3.195 \(\int \frac {(d+e x^2)^2 (a+b \log (c x^n))}{x^8} \, dx\)

Optimal. Leaf size=95 \[ -\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{7 x^7}-\frac {2 d e \left (a+b \log \left (c x^n\right )\right )}{5 x^5}-\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {b d^2 n}{49 x^7}-\frac {2 b d e n}{25 x^5}-\frac {b e^2 n}{9 x^3} \]

[Out]

-1/49*b*d^2*n/x^7-2/25*b*d*e*n/x^5-1/9*b*e^2*n/x^3-1/7*d^2*(a+b*ln(c*x^n))/x^7-2/5*d*e*(a+b*ln(c*x^n))/x^5-1/3
*e^2*(a+b*ln(c*x^n))/x^3

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Rubi [A]  time = 0.08, antiderivative size = 74, normalized size of antiderivative = 0.78, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {270, 2334, 12, 14} \[ -\frac {1}{105} \left (\frac {15 d^2}{x^7}+\frac {42 d e}{x^5}+\frac {35 e^2}{x^3}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {b d^2 n}{49 x^7}-\frac {2 b d e n}{25 x^5}-\frac {b e^2 n}{9 x^3} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)^2*(a + b*Log[c*x^n]))/x^8,x]

[Out]

-(b*d^2*n)/(49*x^7) - (2*b*d*e*n)/(25*x^5) - (b*e^2*n)/(9*x^3) - (((15*d^2)/x^7 + (42*d*e)/x^5 + (35*e^2)/x^3)
*(a + b*Log[c*x^n]))/105

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(d + e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]
] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

Rubi steps

\begin {align*} \int \frac {\left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^8} \, dx &=-\frac {1}{105} \left (\frac {15 d^2}{x^7}+\frac {42 d e}{x^5}+\frac {35 e^2}{x^3}\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \frac {-15 d^2-42 d e x^2-35 e^2 x^4}{105 x^8} \, dx\\ &=-\frac {1}{105} \left (\frac {15 d^2}{x^7}+\frac {42 d e}{x^5}+\frac {35 e^2}{x^3}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{105} (b n) \int \frac {-15 d^2-42 d e x^2-35 e^2 x^4}{x^8} \, dx\\ &=-\frac {1}{105} \left (\frac {15 d^2}{x^7}+\frac {42 d e}{x^5}+\frac {35 e^2}{x^3}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{105} (b n) \int \left (-\frac {15 d^2}{x^8}-\frac {42 d e}{x^6}-\frac {35 e^2}{x^4}\right ) \, dx\\ &=-\frac {b d^2 n}{49 x^7}-\frac {2 b d e n}{25 x^5}-\frac {b e^2 n}{9 x^3}-\frac {1}{105} \left (\frac {15 d^2}{x^7}+\frac {42 d e}{x^5}+\frac {35 e^2}{x^3}\right ) \left (a+b \log \left (c x^n\right )\right )\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 95, normalized size = 1.00 \[ -\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{7 x^7}-\frac {2 d e \left (a+b \log \left (c x^n\right )\right )}{5 x^5}-\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {b d^2 n}{49 x^7}-\frac {2 b d e n}{25 x^5}-\frac {b e^2 n}{9 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)^2*(a + b*Log[c*x^n]))/x^8,x]

[Out]

-1/49*(b*d^2*n)/x^7 - (2*b*d*e*n)/(25*x^5) - (b*e^2*n)/(9*x^3) - (d^2*(a + b*Log[c*x^n]))/(7*x^7) - (2*d*e*(a
+ b*Log[c*x^n]))/(5*x^5) - (e^2*(a + b*Log[c*x^n]))/(3*x^3)

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fricas [A]  time = 0.55, size = 112, normalized size = 1.18 \[ -\frac {1225 \, {\left (b e^{2} n + 3 \, a e^{2}\right )} x^{4} + 225 \, b d^{2} n + 1575 \, a d^{2} + 882 \, {\left (b d e n + 5 \, a d e\right )} x^{2} + 105 \, {\left (35 \, b e^{2} x^{4} + 42 \, b d e x^{2} + 15 \, b d^{2}\right )} \log \relax (c) + 105 \, {\left (35 \, b e^{2} n x^{4} + 42 \, b d e n x^{2} + 15 \, b d^{2} n\right )} \log \relax (x)}{11025 \, x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^2*(a+b*log(c*x^n))/x^8,x, algorithm="fricas")

