3.191 \(\int (d+e x^2)^2 (a+b \log (c x^n)) \, dx\)

Optimal. Leaf size=86 \[ d^2 x \left (a+b \log \left (c x^n\right )\right )+\frac {2}{3} d e x^3 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{5} e^2 x^5 \left (a+b \log \left (c x^n\right )\right )-b d^2 n x-\frac {2}{9} b d e n x^3-\frac {1}{25} b e^2 n x^5 \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 68, normalized size of antiderivative = 0.79, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {194, 2313} \[ \frac {1}{15} \left (15 d^2 x+10 d e x^3+3 e^2 x^5\right ) \left (a+b \log \left (c x^n\right )\right )-b d^2 n x-\frac {2}{9} b d e n x^3-\frac {1}{25} b e^2 n x^5 \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*d^2*n*x) - (2*b*d*e*n*x^3)/9 - (b*e^2*n*x^5)/25 + ((15*d^2*x + 10*d*e*x^3 + 3*e^2*x^5)*(a + b*Log[c*x^n]))
/15

Rule 194

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

Rule 2313

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(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]

Rubi steps

\begin {align*} \int \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac {1}{15} \left (15 d^2 x+10 d e x^3+3 e^2 x^5\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \left (d^2+\frac {2}{3} d e x^2+\frac {e^2 x^4}{5}\right ) \, dx\\ &=-b d^2 n x-\frac {2}{9} b d e n x^3-\frac {1}{25} b e^2 n x^5+\frac {1}{15} \left (15 d^2 x+10 d e x^3+3 e^2 x^5\right ) \left (a+b \log \left (c x^n\right )\right )\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 89, normalized size = 1.03 \[ \frac {2}{3} d e x^3 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{5} e^2 x^5 \left (a+b \log \left (c x^n\right )\right )+a d^2 x+b d^2 x \log \left (c x^n\right )-b d^2 n x-\frac {2}{9} b d e n x^3-\frac {1}{25} b e^2 n x^5 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 112, normalized size = 1.30 \[ -\frac {1}{25} \, {\left (b e^{2} n - 5 \, a e^{2}\right )} x^{5} - \frac {2}{9} \, {\left (b d e n - 3 \, a d e\right )} x^{3} - {\left (b d^{2} n - a d^{2}\right )} x + \frac {1}{15} \, {\left (3 \, b e^{2} x^{5} + 10 \, b d e x^{3} + 15 \, b d^{2} x\right )} \log \relax (c) + \frac {1}{15} \, {\left (3 \, b e^{2} n x^{5} + 10 \, b d e n x^{3} + 15 \, b d^{2} n x\right )} \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/25*(b*e^2*n - 5*a*e^2)*x^5 - 2/9*(b*d*e*n - 3*a*d*e)*x^3 - (b*d^2*n - a*d^2)*x + 1/15*(3*b*e^2*x^5 + 10*b*d
*e*x^3 + 15*b*d^2*x)*log(c) + 1/15*(3*b*e^2*n*x^5 + 10*b*d*e*n*x^3 + 15*b*d^2*n*x)*log(x)

