3.322 \(\int \frac{(f+g x^2) \log (c (d+e x^2)^p)}{x^6} \, dx\)

Optimal. Leaf size=140 \[ -\frac{f \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac{g \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}+\frac{2 e^2 f p}{5 d^2 x}+\frac{2 e^{5/2} f p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{5 d^{5/2}}-\frac{2 e^{3/2} g p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{3 d^{3/2}}-\frac{2 e f p}{15 d x^3}-\frac{2 e g p}{3 d x} \]

[Out]

(-2*e*f*p)/(15*d*x^3) + (2*e^2*f*p)/(5*d^2*x) - (2*e*g*p)/(3*d*x) + (2*e^(5/2)*f*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]]
)/(5*d^(5/2)) - (2*e^(3/2)*g*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(3*d^(3/2)) - (f*Log[c*(d + e*x^2)^p])/(5*x^5) - (
g*Log[c*(d + e*x^2)^p])/(3*x^3)

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Rubi [A]  time = 0.121402, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2476, 2455, 325, 205} \[ -\frac{f \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac{g \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}+\frac{2 e^2 f p}{5 d^2 x}+\frac{2 e^{5/2} f p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{5 d^{5/2}}-\frac{2 e^{3/2} g p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{3 d^{3/2}}-\frac{2 e f p}{15 d x^3}-\frac{2 e g p}{3 d x} \]

Antiderivative was successfully verified.

[In]

Int[((f + g*x^2)*Log[c*(d + e*x^2)^p])/x^6,x]

[Out]

(-2*e*f*p)/(15*d*x^3) + (2*e^2*f*p)/(5*d^2*x) - (2*e*g*p)/(3*d*x) + (2*e^(5/2)*f*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]]
)/(5*d^(5/2)) - (2*e^(3/2)*g*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(3*d^(3/2)) - (f*Log[c*(d + e*x^2)^p])/(5*x^5) - (
g*Log[c*(d + e*x^2)^p])/(3*x^3)

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]

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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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]

Rubi steps

\begin{align*} \int \frac{\left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{x^6} \, dx &=\int \left (\frac{f \log \left (c \left (d+e x^2\right )^p\right )}{x^6}+\frac{g \log \left (c \left (d+e x^2\right )^p\right )}{x^4}\right ) \, dx\\ &=f \int \frac{\log \left (c \left (d+e x^2\right )^p\right )}{x^6} \, dx+g \int \frac{\log \left (c \left (d+e x^2\right )^p\right )}{x^4} \, dx\\ &=-\frac{f \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac{g \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}+\frac{1}{5} (2 e f p) \int \frac{1}{x^4 \left (d+e x^2\right )} \, dx+\frac{1}{3} (2 e g p) \int \frac{1}{x^2 \left (d+e x^2\right )} \, dx\\ &=-\frac{2 e f p}{15 d x^3}-\frac{2 e g p}{3 d x}-\frac{f \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac{g \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}-\frac{\left (2 e^2 f p\right ) \int \frac{1}{x^2 \left (d+e x^2\right )} \, dx}{5 d}-\frac{\left (2 e^2 g p\right ) \int \frac{1}{d+e x^2} \, dx}{3 d}\\ &=-\frac{2 e f p}{15 d x^3}+\frac{2 e^2 f p}{5 d^2 x}-\frac{2 e g p}{3 d x}-\frac{2 e^{3/2} g p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{3 d^{3/2}}-\frac{f \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac{g \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}+\frac{\left (2 e^3 f p\right ) \int \frac{1}{d+e x^2} \, dx}{5 d^2}\\ &=-\frac{2 e f p}{15 d x^3}+\frac{2 e^2 f p}{5 d^2 x}-\frac{2 e g p}{3 d x}+\frac{2 e^{5/2} f p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{5 d^{5/2}}-\frac{2 e^{3/2} g p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{3 d^{3/2}}-\frac{f \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac{g \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}\\ \end{align*}

Mathematica [C]  time = 0.0055823, size = 101, normalized size = 0.72 \[ -\frac{f \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac{g \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}-\frac{2 e f p \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};-\frac{e x^2}{d}\right )}{15 d x^3}-\frac{2 e g p \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};-\frac{e x^2}{d}\right )}{3 d x} \]

Antiderivative was successfully verified.

