3.4.22 \(\int \frac {(f+g x^2) \log (c (d+e x^2)^p)}{x^6} \, dx\) [322]

Optimal. Leaf size=140 \[ -\frac {2 e f p}{15 d x^3}+\frac {2 e^2 f p}{5 d^2 x}-\frac {2 e g p}{3 d x}+\frac {2 e^{5/2} f p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 d^{5/2}}-\frac {2 e^{3/2} g p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 d^{3/2}}-\frac {f \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac {g \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3} \]

[Out]

-2/15*e*f*p/d/x^3+2/5*e^2*f*p/d^2/x-2/3*e*g*p/d/x+2/5*e^(5/2)*f*p*arctan(x*e^(1/2)/d^(1/2))/d^(5/2)-2/3*e^(3/2
)*g*p*arctan(x*e^(1/2)/d^(1/2))/d^(3/2)-1/5*f*ln(c*(e*x^2+d)^p)/x^5-1/3*g*ln(c*(e*x^2+d)^p)/x^3

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Rubi [A]
time = 0.08, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2526, 2505, 331, 211} \begin {gather*} \frac {2 e^{5/2} f p \text {ArcTan}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 d^{5/2}}-\frac {2 e^{3/2} g p \text {ArcTan}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 d^{3/2}}-\frac {f \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac {g \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}+\frac {2 e^2 f p}{5 d^2 x}-\frac {2 e f p}{15 d x^3}-\frac {2 e g p}{3 d x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 331

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 2505

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 2526

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]

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.01, size = 101, normalized size = 0.72 \begin {gather*} -\frac {2 e f 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 p \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-\frac {e x^2}{d}\right )}{3 d x}-\frac {f \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac {g \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.10, size = 474, normalized size = 3.39

method	result	size
risch	\(-\frac {\left (5 g \,x^{2}+3 f \right ) \ln \left (\left (e \,x^{2}+d \right )^{p}\right )}{15 x^{5}}-\frac {-3 i \pi \,d^{3} f \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}-5 i \pi \,d^{3} g \,x^{2} \mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \right )+3 i \pi \,d^{3} f \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right )-3 i \pi \,d^{3} f \,\mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \right )+10 \sqrt {-e d}\, p e \ln \left (-e x -\sqrt {-e d}\right ) g d \,x^{5}-6 \sqrt {-e d}\, p \,e^{2} \ln \left (-e x -\sqrt {-e d}\right ) f \,x^{5}-10 \sqrt {-e d}\, p e \ln \left (-e x +\sqrt {-e d}\right ) g d \,x^{5}+6 \sqrt {-e d}\, p \,e^{2} \ln \left (-e x +\sqrt {-e d}\right ) f \,x^{5}+5 i \pi \,d^{3} g \,x^{2} \mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}+3 i \pi \,d^{3} f \,\mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}+5 i \pi \,d^{3} g \,x^{2} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right )-5 i \pi \,d^{3} g \,x^{2} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}+20 d^{2} e g p \,x^{4}-12 d \,e^{2} f p \,x^{4}+10 \ln \left (c \right ) d^{3} g \,x^{2}+4 d^{2} e f p \,x^{2}+6 \ln \left (c \right ) d^{3} f}{30 d^{3} x^{5}}\)	\(474\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x^2+f)*ln(c*(e*x^2+d)^p)/x^6,x,method=_RETURNVERBOSE)

[Out]

-1/15*(5*g*x^2+3*f)/x^5*ln((e*x^2+d)^p)-1/30*(-3*I*Pi*d^3*f*csgn(I*c*(e*x^2+d)^p)^3-5*I*Pi*d^3*g*x^2*csgn(I*(e
*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)+3*I*Pi*d^3*f*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)-3*I*Pi*d^3*f*csgn(I*
(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)+10*(-e*d)^(1/2)*p*e*ln(-e*x-(-e*d)^(1/2))*g*d*x^5-6*(-e*d)^(1/2)*
p*e^2*ln(-e*x-(-e*d)^(1/2))*f*x^5-10*(-e*d)^(1/2)*p*e*ln(-e*x+(-e*d)^(1/2))*g*d*x^5+6*(-e*d)^(1/2)*p*e^2*ln(-e
*x+(-e*d)^(1/2))*f*x^5+5*I*Pi*d^3*g*x^2*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2+3*I*Pi*d^3*f*csgn(I*(e*x^2
+d)^p)*csgn(I*c*(e*x^2+d)^p)^2+5*I*Pi*d^3*g*x^2*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)-5*I*Pi*d^3*g*x^2*csgn(I*c*(e
*x^2+d)^p)^3+20*d^2*e*g*p*x^4-12*d*e^2*f*p*x^4+10*ln(c)*d^3*g*x^2+4*d^2*e*f*p*x^2+6*ln(c)*d^3*f)/d^3/x^5

