3.335 \(\int \frac{(f+g x^2)^2 \log (c (d+e x^2)^p)}{x^4} \, dx\)

Optimal. Leaf size=169 \[ -\frac{f^2 \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}-\frac{2 f g \log \left (c \left (d+e x^2\right )^p\right )}{x}+g^2 x \log \left (c \left (d+e x^2\right )^p\right )-\frac{2 e^{3/2} f^2 p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{3 d^{3/2}}-\frac{2 e f^2 p}{3 d x}+\frac{4 \sqrt{e} f g p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{\sqrt{d}}+\frac{2 \sqrt{d} g^2 p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{\sqrt{e}}-2 g^2 p x \]

[Out]

(-2*e*f^2*p)/(3*d*x) - 2*g^2*p*x - (2*e^(3/2)*f^2*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(3*d^(3/2)) + (4*Sqrt[e]*f*g*
p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/Sqrt[d] + (2*Sqrt[d]*g^2*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/Sqrt[e] - (f^2*Log[c*(d
+ e*x^2)^p])/(3*x^3) - (2*f*g*Log[c*(d + e*x^2)^p])/x + g^2*x*Log[c*(d + e*x^2)^p]

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Rubi [A]  time = 0.146754, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {2476, 2448, 321, 205, 2455, 325} \[ -\frac{f^2 \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}-\frac{2 f g \log \left (c \left (d+e x^2\right )^p\right )}{x}+g^2 x \log \left (c \left (d+e x^2\right )^p\right )-\frac{2 e^{3/2} f^2 p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{3 d^{3/2}}-\frac{2 e f^2 p}{3 d x}+\frac{4 \sqrt{e} f g p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{\sqrt{d}}+\frac{2 \sqrt{d} g^2 p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{\sqrt{e}}-2 g^2 p x \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*e*f^2*p)/(3*d*x) - 2*g^2*p*x - (2*e^(3/2)*f^2*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(3*d^(3/2)) + (4*Sqrt[e]*f*g*
p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/Sqrt[d] + (2*Sqrt[d]*g^2*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/Sqrt[e] - (f^2*Log[c*(d
+ e*x^2)^p])/(3*x^3) - (2*f*g*Log[c*(d + e*x^2)^p])/x + g^2*x*Log[c*(d + e*x^2)^p]

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 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 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 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 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]

Rubi steps

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

Mathematica [C]  time = 0.133533, size = 113, normalized size = 0.67 \[ -\frac{\left (f^2+6 f g x^2-3 g^2 x^4\right ) \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}-\frac{2 e f^2 p \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};-\frac{e x^2}{d}\right )}{3 d x}+\frac{2 g p (d g+2 e f) \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{\sqrt{d} \sqrt{e}}-2 g^2 p x \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2*g^2*p*x + (2*g*(2*e*f + d*g)*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(Sqrt[d]*Sqrt[e]) - (2*e*f^2*p*Hypergeometric2F
1[-1/2, 1, 1/2, -((e*x^2)/d)])/(3*d*x) - ((f^2 + 6*f*g*x^2 - 3*g^2*x^4)*Log[c*(d + e*x^2)^p])/(3*x^3)

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Maple [C]  time = 0.805, size = 700, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(-3*g^2*x^4+6*f*g*x^2+f^2)/x^3*ln((e*x^2+d)^p)+1/6*(6*I*Pi*d*f*g*x^2*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+
d)^p)*csgn(I*c)-6*I*Pi*d*f*g*x^2*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2-I*Pi*d*f^2*csgn(I*c*(e*x^2+d)^p)^
2*csgn(I*c)+6*I*Pi*d*f*g*x^2*csgn(I*c*(e*x^2+d)^p)^3+3*I*Pi*d*g^2*x^4*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p
)^2-I*Pi*d*f^2*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2+3*I*Pi*d*g^2*x^4*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)-
6*I*Pi*d*f*g*x^2*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)+6*ln(c)*d*g^2*x^4-3*I*Pi*d*g^2*x^4*csgn(I*c*(e*x^2+d)^p)^3-
3*I*Pi*d*g^2*x^4*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)+I*Pi*d*f^2*csgn(I*c*(e*x^2+d)^p)^3+I*Pi*d
*f^2*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)-12*d*g^2*p*x^4-12*ln(c)*d*f*g*x^2-4*e*f^2*p*x^2+2*sum
(_R*ln((18*d^4*g^4*p^2+72*d^3*e*f*g^3*p^2+60*d^2*e^2*f^2*g^2*p^2-24*d*e^3*f^3*g*p^2+2*e^4*f^4*p^2+3*_R^2*d^3*e
)*x+(-3*d^4*g^2*p-6*d^3*e*f*g*p+d^2*e^2*f^2*p)*_R),_R=RootOf(9*d^4*g^4*p^2+36*d^3*e*f*g^3*p^2+30*d^2*e^2*f^2*g
^2*p^2-12*d*e^3*f^3*g*p^2+e^4*f^4*p^2+_Z^2*d^3*e))*d*x^3-2*ln(c)*d*f^2)/d/x^3