[In]

Integrate[((f + g*x^2)*Log[c*(d + e*x^2)^p])/x^6,x]

[Out]

(-2*e*f*p*Hypergeometric2F1[-3/2, 1, -1/2, -((e*x^2)/d)])/(15*d*x^3) - (2*e*g*p*Hypergeometric2F1[-1/2, 1, 1/2
, -((e*x^2)/d)])/(3*d*x) - (f*Log[c*(d + e*x^2)^p])/(5*x^5) - (g*Log[c*(d + e*x^2)^p])/(3*x^3)

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Maple [C]  time = 0.574, size = 483, normalized size = 3.5 \begin{align*} -{\frac{ \left ( 5\,g{x}^{2}+3\,f \right ) \ln \left ( \left ( e{x}^{2}+d \right ) ^{p} \right ) }{15\,{x}^{5}}}+{\frac{-5\,i\pi \,{d}^{2}g{x}^{2}{\it csgn} \left ( i \left ( e{x}^{2}+d \right ) ^{p} \right ) \left ({\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{2}+5\,i\pi \,{d}^{2}g{x}^{2}{\it csgn} \left ( i \left ( e{x}^{2}+d \right ) ^{p} \right ){\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ){\it csgn} \left ( ic \right ) +5\,i\pi \,{d}^{2}g{x}^{2} \left ({\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{3}-5\,i\pi \,{d}^{2}g{x}^{2} \left ({\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) -3\,i\pi \,f{\it csgn} \left ( i \left ( e{x}^{2}+d \right ) ^{p} \right ) \left ({\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{2}{d}^{2}+3\,i\pi \,{d}^{2}f{\it csgn} \left ( i \left ( e{x}^{2}+d \right ) ^{p} \right ){\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ){\it csgn} \left ( ic \right ) +3\,i\pi \,{d}^{2}f \left ({\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{3}-3\,i\pi \,f \left ({\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ){d}^{2}+2\,\sum _{{\it \_R}={\it RootOf} \left ( 25\,{d}^{2}{e}^{3}{g}^{2}{p}^{2}-30\,d{e}^{4}fg{p}^{2}+9\,{e}^{5}{f}^{2}{p}^{2}+{d}^{5}{{\it \_Z}}^{2} \right ) }{\it \_R}\,\ln \left ( \left ( 50\,{d}^{2}{e}^{3}{g}^{2}{p}^{2}-60\,d{e}^{4}fg{p}^{2}+18\,{e}^{5}{f}^{2}{p}^{2}+3\,{{\it \_R}}^{2}{d}^{5} \right ) x+ \left ( 5\,{d}^{4}gpe-3\,{d}^{3}fp{e}^{2} \right ){\it \_R} \right ){d}^{2}{x}^{5}-20\,degp{x}^{4}+12\,{e}^{2}fp{x}^{4}-10\,\ln \left ( c \right ){d}^{2}g{x}^{2}-4\,defp{x}^{2}-6\,\ln \left ( c \right ){d}^{2}f}{30\,{d}^{2}{x}^{5}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x^2+f)*ln(c*(e*x^2+d)^p)/x^6,x)

[Out]

-1/15*(5*g*x^2+3*f)/x^5*ln((e*x^2+d)^p)+1/30*(-5*I*Pi*d^2*g*x^2*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2+5*
I*Pi*d^2*g*x^2*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)+5*I*Pi*d^2*g*x^2*csgn(I*c*(e*x^2+d)^p)^3-5*
I*Pi*d^2*g*x^2*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)-3*I*Pi*f*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2*d^2+3*I*
Pi*d^2*f*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)+3*I*Pi*d^2*f*csgn(I*c*(e*x^2+d)^p)^3-3*I*Pi*f*csg
n(I*c*(e*x^2+d)^p)^2*csgn(I*c)*d^2+2*sum(_R*ln((50*d^2*e^3*g^2*p^2-60*d*e^4*f*g*p^2+18*e^5*f^2*p^2+3*_R^2*d^5)
*x+(5*d^4*e*g*p-3*d^3*e^2*f*p)*_R),_R=RootOf(25*d^2*e^3*g^2*p^2-30*d*e^4*f*g*p^2+9*e^5*f^2*p^2+_Z^2*d^5))*d^2*
x^5-20*d*e*g*p*x^4+12*e^2*f*p*x^4-10*ln(c)*d^2*g*x^2-4*d*e*f*p*x^2-6*ln(c)*d^2*f)/d^2/x^5

