\(\int (d+e x^2)^2 (a+b \log (c x^n)) \, dx\) [191]

Optimal result

Integrand size = 20, antiderivative size = 86 \[ \int \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=-b d^2 n x-\frac {2}{9} b d e n x^3-\frac {1}{25} b e^2 n x^5+d^2 x \left (a+b \log \left (c x^n\right )\right )+\frac {2}{3} d e x^3 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{5} e^2 x^5 \left (a+b \log \left (c x^n\right )\right ) \]

[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))

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {200, 2350} \[ \int \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=d^2 x \left (a+b \log \left (c x^n\right )\right )+\frac {2}{3} d e x^3 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{5} e^2 x^5 \left (a+b \log \left (c x^n\right )\right )-b d^2 n x-\frac {2}{9} b d e n x^3-\frac {1}{25} b e^2 n x^5 \]

[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 + d^2*x*(a + b*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

Rule 200

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 2350

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(d +
e*x^r)^q, x]}, Dist[a + b*Log[c*x^n], u, 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*} \text {integral}& = 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 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+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 ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.03 \[ \int \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=a d^2 x-b d^2 n x-\frac {2}{9} b d e n x^3-\frac {1}{25} b e^2 n x^5+b d^2 x \log \left (c x^n\right )+\frac {2}{3} d e x^3 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{5} e^2 x^5 \left (a+b \log \left (c x^n\right )\right ) \]

[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

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.08

method	result	size
parallelrisch	\(\frac {x^{5} b \ln \left (c \,x^{n}\right ) e^{2}}{5}-\frac {b \,e^{2} n \,x^{5}}{25}+\frac {x^{5} a \,e^{2}}{5}+\frac {2 x^{3} b \ln \left (c \,x^{n}\right ) d e}{3}-\frac {2 b d e n \,x^{3}}{9}+\frac {2 x^{3} a d e}{3}+x b \ln \left (c \,x^{n}\right ) d^{2}-b \,d^{2} n x +a \,d^{2} x\)	\(93\)
risch	\(\frac {b x \left (3 e^{2} x^{4}+10 d e \,x^{2}+15 d^{2}\right ) \ln \left (x^{n}\right )}{15}+\frac {i \pi b \,e^{2} x^{5} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{10}-\frac {i \pi b d e \,x^{3} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{3}+\frac {i \pi b \,d^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} x}{2}-\frac {i \pi b d e \,x^{3} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{3}-\frac {i \pi b \,e^{2} x^{5} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{10}-\frac {i \pi b \,d^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) x}{2}-\frac {i \pi b \,d^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3} x}{2}+\frac {i \pi b \,d^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} x}{2}+\frac {\ln \left (c \right ) b \,e^{2} x^{5}}{5}+\frac {i \pi b \,e^{2} x^{5} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{10}+\frac {i \pi b d e \,x^{3} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{3}-\frac {i \pi b \,e^{2} x^{5} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{10}+\frac {i \pi b d e \,x^{3} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{3}-\frac {b \,e^{2} n \,x^{5}}{25}+\frac {x^{5} a \,e^{2}}{5}+\frac {2 \ln \left (c \right ) b d e \,x^{3}}{3}-\frac {2 b d e n \,x^{3}}{9}+\frac {2 x^{3} a d e}{3}+\ln \left (c \right ) b \,d^{2} x -b \,d^{2} n x +a \,d^{2} x\)	\(416\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.30 \[ \int \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{25} \, {\left (b e^{2} n - 5 \, a e^{2}\right )} x^{5} - \frac {2}{9} \, {\left (b d e n - 3 \, a d e\right )} x^{3} - {\left (b d^{2} n - a d^{2}\right )} x + \frac {1}{15} \, {\left (3 \, b e^{2} x^{5} + 10 \, b d e x^{3} + 15 \, b d^{2} x\right )} \log \left (c\right ) + \frac {1}{15} \, {\left (3 \, b e^{2} n x^{5} + 10 \, b d e n x^{3} + 15 \, b d^{2} n x\right )} \log \left (x\right ) \]

[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)

Sympy [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.28 \[ \int \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=a d^{2} x + \frac {2 a d e x^{3}}{3} + \frac {a e^{2} x^{5}}{5} - b d^{2} n x + b d^{2} x \log {\left (c x^{n} \right )} - \frac {2 b d e n x^{3}}{9} + \frac {2 b d e x^{3} \log {\left (c x^{n} \right )}}{3} - \frac {b e^{2} n x^{5}}{25} + \frac {b e^{2} x^{5} \log {\left (c x^{n} \right )}}{5} \]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.07 \[ \int \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{25} \, b e^{2} n x^{5} + \frac {1}{5} \, b e^{2} x^{5} \log \left (c x^{n}\right ) + \frac {1}{5} \, a e^{2} x^{5} - \frac {2}{9} \, b d e n x^{3} + \frac {2}{3} \, b d e x^{3} \log \left (c x^{n}\right ) + \frac {2}{3} \, a d e x^{3} - b d^{2} n x + b d^{2} x \log \left (c x^{n}\right ) + a d^{2} x \]

[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

Giac [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.30 \[ \int \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{5} \, b e^{2} n x^{5} \log \left (x\right ) - \frac {1}{25} \, b e^{2} n x^{5} + \frac {1}{5} \, b e^{2} x^{5} \log \left (c\right ) + \frac {1}{5} \, a e^{2} x^{5} + \frac {2}{3} \, b d e n x^{3} \log \left (x\right ) - \frac {2}{9} \, b d e n x^{3} + \frac {2}{3} \, b d e x^{3} \log \left (c\right ) + \frac {2}{3} \, a d e x^{3} + b d^{2} n x \log \left (x\right ) - b d^{2} n x + b d^{2} x \log \left (c\right ) + a d^{2} x \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.86 \[ \int \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=\ln \left (c\,x^n\right )\,\left (b\,d^2\,x+\frac {2\,b\,d\,e\,x^3}{3}+\frac {b\,e^2\,x^5}{5}\right )+\frac {e^2\,x^5\,\left (5\,a-b\,n\right )}{25}+d^2\,x\,\left (a-b\,n\right )+\frac {2\,d\,e\,x^3\,\left (3\,a-b\,n\right )}{9} \]

[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