[Out]

-1/11025*(1225*(b*e^2*n + 3*a*e^2)*x^4 + 225*b*d^2*n + 1575*a*d^2 + 882*(b*d*e*n + 5*a*d*e)*x^2 + 105*(35*b*e^
2*x^4 + 42*b*d*e*x^2 + 15*b*d^2)*log(c) + 105*(35*b*e^2*n*x^4 + 42*b*d*e*n*x^2 + 15*b*d^2*n)*log(x))/x^7

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giac [A]  time = 0.29, size = 116, normalized size = 1.22 \[ -\frac {3675 \, b n x^{4} e^{2} \log \relax (x) + 1225 \, b n x^{4} e^{2} + 3675 \, b x^{4} e^{2} \log \relax (c) + 4410 \, b d n x^{2} e \log \relax (x) + 3675 \, a x^{4} e^{2} + 882 \, b d n x^{2} e + 4410 \, b d x^{2} e \log \relax (c) + 4410 \, a d x^{2} e + 1575 \, b d^{2} n \log \relax (x) + 225 \, b d^{2} n + 1575 \, b d^{2} \log \relax (c) + 1575 \, a d^{2}}{11025 \, x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^2*(a+b*log(c*x^n))/x^8,x, algorithm="giac")

[Out]

-1/11025*(3675*b*n*x^4*e^2*log(x) + 1225*b*n*x^4*e^2 + 3675*b*x^4*e^2*log(c) + 4410*b*d*n*x^2*e*log(x) + 3675*
a*x^4*e^2 + 882*b*d*n*x^2*e + 4410*b*d*x^2*e*log(c) + 4410*a*d*x^2*e + 1575*b*d^2*n*log(x) + 225*b*d^2*n + 157
5*b*d^2*log(c) + 1575*a*d^2)/x^7

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maple [C]  time = 0.17, size = 419, normalized size = 4.41 \[ -\frac {\left (35 e^{2} x^{4}+42 d e \,x^{2}+15 d^{2}\right ) b \ln \left (x^{n}\right )}{105 x^{7}}-\frac {-3675 i \pi b \,e^{2} x^{4} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )+3675 i \pi b \,e^{2} x^{4} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+3675 i \pi b \,e^{2} x^{4} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-3675 i \pi b \,e^{2} x^{4} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}-4410 i \pi b d e \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )+4410 i \pi b d e \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+4410 i \pi b d e \,x^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-4410 i \pi b d e \,x^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}+2450 b \,e^{2} n \,x^{4}+7350 b \,e^{2} x^{4} \ln \relax (c )+7350 a \,e^{2} x^{4}-1575 i \pi b \,d^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )+1575 i \pi b \,d^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+1575 i \pi b \,d^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-1575 i \pi b \,d^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}+1764 b d e n \,x^{2}+8820 b d e \,x^{2} \ln \relax (c )+8820 a d e \,x^{2}+450 b \,d^{2} n +3150 b \,d^{2} \ln \relax (c )+3150 a \,d^{2}}{22050 x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)^2*(b*ln(c*x^n)+a)/x^8,x)

[Out]

-1/105*b*(35*e^2*x^4+42*d*e*x^2+15*d^2)/x^7*ln(x^n)-1/22050*(1575*I*Pi*b*d^2*csgn(I*x^n)*csgn(I*c*x^n)^2-3675*
I*Pi*b*e^2*x^4*csgn(I*c*x^n)^3+3675*I*Pi*b*e^2*x^4*csgn(I*x^n)*csgn(I*c*x^n)^2+4410*I*Pi*b*d*e*x^2*csgn(I*x^n)
*csgn(I*c*x^n)^2+7350*b*e^2*x^4*ln(c)+2450*b*e^2*n*x^4+7350*a*e^2*x^4-3675*I*Pi*b*e^2*x^4*csgn(I*x^n)*csgn(I*c
*x^n)*csgn(I*c)-1575*I*Pi*b*d^2*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+4410*I*Pi*b*d*e*x^2*csgn(I*c*x^n)^2*csgn(I
*c)+1575*I*Pi*b*d^2*csgn(I*c*x^n)^2*csgn(I*c)+8820*b*d*e*x^2*ln(c)+1764*b*d*e*n*x^2+8820*a*d*e*x^2-4410*I*Pi*b
*d*e*x^2*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+3675*I*Pi*b*e^2*x^4*csgn(I*c*x^n)^2*csgn(I*c)-1575*I*Pi*b*d^2*csg
n(I*c*x^n)^3-4410*I*Pi*b*d*e*x^2*csgn(I*c*x^n)^3+3150*b*d^2*ln(c)+450*b*d^2*n+3150*a*d^2)/x^7

