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

Optimal. Leaf size=117 \[ f x \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{3} g x^3 \log \left (c \left (d+e x^2\right )^p\right )-\frac {2 d^{3/2} g p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 e^{3/2}}+\frac {2 \sqrt {d} f p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}+\frac {2 d g p x}{3 e}-2 f p x-\frac {2}{9} g p x^3 \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2471, 2448, 321, 205, 2455, 302} \[ f x \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{3} g x^3 \log \left (c \left (d+e x^2\right )^p\right )-\frac {2 d^{3/2} g p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 e^{3/2}}+\frac {2 \sqrt {d} f p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}+\frac {2 d g p x}{3 e}-2 f p x-\frac {2}{9} g p x^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2*f*p*x + (2*d*g*p*x)/(3*e) - (2*g*p*x^3)/9 + (2*Sqrt[d]*f*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/Sqrt[e] - (2*d^(3/2
)*g*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(3*e^(3/2)) + f*x*Log[c*(d + e*x^2)^p] + (g*x^3*Log[c*(d + e*x^2)^p])/3

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 ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx &=\int \left (f \log \left (c \left (d+e x^2\right )^p\right )+g x^2 \log \left (c \left (d+e x^2\right )^p\right )\right ) \, dx\\ &=f \int \log \left (c \left (d+e x^2\right )^p\right ) \, dx+g \int x^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx\\ &=f x \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{3} g x^3 \log \left (c \left (d+e x^2\right )^p\right )-(2 e f p) \int \frac {x^2}{d+e x^2} \, dx-\frac {1}{3} (2 e g p) \int \frac {x^4}{d+e x^2} \, dx\\ &=-2 f p x+f x \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{3} g x^3 \log \left (c \left (d+e x^2\right )^p\right )+(2 d f p) \int \frac {1}{d+e x^2} \, dx-\frac {1}{3} (2 e g p) \int \left (-\frac {d}{e^2}+\frac {x^2}{e}+\frac {d^2}{e^2 \left (d+e x^2\right )}\right ) \, dx\\ &=-2 f p x+\frac {2 d g p x}{3 e}-\frac {2}{9} g p x^3+\frac {2 \sqrt {d} f p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}+f x \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{3} g x^3 \log \left (c \left (d+e x^2\right )^p\right )-\frac {\left (2 d^2 g p\right ) \int \frac {1}{d+e x^2} \, dx}{3 e}\\ &=-2 f p x+\frac {2 d g p x}{3 e}-\frac {2}{9} g p x^3+\frac {2 \sqrt {d} f p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}-\frac {2 d^{3/2} g p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 e^{3/2}}+f x \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{3} g x^3 \log \left (c \left (d+e x^2\right )^p\right )\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 117, normalized size = 1.00 \[ f x \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{3} g x^3 \log \left (c \left (d+e x^2\right )^p\right )-\frac {2 d^{3/2} g p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 e^{3/2}}+\frac {2 \sqrt {d} f p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}+\frac {2 d g p x}{3 e}-2 f p x-\frac {2}{9} g p x^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2*f*p*x + (2*d*g*p*x)/(3*e) - (2*g*p*x^3)/9 + (2*Sqrt[d]*f*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/Sqrt[e] - (2*d^(3/2
)*g*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(3*e^(3/2)) + f*x*Log[c*(d + e*x^2)^p] + (g*x^3*Log[c*(d + e*x^2)^p])/3

