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

Optimal. Leaf size=200 \[ -\frac{f^2 \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac{2 f g \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}-\frac{g^2 \log \left (c \left (d+e x^2\right )^p\right )}{x}+\frac{2 e^2 f^2 p}{5 d^2 x}+\frac{2 e^{5/2} f^2 p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{5 d^{5/2}}-\frac{4 e^{3/2} f g p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{3 d^{3/2}}-\frac{2 e f^2 p}{15 d x^3}-\frac{4 e f g p}{3 d x}+\frac{2 \sqrt{e} g^2 p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{\sqrt{d}} \]

[Out]

(-2*e*f^2*p)/(15*d*x^3) + (2*e^2*f^2*p)/(5*d^2*x) - (4*e*f*g*p)/(3*d*x) + (2*e^(5/2)*f^2*p*ArcTan[(Sqrt[e]*x)/
Sqrt[d]])/(5*d^(5/2)) - (4*e^(3/2)*f*g*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(3*d^(3/2)) + (2*Sqrt[e]*g^2*p*ArcTan[(S
qrt[e]*x)/Sqrt[d]])/Sqrt[d] - (f^2*Log[c*(d + e*x^2)^p])/(5*x^5) - (2*f*g*Log[c*(d + e*x^2)^p])/(3*x^3) - (g^2
*Log[c*(d + e*x^2)^p])/x

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Rubi [A]  time = 0.170722, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2476, 2455, 325, 205} \[ -\frac{f^2 \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac{2 f g \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}-\frac{g^2 \log \left (c \left (d+e x^2\right )^p\right )}{x}+\frac{2 e^2 f^2 p}{5 d^2 x}+\frac{2 e^{5/2} f^2 p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{5 d^{5/2}}-\frac{4 e^{3/2} f g p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{3 d^{3/2}}-\frac{2 e f^2 p}{15 d x^3}-\frac{4 e f g p}{3 d x}+\frac{2 \sqrt{e} g^2 p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{\sqrt{d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*e*f^2*p)/(15*d*x^3) + (2*e^2*f^2*p)/(5*d^2*x) - (4*e*f*g*p)/(3*d*x) + (2*e^(5/2)*f^2*p*ArcTan[(Sqrt[e]*x)/
Sqrt[d]])/(5*d^(5/2)) - (4*e^(3/2)*f*g*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(3*d^(3/2)) + (2*Sqrt[e]*g^2*p*ArcTan[(S
qrt[e]*x)/Sqrt[d]])/Sqrt[d] - (f^2*Log[c*(d + e*x^2)^p])/(5*x^5) - (2*f*g*Log[c*(d + e*x^2)^p])/(3*x^3) - (g^2
*Log[c*(d + e*x^2)^p])/x

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

Mathematica [C]  time = 0.063699, size = 156, normalized size = 0.78 \[ -\frac{f^2 \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac{2 f g \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}-\frac{g^2 \log \left (c \left (d+e x^2\right )^p\right )}{x}-\frac{2 e f^2 p \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};-\frac{e x^2}{d}\right )}{15 d x^3}-\frac{4 e f g p \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};-\frac{e x^2}{d}\right )}{3 d x}+\frac{2 \sqrt{e} g^2 p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{\sqrt{d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[e]*g^2*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/Sqrt[d] - (2*e*f^2*p*Hypergeometric2F1[-3/2, 1, -1/2, -((e*x^2)/
d)])/(15*d*x^3) - (4*e*f*g*p*Hypergeometric2F1[-1/2, 1, 1/2, -((e*x^2)/d)])/(3*d*x) - (f^2*Log[c*(d + e*x^2)^p
])/(5*x^5) - (2*f*g*Log[c*(d + e*x^2)^p])/(3*x^3) - (g^2*Log[c*(d + e*x^2)^p])/x

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Maple [C]  time = 0.594, size = 753, normalized size = 3.8 \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^6,x)

[Out]