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giac [A]  time = 0.41, size = 112, normalized size = 1.30 \[ \frac {1}{5} \, b n x^{5} e^{2} \log \relax (x) - \frac {1}{25} \, b n x^{5} e^{2} + \frac {1}{5} \, b x^{5} e^{2} \log \relax (c) + \frac {2}{3} \, b d n x^{3} e \log \relax (x) + \frac {1}{5} \, a x^{5} e^{2} - \frac {2}{9} \, b d n x^{3} e + \frac {2}{3} \, b d x^{3} e \log \relax (c) + \frac {2}{3} \, a d x^{3} e + b d^{2} n x \log \relax (x) - b d^{2} n x + b d^{2} x \log \relax (c) + a d^{2} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.22, size = 416, normalized size = 4.84 \[ -\frac {i \pi b \,e^{2} x^{5} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{10}+\frac {i \pi b \,e^{2} x^{5} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{10}+\frac {i \pi b \,e^{2} x^{5} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{10}-\frac {i \pi b \,e^{2} x^{5} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{10}-\frac {i \pi b d e \,x^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{3}+\frac {i \pi b d e \,x^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{3}+\frac {i \pi b d e \,x^{3} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{3}-\frac {i \pi b d e \,x^{3} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{3}-\frac {b \,e^{2} n \,x^{5}}{25}+\frac {b \,e^{2} x^{5} \ln \relax (c )}{5}+\frac {a \,e^{2} x^{5}}{5}-\frac {i \pi b \,d^{2} x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i \pi b \,d^{2} x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i \pi b \,d^{2} x \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i \pi b \,d^{2} x \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{2}-\frac {2 b d e n \,x^{3}}{9}+\frac {2 b d e \,x^{3} \ln \relax (c )}{3}+\frac {2 a d e \,x^{3}}{3}-b \,d^{2} n x +b \,d^{2} x \ln \relax (c )+a \,d^{2} x +\frac {\left (3 e^{2} x^{4}+10 d e \,x^{2}+15 d^{2}\right ) b x \ln \left (x^{n}\right )}{15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*b*x*(3*e^2*x^4+10*d*e*x^2+15*d^2)*ln(x^n)-1/3*I*Pi*b*d*e*x^3*csgn(I*c*x^n)^3+1/3*I*Pi*b*d*e*x^3*csgn(I*x^
n)*csgn(I*c*x^n)^2+1/3*I*Pi*b*d*e*x^3*csgn(I*c*x^n)^2*csgn(I*c)+1/10*I*Pi*b*e^2*x^5*csgn(I*c*x^n)^2*csgn(I*c)+
1/10*I*Pi*b*e^2*x^5*csgn(I*x^n)*csgn(I*c*x^n)^2-1/2*I*Pi*b*d^2*csgn(I*c*x^n)^3*x-1/3*I*Pi*b*d*e*x^3*csgn(I*x^n
)*csgn(I*c*x^n)*csgn(I*c)-1/10*I*Pi*b*e^2*x^5*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+1/2*I*Pi*b*d^2*csgn(I*x^n)*c
sgn(I*c*x^n)^2*x-1/2*I*Pi*b*d^2*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*x+1/2*I*Pi*b*d^2*csgn(I*c*x^n)^2*csgn(I*c)
*x-1/10*I*Pi*b*e^2*x^5*csgn(I*c*x^n)^3+1/5*b*e^2*x^5*ln(c)-1/25*b*e^2*n*x^5+1/5*a*e^2*x^5+2/3*b*d*e*x^3*ln(c)-
2/9*b*d*e*n*x^3+2/3*a*d*e*x^3+ln(c)*b*d^2*x-b*d^2*n*x+a*d^2*x

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maxima [A]  time = 0.47, size = 92, normalized size = 1.07 \[ -\frac {1}{25} \, b e^{2} n x^{5} + \frac {1}{5} \, b e^{2} x^{5} \log \left (c x^{n}\right ) + \frac {1}{5} \, a e^{2} x^{5} - \frac {2}{9} \, b d e n x^{3} + \frac {2}{3} \, b d e x^{3} \log \left (c x^{n}\right ) + \frac {2}{3} \, a d e x^{3} - b d^{2} n x + b d^{2} x \log \left (c x^{n}\right ) + a d^{2} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.44, size = 74, normalized size = 0.86 \[ \ln \left (c\,x^n\right )\,\left (b\,d^2\,x+\frac {2\,b\,d\,e\,x^3}{3}+\frac {b\,e^2\,x^5}{5}\right )+\frac {e^2\,x^5\,\left (5\,a-b\,n\right )}{25}+d^2\,x\,\left (a-b\,n\right )+\frac {2\,d\,e\,x^3\,\left (3\,a-b\,n\right )}{9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.53, size = 144, normalized size = 1.67 \[ a d^{2} x + \frac {2 a d e x^{3}}{3} + \frac {a e^{2} x^{5}}{5} + b d^{2} n x \log {\relax (x )} - b d^{2} n x + b d^{2} x \log {\relax (c )} + \frac {2 b d e n x^{3} \log {\relax (x )}}{3} - \frac {2 b d e n x^{3}}{9} + \frac {2 b d e x^{3} \log {\relax (c )}}{3} + \frac {b e^{2} n x^{5} \log {\relax (x )}}{5} - \frac {b e^{2} n x^{5}}{25} + \frac {b e^{2} x^{5} \log {\relax (c )}}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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