3.4.20 \(\int \frac {(f+g x^2) \log (c (d+e x^2)^p)}{x^2} \, dx\) [320]

Optimal. Leaf size=72 \[ -2 g p x+\frac {2 (e f+d g) p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \sqrt {e}}-\frac {f \log \left (c \left (d+e x^2\right )^p\right )}{x}+g x \log \left (c \left (d+e x^2\right )^p\right ) \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 93, normalized size of antiderivative = 1.29, number of steps used = 7, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2526, 2498, 327, 211, 2505} \begin {gather*} \frac {2 \sqrt {e} f p \text {ArcTan}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d}}+\frac {2 \sqrt {d} g p \text {ArcTan}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}-\frac {f \log \left (c \left (d+e x^2\right )^p\right )}{x}+g x \log \left (c \left (d+e x^2\right )^p\right )-2 g p x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 2526

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x^n)^p])^q, x^m*(f + g*x^s)^r, x], x] /; FreeQ[{a, b, c,
d, e, f, g, m, n, p, q, r, s}, x] && IGtQ[q, 0] && IntegerQ[m] && IntegerQ[r] && IntegerQ[s]

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 62, normalized size = 0.86 \begin {gather*} -2 g p x+\frac {2 (e f+d g) p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \sqrt {e}}+\left (-\frac {f}{x}+g x\right ) \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^2,x]

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.17, size = 427, normalized size = 5.93

method	result	size
risch	\(-\frac {\left (-g \,x^{2}+f \right ) \ln \left (\left (e \,x^{2}+d \right )^{p}\right )}{x}+\frac {i \pi g \,x^{2} \mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2} d e -i \pi g \,x^{2} \mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \right ) d e -i \pi g \,x^{2} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3} d e +i \pi g \,x^{2} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right ) d e -i \pi d f \,\mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2} e +i \pi d f \,\mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \right ) e +i \pi d f \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3} e -i \pi d f \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right ) e +2 \ln \left (c \right ) g \,x^{2} d e +2 \sqrt {-e d}\, p \ln \left (-\sqrt {-e d}\, x +d \right ) g d x +2 \sqrt {-e d}\, p \ln \left (-\sqrt {-e d}\, x +d \right ) f e x -2 \sqrt {-e d}\, p \ln \left (-\sqrt {-e d}\, x -d \right ) g d x -2 \sqrt {-e d}\, p \ln \left (-\sqrt {-e d}\, x -d \right ) f e x -4 x^{2} d e g p -2 \ln \left (c \right ) d e f}{2 d e x}\)	\(427\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-g*x^2+f)/x*ln((e*x^2+d)^p)+1/2*(I*Pi*g*x^2*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2*d*e-I*Pi*g*x^2*csgn(
I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)*d*e-I*Pi*g*x^2*csgn(I*c*(e*x^2+d)^p)^3*d*e+I*Pi*g*x^2*csgn(I*c*
(e*x^2+d)^p)^2*csgn(I*c)*d*e-I*Pi*d*f*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2*e+I*Pi*d*f*csgn(I*(e*x^2+d)^
p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)*e+I*Pi*d*f*csgn(I*c*(e*x^2+d)^p)^3*e-I*Pi*d*f*csgn(I*c*(e*x^2+d)^p)^2*csgn(
I*c)*e+2*ln(c)*g*x^2*d*e+2*(-e*d)^(1/2)*p*ln(-(-e*d)^(1/2)*x+d)*g*d*x+2*(-e*d)^(1/2)*p*ln(-(-e*d)^(1/2)*x+d)*f
*e*x-2*(-e*d)^(1/2)*p*ln(-(-e*d)^(1/2)*x-d)*g*d*x-2*(-e*d)^(1/2)*p*ln(-(-e*d)^(1/2)*x-d)*f*e*x-4*x^2*d*e*g*p-2
*ln(c)*d*e*f)/d/e/x

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Maxima [A]
time = 0.55, size = 59, normalized size = 0.82 \begin {gather*} -2 \, {\left (g x e^{\left (-1\right )} - \frac {{\left (d g + f e\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {3}{2}\right )}}{\sqrt {d}}\right )} p e + {\left (g x - \frac {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^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.39, size = 208, normalized size = 2.89 \begin {gather*} \left [-\frac {{\left (2 \, d g p x^{2} e - {\left (d g p x^{2} - d f p\right )} e \log \left (x^{2} e + d\right ) - {\left (d g x^{2} - d f\right )} e \log \left (c\right ) + {\left (d g p x + f p x e\right )} \sqrt {-d e} \log \left (\frac {x^{2} e - 2 \, \sqrt {-d e} x - d}{x^{2} e + d}\right )\right )} e^{\left (-1\right )}}{d x}, -\frac {{\left (2 \, d g p x^{2} e - 2 \, {\left (d g p x + f p x e\right )} \sqrt {d} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\frac {1}{2}} - {\left (d g p x^{2} - d f p\right )} e \log \left (x^{2} e + d\right ) - {\left (d g x^{2} - d f\right )} e \log \left (c\right )\right )} e^{\left (-1\right )}}{d x}\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^2,x, algorithm="fricas")

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 204 vs. \(2 (71) = 142\).
time = 16.96, size = 204, normalized size = 2.83 \begin {gather*} \begin {cases} \left (- \frac {f}{x} + g x\right ) \log {\left (0^{p} c \right )} & \text {for}\: d = 0 \wedge e = 0 \\\left (- \frac {f}{x} + g x\right ) \log {\left (c d^{p} \right )} & \text {for}\: e = 0 \\- \frac {2 f p}{x} - \frac {f \log {\left (c \left (e x^{2}\right )^{p} \right )}}{x} - 2 g p x + g x \log {\left (c \left (e x^{2}\right )^{p} \right )} & \text {for}\: d = 0 \\\frac {2 d g p \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{e \sqrt {- \frac {d}{e}}} - \frac {d g \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{e \sqrt {- \frac {d}{e}}} + \frac {2 f p \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{\sqrt {- \frac {d}{e}}} - \frac {f \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{\sqrt {- \frac {d}{e}}} - \frac {f \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{x} - 2 g p x + g x \log {\left (c \left (d + e x^{2}\right )^{p} \right )} & \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**2,x)

[Out]

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

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Giac [A]
time = 3.13, size = 78, normalized size = 1.08 \begin {gather*} \frac {2 \, {\left (d g p + f p e\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {1}{2}\right )}}{\sqrt {d}} + \frac {g p x^{2} \log \left (x^{2} e + d\right ) - 2 \, g p x^{2} + g x^{2} \log \left (c\right ) - f p \log \left (x^{2} e + d\right ) - f \log \left (c\right )}{x} \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^2,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.33, size = 83, normalized size = 1.15 \begin {gather*} \ln \left (c\,{\left (e\,x^2+d\right )}^p\right )\,\left (2\,g\,x-\frac {g\,x^2+f}{x}\right )-2\,g\,p\,x+\frac {2\,p\,\mathrm {atan}\left (\frac {2\,\sqrt {e}\,p\,x\,\left (d\,g+e\,f\right )}{\sqrt {d}\,\left (2\,d\,g\,p+2\,e\,f\,p\right )}\right )\,\left (d\,g+e\,f\right )}{\sqrt {d}\,\sqrt {e}} \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^2,x)

[Out]

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

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