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

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

[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 = 0.08, 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 = {2476, 2448, 321, 205, 2455} \[ -\frac {f \log \left (c \left (d+e x^2\right )^p\right )}{x}+g x \log \left (c \left (d+e x^2\right )^p\right )+\frac {2 \sqrt {e} f p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d}}+\frac {2 \sqrt {d} g p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}-2 g p x \]

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 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 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 2476

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 = 0.05, size = 62, normalized size = 0.86 \[ \left (g x-\frac {f}{x}\right ) \log \left (c \left (d+e x^2\right )^p\right )+\frac {2 p (d g+e f) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \sqrt {e}}-2 g p x \]

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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fricas [A]  time = 0.71, size = 199, normalized size = 2.76 \[ \left [-\frac {2 \, d e g p x^{2} + \sqrt {-d e} {\left (e f + d g\right )} p x \log \left (\frac {e x^{2} - 2 \, \sqrt {-d e} x - d}{e x^{2} + d}\right ) - {\left (d e g p x^{2} - d e f p\right )} \log \left (e x^{2} + d\right ) - {\left (d e g x^{2} - d e f\right )} \log \relax (c)}{d e x}, -\frac {2 \, d e g p x^{2} - 2 \, \sqrt {d e} {\left (e f + d g\right )} p x \arctan \left (\frac {\sqrt {d e} x}{d}\right ) - {\left (d e g p x^{2} - d e f p\right )} \log \left (e x^{2} + d\right ) - {\left (d e g x^{2} - d e f\right )} \log \relax (c)}{d e x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-(2*d*e*g*p*x^2 + sqrt(-d*e)*(e*f + d*g)*p*x*log((e*x^2 - 2*sqrt(-d*e)*x - d)/(e*x^2 + d)) - (d*e*g*p*x^2 - d
*e*f*p)*log(e*x^2 + d) - (d*e*g*x^2 - d*e*f)*log(c))/(d*e*x), -(2*d*e*g*p*x^2 - 2*sqrt(d*e)*(e*f + d*g)*p*x*ar
ctan(sqrt(d*e)*x/d) - (d*e*g*p*x^2 - d*e*f*p)*log(e*x^2 + d) - (d*e*g*x^2 - d*e*f)*log(c))/(d*e*x)]

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giac [A]  time = 0.19, size = 78, normalized size = 1.08 \[ \frac {2 \, {\left (d g p + f p e\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {1}{2}\right )}}{\sqrt {d}} + \frac {g p x^{2} \log \left (x^{2} e + d\right ) - 2 \, g p x^{2} + g x^{2} \log \relax (c) - f p \log \left (x^{2} e + d\right ) - f \log \relax (c)}{x} \]

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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maple [C]  time = 0.77, size = 403, normalized size = 5.60 \[ -\frac {\left (-g \,x^{2}+f \right ) \ln \left (\left (e \,x^{2}+d \right )^{p}\right )}{x}+\frac {-i \pi g \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )+i \pi g \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}+i \pi g \,x^{2} \mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}-i \pi g \,x^{2} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}+i \pi f \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )-i \pi f \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}-i \pi f \,\mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}+i \pi f \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}-4 g p \,x^{2}+2 g \,x^{2} \ln \relax (c )-2 f \ln \relax (c )+2 x \RootOf \left (d^{2} g^{2} p^{2}+2 d e f g \,p^{2}+e^{2} f^{2} p^{2}+d \,\textit {\_Z}^{2} e \right ) \ln \left (\left (-d^{2} g p -d e f p \right ) \RootOf \left (d^{2} g^{2} p^{2}+2 d e f g \,p^{2}+e^{2} f^{2} p^{2}+d \,\textit {\_Z}^{2} e \right )+\left (2 d^{2} g^{2} p^{2}+4 d e f g \,p^{2}+2 e^{2} f^{2} p^{2}+3 \RootOf \left (d^{2} g^{2} p^{2}+2 d e f g \,p^{2}+e^{2} f^{2} p^{2}+d \,\textit {\_Z}^{2} e \right )^{2} d e \right ) x \right )}{2 x} \]

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)

[Out]

-(-g*x^2+f)/x*ln((e*x^2+d)^p)+1/2*(I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2*g*x^2-I*Pi*g*x^2*csgn(I*(e
*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)-I*Pi*g*x^2*csgn(I*c*(e*x^2+d)^p)^3+I*Pi*g*x^2*csgn(I*c*(e*x^2+d)^p)
^2*csgn(I*c)-I*Pi*f*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2+I*Pi*f*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^
p)*csgn(I*c)+I*Pi*f*csgn(I*c*(e*x^2+d)^p)^3-I*Pi*f*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)+2*g*x^2*ln(c)-4*g*p*x^2-2
*ln(c)*f+2*sum(_R*ln((2*d^2*g^2*p^2+4*d*e*f*g*p^2+2*e^2*f^2*p^2+3*_R^2*d*e)*x+(-d^2*g*p-d*e*f*p)*_R),_R=RootOf
(d^2*g^2*p^2+2*d*e*f*g*p^2+e^2*f^2*p^2+_Z^2*d*e))*x)/x

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maxima [A]  time = 0.99, size = 61, normalized size = 0.85 \[ -2 \, e p {\left (\frac {g x}{e} - \frac {{\left (e f + d g\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{\sqrt {d e} e}\right )} + {\left (g x - \frac {f}{x}\right )} \log \left ({\left (e x^{2} + d\right )}^{p} c\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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sympy [A]  time = 46.00, size = 262, normalized size = 3.64 \[ \begin {cases} \left (- \frac {f}{x} + g x\right ) \log {\left (0^{p} c \right )} & \text {for}\: d = 0 \wedge e = 0 \\- \frac {f p \log {\relax (e )}}{x} - \frac {2 f p \log {\relax (x )}}{x} - \frac {2 f p}{x} - \frac {f \log {\relax (c )}}{x} + g p x \log {\relax (e )} + 2 g p x \log {\relax (x )} - 2 g p x + g x \log {\relax (c )} & \text {for}\: d = 0 \\\left (- \frac {f}{x} + g x\right ) \log {\left (c d^{p} \right )} & \text {for}\: e = 0 \\\frac {i \sqrt {d} g p \log {\left (d + e x^{2} \right )}}{e \sqrt {\frac {1}{e}}} - \frac {2 i \sqrt {d} g p \log {\left (- i \sqrt {d} \sqrt {\frac {1}{e}} + x \right )}}{e \sqrt {\frac {1}{e}}} - \frac {f p \log {\left (d + e x^{2} \right )}}{x} - \frac {f \log {\relax (c )}}{x} + g p x \log {\left (d + e x^{2} \right )} - 2 g p x + g x \log {\relax (c )} + \frac {i f p \log {\left (d + e x^{2} \right )}}{\sqrt {d} \sqrt {\frac {1}{e}}} - \frac {2 i f p \log {\left (- i \sqrt {d} \sqrt {\frac {1}{e}} + x \right )}}{\sqrt {d} \sqrt {\frac {1}{e}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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