4}+\frac {d f g p \,x^{2}}{2 e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.99, size = 178, normalized size = 0.77 \[ -\frac {1}{2940} \, {\left (\frac {1470 \, d^{2} f g \log \left (e x^{2} + d\right )}{e^{3}} - \frac {840 \, {\left (7 \, d e^{3} f^{2} - d^{4} g^{2}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{\sqrt {d e} e^{4}} + \frac {120 \, e^{3} g^{2} x^{7} - 168 \, d e^{2} g^{2} x^{5} + 735 \, e^{3} f g x^{4} + 280 \, d^{2} e g^{2} x^{3} - 1470 \, d e^{2} f g x^{2} + 840 \, {\left (7 \, e^{3} f^{2} - d^{3} g^{2}\right )} x}{e^{4}}\right )} e p + \frac {1}{14} \, {\left (2 \, g^{2} x^{7} + 7 \, f g x^{4} + 14 \, f^{2} x\right )} \log \left ({\left (e x^{2} + d\right )}^{p} c\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2940*(1470*d^2*f*g*log(e*x^2 + d)/e^3 - 840*(7*d*e^3*f^2 - d^4*g^2)*arctan(e*x/sqrt(d*e))/(sqrt(d*e)*e^4) +
(120*e^3*g^2*x^7 - 168*d*e^2*g^2*x^5 + 735*e^3*f*g*x^4 + 280*d^2*e*g^2*x^3 - 1470*d*e^2*f*g*x^2 + 840*(7*e^3*
f^2 - d^3*g^2)*x)/e^4)*e*p + 1/14*(2*g^2*x^7 + 7*f*g*x^4 + 14*f^2*x)*log((e*x^2 + d)^p*c)

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mupad [B]  time = 2.77, size = 317, normalized size = 1.37 \[ \frac {g^2\,x^7\,\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )}{7}-2\,f^2\,p\,x-\frac {2\,g^2\,p\,x^7}{49}+f^2\,x\,\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )+\frac {f\,g\,x^4\,\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )}{2}-\frac {f\,g\,p\,x^4}{4}+\frac {2\,d\,g^2\,p\,x^5}{35\,e}+\frac {2\,d^3\,g^2\,p\,x}{7\,e^3}-\frac {2\,\sqrt {d}\,f^2\,p\,\mathrm {atan}\left (\frac {7\,\sqrt {d}\,e^{7/2}\,f^2\,p\,x}{d^4\,g^2\,p-7\,d\,e^3\,f^2\,p}-\frac {d^{7/2}\,\sqrt {e}\,g^2\,p\,x}{d^4\,g^2\,p-7\,d\,e^3\,f^2\,p}\right )}{\sqrt {e}}+\frac {2\,d^{7/2}\,g^2\,p\,\mathrm {atan}\left (\frac {7\,\sqrt {d}\,e^{7/2}\,f^2\,p\,x}{d^4\,g^2\,p-7\,d\,e^3\,f^2\,p}-\frac {d^{7/2}\,\sqrt {e}\,g^2\,p\,x}{d^4\,g^2\,p-7\,d\,e^3\,f^2\,p}\right )}{7\,e^{7/2}}-\frac {2\,d^2\,g^2\,p\,x^3}{21\,e^2}+\frac {d\,f\,g\,p\,x^2}{2\,e}-\frac {d^2\,f\,g\,p\,\ln \left (e\,x^2+d\right )}{2\,e^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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