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]

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

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Rubi [A]  time = 0.14, antiderivative size = 93, normalized size of antiderivative = 1.00, 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 12

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

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 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]))

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 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])

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*}

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Mathematica [A]  time = 0.04, 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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fricas [A]  time = 0.68, size = 97, normalized size = 1.04 \[ -\frac {2 \, {\left (e^{2} f - 2 \, d e g\right )} p x^{4} \log \relax (x) + d e f p x^{2} + {\left (2 \, d^{2} g p x^{2} - {\left (e^{2} f - 2 \, d e g\right )} p x^{4} + d^{2} f p\right )} \log \left (e x^{2} + d\right ) + {\left (2 \, d^{2} g x^{2} + d^{2} f\right )} \log \relax (c)}{4 \, d^{2} x^{4}} \]

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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giac [B]  time = 0.20, size = 322, normalized size = 3.46 \[ -\frac {{\left (2 \, {\left (x^{2} e + d\right )}^{2} d g p e^{2} \log \left (x^{2} e + d\right ) - 2 \, {\left (x^{2} e + d\right )} d^{2} g p e^{2} \log \left (x^{2} e + d\right ) - 2 \, {\left (x^{2} e + d\right )}^{2} d g p e^{2} \log \left (x^{2} e\right ) + 4 \, {\left (x^{2} e + d\right )} d^{2} g p e^{2} \log \left (x^{2} e\right ) - 2 \, d^{3} g p e^{2} \log \left (x^{2} e\right ) - {\left (x^{2} e + d\right )}^{2} f p e^{3} \log \left (x^{2} e + d\right ) + 2 \, {\left (x^{2} e + d\right )} d f p e^{3} \log \left (x^{2} e + d\right ) + {\left (x^{2} e + d\right )}^{2} f p e^{3} \log \left (x^{2} e\right ) - 2 \, {\left (x^{2} e + d\right )} d f p e^{3} \log \left (x^{2} e\right ) + d^{2} f p e^{3} \log \left (x^{2} e\right ) + 2 \, {\left (x^{2} e + d\right )} d^{2} g e^{2} \log \relax (c) - 2 \, d^{3} g e^{2} \log \relax (c) + {\left (x^{2} e + d\right )} d f p e^{3} - d^{2} f p e^{3} + d^{2} f e^{3} \log \relax (c)\right )} e^{\left (-1\right )}}{4 \, {\left ({\left (x^{2} e + d\right )}^{2} d^{2} - 2 \, {\left (x^{2} e + d\right )} d^{3} + d^{4}\right )}} \]

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)

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maple [C]  time = 0.40, size = 392, normalized size = 4.22 \[ -\frac {\left (2 g \,x^{2}+f \right ) \ln \left (\left (e \,x^{2}+d \right )^{p}\right )}{4 x^{4}}-\frac {-8 d e g p \,x^{4} \ln \relax (x )+4 d e g p \,x^{4} \ln \left (e \,x^{2}+d \right )+4 e^{2} f p \,x^{4} \ln \relax (x )-2 e^{2} f p \,x^{4} \ln \left (e \,x^{2}+d \right )-2 i \pi \,d^{2} g \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )+2 i \pi \,d^{2} g \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}+2 i \pi \,d^{2} g \,x^{2} \mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}-2 i \pi \,d^{2} g \,x^{2} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}-i \pi \,d^{2} f \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )+i \pi \,d^{2} f \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}+i \pi \,d^{2} f \,\mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}-i \pi \,d^{2} f \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}+4 d^{2} g \,x^{2} \ln \relax (c )+2 d e f p \,x^{2}+2 d^{2} f \ln \relax (c )}{8 d^{2} x^{4}} \]

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 = 0.47, size = 77, normalized size = 0.83 \[ \frac {1}{4} \, e p {\left (\frac {{\left (e f - 2 \, d g\right )} \log \left (e x^{2} + d\right )}{d^{2}} - \frac {{\left (e f - 2 \, d g\right )} \log \left (x^{2}\right )}{d^{2}} - \frac {f}{d x^{2}}\right )} - \frac {{\left (2 \, g x^{2} + f\right )} \log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{4 \, x^{4}} \]

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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mupad [B]  time = 0.37, size = 85, normalized size = 0.91 \[ \frac {\ln \left (e\,x^2+d\right )\,\left (e^2\,f\,p-2\,d\,e\,g\,p\right )}{4\,d^2}-\frac {\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )\,\left (\frac {g\,x^2}{2}+\frac {f}{4}\right )}{x^4}-\frac {\ln \relax (x)\,\left (e^2\,f\,p-2\,d\,e\,g\,p\right )}{2\,d^2}-\frac {e\,f\,p}{4\,d\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(c*(d + e*x^2)^p)*(f + g*x^2))/x^5,x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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