3.3.41 \(\int \frac {(a+b x)^3}{(c+d x)^5 \log (e (\frac {a+b x}{c+d x})^n)} \, dx\) [241]

Optimal. Leaf size=75 \[ \frac {(a+b x)^4 \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-4/n} \text {Ei}\left (\frac {4 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{n}\right )}{(b c-a d) n (c+d x)^4} \]

[Out]

(b*x+a)^4*Ei(4*ln(e*((b*x+a)/(d*x+c))^n)/n)/(-a*d+b*c)/n/((e*((b*x+a)/(d*x+c))^n)^(4/n))/(d*x+c)^4

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Rubi [A]
time = 0.07, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {2561, 2347, 2209} \begin {gather*} \frac {(a+b x)^4 \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-4/n} \text {Ei}\left (\frac {4 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{n}\right )}{n (c+d x)^4 (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^3/((c + d*x)^5*Log[e*((a + b*x)/(c + d*x))^n]),x]

[Out]

((a + b*x)^4*ExpIntegralEi[(4*Log[e*((a + b*x)/(c + d*x))^n])/n])/((b*c - a*d)*n*(e*((a + b*x)/(c + d*x))^n)^(
4/n)*(c + d*x)^4)

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2347

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

Rule 2561

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

Rubi steps

\begin {align*} \int \frac {(a+b x)^3}{(c+d x)^5 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx &=\frac {(a+b x)^4 \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-4/n} \text {Ei}\left (\frac {4 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{n}\right )}{(b c-a d) n (c+d x)^4}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 75, normalized size = 1.00 \begin {gather*} \frac {(a+b x)^4 \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-4/n} \text {Ei}\left (\frac {4 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{n}\right )}{(b c-a d) n (c+d x)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^3/((c + d*x)^5*Log[e*((a + b*x)/(c + d*x))^n]),x]

[Out]

((a + b*x)^4*ExpIntegralEi[(4*Log[e*((a + b*x)/(c + d*x))^n])/n])/((b*c - a*d)*n*(e*((a + b*x)/(c + d*x))^n)^(
4/n)*(c + d*x)^4)

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Maple [F]
time = 0.23, size = 0, normalized size = 0.00 \[\int \frac {\left (b x +a \right )^{3}}{\left (d x +c \right )^{5} \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^3/(d*x+c)^5/ln(e*((b*x+a)/(d*x+c))^n),x)

[Out]

int((b*x+a)^3/(d*x+c)^5/ln(e*((b*x+a)/(d*x+c))^n),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((b*x+a)^3/(d*x+c)^5/log(e*((b*x+a)/(d*x+c))^n),x, algorithm="maxima")

[Out]

integrate((b*x + a)^3/((d*x + c)^5*log(((b*x + a)/(d*x + c))^n*e)), x)

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Fricas [A]
time = 0.35, size = 106, normalized size = 1.41 \begin {gather*} \frac {e^{\left (-\frac {4}{n}\right )} \operatorname {log\_integral}\left (\frac {{\left (b^{4} x^{4} + 4 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x + a^{4}\right )} e^{\frac {4}{n}}}{d^{4} x^{4} + 4 \, c d^{3} x^{3} + 6 \, c^{2} d^{2} x^{2} + 4 \, c^{3} d x + c^{4}}\right )}{{\left (b c - a d\right )} n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3/(d*x+c)^5/log(e*((b*x+a)/(d*x+c))^n),x, algorithm="fricas")

[Out]

e^(-4/n)*log_integral((b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2*x^2 + 4*a^3*b*x + a^4)*e^(4/n)/(d^4*x^4 + 4*c*d^3*x^3
+ 6*c^2*d^2*x^2 + 4*c^3*d*x + c^4))/((b*c - a*d)*n)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**3/(d*x+c)**5/ln(e*((b*x+a)/(d*x+c))**n),x)

[Out]

Timed out

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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((b*x+a)^3/(d*x+c)^5/log(e*((b*x+a)/(d*x+c))^n),x, algorithm="giac")

[Out]

integrate((b*x + a)^3/((d*x + c)^5*log(((b*x + a)/(d*x + c))^n*e)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,x\right )}^3}{\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\,{\left (c+d\,x\right )}^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^3/(log(e*((a + b*x)/(c + d*x))^n)*(c + d*x)^5),x)

[Out]

int((a + b*x)^3/(log(e*((a + b*x)/(c + d*x))^n)*(c + d*x)^5), x)

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