-1/15*(15*g^2*x^4+10*f*g*x^2+3*f^2)/x^5*ln((e*x^2+d)^p)+1/30*(15*I*Pi*d^2*g^2*x^4*csgn(I*c*(e*x^2+d)^p)^3-3*I*
Pi*f^2*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2*d^2+3*I*Pi*f^2*csgn(I*c*(e*x^2+d)^p)^3*d^2+10*I*Pi*d^2*f*g*
x^2*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)-15*I*Pi*d^2*g^2*x^4*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^
2+d)^p)^2+10*I*Pi*d^2*f*g*x^2*csgn(I*c*(e*x^2+d)^p)^3-10*I*Pi*d^2*f*g*x^2*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)-10
*I*Pi*d^2*f*g*x^2*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2-30*ln(c)*d^2*g^2*x^4-3*I*Pi*f^2*csgn(I*c*(e*x^2+
d)^p)^2*csgn(I*c)*d^2+15*I*Pi*d^2*g^2*x^4*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)-15*I*Pi*d^2*g^2*
x^4*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)+3*I*Pi*f^2*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)*d^2-40*d*
e*f*g*p*x^4+12*e^2*f^2*p*x^4+2*sum(_R*ln((450*d^4*e*g^4*p^2-600*d^3*e^2*f*g^3*p^2+380*d^2*e^3*f^2*g^2*p^2-120*
d*e^4*f^3*g*p^2+18*e^5*f^4*p^2+3*_R^2*d^5)*x+(-15*d^5*g^2*p+10*d^4*e*f*g*p-3*d^3*e^2*f^2*p)*_R),_R=RootOf(225*
d^4*e*g^4*p^2-300*d^3*e^2*f*g^3*p^2+190*d^2*e^3*f^2*g^2*p^2-60*d*e^4*f^3*g*p^2+9*e^5*f^4*p^2+_Z^2*d^5))*d^2*x^
5-20*ln(c)*d^2*f*g*x^2-4*d*e*f^2*p*x^2-6*ln(c)*f^2*d^2)/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)^2*log(c*(e*x^2+d)^p)/x^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.88557, size = 774, normalized size = 3.87 \begin{align*} \left [\frac{{\left (3 \, e^{2} f^{2} - 10 \, d e f g + 15 \, d^{2} g^{2}\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^{2} p x^{2} + 2 \,{\left (3 \, e^{2} f^{2} - 10 \, d e f g\right )} p x^{4} -{\left (15 \, d^{2} g^{2} p x^{4} + 10 \, d^{2} f g p x^{2} + 3 \, d^{2} f^{2} p\right )} \log \left (e x^{2} + d\right ) -{\left (15 \, d^{2} g^{2} x^{4} + 10 \, d^{2} f g x^{2} + 3 \, d^{2} f^{2}\right )} \log \left (c\right )}{15 \, d^{2} x^{5}}, \frac{2 \,{\left (3 \, e^{2} f^{2} - 10 \, d e f g + 15 \, d^{2} g^{2}\right )} p x^{5} \sqrt{\frac{e}{d}} \arctan \left (x \sqrt{\frac{e}{d}}\right ) - 2 \, d e f^{2} p x^{2} + 2 \,{\left (3 \, e^{2} f^{2} - 10 \, d e f g\right )} p x^{4} -{\left (15 \, d^{2} g^{2} p x^{4} + 10 \, d^{2} f g p x^{2} + 3 \, d^{2} f^{2} p\right )} \log \left (e x^{2} + d\right ) -{\left (15 \, d^{2} g^{2} x^{4} + 10 \, d^{2} f g x^{2} + 3 \, d^{2} f^{2}\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)^2*log(c*(e*x^2+d)^p)/x^6,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.31353, size = 244, normalized size = 1.22 \begin{align*} \frac{2 \,{\left (15 \, d^{2} g^{2} p e - 10 \, d f g p e^{2} + 3 \, f^{2} 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{15 \, d^{2} g^{2} p x^{4} \log \left (x^{2} e + d\right ) + 20 \, d f g p x^{4} e + 15 \, d^{2} g^{2} x^{4} \log \left (c\right ) - 6 \, f^{2} p x^{4} e^{2} + 10 \, d^{2} f g p x^{2} \log \left (x^{2} e + d\right ) + 2 \, d f^{2} p x^{2} e + 10 \, d^{2} f g x^{2} \log \left (c\right ) + 3 \, d^{2} f^{2} p \log \left (x^{2} e + d\right ) + 3 \, d^{2} f^{2} \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)^2*log(c*(e*x^2+d)^p)/x^6,x, algorithm="giac")

[Out]

2/15*(15*d^2*g^2*p*e - 10*d*f*g*p*e^2 + 3*f^2*p*e^3)*arctan(x*e^(1/2)/sqrt(d))*e^(-1/2)/d^(5/2) - 1/15*(15*d^2
*g^2*p*x^4*log(x^2*e + d) + 20*d*f*g*p*x^4*e + 15*d^2*g^2*x^4*log(c) - 6*f^2*p*x^4*e^2 + 10*d^2*f*g*p*x^2*log(
x^2*e + d) + 2*d*f^2*p*x^2*e + 10*d^2*f*g*x^2*log(c) + 3*d^2*f^2*p*log(x^2*e + d) + 3*d^2*f^2*log(c))/(d^2*x^5
)