3.162 \(\int (f x)^m (d+e x)^3 (a+b \log (c x^n)) \, dx\)

Optimal. Leaf size=211 \[ \frac{3 d^2 e (f x)^{m+2} \left (a+b \log \left (c x^n\right )\right )}{f^2 (m+2)}+\frac{d^3 (f x)^{m+1} \left (a+b \log \left (c x^n\right )\right )}{f (m+1)}+\frac{3 d e^2 (f x)^{m+3} \left (a+b \log \left (c x^n\right )\right )}{f^3 (m+3)}+\frac{e^3 (f x)^{m+4} \left (a+b \log \left (c x^n\right )\right )}{f^4 (m+4)}-\frac{3 b d^2 e n (f x)^{m+2}}{f^2 (m+2)^2}-\frac{b d^3 n (f x)^{m+1}}{f (m+1)^2}-\frac{3 b d e^2 n (f x)^{m+3}}{f^3 (m+3)^2}-\frac{b e^3 n (f x)^{m+4}}{f^4 (m+4)^2} \]

[Out]

-((b*d^3*n*(f*x)^(1 + m))/(f*(1 + m)^2)) - (3*b*d^2*e*n*(f*x)^(2 + m))/(f^2*(2 + m)^2) - (3*b*d*e^2*n*(f*x)^(3
+ m))/(f^3*(3 + m)^2) - (b*e^3*n*(f*x)^(4 + m))/(f^4*(4 + m)^2) + (d^3*(f*x)^(1 + m)*(a + b*Log[c*x^n]))/(f*(
1 + m)) + (3*d^2*e*(f*x)^(2 + m)*(a + b*Log[c*x^n]))/(f^2*(2 + m)) + (3*d*e^2*(f*x)^(3 + m)*(a + b*Log[c*x^n])
)/(f^3*(3 + m)) + (e^3*(f*x)^(4 + m)*(a + b*Log[c*x^n]))/(f^4*(4 + m))

________________________________________________________________________________________

Rubi [A]  time = 0.228266, antiderivative size = 211, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {43, 2350, 14} \[ \frac{3 d^2 e (f x)^{m+2} \left (a+b \log \left (c x^n\right )\right )}{f^2 (m+2)}+\frac{d^3 (f x)^{m+1} \left (a+b \log \left (c x^n\right )\right )}{f (m+1)}+\frac{3 d e^2 (f x)^{m+3} \left (a+b \log \left (c x^n\right )\right )}{f^3 (m+3)}+\frac{e^3 (f x)^{m+4} \left (a+b \log \left (c x^n\right )\right )}{f^4 (m+4)}-\frac{3 b d^2 e n (f x)^{m+2}}{f^2 (m+2)^2}-\frac{b d^3 n (f x)^{m+1}}{f (m+1)^2}-\frac{3 b d e^2 n (f x)^{m+3}}{f^3 (m+3)^2}-\frac{b e^3 n (f x)^{m+4}}{f^4 (m+4)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*d^3*n*(f*x)^(1 + m))/(f*(1 + m)^2)) - (3*b*d^2*e*n*(f*x)^(2 + m))/(f^2*(2 + m)^2) - (3*b*d*e^2*n*(f*x)^(3
+ m))/(f^3*(3 + m)^2) - (b*e^3*n*(f*x)^(4 + m))/(f^4*(4 + m)^2) + (d^3*(f*x)^(1 + m)*(a + b*Log[c*x^n]))/(f*(
1 + m)) + (3*d^2*e*(f*x)^(2 + m)*(a + b*Log[c*x^n]))/(f^2*(2 + m)) + (3*d*e^2*(f*x)^(3 + m)*(a + b*Log[c*x^n])
)/(f^3*(3 + m)) + (e^3*(f*x)^(4 + m)*(a + b*Log[c*x^n]))/(f^4*(4 + m))