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x^2+f)*log(c*(e*x^2+d)^p)/x^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.93368, size = 583, normalized size = 4.16 \begin{align*} \left [-\frac{{\left (3 \, e^{2} f - 5 \, d e g\right )} p x^{5} \sqrt{-\frac{e}{d}} \log \left (\frac{e x^{2} - 2 \, d x \sqrt{-\frac{e}{d}} - d}{e x^{2} + d}\right ) + 2 \, d e f p x^{2} - 2 \,{\left (3 \, e^{2} f - 5 \, d e g\right )} p x^{4} +{\left (5 \, d^{2} g p x^{2} + 3 \, d^{2} f p\right )} \log \left (e x^{2} + d\right ) +{\left (5 \, d^{2} g x^{2} + 3 \, d^{2} f\right )} \log \left (c\right )}{15 \, d^{2} x^{5}}, \frac{2 \,{\left (3 \, e^{2} f - 5 \, d e g\right )} p x^{5} \sqrt{\frac{e}{d}} \arctan \left (x \sqrt{\frac{e}{d}}\right ) - 2 \, d e f p x^{2} + 2 \,{\left (3 \, e^{2} f - 5 \, d e g\right )} p x^{4} -{\left (5 \, d^{2} g p x^{2} + 3 \, d^{2} f p\right )} \log \left (e x^{2} + d\right ) -{\left (5 \, d^{2} g x^{2} + 3 \, d^{2} f\right )} \log \left (c\right )}{15 \, d^{2} x^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x^2+f)*log(c*(e*x^2+d)^p)/x^6,x, algorithm="fricas")

[Out]

[-1/15*((3*e^2*f - 5*d*e*g)*p*x^5*sqrt(-e/d)*log((e*x^2 - 2*d*x*sqrt(-e/d) - d)/(e*x^2 + d)) + 2*d*e*f*p*x^2 -
2*(3*e^2*f - 5*d*e*g)*p*x^4 + (5*d^2*g*p*x^2 + 3*d^2*f*p)*log(e*x^2 + d) + (5*d^2*g*x^2 + 3*d^2*f)*log(c))/(d
^2*x^5), 1/15*(2*(3*e^2*f - 5*d*e*g)*p*x^5*sqrt(e/d)*arctan(x*sqrt(e/d)) - 2*d*e*f*p*x^2 + 2*(3*e^2*f - 5*d*e*
g)*p*x^4 - (5*d^2*g*p*x^2 + 3*d^2*f*p)*log(e*x^2 + d) - (5*d^2*g*x^2 + 3*d^2*f)*log(c))/(d^2*x^5)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x**2+f)*ln(c*(e*x**2+d)**p)/x**6,x)

[Out]

Timed out

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Giac [A]  time = 1.32981, size = 165, normalized size = 1.18 \begin{align*} -\frac{2 \,{\left (5 \, d g p e^{2} - 3 \, f p e^{3}\right )} \arctan \left (\frac{x e^{\frac{1}{2}}}{\sqrt{d}}\right ) e^{\left (-\frac{1}{2}\right )}}{15 \, d^{\frac{5}{2}}} - \frac{10 \, d g p x^{4} e - 6 \, f p x^{4} e^{2} + 5 \, d^{2} g p x^{2} \log \left (x^{2} e + d\right ) + 2 \, d f p x^{2} e + 5 \, d^{2} g x^{2} \log \left (c\right ) + 3 \, d^{2} f p \log \left (x^{2} e + d\right ) + 3 \, d^{2} f \log \left (c\right )}{15 \, d^{2} x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x^2+f)*log(c*(e*x^2+d)^p)/x^6,x, algorithm="giac")

[Out]

-2/15*(5*d*g*p*e^2 - 3*f*p*e^3)*arctan(x*e^(1/2)/sqrt(d))*e^(-1/2)/d^(5/2) - 1/15*(10*d*g*p*x^4*e - 6*f*p*x^4*
e^2 + 5*d^2*g*p*x^2*log(x^2*e + d) + 2*d*f*p*x^2*e + 5*d^2*g*x^2*log(c) + 3*d^2*f*p*log(x^2*e + d) + 3*d^2*f*l
og(c))/(d^2*x^5)