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Maxima [A]
time = 0.49, size = 86, normalized size = 0.61 \begin {gather*} -\frac {2}{15} \, p {\left (\frac {{\left (5 \, d g e - 3 \, f e^{2}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {1}{2}\right )}}{d^{\frac {5}{2}}} + \frac {{\left (5 \, d g - 3 \, f e\right )} x^{2} + d f}{d^{2} x^{3}}\right )} e - \frac {{\left (5 \, g x^{2} + 3 \, f\right )} \log \left ({\left (x^{2} e + d\right )}^{p} c\right )}{15 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x^2+f)*log(c*(e*x^2+d)^p)/x^6,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.36, size = 276, normalized size = 1.97 \begin {gather*} \left [\frac {6 \, f p x^{4} e^{2} - {\left (5 \, d g p x^{5} e - 3 \, f p x^{5} e^{2}\right )} \sqrt {-\frac {e}{d}} \log \left (\frac {x^{2} e + 2 \, d x \sqrt {-\frac {e}{d}} - d}{x^{2} e + d}\right ) - 2 \, {\left (5 \, d g p x^{4} + d f p x^{2}\right )} e - {\left (5 \, d^{2} g p x^{2} + 3 \, d^{2} f p\right )} \log \left (x^{2} e + d\right ) - {\left (5 \, d^{2} g x^{2} + 3 \, d^{2} f\right )} \log \left (c\right )}{15 \, d^{2} x^{5}}, \frac {6 \, f p x^{4} e^{2} - \frac {2 \, {\left (5 \, d g p x^{5} e - 3 \, f p x^{5} e^{2}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\frac {1}{2}}}{\sqrt {d}} - 2 \, {\left (5 \, d g p x^{4} + d f p x^{2}\right )} e - {\left (5 \, d^{2} g p x^{2} + 3 \, d^{2} f p\right )} \log \left (x^{2} e + d\right ) - {\left (5 \, d^{2} g x^{2} + 3 \, d^{2} f\right )} \log \left (c\right )}{15 \, d^{2} x^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x^2+f)*log(c*(e*x^2+d)^p)/x^6,x, algorithm="fricas")

