3.337 \(\int \frac{(f+g x^2)^2 \log (c (d+e x^2)^p)}{x^8} \, dx\)

Optimal. Leaf size=252 \[ -\frac{f^2 \log \left (c \left (d+e x^2\right )^p\right )}{7 x^7}-\frac{2 f g \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac{g^2 \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}+\frac{2 e^2 f^2 p}{21 d^2 x^3}-\frac{2 e^3 f^2 p}{7 d^3 x}-\frac{2 e^{7/2} f^2 p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{7 d^{7/2}}+\frac{4 e^2 f g p}{5 d^2 x}+\frac{4 e^{5/2} f g p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{5 d^{5/2}}-\frac{2 e^{3/2} g^2 p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{3 d^{3/2}}-\frac{2 e f^2 p}{35 d x^5}-\frac{4 e f g p}{15 d x^3}-\frac{2 e g^2 p}{3 d x} \]

[Out]

(-2*e*f^2*p)/(35*d*x^5) + (2*e^2*f^2*p)/(21*d^2*x^3) - (4*e*f*g*p)/(15*d*x^3) - (2*e^3*f^2*p)/(7*d^3*x) + (4*e
^2*f*g*p)/(5*d^2*x) - (2*e*g^2*p)/(3*d*x) - (2*e^(7/2)*f^2*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(7*d^(7/2)) + (4*e^(
5/2)*f*g*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(5*d^(5/2)) - (2*e^(3/2)*g^2*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(3*d^(3/2)
) - (f^2*Log[c*(d + e*x^2)^p])/(7*x^7) - (2*f*g*Log[c*(d + e*x^2)^p])/(5*x^5) - (g^2*Log[c*(d + e*x^2)^p])/(3*
x^3)

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Rubi [A]  time = 0.206453, antiderivative size = 252, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2476, 2455, 325, 205} \[ -\frac{f^2 \log \left (c \left (d+e x^2\right )^p\right )}{7 x^7}-\frac{2 f g \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac{g^2 \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}+\frac{2 e^2 f^2 p}{21 d^2 x^3}-\frac{2 e^3 f^2 p}{7 d^3 x}-\frac{2 e^{7/2} f^2 p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{7 d^{7/2}}+\frac{4 e^2 f g p}{5 d^2 x}+\frac{4 e^{5/2} f g p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{5 d^{5/2}}-\frac{2 e^{3/2} g^2 p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{3 d^{3/2}}-\frac{2 e f^2 p}{35 d x^5}-\frac{4 e f g p}{15 d x^3}-\frac{2 e g^2 p}{3 d x} \]

Antiderivative was successfully verified.

[In]

Int[((f + g*x^2)^2*Log[c*(d + e*x^2)^p])/x^8,x]

[Out]

