\(\int x^m \log (d (b x+c x^2)^n) \, dx\) [59]

Optimal result

Integrand size = 18, antiderivative size = 66 \[ \int x^m \log \left (d \left (b x+c x^2\right )^n\right ) \, dx=-\frac {2 n x^{1+m}}{(1+m)^2}+\frac {n x^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {c x}{b}\right )}{(1+m)^2}+\frac {x^{1+m} \log \left (d \left (b x+c x^2\right )^n\right )}{1+m} \]

[Out]

-2*n*x^(1+m)/(1+m)^2+n*x^(1+m)*hypergeom([1, 1+m],[2+m],-c*x/b)/(1+m)^2+x^(1+m)*ln(d*(c*x^2+b*x)^n)/(1+m)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2605, 81, 66} \[ \int x^m \log \left (d \left (b x+c x^2\right )^n\right ) \, dx=\frac {x^{m+1} \log \left (d \left (b x+c x^2\right )^n\right )}{m+1}+\frac {n x^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {c x}{b}\right )}{(m+1)^2}-\frac {2 n x^{m+1}}{(m+1)^2} \]

[In]

Int[x^m*Log[d*(b*x + c*x^2)^n],x]

[Out]

(-2*n*x^(1 + m))/(1 + m)^2 + (n*x^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, -((c*x)/b)])/(1 + m)^2 + (x^(1 +
m)*Log[d*(b*x + c*x^2)^n])/(1 + m)

Rule 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))

Rule 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 2605

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[(d + e*x)^(m +
1)*((a + b*Log[c*RFx^p])^n/(e*(m + 1))), x] - Dist[b*n*(p/(e*(m + 1))), Int[SimplifyIntegrand[(d + e*x)^(m +
1)*(a + b*Log[c*RFx^p])^(n - 1)*(D[RFx, x]/RFx), x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && RationalFunc
tionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || IntegerQ[m]) && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {x^{1+m} \log \left (d \left (b x+c x^2\right )^n\right )}{1+m}-\frac {n \int \frac {x^m (b+2 c x)}{b+c x} \, dx}{1+m} \\ & = -\frac {2 n x^{1+m}}{(1+m)^2}+\frac {x^{1+m} \log \left (d \left (b x+c x^2\right )^n\right )}{1+m}+\frac {(b n) \int \frac {x^m}{b+c x} \, dx}{1+m} \\ & = -\frac {2 n x^{1+m}}{(1+m)^2}+\frac {n x^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {c x}{b}\right )}{(1+m)^2}+\frac {x^{1+m} \log \left (d \left (b x+c x^2\right )^n\right )}{1+m} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.73 \[ \int x^m \log \left (d \left (b x+c x^2\right )^n\right ) \, dx=\frac {x^{1+m} \left (-2 n+n \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {c x}{b}\right )+(1+m) \log \left (d (x (b+c x))^n\right )\right )}{(1+m)^2} \]

[In]

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

[Out]

(x^(1 + m)*(-2*n + n*Hypergeometric2F1[1, 1 + m, 2 + m, -((c*x)/b)] + (1 + m)*Log[d*(x*(b + c*x))^n]))/(1 + m)
^2

Maple [F]

\[\int x^{m} \ln \left (d \left (c \,x^{2}+b x \right )^{n}\right )d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int x^m \log \left (d \left (b x+c x^2\right )^n\right ) \, dx=\int { x^{m} \log \left ({\left (c x^{2} + b x\right )}^{n} d\right ) \,d x } \]

[In]

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

[Out]

integral(x^m*log((c*x^2 + b*x)^n*d), x)

Sympy [F]

\[ \int x^m \log \left (d \left (b x+c x^2\right )^n\right ) \, dx=\int x^{m} \log {\left (d \left (b x + c x^{2}\right )^{n} \right )}\, dx \]

[In]

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

[Out]

Integral(x**m*log(d*(b*x + c*x**2)**n), x)

Maxima [F]

\[ \int x^m \log \left (d \left (b x+c x^2\right )^n\right ) \, dx=\int { x^{m} \log \left ({\left (c x^{2} + b x\right )}^{n} d\right ) \,d x } \]

[In]

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

[Out]

(x*x^m*log((c*x + b)^n) + x*x^m*log(x^n))/(m + 1) + integrate((((m + 1)*log(d) - 2*n)*c*x + ((m + 1)*log(d) -
n)*b)*x^m/(c*(m + 1)*x + b*(m + 1)), x)

Giac [F]

\[ \int x^m \log \left (d \left (b x+c x^2\right )^n\right ) \, dx=\int { x^{m} \log \left ({\left (c x^{2} + b x\right )}^{n} d\right ) \,d x } \]

[In]

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

[Out]

integrate(x^m*log((c*x^2 + b*x)^n*d), x)

Mupad [F(-1)]

Timed out. \[ \int x^m \log \left (d \left (b x+c x^2\right )^n\right ) \, dx=\int x^m\,\ln \left (d\,{\left (c\,x^2+b\,x\right )}^n\right ) \,d x \]

[In]

int(x^m*log(d*(b*x + c*x^2)^n),x)

[Out]

int(x^m*log(d*(b*x + c*x^2)^n), x)