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 1108 vs. \(2 (138) = 276\).
time = 164.84, size = 1108, normalized size = 7.91 \begin {gather*} \begin {cases} \left (- \frac {f}{5 x^{5}} - \frac {g}{3 x^{3}}\right ) \log {\left (0^{p} c \right )} & \text {for}\: d = 0 \wedge e = 0 \\\left (- \frac {f}{5 x^{5}} - \frac {g}{3 x^{3}}\right ) \log {\left (c d^{p} \right )} & \text {for}\: e = 0 \\- \frac {2 f p}{25 x^{5}} - \frac {f \log {\left (c \left (e x^{2}\right )^{p} \right )}}{5 x^{5}} - \frac {2 g p}{9 x^{3}} - \frac {g \log {\left (c \left (e x^{2}\right )^{p} \right )}}{3 x^{3}} & \text {for}\: d = 0 \\- \frac {3 d^{3} f \sqrt {- \frac {d}{e}} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{15 d^{3} x^{5} \sqrt {- \frac {d}{e}} + 15 d^{2} e x^{7} \sqrt {- \frac {d}{e}}} - \frac {5 d^{3} g x^{2} \sqrt {- \frac {d}{e}} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{15 d^{3} x^{5} \sqrt {- \frac {d}{e}} + 15 d^{2} e x^{7} \sqrt {- \frac {d}{e}}} - \frac {2 d^{2} f p x^{2} \sqrt {- \frac {d}{e}}}{\frac {15 d^{3} x^{5} \sqrt {- \frac {d}{e}}}{e} + 15 d^{2} x^{7} \sqrt {- \frac {d}{e}}} - \frac {3 d^{2} f x^{2} \sqrt {- \frac {d}{e}} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{\frac {15 d^{3} x^{5} \sqrt {- \frac {d}{e}}}{e} + 15 d^{2} x^{7} \sqrt {- \frac {d}{e}}} - \frac {10 d^{2} g p x^{5} \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{\frac {15 d^{3} x^{5} \sqrt {- \frac {d}{e}}}{e} + 15 d^{2} x^{7} \sqrt {- \frac {d}{e}}} - \frac {10 d^{2} g p x^{4} \sqrt {- \frac {d}{e}}}{\frac {15 d^{3} x^{5} \sqrt {- \frac {d}{e}}}{e} + 15 d^{2} x^{7} \sqrt {- \frac {d}{e}}} + \frac {5 d^{2} g x^{5} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{\frac {15 d^{3} x^{5} \sqrt {- \frac {d}{e}}}{e} + 15 d^{2} x^{7} \sqrt {- \frac {d}{e}}} - \frac {5 d^{2} g x^{4} \sqrt {- \frac {d}{e}} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{\frac {15 d^{3} x^{5} \sqrt {- \frac {d}{e}}}{e} + 15 d^{2} x^{7} \sqrt {- \frac {d}{e}}} + \frac {6 d e f p x^{5} \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{\frac {15 d^{3} x^{5} \sqrt {- \frac {d}{e}}}{e} + 15 d^{2} x^{7} \sqrt {- \frac {d}{e}}} + \frac {4 d e f p x^{4} \sqrt {- \frac {d}{e}}}{\frac {15 d^{3} x^{5} \sqrt {- \frac {d}{e}}}{e} + 15 d^{2} x^{7} \sqrt {- \frac {d}{e}}} - \frac {3 d e f x^{5} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{\frac {15 d^{3} x^{5} \sqrt {- \frac {d}{e}}}{e} + 15 d^{2} x^{7} \sqrt {- \frac {d}{e}}} - \frac {10 d e g p x^{7} \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{\frac {15 d^{3} x^{5} \sqrt {- \frac {d}{e}}}{e} + 15 d^{2} x^{7} \sqrt {- \frac {d}{e}}} - \frac {10 d e g p x^{6} \sqrt {- \frac {d}{e}}}{\frac {15 d^{3} x^{5} \sqrt {- \frac {d}{e}}}{e} + 15 d^{2} x^{7} \sqrt {- \frac {d}{e}}} + \frac {5 d e g x^{7} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{\frac {15 d^{3} x^{5} \sqrt {- \frac {d}{e}}}{e} + 15 d^{2} x^{7} \sqrt {- \frac {d}{e}}} + \frac {6 e^{2} f p x^{7} \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{\frac {15 d^{3} x^{5} \sqrt {- \frac {d}{e}}}{e} + 15 d^{2} x^{7} \sqrt {- \frac {d}{e}}} + \frac {6 e^{2} f p x^{6} \sqrt {- \frac {d}{e}}}{\frac {15 d^{3} x^{5} \sqrt {- \frac {d}{e}}}{e} + 15 d^{2} x^{7} \sqrt {- \frac {d}{e}}} - \frac {3 e^{2} f x^{7} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{\frac {15 d^{3} x^{5} \sqrt {- \frac {d}{e}}}{e} + 15 d^{2} x^{7} \sqrt {- \frac {d}{e}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x**2+f)*ln(c*(e*x**2+d)**p)/x**6,x)

[Out]