(-2*e*f^2*p)/(35*d*x^5) + (2*e^2*f^2*p)/(21*d^2*x^3) - (4*e*f*g*p)/(15*d*x^3) - (2*e^3*f^2*p)/(7*d^3*x) + (4*e
^2*f*g*p)/(5*d^2*x) - (2*e*g^2*p)/(3*d*x) - (2*e^(7/2)*f^2*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(7*d^(7/2)) + (4*e^(
5/2)*f*g*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(5*d^(5/2)) - (2*e^(3/2)*g^2*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(3*d^(3/2)
) - (f^2*Log[c*(d + e*x^2)^p])/(7*x^7) - (2*f*g*Log[c*(d + e*x^2)^p])/(5*x^5) - (g^2*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 )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^8} \, dx &=\int \left (\frac{f^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^8}+\frac{2 f g \log \left (c \left (d+e x^2\right )^p\right )}{x^6}+\frac{g^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^4}\right ) \, dx\\ &=f^2 \int \frac{\log \left (c \left (d+e x^2\right )^p\right )}{x^8} \, dx+(2 f g) \int \frac{\log \left (c \left (d+e x^2\right )^p\right )}{x^6} \, dx+g^2 \int \frac{\log \left (c \left (d+e x^2\right )^p\right )}{x^4} \, dx\\ &=-\frac{f^2 \log \left (c \left (d+e x^2\right )^p\right )}{7 x^7}-\frac{2 f g \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac{g^2 \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}+\frac{1}{7} \left (2 e f^2 p\right ) \int \frac{1}{x^6 \left (d+e x^2\right )} \, dx+\frac{1}{5} (4 e f g p) \int \frac{1}{x^4 \left (d+e x^2\right )} \, dx+\frac{1}{3} \left (2 e g^2 p\right ) \int \frac{1}{x^2 \left (d+e x^2\right )} \, dx\\ &=-\frac{2 e f^2 p}{35 d x^5}-\frac{4 e f g p}{15 d x^3}-\frac{2 e g^2 p}{3 d x}-\frac{f^2 \log \left (c \left (d+e x^2\right )^p\right )}{7 x^7}-\frac{2 f g \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac{g^2 \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}-\frac{\left (2 e^2 f^2 p\right ) \int \frac{1}{x^4 \left (d+e x^2\right )} \, dx}{7 d}-\frac{\left (4 e^2 f g p\right ) \int \frac{1}{x^2 \left (d+e x^2\right )} \, dx}{5 d}-\frac{\left (2 e^2 g^2 p\right ) \int \frac{1}{d+e x^2} \, dx}{3 d}\\ &=-\frac{2 e f^2 p}{35 d x^5}+\frac{2 e^2 f^2 p}{21 d^2 x^3}-\frac{4 e f g p}{15 d x^3}+\frac{4 e^2 f g p}{5 d^2 x}-\frac{2 e g^2 p}{3 d x}-\frac{2 e^{3/2} g^2 p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{3 d^{3/2}}-\frac{f^2 \log \left (c \left (d+e x^2\right )^p\right )}{7 x^7}-\frac{2 f g \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac{g^2 \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}+\frac{\left (2 e^3 f^2 p\right ) \int \frac{1}{x^2 \left (d+e x^2\right )} \, dx}{7 d^2}+\frac{\left (4 e^3 f g p\right ) \int \frac{1}{d+e x^2} \, dx}{5 d^2}\\ &=-\frac{2 e f^2 p}{35 d x^5}+\frac{2 e^2 f^2 p}{21 d^2 x^3}-\frac{4 e f g p}{15 d x^3}-\frac{2 e^3 f^2 p}{7 d^3 x}+\frac{4 e^2 f g p}{5 d^2 x}-\frac{2 e g^2 p}{3 d x}+\frac{4 e^{5/2} f g p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{5 d^{5/2}}-\frac{2 e^{3/2} g^2 p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{3 d^{3/2}}-\frac{f^2 \log \left (c \left (d+e x^2\right )^p\right )}{7 x^7}-\frac{2 f g \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac{g^2 \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}-\frac{\left (2 e^4 f^2 p\right ) \int \frac{1}{d+e x^2} \, dx}{7 d^3}\\ &=-\frac{2 e f^2 p}{35 d x^5}+\frac{2 e^2 f^2 p}{21 d^2 x^3}-\frac{4 e f g p}{15 d x^3}-\frac{2 e^3 f^2 p}{7 d^3 x}+\frac{4 e^2 f g p}{5 d^2 x}-\frac{2 e g^2 p}{3 d x}-\frac{2 e^{7/2} f^2 p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{7 d^{7/2}}+\frac{4 e^{5/2} f g p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{5 d^{5/2}}-\frac{2 e^{3/2} g^2 p \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{3 d^{3/2}}-\frac{f^2 \log \left (c \left (d+e x^2\right )^p\right )}{7 x^7}-\frac{2 f g \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac{g^2 \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}\\ \end{align*}

Mathematica [C]  time = 0.0293723, size = 161, normalized size = 0.64 \[ -\frac{f^2 \log \left (c \left (d+e x^2\right )^p\right )}{7 x^7}-\frac{2 f g \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac{g^2 \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}-\frac{2 e f^2 p \, _2F_1\left (-\frac{5}{2},1;-\frac{3}{2};-\frac{e x^2}{d}\right )}{35 d x^5}-\frac{4 e f g p \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};-\frac{e x^2}{d}\right )}{15 d x^3}-\frac{2 e g^2 p \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};-\frac{e x^2}{d}\right )}{3 d x} \]

Antiderivative was successfully verified.