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2350

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Wit
h[{u = IntHide[(f*x)^m*(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] /; ((EqQ[r, 1] || EqQ[r, 2]) && IntegerQ[m] && IntegerQ[q - 1/2]) || InverseFunctionFreeQ[u, x]] /
; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && IntegerQ[2*q] && ((IntegerQ[m] && IntegerQ[r]) || IGtQ[q, 0])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int (f x)^m (d+e x)^3 \left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac{d^3 (f x)^{1+m} \left (a+b \log \left (c x^n\right )\right )}{f (1+m)}+\frac{3 d^2 e (f x)^{2+m} \left (a+b \log \left (c x^n\right )\right )}{f^2 (2+m)}+\frac{3 d e^2 (f x)^{3+m} \left (a+b \log \left (c x^n\right )\right )}{f^3 (3+m)}+\frac{e^3 (f x)^{4+m} \left (a+b \log \left (c x^n\right )\right )}{f^4 (4+m)}-(b n) \int (f x)^m \left (\frac{d^3}{1+m}+\frac{3 d^2 e x}{2+m}+\frac{3 d e^2 x^2}{3+m}+\frac{e^3 x^3}{4+m}\right ) \, dx\\ &=\frac{d^3 (f x)^{1+m} \left (a+b \log \left (c x^n\right )\right )}{f (1+m)}+\frac{3 d^2 e (f x)^{2+m} \left (a+b \log \left (c x^n\right )\right )}{f^2 (2+m)}+\frac{3 d e^2 (f x)^{3+m} \left (a+b \log \left (c x^n\right )\right )}{f^3 (3+m)}+\frac{e^3 (f x)^{4+m} \left (a+b \log \left (c x^n\right )\right )}{f^4 (4+m)}-(b n) \int \left (\frac{d^3 (f x)^m}{1+m}+\frac{3 d^2 e (f x)^{1+m}}{f (2+m)}+\frac{3 d e^2 (f x)^{2+m}}{f^2 (3+m)}+\frac{e^3 (f x)^{3+m}}{f^3 (4+m)}\right ) \, dx\\ &=-\frac{b d^3 n (f x)^{1+m}}{f (1+m)^2}-\frac{3 b d^2 e n (f x)^{2+m}}{f^2 (2+m)^2}-\frac{3 b d e^2 n (f x)^{3+m}}{f^3 (3+m)^2}-\frac{b e^3 n (f x)^{4+m}}{f^4 (4+m)^2}+\frac{d^3 (f x)^{1+m} \left (a+b \log \left (c x^n\right )\right )}{f (1+m)}+\frac{3 d^2 e (f x)^{2+m} \left (a+b \log \left (c x^n\right )\right )}{f^2 (2+m)}+\frac{3 d e^2 (f x)^{3+m} \left (a+b \log \left (c x^n\right )\right )}{f^3 (3+m)}+\frac{e^3 (f x)^{4+m} \left (a+b \log \left (c x^n\right )\right )}{f^4 (4+m)}\\ \end{align*}

Mathematica [A]  time = 0.224841, size = 152, normalized size = 0.72 \[ x (f x)^m \left (\frac{3 d^2 e x \left (a+b \log \left (c x^n\right )\right )}{m+2}+\frac{d^3 \left (a+b \log \left (c x^n\right )\right )}{m+1}+\frac{3 d e^2 x^2 \left (a+b \log \left (c x^n\right )\right )}{m+3}+\frac{e^3 x^3 \left (a+b \log \left (c x^n\right )\right )}{m+4}-\frac{3 b d^2 e n x}{(m+2)^2}-\frac{b d^3 n}{(m+1)^2}-\frac{3 b d e^2 n x^2}{(m+3)^2}-\frac{b e^3 n x^3}{(m+4)^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

x*(f*x)^m*(-((b*d^3*n)/(1 + m)^2) - (3*b*d^2*e*n*x)/(2 + m)^2 - (3*b*d*e^2*n*x^2)/(3 + m)^2 - (b*e^3*n*x^3)/(4
+ m)^2 + (d^3*(a + b*Log[c*x^n]))/(1 + m) + (3*d^2*e*x*(a + b*Log[c*x^n]))/(2 + m) + (3*d*e^2*x^2*(a + b*Log[
c*x^n]))/(3 + m) + (e^3*x^3*(a + b*Log[c*x^n]))/(4 + m))

