3.315 \(\int \frac{(f+g x^2) \log (c (d+e x^2)^p)}{x^5} \, dx\)

Optimal. Leaf size=93 \[ -\frac{\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{4 f x^4}+\frac{p (e f-d g)^2 \log \left (d+e x^2\right )}{4 d^2 f}-\frac{e p \log (x) (e f-2 d g)}{2 d^2}-\frac{e f p}{4 d x^2} \]

[Out]

-(e*f*p)/(4*d*x^2) - (e*(e*f - 2*d*g)*p*Log[x])/(2*d^2) + ((e*f - d*g)^2*p*Log[d + e*x^2])/(4*d^2*f) - ((f + g
*x^2)^2*Log[c*(d + e*x^2)^p])/(4*f*x^4)

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Rubi [A]  time = 0.137383, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {2475, 37, 2414, 12, 88} \[ -\frac{\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{4 f x^4}+\frac{p (e f-d g)^2 \log \left (d+e x^2\right )}{4 d^2 f}-\frac{e p \log (x) (e f-2 d g)}{2 d^2}-\frac{e f p}{4 d x^2} \]

Antiderivative was successfully verified.

[In]

Int[((f + g*x^2)*Log[c*(d + e*x^2)^p])/x^5,x]

[Out]

-(e*f*p)/(4*d*x^2) - (e*(e*f - 2*d*g)*p*Log[x])/(2*d^2) + ((e*f - d*g)^2*p*Log[d + e*x^2])/(4*d^2*f) - ((f + g
*x^2)^2*Log[c*(d + e*x^2)^p])/(4*f*x^4)

Rule 2475

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q,
x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && IntegerQ[r] && IntegerQ[s/n] && Intege
rQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0])

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 2414

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*(x_)^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = IntHide[x^m*(f + g*x^r)^q, x]}, Dist[a + b*Log[c*(d + e*x)^n], u, x] - Dist[b*e*n, Int[SimplifyI
ntegrand[u/(d + e*x), x], x], x] /; InverseFunctionFreeQ[u, x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, q, r}, x]
&& IntegerQ[m] && IntegerQ[q] && IntegerQ[r]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

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

Mathematica [A]  time = 0.0378179, size = 105, normalized size = 1.13 \[ -\frac{f \log \left (c \left (d+e x^2\right )^p\right )}{4 x^4}-\frac{g \log \left (c \left (d+e x^2\right )^p\right )}{2 x^2}+\frac{1}{4} e f p \left (\frac{e \log \left (d+e x^2\right )}{d^2}-\frac{2 e \log (x)}{d^2}-\frac{1}{d x^2}\right )-\frac{e g p \log \left (d+e x^2\right )}{2 d}+\frac{e g p \log (x)}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[((f + g*x^2)*Log[c*(d + e*x^2)^p])/x^5,x]

[Out]

(e*g*p*Log[x])/d - (e*g*p*Log[d + e*x^2])/(2*d) + (e*f*p*(-(1/(d*x^2)) - (2*e*Log[x])/d^2 + (e*Log[d + e*x^2])
/d^2))/4 - (f*Log[c*(d + e*x^2)^p])/(4*x^4) - (g*Log[c*(d + e*x^2)^p])/(2*x^2)

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Maple [C]  time = 0.373, size = 392, normalized size = 4.2 \begin{align*} -{\frac{ \left ( 2\,g{x}^{2}+f \right ) \ln \left ( \left ( e{x}^{2}+d \right ) ^{p} \right ) }{4\,{x}^{4}}}-{\frac{2\,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}-2\,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 ) -2\,i\pi \,{d}^{2}g{x}^{2} \left ({\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{3}+2\,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 ) -8\,\ln \left ( x \right ) degp{x}^{4}+4\,\ln \left ( x \right ){e}^{2}fp{x}^{4}+4\,\ln \left ( e{x}^{2}+d \right ) degp{x}^{4}-2\,\ln \left ( e{x}^{2}+d \right ){e}^{2}fp{x}^{4}+i\pi \,{d}^{2}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}-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 ) -i\pi \,{d}^{2}f \left ({\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{3}+i\pi \,{d}^{2}f \left ({\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) +4\,\ln \left ( c \right ){d}^{2}g{x}^{2}+2\,defp{x}^{2}+2\,\ln \left ( c \right ){d}^{2}f}{8\,{d}^{2}{x}^{4}}} \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^5,x)

[Out]

-1/4*(2*g*x^2+f)/x^4*ln((e*x^2+d)^p)-1/8*(2*I*Pi*d^2*g*x^2*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2-2*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)-2*I*Pi*d^2*g*x^2*csgn(I*c*(e*x^2+d)^p)^3+2*I*Pi*
d^2*g*x^2*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)-8*ln(x)*d*e*g*p*x^4+4*ln(x)*e^2*f*p*x^4+4*ln(e*x^2+d)*d*e*g*p*x^4-
2*ln(e*x^2+d)*e^2*f*p*x^4+I*Pi*d^2*f*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2-I*Pi*d^2*f*csgn(I*(e*x^2+d)^p
)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)-I*Pi*d^2*f*csgn(I*c*(e*x^2+d)^p)^3+I*Pi*d^2*f*csgn(I*c*(e*x^2+d)^p)^2*csgn(I
*c)+4*ln(c)*d^2*g*x^2+2*d*e*f*p*x^2+2*ln(c)*d^2*f)/d^2/x^4