[In]

Integrate[((f + g*x^2)^2*Log[c*(d + e*x^2)^p])/x^8,x]

[Out]

(-2*e*f^2*p*Hypergeometric2F1[-5/2, 1, -3/2, -((e*x^2)/d)])/(35*d*x^5) - (4*e*f*g*p*Hypergeometric2F1[-3/2, 1,
-1/2, -((e*x^2)/d)])/(15*d*x^3) - (2*e*g^2*p*Hypergeometric2F1[-1/2, 1, 1/2, -((e*x^2)/d)])/(3*d*x) - (f^2*Lo
g[c*(d + e*x^2)^p])/(7*x^7) - (2*f*g*Log[c*(d + e*x^2)^p])/(5*x^5) - (g^2*Log[c*(d + e*x^2)^p])/(3*x^3)

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Maple [C]  time = 0.424, size = 784, normalized size = 3.1 \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^8,x)

[Out]

-1/105*(35*g^2*x^4+42*f*g*x^2+15*f^2)/x^7*ln((e*x^2+d)^p)-1/210*(-42*I*Pi*d^4*f*g*x^2*csgn(I*c*(e*x^2+d)^p)^3-
15*I*Pi*d^4*f^2*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)+30*ln(c)*d^4*f^2+84*(-d*e)^(1/2)*p*e^2*ln(
-e*x+(-d*e)^(1/2))*f*g*d*x^7-84*(-d*e)^(1/2)*p*e^2*ln(-e*x-(-d*e)^(1/2))*f*g*d*x^7-15*I*Pi*d^4*f^2*csgn(I*c*(e
*x^2+d)^p)^3+70*ln(c)*d^4*g^2*x^4+42*I*Pi*d^4*f*g*x^2*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2-35*I*Pi*d^4*
g^2*x^4*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)+42*I*Pi*d^4*f*g*x^2*csgn(I*c*(e*x^2+d)^p)^2*csgn(I
*c)+140*d^3*e*g^2*p*x^6+60*d*e^3*f^2*p*x^6-20*d^2*e^2*f^2*p*x^4+12*d^3*e*f^2*p*x^2+35*I*Pi*d^4*g^2*x^4*csgn(I*
(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2+35*I*Pi*d^4*g^2*x^4*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)-70*(-d*e)^(1/2)*p*e
*ln(-e*x+(-d*e)^(1/2))*g^2*d^2*x^7+70*(-d*e)^(1/2)*p*e*ln(-e*x-(-d*e)^(1/2))*g^2*d^2*x^7+84*ln(c)*d^4*f*g*x^2+
15*I*Pi*d^4*f^2*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2+15*I*Pi*d^4*f^2*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)-
35*I*Pi*d^4*g^2*x^4*csgn(I*c*(e*x^2+d)^p)^3-30*(-d*e)^(1/2)*p*e^3*ln(-e*x+(-d*e)^(1/2))*f^2*x^7+30*(-d*e)^(1/2
)*p*e^3*ln(-e*x-(-d*e)^(1/2))*f^2*x^7-168*d^2*e^2*f*g*p*x^6+56*d^3*e*f*g*p*x^4-42*I*Pi*d^4*f*g*x^2*csgn(I*(e*x
^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c))/d^4/x^7

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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^8,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.81863, size = 956, normalized size = 3.79 \begin{align*} \left [\frac{{\left (15 \, e^{3} f^{2} - 42 \, d e^{2} f g + 35 \, d^{2} e g^{2}\right )} p x^{7} \sqrt{-\frac{e}{d}} \log \left (\frac{e x^{2} - 2 \, d x \sqrt{-\frac{e}{d}} - d}{e x^{2} + d}\right ) - 6 \, d^{2} e f^{2} p x^{2} - 2 \,{\left (15 \, e^{3} f^{2} - 42 \, d e^{2} f g + 35 \, d^{2} e g^{2}\right )} p x^{6} + 2 \,{\left (5 \, d e^{2} f^{2} - 14 \, d^{2} e f g\right )} p x^{4} -{\left (35 \, d^{3} g^{2} p x^{4} + 42 \, d^{3} f g p x^{2} + 15 \, d^{3} f^{2} p\right )} \log \left (e x^{2} + d\right ) -{\left (35 \, d^{3} g^{2} x^{4} + 42 \, d^{3} f g x^{2} + 15 \, d^{3} f^{2}\right )} \log \left (c\right )}{105 \, d^{3} x^{7}}, -\frac{2 \,{\left (15 \, e^{3} f^{2} - 42 \, d e^{2} f g + 35 \, d^{2} e g^{2}\right )} p x^{7} \sqrt{\frac{e}{d}} \arctan \left (x \sqrt{\frac{e}{d}}\right ) + 6 \, d^{2} e f^{2} p x^{2} + 2 \,{\left (15 \, e^{3} f^{2} - 42 \, d e^{2} f g + 35 \, d^{2} e g^{2}\right )} p x^{6} - 2 \,{\left (5 \, d e^{2} f^{2} - 14 \, d^{2} e f g\right )} p x^{4} +{\left (35 \, d^{3} g^{2} p x^{4} + 42 \, d^{3} f g p x^{2} + 15 \, d^{3} f^{2} p\right )} \log \left (e x^{2} + d\right ) +{\left (35 \, d^{3} g^{2} x^{4} + 42 \, d^{3} f g x^{2} + 15 \, d^{3} f^{2}\right )} \log \left (c\right )}{105 \, d^{3} x^{7}}\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^8,x, algorithm="fricas")