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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)^2*log(c*(e*x^2+d)^p)/x^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.67833, size = 744, normalized size = 4.4 \begin{align*} \left [-\frac{6 \, d^{2} e g^{2} p x^{4} + 2 \, d e^{2} f^{2} p x^{2} -{\left (e^{2} f^{2} - 6 \, d e f g - 3 \, d^{2} g^{2}\right )} \sqrt{-d e} p x^{3} \log \left (\frac{e x^{2} - 2 \, \sqrt{-d e} x - d}{e x^{2} + d}\right ) -{\left (3 \, d^{2} e g^{2} p x^{4} - 6 \, d^{2} e f g p x^{2} - d^{2} e f^{2} p\right )} \log \left (e x^{2} + d\right ) -{\left (3 \, d^{2} e g^{2} x^{4} - 6 \, d^{2} e f g x^{2} - d^{2} e f^{2}\right )} \log \left (c\right )}{3 \, d^{2} e x^{3}}, -\frac{6 \, d^{2} e g^{2} p x^{4} + 2 \, d e^{2} f^{2} p x^{2} + 2 \,{\left (e^{2} f^{2} - 6 \, d e f g - 3 \, d^{2} g^{2}\right )} \sqrt{d e} p x^{3} \arctan \left (\frac{\sqrt{d e} x}{d}\right ) -{\left (3 \, d^{2} e g^{2} p x^{4} - 6 \, d^{2} e f g p x^{2} - d^{2} e f^{2} p\right )} \log \left (e x^{2} + d\right ) -{\left (3 \, d^{2} e g^{2} x^{4} - 6 \, d^{2} e f g x^{2} - d^{2} e f^{2}\right )} \log \left (c\right )}{3 \, d^{2} e x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x^2+f)^2*log(c*(e*x^2+d)^p)/x^4,x, algorithm="fricas")

[Out]

[-1/3*(6*d^2*e*g^2*p*x^4 + 2*d*e^2*f^2*p*x^2 - (e^2*f^2 - 6*d*e*f*g - 3*d^2*g^2)*sqrt(-d*e)*p*x^3*log((e*x^2 -
2*sqrt(-d*e)*x - d)/(e*x^2 + d)) - (3*d^2*e*g^2*p*x^4 - 6*d^2*e*f*g*p*x^2 - d^2*e*f^2*p)*log(e*x^2 + d) - (3*
d^2*e*g^2*x^4 - 6*d^2*e*f*g*x^2 - d^2*e*f^2)*log(c))/(d^2*e*x^3), -1/3*(6*d^2*e*g^2*p*x^4 + 2*d*e^2*f^2*p*x^2
+ 2*(e^2*f^2 - 6*d*e*f*g - 3*d^2*g^2)*sqrt(d*e)*p*x^3*arctan(sqrt(d*e)*x/d) - (3*d^2*e*g^2*p*x^4 - 6*d^2*e*f*g
*p*x^2 - d^2*e*f^2*p)*log(e*x^2 + d) - (3*d^2*e*g^2*x^4 - 6*d^2*e*f*g*x^2 - d^2*e*f^2)*log(c))/(d^2*e*x^3)]

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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)**2*ln(c*(e*x**2+d)**p)/x**4,x)

[Out]

Timed out

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Giac [A]  time = 1.35648, size = 208, normalized size = 1.23 \begin{align*} \frac{2 \,{\left (3 \, d^{2} g^{2} p + 6 \, d f g p e - f^{2} p e^{2}\right )} \arctan \left (\frac{x e^{\frac{1}{2}}}{\sqrt{d}}\right ) e^{\left (-\frac{1}{2}\right )}}{3 \, d^{\frac{3}{2}}} + \frac{3 \, d g^{2} p x^{4} \log \left (x^{2} e + d\right ) - 6 \, d g^{2} p x^{4} + 3 \, d g^{2} x^{4} \log \left (c\right ) - 6 \, d f g p x^{2} \log \left (x^{2} e + d\right ) - 2 \, f^{2} p x^{2} e - 6 \, d f g x^{2} \log \left (c\right ) - d f^{2} p \log \left (x^{2} e + d\right ) - d f^{2} \log \left (c\right )}{3 \, d x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x^2+f)^2*log(c*(e*x^2+d)^p)/x^4,x, algorithm="giac")

[Out]

2/3*(3*d^2*g^2*p + 6*d*f*g*p*e - f^2*p*e^2)*arctan(x*e^(1/2)/sqrt(d))*e^(-1/2)/d^(3/2) + 1/3*(3*d*g^2*p*x^4*lo
g(x^2*e + d) - 6*d*g^2*p*x^4 + 3*d*g^2*x^4*log(c) - 6*d*f*g*p*x^2*log(x^2*e + d) - 2*f^2*p*x^2*e - 6*d*f*g*x^2
*log(c) - d*f^2*p*log(x^2*e + d) - d*f^2*log(c))/(d*x^3)