________________________________________________________________________________________

Maple [C]  time = 0.482, size = 5021, normalized size = 23.8 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*(e*x+d)^3*(a+b*log(c*x^n)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.43351, size = 2942, normalized size = 13.94 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a*e^3*m^7 + 16*a*e^3*m^6 + 106*a*e^3*m^5 + 376*a*e^3*m^4 + 769*a*e^3*m^3 + 904*a*e^3*m^2 + 564*a*e^3*m + 144
*a*e^3 - (b*e^3*m^6 + 12*b*e^3*m^5 + 58*b*e^3*m^4 + 144*b*e^3*m^3 + 193*b*e^3*m^2 + 132*b*e^3*m + 36*b*e^3)*n)
*x^4 + 3*(a*d*e^2*m^7 + 17*a*d*e^2*m^6 + 119*a*d*e^2*m^5 + 443*a*d*e^2*m^4 + 944*a*d*e^2*m^3 + 1148*a*d*e^2*m^
2 + 736*a*d*e^2*m + 192*a*d*e^2 - (b*d*e^2*m^6 + 14*b*d*e^2*m^5 + 77*b*d*e^2*m^4 + 212*b*d*e^2*m^3 + 308*b*d*e
^2*m^2 + 224*b*d*e^2*m + 64*b*d*e^2)*n)*x^3 + 3*(a*d^2*e*m^7 + 18*a*d^2*e*m^6 + 134*a*d^2*e*m^5 + 532*a*d^2*e*
m^4 + 1209*a*d^2*e*m^3 + 1562*a*d^2*e*m^2 + 1056*a*d^2*e*m + 288*a*d^2*e - (b*d^2*e*m^6 + 16*b*d^2*e*m^5 + 102
*b*d^2*e*m^4 + 328*b*d^2*e*m^3 + 553*b*d^2*e*m^2 + 456*b*d^2*e*m + 144*b*d^2*e)*n)*x^2 + (a*d^3*m^7 + 19*a*d^3
*m^6 + 151*a*d^3*m^5 + 649*a*d^3*m^4 + 1624*a*d^3*m^3 + 2356*a*d^3*m^2 + 1824*a*d^3*m + 576*a*d^3 - (b*d^3*m^6
+ 18*b*d^3*m^5 + 133*b*d^3*m^4 + 516*b*d^3*m^3 + 1108*b*d^3*m^2 + 1248*b*d^3*m + 576*b*d^3)*n)*x + ((b*e^3*m^
7 + 16*b*e^3*m^6 + 106*b*e^3*m^5 + 376*b*e^3*m^4 + 769*b*e^3*m^3 + 904*b*e^3*m^2 + 564*b*e^3*m + 144*b*e^3)*x^
4 + 3*(b*d*e^2*m^7 + 17*b*d*e^2*m^6 + 119*b*d*e^2*m^5 + 443*b*d*e^2*m^4 + 944*b*d*e^2*m^3 + 1148*b*d*e^2*m^2 +
736*b*d*e^2*m + 192*b*d*e^2)*x^3 + 3*(b*d^2*e*m^7 + 18*b*d^2*e*m^6 + 134*b*d^2*e*m^5 + 532*b*d^2*e*m^4 + 1209
*b*d^2*e*m^3 + 1562*b*d^2*e*m^2 + 1056*b*d^2*e*m + 288*b*d^2*e)*x^2 + (b*d^3*m^7 + 19*b*d^3*m^6 + 151*b*d^3*m^
5 + 649*b*d^3*m^4 + 1624*b*d^3*m^3 + 2356*b*d^3*m^2 + 1824*b*d^3*m + 576*b*d^3)*x)*log(c) + ((b*e^3*m^7 + 16*b
*e^3*m^6 + 106*b*e^3*m^5 + 376*b*e^3*m^4 + 769*b*e^3*m^3 + 904*b*e^3*m^2 + 564*b*e^3*m + 144*b*e^3)*n*x^4 + 3*
(b*d*e^2*m^7 + 17*b*d*e^2*m^6 + 119*b*d*e^2*m^5 + 443*b*d*e^2*m^4 + 944*b*d*e^2*m^3 + 1148*b*d*e^2*m^2 + 736*b
*d*e^2*m + 192*b*d*e^2)*n*x^3 + 3*(b*d^2*e*m^7 + 18*b*d^2*e*m^6 + 134*b*d^2*e*m^5 + 532*b*d^2*e*m^4 + 1209*b*d
^2*e*m^3 + 1562*b*d^2*e*m^2 + 1056*b*d^2*e*m + 288*b*d^2*e)*n*x^2 + (b*d^3*m^7 + 19*b*d^3*m^6 + 151*b*d^3*m^5
+ 649*b*d^3*m^4 + 1624*b*d^3*m^3 + 2356*b*d^3*m^2 + 1824*b*d^3*m + 576*b*d^3)*n*x)*log(x))*e^(m*log(f) + m*log
(x))/(m^8 + 20*m^7 + 170*m^6 + 800*m^5 + 2273*m^4 + 3980*m^3 + 4180*m^2 + 2400*m + 576)

________________________________________________________________________________________

Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)**m*(e*x+d)**3*(a+b*ln(c*x**n)),x)

[Out]

