3.3.11 \(\int \frac {(h+i x)^2 (a+b \log (c (e+f x)))^p}{d e+d f x} \, dx\) [211]

Optimal. Leaf size=210 \[ \frac {(f h-e i)^2 (a+b \log (c (e+f x)))^{1+p}}{b d f^3 (1+p)}+\frac {2^{-1-p} e^{-\frac {2 a}{b}} i^2 \Gamma \left (1+p,-\frac {2 (a+b \log (c (e+f x)))}{b}\right ) (a+b \log (c (e+f x)))^p \left (-\frac {a+b \log (c (e+f x))}{b}\right )^{-p}}{c^2 d f^3}+\frac {2 e^{-\frac {a}{b}} i (f h-e i) \Gamma \left (1+p,-\frac {a+b \log (c (e+f x))}{b}\right ) (a+b \log (c (e+f x)))^p \left (-\frac {a+b \log (c (e+f x))}{b}\right )^{-p}}{c d f^3} \]

[Out]

(-e*i+f*h)^2*(a+b*ln(c*(f*x+e)))^(1+p)/b/d/f^3/(1+p)+2^(-1-p)*i^2*GAMMA(1+p,-2*(a+b*ln(c*(f*x+e)))/b)*(a+b*ln(
c*(f*x+e)))^p/c^2/d/exp(2*a/b)/f^3/(((-a-b*ln(c*(f*x+e)))/b)^p)+2*i*(-e*i+f*h)*GAMMA(1+p,(-a-b*ln(c*(f*x+e)))/
b)*(a+b*ln(c*(f*x+e)))^p/c/d/exp(a/b)/f^3/(((-a-b*ln(c*(f*x+e)))/b)^p)

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Rubi [A]
time = 0.34, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2458, 12, 2395, 2336, 2212, 2339, 30, 2346} \begin {gather*} \frac {i^2 2^{-p-1} e^{-\frac {2 a}{b}} (a+b \log (c (e+f x)))^p \left (-\frac {a+b \log (c (e+f x))}{b}\right )^{-p} \text {Gamma}\left (p+1,-\frac {2 (a+b \log (c (e+f x)))}{b}\right )}{c^2 d f^3}+\frac {2 i e^{-\frac {a}{b}} (f h-e i) (a+b \log (c (e+f x)))^p \left (-\frac {a+b \log (c (e+f x))}{b}\right )^{-p} \text {Gamma}\left (p+1,-\frac {a+b \log (c (e+f x))}{b}\right )}{c d f^3}+\frac {(f h-e i)^2 (a+b \log (c (e+f x)))^{p+1}}{b d f^3 (p+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((h + i*x)^2*(a + b*Log[c*(e + f*x)])^p)/(d*e + d*f*x),x]

[Out]

((f*h - e*i)^2*(a + b*Log[c*(e + f*x)])^(1 + p))/(b*d*f^3*(1 + p)) + (2^(-1 - p)*i^2*Gamma[1 + p, (-2*(a + b*L
og[c*(e + f*x)]))/b]*(a + b*Log[c*(e + f*x)])^p)/(c^2*d*E^((2*a)/b)*f^3*(-((a + b*Log[c*(e + f*x)])/b))^p) + (
2*i*(f*h - e*i)*Gamma[1 + p, -((a + b*Log[c*(e + f*x)])/b)]*(a + b*Log[c*(e + f*x)])^p)/(c*d*E^(a/b)*f^3*(-((a
+ b*Log[c*(e + f*x)])/b))^p)

Rule 12

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2212

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-F^(g*(e - c*(f/d))))*((c
+ d*x)^FracPart[m]/(d*((-f)*g*(Log[F]/d))^(IntPart[m] + 1)*((-f)*g*Log[F]*((c + d*x)/d))^FracPart[m]))*Gamma[m
+ 1, ((-f)*g*(Log[F]/d))*(c + d*x)], x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rule 2336

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[1/(n*c^(1/n)), Subst[Int[E^(x/n)*(a + b*x)^p
, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[1/n]

Rule 2339

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2346

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Rule 2395

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

Rule 2458

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[(g*(x/e))^q*((e*h - d*i)/e + i*(x/e))^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rubi steps

\begin {align*} \int \frac {(h+211 x)^2 (a+b \log (c (e+f x)))^p}{d e+d f x} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (\frac {-211 e+f h}{f}+\frac {211 x}{f}\right )^2 (a+b \log (c x))^p}{d x} \, dx,x,e+f x\right )}{f}\\ &=\frac {\text {Subst}\left (\int \frac {\left (\frac {-211 e+f h}{f}+\frac {211 x}{f}\right )^2 (a+b \log (c x))^p}{x} \, dx,x,e+f x\right )}{d f}\\ &=\frac {\text {Subst}\left (\int \left (-\frac {422 (211 e-f h) (a+b \log (c x))^p}{f^2}+\frac {(211 e-f h)^2 (a+b \log (c x))^p}{f^2 x}+\frac {44521 x (a+b \log (c x))^p}{f^2}\right ) \, dx,x,e+f x\right )}{d f}\\ &=\frac {44521 \text {Subst}\left (\int x (a+b \log (c x))^p \, dx,x,e+f x\right )}{d f^3}-\frac {(422 (211 e-f h)) \text {Subst}\left (\int (a+b \log (c x))^p \, dx,x,e+f x\right )}{d f^3}+\frac {(211 e-f h)^2 \text {Subst}\left (\int \frac {(a+b \log (c x))^p}{x} \, dx,x,e+f x\right )}{d f^3}\\ &=\frac {44521 \text {Subst}\left (\int e^{2 x} (a+b x)^p \, dx,x,\log (c (e+f x))\right )}{c^2 d f^3}-\frac {(422 (211 e-f h)) \text {Subst}\left (\int e^x (a+b x)^p \, dx,x,\log (c (e+f x))\right )}{c d f^3}+\frac {(211 e-f h)^2 \text {Subst}\left (\int x^p \, dx,x,a+b \log (c (e+f x))\right )}{b d f^3}\\ &=\frac {(211 e-f h)^2 (a+b \log (c (e+f x)))^{1+p}}{b d f^3 (1+p)}+\frac {44521\ 2^{-1-p} e^{-\frac {2 a}{b}} \Gamma \left (1+p,-\frac {2 (a+b \log (c (e+f x)))}{b}\right ) (a+b \log (c (e+f x)))^p \left (-\frac {a+b \log (c (e+f x))}{b}\right )^{-p}}{c^2 d f^3}-\frac {422 e^{-\frac {a}{b}} (211 e-f h) \Gamma \left (1+p,-\frac {a+b \log (c (e+f x))}{b}\right ) (a+b \log (c (e+f x)))^p \left (-\frac {a+b \log (c (e+f x))}{b}\right )^{-p}}{c d f^3}\\ \end {align*}