Piecewise(((-f/(5*x**5) - g/(3*x**3))*log(0**p*c), Eq(d, 0) & Eq(e, 0)), ((-f/(5*x**5) - g/(3*x**3))*log(c*d**
p), Eq(e, 0)), (-2*f*p/(25*x**5) - f*log(c*(e*x**2)**p)/(5*x**5) - 2*g*p/(9*x**3) - g*log(c*(e*x**2)**p)/(3*x*
*3), Eq(d, 0)), (-3*d**3*f*sqrt(-d/e)*log(c*(d + e*x**2)**p)/(15*d**3*x**5*sqrt(-d/e) + 15*d**2*e*x**7*sqrt(-d
/e)) - 5*d**3*g*x**2*sqrt(-d/e)*log(c*(d + e*x**2)**p)/(15*d**3*x**5*sqrt(-d/e) + 15*d**2*e*x**7*sqrt(-d/e)) -
2*d**2*f*p*x**2*sqrt(-d/e)/(15*d**3*x**5*sqrt(-d/e)/e + 15*d**2*x**7*sqrt(-d/e)) - 3*d**2*f*x**2*sqrt(-d/e)*l
og(c*(d + e*x**2)**p)/(15*d**3*x**5*sqrt(-d/e)/e + 15*d**2*x**7*sqrt(-d/e)) - 10*d**2*g*p*x**5*log(x - sqrt(-d
/e))/(15*d**3*x**5*sqrt(-d/e)/e + 15*d**2*x**7*sqrt(-d/e)) - 10*d**2*g*p*x**4*sqrt(-d/e)/(15*d**3*x**5*sqrt(-d
/e)/e + 15*d**2*x**7*sqrt(-d/e)) + 5*d**2*g*x**5*log(c*(d + e*x**2)**p)/(15*d**3*x**5*sqrt(-d/e)/e + 15*d**2*x
**7*sqrt(-d/e)) - 5*d**2*g*x**4*sqrt(-d/e)*log(c*(d + e*x**2)**p)/(15*d**3*x**5*sqrt(-d/e)/e + 15*d**2*x**7*sq
rt(-d/e)) + 6*d*e*f*p*x**5*log(x - sqrt(-d/e))/(15*d**3*x**5*sqrt(-d/e)/e + 15*d**2*x**7*sqrt(-d/e)) + 4*d*e*f
*p*x**4*sqrt(-d/e)/(15*d**3*x**5*sqrt(-d/e)/e + 15*d**2*x**7*sqrt(-d/e)) - 3*d*e*f*x**5*log(c*(d + e*x**2)**p)
/(15*d**3*x**5*sqrt(-d/e)/e + 15*d**2*x**7*sqrt(-d/e)) - 10*d*e*g*p*x**7*log(x - sqrt(-d/e))/(15*d**3*x**5*sqr
t(-d/e)/e + 15*d**2*x**7*sqrt(-d/e)) - 10*d*e*g*p*x**6*sqrt(-d/e)/(15*d**3*x**5*sqrt(-d/e)/e + 15*d**2*x**7*sq
rt(-d/e)) + 5*d*e*g*x**7*log(c*(d + e*x**2)**p)/(15*d**3*x**5*sqrt(-d/e)/e + 15*d**2*x**7*sqrt(-d/e)) + 6*e**2
*f*p*x**7*log(x - sqrt(-d/e))/(15*d**3*x**5*sqrt(-d/e)/e + 15*d**2*x**7*sqrt(-d/e)) + 6*e**2*f*p*x**6*sqrt(-d/
e)/(15*d**3*x**5*sqrt(-d/e)/e + 15*d**2*x**7*sqrt(-d/e)) - 3*e**2*f*x**7*log(c*(d + e*x**2)**p)/(15*d**3*x**5*
sqrt(-d/e)/e + 15*d**2*x**7*sqrt(-d/e)), True))

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Giac [A]
time = 3.91, size = 122, normalized size = 0.87 \begin {gather*} -\frac {2 \, {\left (5 \, d g p e^{2} - 3 \, f 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 {10 \, d g p x^{4} e - 6 \, f p x^{4} e^{2} + 5 \, d^{2} g p x^{2} \log \left (x^{2} e + d\right ) + 2 \, d f p x^{2} e + 5 \, d^{2} g x^{2} \log \left (c\right ) + 3 \, d^{2} f p \log \left (x^{2} e + d\right ) + 3 \, d^{2} f \log \left (c\right )}{15 \, d^{2} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x^2+f)*log(c*(e*x^2+d)^p)/x^6,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.38, size = 88, normalized size = 0.63 \begin {gather*} -\frac {\frac {2\,e\,f\,p}{d}+\frac {2\,e\,p\,x^2\,\left (5\,d\,g-3\,e\,f\right )}{d^2}}{15\,x^3}-\frac {\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )\,\left (\frac {g\,x^2}{3}+\frac {f}{5}\right )}{x^5}-\frac {2\,e^{3/2}\,p\,\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (5\,d\,g-3\,e\,f\right )}{15\,d^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(c*(d + e*x^2)^p)*(f + g*x^2))/x^6,x)

[Out]

- ((2*e*f*p)/d + (2*e*p*x^2*(5*d*g - 3*e*f))/d^2)/(15*x^3) - (log(c*(d + e*x^2)^p)*(f/5 + (g*x^2)/3))/x^5 - (2
*e^(3/2)*p*atan((e^(1/2)*x)/d^(1/2))*(5*d*g - 3*e*f))/(15*d^(5/2))

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