3.320 \(\int (f x)^m (d+e x^2) (a+b \log (c x^n)) \, dx\)

Optimal. Leaf size=95 \[ \frac{d (f x)^{m+1} \left (a+b \log \left (c x^n\right )\right )}{f (m+1)}+\frac{e (f x)^{m+3} \left (a+b \log \left (c x^n\right )\right )}{f^3 (m+3)}-\frac{b d n (f x)^{m+1}}{f (m+1)^2}-\frac{b e n (f x)^{m+3}}{f^3 (m+3)^2} \]

[Out]

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

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Rubi [A]  time = 0.085458, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {14, 2350} \[ \frac{d (f x)^{m+1} \left (a+b \log \left (c x^n\right )\right )}{f (m+1)}+\frac{e (f x)^{m+3} \left (a+b \log \left (c x^n\right )\right )}{f^3 (m+3)}-\frac{b d n (f x)^{m+1}}{f (m+1)^2}-\frac{b e n (f x)^{m+3}}{f^3 (m+3)^2} \]

Antiderivative was successfully verified.

[In]

Int[(f*x)^m*(d + e*x^2)*(a + b*Log[c*x^n]),x]

[Out]

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

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]]

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])

Rubi steps

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

Mathematica [A]  time = 0.0723681, size = 68, normalized size = 0.72 \[ x (f x)^m \left (\frac{d \left (a+b \log \left (c x^n\right )\right )}{m+1}+\frac{e x^2 \left (a+b \log \left (c x^n\right )\right )}{m+3}-\frac{b d n}{(m+1)^2}-\frac{b e n x^2}{(m+3)^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.175, size = 1180, normalized size = 12.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*x*(e*m*x^2+e*x^2+d*m+3*d)/(1+m)/(3+m)*exp(1/2*m*(-I*Pi*csgn(I*f*x)^3+I*Pi*csgn(I*f*x)^2*csgn(I*f)+I*Pi*csgn(
I*f*x)^2*csgn(I*x)-I*Pi*csgn(I*f*x)*csgn(I*f)*csgn(I*x)+2*ln(f)+2*ln(x)))*ln(x^n)-1/2*x*(-18*a*d-2*a*d*m^3-14*
a*e*m*x^2+18*b*d*n-30*a*d*m-I*Pi*b*e*m^3*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2-I*Pi*b*e*m^3*x^2*csgn(I*c*x^n)^2*csgn
(I*c)+7*I*Pi*b*e*m*x^2*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+I*Pi*b*e*m^3*x^2*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c
)-2*a*e*m^3*x^2-10*a*e*m^2*x^2-6*a*e*x^2-14*ln(c)*b*d*m^2-30*ln(c)*b*d*m-2*ln(c)*b*d*m^3+7*I*Pi*b*d*m^2*csgn(I
*x^n)*csgn(I*c*x^n)*csgn(I*c)+3*I*Pi*b*e*x^2*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+5*I*Pi*b*e*m^2*x^2*csgn(I*x^n
)*csgn(I*c*x^n)*csgn(I*c)-5*I*Pi*b*e*m^2*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2+I*Pi*b*d*m^3*csgn(I*x^n)*csgn(I*c*x^n
)*csgn(I*c)-7*I*Pi*b*e*m*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2+2*b*e*m^2*n*x^2+2*b*d*m^2*n+9*I*Pi*b*d*csgn(I*c*x^n)^
3-18*ln(c)*b*d+15*I*Pi*b*d*m*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-2*ln(c)*b*e*m^3*x^2-10*ln(c)*b*e*m^2*x^2-14*l
n(c)*b*e*m*x^2+12*b*d*m*n-5*I*Pi*b*e*m^2*x^2*csgn(I*c*x^n)^2*csgn(I*c)-7*I*Pi*b*e*m*x^2*csgn(I*c*x^n)^2*csgn(I
*c)+I*Pi*b*d*m^3*csgn(I*c*x^n)^3+15*I*Pi*b*d*m*csgn(I*c*x^n)^3-3*I*Pi*b*e*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2-6*ln
(c)*b*e*x^2+4*b*e*m*n*x^2-14*a*d*m^2+7*I*Pi*b*e*m*x^2*csgn(I*c*x^n)^3-7*I*Pi*b*d*m^2*csgn(I*x^n)*csgn(I*c*x^n)
^2-7*I*Pi*b*d*m^2*csgn(I*c*x^n)^2*csgn(I*c)+I*Pi*b*e*m^3*x^2*csgn(I*c*x^n)^3-9*I*Pi*b*d*csgn(I*x^n)*csgn(I*c*x
^n)^2-9*I*Pi*b*d*csgn(I*c*x^n)^2*csgn(I*c)+7*I*Pi*b*d*m^2*csgn(I*c*x^n)^3+3*I*Pi*b*e*x^2*csgn(I*c*x^n)^3+9*I*P
i*b*d*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+5*I*Pi*b*e*m^2*x^2*csgn(I*c*x^n)^3-3*I*Pi*b*e*x^2*csgn(I*c*x^n)^2*cs
gn(I*c)-15*I*Pi*b*d*m*csgn(I*x^n)*csgn(I*c*x^n)^2-15*I*Pi*b*d*m*csgn(I*c*x^n)^2*csgn(I*c)-I*Pi*b*d*m^3*csgn(I*
x^n)*csgn(I*c*x^n)^2-I*Pi*b*d*m^3*csgn(I*c*x^n)^2*csgn(I*c)+2*b*e*n*x^2)/(3+m)^2/(1+m)^2*exp(1/2*m*(-I*Pi*csgn
(I*f*x)^3+I*Pi*csgn(I*f*x)^2*csgn(I*f)+I*Pi*csgn(I*f*x)^2*csgn(I*x)-I*Pi*csgn(I*f*x)*csgn(I*f)*csgn(I*x)+2*ln(
f)+2*ln(x)))