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Mathematica [A]
time = 0.27, size = 189, normalized size = 0.90 \begin {gather*} \frac {2^{-1-p} e^{-\frac {2 a}{b}} (a+b \log (c (e+f x)))^p \left (-\frac {a+b \log (c (e+f x))}{b}\right )^{-p} \left (b i^2 (1+p) \Gamma \left (1+p,-\frac {2 (a+b \log (c (e+f x)))}{b}\right )+2^{1+p} c e^{a/b} (f h-e i) \left (2 b i (1+p) \Gamma \left (1+p,-\frac {a+b \log (c (e+f x))}{b}\right )-b c e^{a/b} (f h-e i) \left (-\frac {a+b \log (c (e+f x))}{b}\right )^{1+p}\right )\right )}{b c^2 d f^3 (1+p)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((h + i*x)^2*(a + b*Log[c*(e + f*x)])^p)/(d*e + d*f*x),x]

[Out]

(2^(-1 - p)*(a + b*Log[c*(e + f*x)])^p*(b*i^2*(1 + p)*Gamma[1 + p, (-2*(a + b*Log[c*(e + f*x)]))/b] + 2^(1 + p
)*c*E^(a/b)*(f*h - e*i)*(2*b*i*(1 + p)*Gamma[1 + p, -((a + b*Log[c*(e + f*x)])/b)] - b*c*E^(a/b)*(f*h - e*i)*(
-((a + b*Log[c*(e + f*x)])/b))^(1 + p))))/(b*c^2*d*E^((2*a)/b)*f^3*(1 + p)*(-((a + b*Log[c*(e + f*x)])/b))^p)

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Maple [F]
time = 0.34, size = 0, normalized size = 0.00 \[\int \frac {\left (i x +h \right )^{2} \left (a +b \ln \left (c \left (f x +e \right )\right )\right )^{p}}{d f x +e d}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((i*x+h)^2*(a+b*ln(c*(f*x+e)))^p/(d*f*x+d*e),x)

[Out]

int((i*x+h)^2*(a+b*ln(c*(f*x+e)))^p/(d*f*x+d*e),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x+h)^2*(a+b*log(c*(f*x+e)))^p/(d*f*x+d*e),x, algorithm="maxima")

[Out]

(b*c*log(c*f*x + c*e) + a*c)*(b*log(c*f*x + c*e) + a)^p*h^2/(b*c*d*f*(p + 1)) - integrate((-2*I*h*x + x^2)*(b*
log(f*x + e) + b*log(c) + a)^p/(d*f*x + d*e), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x+h)^2*(a+b*log(c*(f*x+e)))^p/(d*f*x+d*e),x, algorithm="fricas")

[Out]

integral((h^2 + 2*I*h*x - x^2)*(b*log(c*f*x + c*e) + a)^p/(d*f*x + d*e), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {h^{2} \left (a + b \log {\left (c e + c f x \right )}\right )^{p}}{e + f x}\, dx + \int \frac {i^{2} x^{2} \left (a + b \log {\left (c e + c f x \right )}\right )^{p}}{e + f x}\, dx + \int \frac {2 h i x \left (a + b \log {\left (c e + c f x \right )}\right )^{p}}{e + f x}\, dx}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x+h)**2*(a+b*ln(c*(f*x+e)))**p/(d*f*x+d*e),x)

[Out]

(Integral(h**2*(a + b*log(c*e + c*f*x))**p/(e + f*x), x) + Integral(i**2*x**2*(a + b*log(c*e + c*f*x))**p/(e +
f*x), x) + Integral(2*h*i*x*(a + b*log(c*e + c*f*x))**p/(e + f*x), x))/d

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x+h)^2*(a+b*log(c*(f*x+e)))^p/(d*f*x+d*e),x, algorithm="giac")

[Out]

integrate((b*log((f*x + e)*c) + a)^p*(h + I*x)^2/(d*f*x + d*e), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (h+i\,x\right )}^2\,{\left (a+b\,\ln \left (c\,\left (e+f\,x\right )\right )\right )}^p}{d\,e+d\,f\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((h + i*x)^2*(a + b*log(c*(e + f*x)))^p)/(d*e + d*f*x),x)

[Out]

int(((h + i*x)^2*(a + b*log(c*(e + f*x)))^p)/(d*e + d*f*x), x)

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