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maxima [A]  time = 0.46, size = 100, normalized size = 1.05 \[ -\frac {b e^{2} n}{9 \, x^{3}} - \frac {b e^{2} \log \left (c x^{n}\right )}{3 \, x^{3}} - \frac {a e^{2}}{3 \, x^{3}} - \frac {2 \, b d e n}{25 \, x^{5}} - \frac {2 \, b d e \log \left (c x^{n}\right )}{5 \, x^{5}} - \frac {2 \, a d e}{5 \, x^{5}} - \frac {b d^{2} n}{49 \, x^{7}} - \frac {b d^{2} \log \left (c x^{n}\right )}{7 \, x^{7}} - \frac {a d^{2}}{7 \, x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^2*(a+b*log(c*x^n))/x^8,x, algorithm="maxima")

[Out]

-1/9*b*e^2*n/x^3 - 1/3*b*e^2*log(c*x^n)/x^3 - 1/3*a*e^2/x^3 - 2/25*b*d*e*n/x^5 - 2/5*b*d*e*log(c*x^n)/x^5 - 2/
5*a*d*e/x^5 - 1/49*b*d^2*n/x^7 - 1/7*b*d^2*log(c*x^n)/x^7 - 1/7*a*d^2/x^7

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mupad [B]  time = 3.55, size = 89, normalized size = 0.94 \[ -\frac {x^4\,\left (35\,a\,e^2+\frac {35\,b\,e^2\,n}{3}\right )+x^2\,\left (42\,a\,d\,e+\frac {42\,b\,d\,e\,n}{5}\right )+15\,a\,d^2+\frac {15\,b\,d^2\,n}{7}}{105\,x^7}-\frac {\ln \left (c\,x^n\right )\,\left (\frac {b\,d^2}{7}+\frac {2\,b\,d\,e\,x^2}{5}+\frac {b\,e^2\,x^4}{3}\right )}{x^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((d + e*x^2)^2*(a + b*log(c*x^n)))/x^8,x)

[Out]

- (x^4*(35*a*e^2 + (35*b*e^2*n)/3) + x^2*(42*a*d*e + (42*b*d*e*n)/5) + 15*a*d^2 + (15*b*d^2*n)/7)/(105*x^7) -
(log(c*x^n)*((b*d^2)/7 + (b*e^2*x^4)/3 + (2*b*d*e*x^2)/5))/x^7

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sympy [A]  time = 10.08, size = 160, normalized size = 1.68 \[ - \frac {a d^{2}}{7 x^{7}} - \frac {2 a d e}{5 x^{5}} - \frac {a e^{2}}{3 x^{3}} - \frac {b d^{2} n \log {\relax (x )}}{7 x^{7}} - \frac {b d^{2} n}{49 x^{7}} - \frac {b d^{2} \log {\relax (c )}}{7 x^{7}} - \frac {2 b d e n \log {\relax (x )}}{5 x^{5}} - \frac {2 b d e n}{25 x^{5}} - \frac {2 b d e \log {\relax (c )}}{5 x^{5}} - \frac {b e^{2} n \log {\relax (x )}}{3 x^{3}} - \frac {b e^{2} n}{9 x^{3}} - \frac {b e^{2} \log {\relax (c )}}{3 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)**2*(a+b*ln(c*x**n))/x**8,x)

[Out]

-a*d**2/(7*x**7) - 2*a*d*e/(5*x**5) - a*e**2/(3*x**3) - b*d**2*n*log(x)/(7*x**7) - b*d**2*n/(49*x**7) - b*d**2
*log(c)/(7*x**7) - 2*b*d*e*n*log(x)/(5*x**5) - 2*b*d*e*n/(25*x**5) - 2*b*d*e*log(c)/(5*x**5) - b*e**2*n*log(x)
/(3*x**3) - b*e**2*n/(9*x**3) - b*e**2*log(c)/(3*x**3)

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