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.36516, size = 562, normalized size = 5.92 \begin{align*} \frac{{\left ({\left (a e m^{3} + 5 \, a e m^{2} + 7 \, a e m + 3 \, a e -{\left (b e m^{2} + 2 \, b e m + b e\right )} n\right )} x^{3} +{\left (a d m^{3} + 7 \, a d m^{2} + 15 \, a d m + 9 \, a d -{\left (b d m^{2} + 6 \, b d m + 9 \, b d\right )} n\right )} x +{\left ({\left (b e m^{3} + 5 \, b e m^{2} + 7 \, b e m + 3 \, b e\right )} x^{3} +{\left (b d m^{3} + 7 \, b d m^{2} + 15 \, b d m + 9 \, b d\right )} x\right )} \log \left (c\right ) +{\left ({\left (b e m^{3} + 5 \, b e m^{2} + 7 \, b e m + 3 \, b e\right )} n x^{3} +{\left (b d m^{3} + 7 \, b d m^{2} + 15 \, b d m + 9 \, b d\right )} n x\right )} \log \left (x\right )\right )} e^{\left (m \log \left (f\right ) + m \log \left (x\right )\right )}}{m^{4} + 8 \, m^{3} + 22 \, m^{2} + 24 \, m + 9} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a*e*m^3 + 5*a*e*m^2 + 7*a*e*m + 3*a*e - (b*e*m^2 + 2*b*e*m + b*e)*n)*x^3 + (a*d*m^3 + 7*a*d*m^2 + 15*a*d*m +
9*a*d - (b*d*m^2 + 6*b*d*m + 9*b*d)*n)*x + ((b*e*m^3 + 5*b*e*m^2 + 7*b*e*m + 3*b*e)*x^3 + (b*d*m^3 + 7*b*d*m^
2 + 15*b*d*m + 9*b*d)*x)*log(c) + ((b*e*m^3 + 5*b*e*m^2 + 7*b*e*m + 3*b*e)*n*x^3 + (b*d*m^3 + 7*b*d*m^2 + 15*b
*d*m + 9*b*d)*n*x)*log(x))*e^(m*log(f) + m*log(x))/(m^4 + 8*m^3 + 22*m^2 + 24*m + 9)

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

[Out]

Exception raised: TypeError

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Giac [B]  time = 1.34681, size = 323, normalized size = 3.4 \begin{align*} \frac{b f^{2} f^{m} x^{3} x^{m} e \log \left (c\right )}{f^{2} m + 3 \, f^{2}} + \frac{b f^{m} m n x^{3} x^{m} e \log \left (x\right )}{m^{2} + 6 \, m + 9} + \frac{a f^{2} f^{m} x^{3} x^{m} e}{f^{2} m + 3 \, f^{2}} + \frac{3 \, b f^{m} n x^{3} x^{m} e \log \left (x\right )}{m^{2} + 6 \, m + 9} - \frac{b f^{m} n x^{3} x^{m} e}{m^{2} + 6 \, m + 9} + \frac{b d f^{m} m n x x^{m} \log \left (x\right )}{m^{2} + 2 \, m + 1} + \frac{b d f^{m} n x x^{m} \log \left (x\right )}{m^{2} + 2 \, m + 1} - \frac{b d f^{m} n x x^{m}}{m^{2} + 2 \, m + 1} + \frac{\left (f x\right )^{m} b d x \log \left (c\right )}{m + 1} + \frac{\left (f x\right )^{m} a d x}{m + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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