Exception raised: TypeError

________________________________________________________________________________________

Giac [B]  time = 1.40079, size = 717, normalized size = 3.4 \begin{align*} \frac{b f^{3} f^{m} x^{4} x^{m} e^{3} \log \left (c\right )}{f^{3} m + 4 \, f^{3}} + \frac{a f^{3} f^{m} x^{4} x^{m} e^{3}}{f^{3} m + 4 \, f^{3}} + \frac{3 \, b d f^{2} f^{m} x^{3} x^{m} e^{2} \log \left (c\right )}{f^{2} m + 3 \, f^{2}} + \frac{b f^{m} m n x^{4} x^{m} e^{3} \log \left (x\right )}{m^{2} + 8 \, m + 16} + \frac{3 \, b d f^{m} m n x^{3} x^{m} e^{2} \log \left (x\right )}{m^{2} + 6 \, m + 9} + \frac{3 \, b d^{2} f^{m} m n x^{2} x^{m} e \log \left (x\right )}{m^{2} + 4 \, m + 4} + \frac{3 \, a d f^{2} f^{m} x^{3} x^{m} e^{2}}{f^{2} m + 3 \, f^{2}} + \frac{b d^{3} f^{m} m n x x^{m} \log \left (x\right )}{m^{2} + 2 \, m + 1} + \frac{4 \, b f^{m} n x^{4} x^{m} e^{3} \log \left (x\right )}{m^{2} + 8 \, m + 16} + \frac{9 \, b d f^{m} n x^{3} x^{m} e^{2} \log \left (x\right )}{m^{2} + 6 \, m + 9} + \frac{6 \, b d^{2} f^{m} n x^{2} x^{m} e \log \left (x\right )}{m^{2} + 4 \, m + 4} - \frac{b f^{m} n x^{4} x^{m} e^{3}}{m^{2} + 8 \, m + 16} - \frac{3 \, b d f^{m} n x^{3} x^{m} e^{2}}{m^{2} + 6 \, m + 9} - \frac{3 \, b d^{2} f^{m} n x^{2} x^{m} e}{m^{2} + 4 \, m + 4} + \frac{3 \, b d^{2} f^{m} x^{2} x^{m} e \log \left (c\right )}{m + 2} + \frac{b d^{3} f^{m} n x x^{m} \log \left (x\right )}{m^{2} + 2 \, m + 1} - \frac{b d^{3} f^{m} n x x^{m}}{m^{2} + 2 \, m + 1} + \frac{3 \, a d^{2} f^{m} x^{2} x^{m} e}{m + 2} + \frac{\left (f x\right )^{m} b d^{3} x \log \left (c\right )}{m + 1} + \frac{\left (f x\right )^{m} a d^{3} x}{m + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*(e*x+d)^3*(a+b*log(c*x^n)),x, algorithm="giac")

[Out]

b*f^3*f^m*x^4*x^m*e^3*log(c)/(f^3*m + 4*f^3) + a*f^3*f^m*x^4*x^m*e^3/(f^3*m + 4*f^3) + 3*b*d*f^2*f^m*x^3*x^m*e
^2*log(c)/(f^2*m + 3*f^2) + b*f^m*m*n*x^4*x^m*e^3*log(x)/(m^2 + 8*m + 16) + 3*b*d*f^m*m*n*x^3*x^m*e^2*log(x)/(
m^2 + 6*m + 9) + 3*b*d^2*f^m*m*n*x^2*x^m*e*log(x)/(m^2 + 4*m + 4) + 3*a*d*f^2*f^m*x^3*x^m*e^2/(f^2*m + 3*f^2)
+ b*d^3*f^m*m*n*x*x^m*log(x)/(m^2 + 2*m + 1) + 4*b*f^m*n*x^4*x^m*e^3*log(x)/(m^2 + 8*m + 16) + 9*b*d*f^m*n*x^3
*x^m*e^2*log(x)/(m^2 + 6*m + 9) + 6*b*d^2*f^m*n*x^2*x^m*e*log(x)/(m^2 + 4*m + 4) - b*f^m*n*x^4*x^m*e^3/(m^2 +
8*m + 16) - 3*b*d*f^m*n*x^3*x^m*e^2/(m^2 + 6*m + 9) - 3*b*d^2*f^m*n*x^2*x^m*e/(m^2 + 4*m + 4) + 3*b*d^2*f^m*x^
2*x^m*e*log(c)/(m + 2) + b*d^3*f^m*n*x*x^m*log(x)/(m^2 + 2*m + 1) - b*d^3*f^m*n*x*x^m/(m^2 + 2*m + 1) + 3*a*d^
2*f^m*x^2*x^m*e/(m + 2) + (f*x)^m*b*d^3*x*log(c)/(m + 1) + (f*x)^m*a*d^3*x/(m + 1)