[Out]

[1/105*((15*e^3*f^2 - 42*d*e^2*f*g + 35*d^2*e*g^2)*p*x^7*sqrt(-e/d)*log((e*x^2 - 2*d*x*sqrt(-e/d) - d)/(e*x^2
+ d)) - 6*d^2*e*f^2*p*x^2 - 2*(15*e^3*f^2 - 42*d*e^2*f*g + 35*d^2*e*g^2)*p*x^6 + 2*(5*d*e^2*f^2 - 14*d^2*e*f*g
)*p*x^4 - (35*d^3*g^2*p*x^4 + 42*d^3*f*g*p*x^2 + 15*d^3*f^2*p)*log(e*x^2 + d) - (35*d^3*g^2*x^4 + 42*d^3*f*g*x
^2 + 15*d^3*f^2)*log(c))/(d^3*x^7), -1/105*(2*(15*e^3*f^2 - 42*d*e^2*f*g + 35*d^2*e*g^2)*p*x^7*sqrt(e/d)*arcta
n(x*sqrt(e/d)) + 6*d^2*e*f^2*p*x^2 + 2*(15*e^3*f^2 - 42*d*e^2*f*g + 35*d^2*e*g^2)*p*x^6 - 2*(5*d*e^2*f^2 - 14*
d^2*e*f*g)*p*x^4 + (35*d^3*g^2*p*x^4 + 42*d^3*f*g*p*x^2 + 15*d^3*f^2*p)*log(e*x^2 + d) + (35*d^3*g^2*x^4 + 42*
d^3*f*g*x^2 + 15*d^3*f^2)*log(c))/(d^3*x^7)]

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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**8,x)

[Out]

Timed out

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Giac [A]  time = 1.36621, size = 300, normalized size = 1.19 \begin{align*} -\frac{2 \,{\left (35 \, d^{2} g^{2} p e^{2} - 42 \, d f g p e^{3} + 15 \, f^{2} p e^{4}\right )} \arctan \left (\frac{x e^{\frac{1}{2}}}{\sqrt{d}}\right ) e^{\left (-\frac{1}{2}\right )}}{105 \, d^{\frac{7}{2}}} - \frac{70 \, d^{2} g^{2} p x^{6} e - 84 \, d f g p x^{6} e^{2} + 35 \, d^{3} g^{2} p x^{4} \log \left (x^{2} e + d\right ) + 30 \, f^{2} p x^{6} e^{3} + 28 \, d^{2} f g p x^{4} e + 35 \, d^{3} g^{2} x^{4} \log \left (c\right ) - 10 \, d f^{2} p x^{4} e^{2} + 42 \, d^{3} f g p x^{2} \log \left (x^{2} e + d\right ) + 6 \, d^{2} f^{2} p x^{2} e + 42 \, d^{3} f g x^{2} \log \left (c\right ) + 15 \, d^{3} f^{2} p \log \left (x^{2} e + d\right ) + 15 \, d^{3} f^{2} \log \left (c\right )}{105 \, d^{3} x^{7}} \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^8,x, algorithm="giac")

[Out]

-2/105*(35*d^2*g^2*p*e^2 - 42*d*f*g*p*e^3 + 15*f^2*p*e^4)*arctan(x*e^(1/2)/sqrt(d))*e^(-1/2)/d^(7/2) - 1/105*(
70*d^2*g^2*p*x^6*e - 84*d*f*g*p*x^6*e^2 + 35*d^3*g^2*p*x^4*log(x^2*e + d) + 30*f^2*p*x^6*e^3 + 28*d^2*f*g*p*x^
4*e + 35*d^3*g^2*x^4*log(c) - 10*d*f^2*p*x^4*e^2 + 42*d^3*f*g*p*x^2*log(x^2*e + d) + 6*d^2*f^2*p*x^2*e + 42*d^
3*f*g*x^2*log(c) + 15*d^3*f^2*p*log(x^2*e + d) + 15*d^3*f^2*log(c))/(d^3*x^7)