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fricas [A]  time = 0.78, size = 220, normalized size = 1.88 \[ \left [-\frac {2 \, e g p x^{3} + 3 \, {\left (3 \, e f - d g\right )} p \sqrt {-\frac {d}{e}} \log \left (\frac {e x^{2} - 2 \, e x \sqrt {-\frac {d}{e}} - d}{e x^{2} + d}\right ) + 6 \, {\left (3 \, e f - d g\right )} p x - 3 \, {\left (e g p x^{3} + 3 \, e f p x\right )} \log \left (e x^{2} + d\right ) - 3 \, {\left (e g x^{3} + 3 \, e f x\right )} \log \relax (c)}{9 \, e}, -\frac {2 \, e g p x^{3} - 6 \, {\left (3 \, e f - d g\right )} p \sqrt {\frac {d}{e}} \arctan \left (\frac {e x \sqrt {\frac {d}{e}}}{d}\right ) + 6 \, {\left (3 \, e f - d g\right )} p x - 3 \, {\left (e g p x^{3} + 3 \, e f p x\right )} \log \left (e x^{2} + d\right ) - 3 \, {\left (e g x^{3} + 3 \, e f x\right )} \log \relax (c)}{9 \, e}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/9*(2*e*g*p*x^3 + 3*(3*e*f - d*g)*p*sqrt(-d/e)*log((e*x^2 - 2*e*x*sqrt(-d/e) - d)/(e*x^2 + d)) + 6*(3*e*f -
d*g)*p*x - 3*(e*g*p*x^3 + 3*e*f*p*x)*log(e*x^2 + d) - 3*(e*g*x^3 + 3*e*f*x)*log(c))/e, -1/9*(2*e*g*p*x^3 - 6*
(3*e*f - d*g)*p*sqrt(d/e)*arctan(e*x*sqrt(d/e)/d) + 6*(3*e*f - d*g)*p*x - 3*(e*g*p*x^3 + 3*e*f*p*x)*log(e*x^2
+ d) - 3*(e*g*x^3 + 3*e*f*x)*log(c))/e]

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giac [A]  time = 0.18, size = 109, normalized size = 0.93 \[ -\frac {2 \, {\left (d^{2} g p - 3 \, d f p e\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {3}{2}\right )}}{3 \, \sqrt {d}} + \frac {1}{9} \, {\left (3 \, g p x^{3} e \log \left (x^{2} e + d\right ) - 2 \, g p x^{3} e + 3 \, g x^{3} e \log \relax (c) + 9 \, f p x e \log \left (x^{2} e + d\right ) + 6 \, d g p x - 18 \, f p x e + 9 \, f x e \log \relax (c)\right )} e^{\left (-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(d^2*g*p - 3*d*f*p*e)*arctan(x*e^(1/2)/sqrt(d))*e^(-3/2)/sqrt(d) + 1/9*(3*g*p*x^3*e*log(x^2*e + d) - 2*g*
p*x^3*e + 3*g*x^3*e*log(c) + 9*f*p*x*e*log(x^2*e + d) + 6*d*g*p*x - 18*f*p*x*e + 9*f*x*e*log(c))*e^(-1)

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maple [C]  time = 0.49, size = 416, normalized size = 3.56 \[ -\frac {i \pi g \,x^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}{6}+\frac {i \pi g \,x^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}}{6}+\frac {i \pi g \,x^{3} \mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}}{6}-\frac {i \pi g \,x^{3} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}}{6}-\frac {i \pi f x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}{2}+\frac {i \pi f x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}}{2}+\frac {i \pi f x \,\mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}}{2}-\frac {i \pi f x \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}}{2}-\frac {2 g p \,x^{3}}{9}+\frac {g \,x^{3} \ln \relax (c )}{3}+\frac {2 d g p x}{3 e}-2 f p x +f x \ln \relax (c )+\frac {\sqrt {-d e}\, d g p \ln \left (-d -\sqrt {-d e}\, x \right )}{3 e^{2}}-\frac {\sqrt {-d e}\, d g p \ln \left (-d +\sqrt {-d e}\, x \right )}{3 e^{2}}-\frac {\sqrt {-d e}\, f p \ln \left (-d -\sqrt {-d e}\, x \right )}{e}+\frac {\sqrt {-d e}\, f p \ln \left (-d +\sqrt {-d e}\, x \right )}{e}+\left (\frac {1}{3} g \,x^{3}+f x \right ) \ln \left (\left (e \,x^{2}+d \right )^{p}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1/3*g*x^3+f*x)*ln((e*x^2+d)^p)-1/6*I*Pi*g*x^3*csgn(I*c*(e*x^2+d)^p)^3-1/2*I*Pi*f*csgn(I*c*(e*x^2+d)^p)^3*x-1/
6*I*Pi*g*x^3*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)+1/2*I*Pi*f*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)*
x+1/6*I*Pi*g*x^3*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2-1/2*I*Pi*f*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)
^p)*csgn(I*c)*x+1/2*I*Pi*f*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2*x+1/6*I*Pi*g*x^3*csgn(I*c*(e*x^2+d)^p)^
2*csgn(I*c)+1/3*ln(c)*g*x^3-2/9*g*p*x^3+ln(c)*f*x+1/3/e^2*(-d*e)^(1/2)*p*ln(-d-(-d*e)^(1/2)*x)*d*g-1/e*(-d*e)^
(1/2)*p*ln(-d-(-d*e)^(1/2)*x)*f-1/3/e^2*(-d*e)^(1/2)*p*ln(-d+(-d*e)^(1/2)*x)*d*g+1/e*(-d*e)^(1/2)*p*ln(-d+(-d*
e)^(1/2)*x)*f+2/3*d*g*p*x/e-2*f*p*x

