\(\int (f x)^{-1-(1+q) r} (d+e x^r)^q (a+b \log (c x^n)) \, dx\) [447]

Optimal result

Integrand size = 32, antiderivative size = 119 \[ \int (f x)^{-1-(1+q) r} \left (d+e x^r\right )^q \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {b n (f x)^{-((1+q) r)} \left (d+e x^r\right )^q \left (1+\frac {e x^r}{d}\right )^{-q} \operatorname {Hypergeometric2F1}\left (-1-q,-1-q,-q,-\frac {e x^r}{d}\right )}{f (1+q)^2 r^2}-\frac {(f x)^{-((1+q) r)} \left (d+e x^r\right )^{1+q} \left (a+b \log \left (c x^n\right )\right )}{d f (1+q) r} \]

[Out]

-b*n*(d+e*x^r)^q*hypergeom([-1-q, -1-q],[-q],-e*x^r/d)/f/(1+q)^2/r^2/((f*x)^((1+q)*r))/((1+e*x^r/d)^q)-(d+e*x^
r)^(1+q)*(a+b*ln(c*x^n))/d/f/(1+q)/r/((f*x)^((1+q)*r))

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {2373, 372, 371} \[ \int (f x)^{-1-(1+q) r} \left (d+e x^r\right )^q \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {(f x)^{-((q+1) r)} \left (d+e x^r\right )^{q+1} \left (a+b \log \left (c x^n\right )\right )}{d f (q+1) r}-\frac {b n (f x)^{-((q+1) r)} \left (d+e x^r\right )^q \left (\frac {e x^r}{d}+1\right )^{-q} \operatorname {Hypergeometric2F1}\left (-q-1,-q-1,-q,-\frac {e x^r}{d}\right )}{f (q+1)^2 r^2} \]

[In]

Int[(f*x)^(-1 - (1 + q)*r)*(d + e*x^r)^q*(a + b*Log[c*x^n]),x]

[Out]

-((b*n*(d + e*x^r)^q*Hypergeometric2F1[-1 - q, -1 - q, -q, -((e*x^r)/d)])/(f*(1 + q)^2*r^2*(f*x)^((1 + q)*r)*(
1 + (e*x^r)/d)^q)) - ((d + e*x^r)^(1 + q)*(a + b*Log[c*x^n]))/(d*f*(1 + q)*r*(f*x)^((1 + q)*r))

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/
(1 + b*(x^n/a))^FracPart[p]), Int[(c*x)^m*(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 2373

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp
[(f*x)^(m + 1)*(d + e*x^r)^(q + 1)*((a + b*Log[c*x^n])/(d*f*(m + 1))), x] - Dist[b*(n/(d*(m + 1))), Int[(f*x)^
m*(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && EqQ[m + r*(q + 1) + 1, 0] && NeQ[
m, -1]

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

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.82 \[ \int (f x)^{-1-(1+q) r} \left (d+e x^r\right )^q \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {(f x)^{-((1+q) r)} \left (d+e x^r\right )^q \left (b n \left (1+\frac {e x^r}{d}\right )^{-q} \operatorname {Hypergeometric2F1}\left (-1-q,-1-q,-q,-\frac {e x^r}{d}\right )+\frac {(1+q) r \left (d+e x^r\right ) \left (a+b \log \left (c x^n\right )\right )}{d}\right )}{f (1+q)^2 r^2} \]

[In]

Integrate[(f*x)^(-1 - (1 + q)*r)*(d + e*x^r)^q*(a + b*Log[c*x^n]),x]

[Out]

-(((d + e*x^r)^q*((b*n*Hypergeometric2F1[-1 - q, -1 - q, -q, -((e*x^r)/d)])/(1 + (e*x^r)/d)^q + ((1 + q)*r*(d
+ e*x^r)*(a + b*Log[c*x^n]))/d))/(f*(1 + q)^2*r^2*(f*x)^((1 + q)*r)))

Maple [F]

\[\int \left (f x \right )^{-1-\left (1+q \right ) r} \left (d +e \,x^{r}\right )^{q} \left (a +b \ln \left (c \,x^{n}\right )\right )d x\]

[In]

int((f*x)^(-1-(1+q)*r)*(d+e*x^r)^q*(a+b*ln(c*x^n)),x)

[Out]

int((f*x)^(-1-(1+q)*r)*(d+e*x^r)^q*(a+b*ln(c*x^n)),x)

Fricas [F]

\[ \int (f x)^{-1-(1+q) r} \left (d+e x^r\right )^q \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} {\left (e x^{r} + d\right )}^{q} \left (f x\right )^{-{\left (q + 1\right )} r - 1} \,d x } \]

[In]

integrate((f*x)^(-1-(1+q)*r)*(d+e*x^r)^q*(a+b*log(c*x^n)),x, algorithm="fricas")

[Out]

integral(((f*x)^(-(q + 1)*r - 1)*b*log(c*x^n) + (f*x)^(-(q + 1)*r - 1)*a)*(e*x^r + d)^q, x)

Sympy [F(-1)]

Timed out. \[ \int (f x)^{-1-(1+q) r} \left (d+e x^r\right )^q \left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Timed out} \]

[In]

integrate((f*x)**(-1-(1+q)*r)*(d+e*x**r)**q*(a+b*ln(c*x**n)),x)

[Out]

Timed out

Maxima [F]

\[ \int (f x)^{-1-(1+q) r} \left (d+e x^r\right )^q \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} {\left (e x^{r} + d\right )}^{q} \left (f x\right )^{-{\left (q + 1\right )} r - 1} \,d x } \]

[In]

integrate((f*x)^(-1-(1+q)*r)*(d+e*x^r)^q*(a+b*log(c*x^n)),x, algorithm="maxima")

[Out]

integrate((b*log(c*x^n) + a)*(e*x^r + d)^q*(f*x)^(-(q + 1)*r - 1), x)

Giac [F]

\[ \int (f x)^{-1-(1+q) r} \left (d+e x^r\right )^q \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} {\left (e x^{r} + d\right )}^{q} \left (f x\right )^{-{\left (q + 1\right )} r - 1} \,d x } \]

[In]

integrate((f*x)^(-1-(1+q)*r)*(d+e*x^r)^q*(a+b*log(c*x^n)),x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)*(e*x^r + d)^q*(f*x)^(-(q + 1)*r - 1), x)

Mupad [F(-1)]

Timed out. \[ \int (f x)^{-1-(1+q) r} \left (d+e x^r\right )^q \left (a+b \log \left (c x^n\right )\right ) \, dx=\int \frac {{\left (d+e\,x^r\right )}^q\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (f\,x\right )}^{r\,\left (q+1\right )+1}} \,d x \]

[In]

int(((d + e*x^r)^q*(a + b*log(c*x^n)))/(f*x)^(r*(q + 1) + 1),x)

[Out]

int(((d + e*x^r)^q*(a + b*log(c*x^n)))/(f*x)^(r*(q + 1) + 1), x)