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Maxima [A]  time = 1.01961, size = 104, normalized size = 1.12 \begin{align*} \frac{1}{4} \, e p{\left (\frac{{\left (e f - 2 \, d g\right )} \log \left (e x^{2} + d\right )}{d^{2}} - \frac{{\left (e f - 2 \, d g\right )} \log \left (x^{2}\right )}{d^{2}} - \frac{f}{d x^{2}}\right )} - \frac{{\left (2 \, g x^{2} + f\right )} \log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{4 \, x^{4}} \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^5,x, algorithm="maxima")

[Out]

1/4*e*p*((e*f - 2*d*g)*log(e*x^2 + d)/d^2 - (e*f - 2*d*g)*log(x^2)/d^2 - f/(d*x^2)) - 1/4*(2*g*x^2 + f)*log((e
*x^2 + d)^p*c)/x^4

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Fricas [A]  time = 2.09917, size = 223, normalized size = 2.4 \begin{align*} -\frac{2 \,{\left (e^{2} f - 2 \, d e g\right )} p x^{4} \log \left (x\right ) + d e f p x^{2} +{\left (2 \, d^{2} g p x^{2} -{\left (e^{2} f - 2 \, d e g\right )} p x^{4} + d^{2} f p\right )} \log \left (e x^{2} + d\right ) +{\left (2 \, d^{2} g x^{2} + d^{2} f\right )} \log \left (c\right )}{4 \, d^{2} x^{4}} \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^5,x, algorithm="fricas")

[Out]

-1/4*(2*(e^2*f - 2*d*e*g)*p*x^4*log(x) + d*e*f*p*x^2 + (2*d^2*g*p*x^2 - (e^2*f - 2*d*e*g)*p*x^4 + d^2*f*p)*log
(e*x^2 + d) + (2*d^2*g*x^2 + d^2*f)*log(c))/(d^2*x^4)

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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**5,x)

[Out]

Timed out

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Giac [B]  time = 1.26617, size = 435, normalized size = 4.68 \begin{align*} -\frac{{\left (2 \,{\left (x^{2} e + d\right )}^{2} d g p e^{2} \log \left (x^{2} e + d\right ) - 2 \,{\left (x^{2} e + d\right )} d^{2} g p e^{2} \log \left (x^{2} e + d\right ) - 2 \,{\left (x^{2} e + d\right )}^{2} d g p e^{2} \log \left (x^{2} e\right ) + 4 \,{\left (x^{2} e + d\right )} d^{2} g p e^{2} \log \left (x^{2} e\right ) - 2 \, d^{3} g p e^{2} \log \left (x^{2} e\right ) -{\left (x^{2} e + d\right )}^{2} f p e^{3} \log \left (x^{2} e + d\right ) + 2 \,{\left (x^{2} e + d\right )} d f p e^{3} \log \left (x^{2} e + d\right ) +{\left (x^{2} e + d\right )}^{2} f p e^{3} \log \left (x^{2} e\right ) - 2 \,{\left (x^{2} e + d\right )} d f p e^{3} \log \left (x^{2} e\right ) + d^{2} f p e^{3} \log \left (x^{2} e\right ) + 2 \,{\left (x^{2} e + d\right )} d^{2} g e^{2} \log \left (c\right ) - 2 \, d^{3} g e^{2} \log \left (c\right ) +{\left (x^{2} e + d\right )} d f p e^{3} - d^{2} f p e^{3} + d^{2} f e^{3} \log \left (c\right )\right )} e^{\left (-1\right )}}{4 \,{\left ({\left (x^{2} e + d\right )}^{2} d^{2} - 2 \,{\left (x^{2} e + d\right )} d^{3} + d^{4}\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^5,x, algorithm="giac")

[Out]

-1/4*(2*(x^2*e + d)^2*d*g*p*e^2*log(x^2*e + d) - 2*(x^2*e + d)*d^2*g*p*e^2*log(x^2*e + d) - 2*(x^2*e + d)^2*d*
g*p*e^2*log(x^2*e) + 4*(x^2*e + d)*d^2*g*p*e^2*log(x^2*e) - 2*d^3*g*p*e^2*log(x^2*e) - (x^2*e + d)^2*f*p*e^3*l
og(x^2*e + d) + 2*(x^2*e + d)*d*f*p*e^3*log(x^2*e + d) + (x^2*e + d)^2*f*p*e^3*log(x^2*e) - 2*(x^2*e + d)*d*f*
p*e^3*log(x^2*e) + d^2*f*p*e^3*log(x^2*e) + 2*(x^2*e + d)*d^2*g*e^2*log(c) - 2*d^3*g*e^2*log(c) + (x^2*e + d)*
d*f*p*e^3 - d^2*f*p*e^3 + d^2*f*e^3*log(c))*e^(-1)/((x^2*e + d)^2*d^2 - 2*(x^2*e + d)*d^3 + d^4)