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maxima [A]  time = 1.04, size = 85, normalized size = 0.73 \[ \frac {2}{9} \, e p {\left (\frac {3 \, {\left (3 \, d e f - d^{2} g\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{\sqrt {d e} e^{2}} - \frac {e g x^{3} + 3 \, {\left (3 \, e f - d g\right )} x}{e^{2}}\right )} + \frac {1}{3} \, {\left (g x^{3} + 3 \, f x\right )} \log \left ({\left (e x^{2} + d\right )}^{p} c\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.32, size = 97, normalized size = 0.83 \[ \ln \left (c\,{\left (e\,x^2+d\right )}^p\right )\,\left (\frac {g\,x^3}{3}+f\,x\right )-x\,\left (2\,f\,p-\frac {2\,d\,g\,p}{3\,e}\right )-\frac {2\,g\,p\,x^3}{9}-\frac {2\,\sqrt {d}\,p\,\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e}\,p\,x\,\left (d\,g-3\,e\,f\right )}{d^2\,g\,p-3\,d\,e\,f\,p}\right )\,\left (d\,g-3\,e\,f\right )}{3\,e^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 23.52, size = 228, normalized size = 1.95 \[ \begin {cases} - \frac {i d^{\frac {3}{2}} g p \log {\left (d + e x^{2} \right )}}{3 e^{2} \sqrt {\frac {1}{e}}} + \frac {2 i d^{\frac {3}{2}} g p \log {\left (- i \sqrt {d} \sqrt {\frac {1}{e}} + x \right )}}{3 e^{2} \sqrt {\frac {1}{e}}} + \frac {i \sqrt {d} f p \log {\left (d + e x^{2} \right )}}{e \sqrt {\frac {1}{e}}} - \frac {2 i \sqrt {d} f p \log {\left (- i \sqrt {d} \sqrt {\frac {1}{e}} + x \right )}}{e \sqrt {\frac {1}{e}}} + \frac {2 d g p x}{3 e} + f p x \log {\left (d + e x^{2} \right )} - 2 f p x + f x \log {\relax (c )} + \frac {g p x^{3} \log {\left (d + e x^{2} \right )}}{3} - \frac {2 g p x^{3}}{9} + \frac {g x^{3} \log {\relax (c )}}{3} & \text {for}\: e \neq 0 \\\left (f x + \frac {g x^{3}}{3}\right ) \log {\left (c d^{p} \right )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-I*d**(3/2)*g*p*log(d + e*x**2)/(3*e**2*sqrt(1/e)) + 2*I*d**(3/2)*g*p*log(-I*sqrt(d)*sqrt(1/e) + x)
/(3*e**2*sqrt(1/e)) + I*sqrt(d)*f*p*log(d + e*x**2)/(e*sqrt(1/e)) - 2*I*sqrt(d)*f*p*log(-I*sqrt(d)*sqrt(1/e) +
x)/(e*sqrt(1/e)) + 2*d*g*p*x/(3*e) + f*p*x*log(d + e*x**2) - 2*f*p*x + f*x*log(c) + g*p*x**3*log(d + e*x**2)/
3 - 2*g*p*x**3/9 + g*x**3*log(c)/3, Ne(e, 0)), ((f*x + g*x**3/3)*log(c*d**p), True))

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