3.4.19 \(\int (f+g x^2) \log (c (d+e x^2)^p) \, dx\) [319]

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

[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)

________________________________________________________________________________________

Rubi [A]
time = 0.06, 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 = {2521, 2498, 327, 211, 2505, 308} \begin {gather*} -\frac {2 d^{3/2} g p \text {ArcTan}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 e^{3/2}}+\frac {2 \sqrt {d} f p \text {ArcTan}\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 {2 d g p x}{3 e}-2 f p x-\frac {2}{9} g p x^3 \end {gather*}

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 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*} \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*}

________________________________________________________________________________________

Mathematica [A]
time = 0.03, size = 117, normalized size = 1.00 \begin {gather*} -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 {gather*}

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

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.04, size = 416, normalized size = 3.56

method	result	size
risch	\(\left (\frac {1}{3} g \,x^{3}+f x \right ) \ln \left (\left (e \,x^{2}+d \right )^{p}\right )+\frac {i \pi g \,x^{3} \mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}}{6}+\frac {i \pi g \,x^{3} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right )}{6}-\frac {i \pi f \,\mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \right ) x}{2}-\frac {i \pi g \,x^{3} \mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \right )}{6}-\frac {i \pi f \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3} x}{2}-\frac {i \pi g \,x^{3} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}}{6}+\frac {i \pi f \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right ) x}{2}+\frac {i \pi f \,\mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2} x}{2}+\frac {\ln \left (c \right ) g \,x^{3}}{3}-\frac {2 g p \,x^{3}}{9}+\frac {\sqrt {-e d}\, p \ln \left (-\sqrt {-e d}\, x -d \right ) d g}{3 e^{2}}-\frac {\sqrt {-e d}\, p \ln \left (-\sqrt {-e d}\, x -d \right ) f}{e}-\frac {\sqrt {-e d}\, p \ln \left (\sqrt {-e d}\, x -d \right ) d g}{3 e^{2}}+\frac {\sqrt {-e d}\, p \ln \left (\sqrt {-e d}\, x -d \right ) f}{e}+\ln \left (c \right ) f x +\frac {2 d g p x}{3 e}-2 f p x\)	\(416\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.53, size = 82, normalized size = 0.70 \begin {gather*} -\frac {2}{9} \, {\left (\frac {3 \, {\left (d^{2} g - 3 \, d f e\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {5}{2}\right )}}{\sqrt {d}} + {\left (g x^{3} e - 3 \, {\left (d g - 3 \, f e\right )} x\right )} e^{\left (-2\right )}\right )} p e + \frac {1}{3} \, {\left (g x^{3} + 3 \, f x\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((g*x^2+f)*log(c*(e*x^2+d)^p),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.36, size = 217, normalized size = 1.85 \begin {gather*} \left [\frac {1}{9} \, {\left (6 \, d g p x + 3 \, {\left (g p x^{3} + 3 \, f p x\right )} e \log \left (x^{2} e + d\right ) + 3 \, {\left (g x^{3} + 3 \, f x\right )} e \log \left (c\right ) - 3 \, {\left (d g p - 3 \, 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 (g p x^{3} + 9 \, f p x\right )} e\right )} e^{\left (-1\right )}, \frac {1}{9} \, {\left (6 \, d g p x - 6 \, {\left (d g p - 3 \, f p e\right )} \sqrt {d} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {1}{2}\right )} + 3 \, {\left (g p x^{3} + 3 \, f p x\right )} e \log \left (x^{2} e + d\right ) + 3 \, {\left (g x^{3} + 3 \, f x\right )} e \log \left (c\right ) - 2 \, {\left (g p x^{3} + 9 \, f p x\right )} e\right )} e^{\left (-1\right )}\right ] \end {gather*}

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

________________________________________________________________________________________

Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 260 vs. \(2 (121) = 242\).
time = 8.83, size = 260, normalized size = 2.22 \begin {gather*} \begin {cases} \left (f x + \frac {g x^{3}}{3}\right ) \log {\left (0^{p} c \right )} & \text {for}\: d = 0 \wedge e = 0 \\- 2 f p x + f x \log {\left (c \left (e x^{2}\right )^{p} \right )} - \frac {2 g p x^{3}}{9} + \frac {g x^{3} \log {\left (c \left (e x^{2}\right )^{p} \right )}}{3} & \text {for}\: d = 0 \\\left (f x + \frac {g x^{3}}{3}\right ) \log {\left (c d^{p} \right )} & \text {for}\: e = 0 \\- \frac {2 d^{2} g p \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{3 e^{2} \sqrt {- \frac {d}{e}}} + \frac {d^{2} g \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{3 e^{2} \sqrt {- \frac {d}{e}}} + \frac {2 d f p \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{e \sqrt {- \frac {d}{e}}} - \frac {d f \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{e \sqrt {- \frac {d}{e}}} + \frac {2 d g p x}{3 e} - 2 f p x + f x \log {\left (c \left (d + e x^{2}\right )^{p} \right )} - \frac {2 g p x^{3}}{9} + \frac {g x^{3} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{3} & \text {otherwise} \end {cases} \end {gather*}

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(((f*x + g*x**3/3)*log(0**p*c), Eq(d, 0) & Eq(e, 0)), (-2*f*p*x + f*x*log(c*(e*x**2)**p) - 2*g*p*x**3
/9 + g*x**3*log(c*(e*x**2)**p)/3, Eq(d, 0)), ((f*x + g*x**3/3)*log(c*d**p), Eq(e, 0)), (-2*d**2*g*p*log(x - sq
rt(-d/e))/(3*e**2*sqrt(-d/e)) + d**2*g*log(c*(d + e*x**2)**p)/(3*e**2*sqrt(-d/e)) + 2*d*f*p*log(x - sqrt(-d/e)
)/(e*sqrt(-d/e)) - d*f*log(c*(d + e*x**2)**p)/(e*sqrt(-d/e)) + 2*d*g*p*x/(3*e) - 2*f*p*x + f*x*log(c*(d + e*x*
*2)**p) - 2*g*p*x**3/9 + g*x**3*log(c*(d + e*x**2)**p)/3, True))

________________________________________________________________________________________

Giac [A]
time = 5.91, size = 109, normalized size = 0.93 \begin {gather*} -\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 \left (c\right ) + 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 \left (c\right )\right )} e^{\left (-1\right )} \end {gather*}

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)

________________________________________________________________________________________

Mupad [B]
time = 0.00, size = 97, normalized size = 0.83 \begin {gather*} \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}} \end {gather*}

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))

________________________________________________________________________________________