3.4.88 \(\int \frac {(d+e x^r)^2 (a+b \log (c x^n))}{x^2} \, dx\) [388]

Optimal. Leaf size=123 \[ -\frac {b d^2 n}{x}-\frac {2 b d e n x^{-1+r}}{(1-r)^2}-\frac {b e^2 n x^{-1+2 r}}{(1-2 r)^2}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {2 d e x^{-1+r} \left (a+b \log \left (c x^n\right )\right )}{1-r}-\frac {e^2 x^{-1+2 r} \left (a+b \log \left (c x^n\right )\right )}{1-2 r} \]

[Out]

-b*d^2*n/x-2*b*d*e*n*x^(-1+r)/(1-r)^2-b*e^2*n*x^(-1+2*r)/(1-2*r)^2-d^2*(a+b*ln(c*x^n))/x-2*d*e*x^(-1+r)*(a+b*l
n(c*x^n))/(1-r)-e^2*x^(-1+2*r)*(a+b*ln(c*x^n))/(1-2*r)

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Rubi [A]
time = 0.11, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {276, 2372, 14} \begin {gather*} -\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {2 d e x^{r-1} \left (a+b \log \left (c x^n\right )\right )}{1-r}-\frac {e^2 x^{2 r-1} \left (a+b \log \left (c x^n\right )\right )}{1-2 r}-\frac {b d^2 n}{x}-\frac {2 b d e n x^{r-1}}{(1-r)^2}-\frac {b e^2 n x^{2 r-1}}{(1-2 r)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2372

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

Rubi steps

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

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Mathematica [A]
time = 0.17, size = 112, normalized size = 0.91 \begin {gather*} \frac {-b d^2 n \log (x)-d^2 \left (a+b n-b n \log (x)+b \log \left (c x^n\right )\right )+\frac {2 d e x^r \left (-b n+a (-1+r)+b (-1+r) \log \left (c x^n\right )\right )}{(-1+r)^2}+\frac {e^2 x^{2 r} \left (-b n+a (-1+2 r)+b (-1+2 r) \log \left (c x^n\right )\right )}{(1-2 r)^2}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.22, size = 1927, normalized size = 15.67

method	result	size
risch	\(\text {Expression too large to display}\)	\(1927\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d+e*x^r)^2*(a+b*ln(c*x^n))/x^2,x,method=_RETURNVERBOSE)

[Out]

-b*(-e^2*(x^r)^2*r+2*d^2*r^2-4*d*e*x^r*r+e^2*(x^r)^2-3*d^2*r+2*d*e*x^r+d^2)/x/(-1+2*r)/(-1+r)*ln(x^n)-1/2*(2*e
^2*(x^r)^2*a-2*I*Pi*b*e^2*r^3*csgn(I*c)*csgn(I*c*x^n)^2*(x^r)^2+4*d*e*x^r*a+4*I*Pi*b*e^2*r*csgn(I*c*x^n)^3*(x^
r)^2+26*b*d^2*n*r^2-12*b*d^2*n*r+26*ln(c)*b*d^2*r^2-12*ln(c)*b*d^2*r+8*ln(c)*b*d^2*r^4-24*ln(c)*b*d^2*r^3-I*Pi
*b*d^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+2*d^2*b*ln(c)+2*b*d^2*n+2*a*d^2+8*b*d^2*n*r^4-24*b*d^2*n*r^3+5*I*Pi
*b*e^2*r^2*csgn(I*c)*csgn(I*c*x^n)^2*(x^r)^2+8*a*d^2*r^4-24*a*d^2*r^3-I*Pi*b*d^2*csgn(I*c*x^n)^3+26*a*d^2*r^2-
12*a*d^2*r+2*ln(c)*b*e^2*(x^r)^2+4*I*Pi*b*d^2*r^4*csgn(I*c)*csgn(I*c*x^n)^2+4*I*Pi*b*d^2*r^4*csgn(I*x^n)*csgn(
I*c*x^n)^2+13*I*Pi*b*d^2*r^2*csgn(I*c)*csgn(I*c*x^n)^2-16*I*Pi*b*d*e*r^2*csgn(I*c*x^n)^3*x^r-5*I*Pi*b*e^2*r^2*
csgn(I*c*x^n)^3*(x^r)^2+13*I*Pi*b*d^2*r^2*csgn(I*x^n)*csgn(I*c*x^n)^2-6*I*Pi*b*d^2*r*csgn(I*x^n)*csgn(I*c*x^n)
^2-4*a*e^2*r^3*(x^r)^2+10*a*e^2*r^2*(x^r)^2-8*a*e^2*r*(x^r)^2+2*b*e^2*n*(x^r)^2+32*a*d*e*r^2*x^r-20*a*d*e*r*x^
r-4*b*e^2*n*r*(x^r)^2+4*b*d*e*n*x^r+2*b*e^2*n*r^2*(x^r)^2-16*a*d*e*r^3*x^r+10*ln(c)*b*e^2*r^2*(x^r)^2-8*ln(c)*
b*e^2*r*(x^r)^2-4*ln(c)*b*e^2*r^3*(x^r)^2+4*ln(c)*b*d*e*x^r-10*I*Pi*b*d*e*r*csgn(I*x^n)*csgn(I*c*x^n)^2*x^r-2*
I*Pi*b*d*e*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)*x^r+I*Pi*b*e^2*csgn(I*c)*csgn(I*c*x^n)^2*(x^r)^2+12*I*Pi*b*d^2*
r^3*csgn(I*c*x^n)^3-5*I*Pi*b*e^2*r^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)*(x^r)^2+16*I*Pi*b*d*e*r^2*csgn(I*x^n)
*csgn(I*c*x^n)^2*x^r+16*I*Pi*b*d*e*r^2*csgn(I*c)*csgn(I*c*x^n)^2*x^r-4*I*Pi*b*e^2*r*csgn(I*x^n)*csgn(I*c*x^n)^
2*(x^r)^2-8*I*Pi*b*d*e*r^3*csgn(I*x^n)*csgn(I*c*x^n)^2*x^r+4*I*Pi*b*e^2*r*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)*
(x^r)^2-10*I*Pi*b*d*e*r*csgn(I*c)*csgn(I*c*x^n)^2*x^r-12*I*Pi*b*d^2*r^3*csgn(I*c)*csgn(I*c*x^n)^2-12*I*Pi*b*d^
2*r^3*csgn(I*x^n)*csgn(I*c*x^n)^2-13*I*Pi*b*d^2*r^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-16*b*d*e*n*r*x^r+32*ln
(c)*b*d*e*r^2*x^r+6*I*Pi*b*d^2*r*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+8*I*Pi*b*d*e*r^3*csgn(I*c)*csgn(I*x^n)*cs
gn(I*c*x^n)*x^r-6*I*Pi*b*d^2*r*csgn(I*c)*csgn(I*c*x^n)^2+6*I*Pi*b*d^2*r*csgn(I*c*x^n)^3+I*Pi*b*e^2*csgn(I*x^n)
*csgn(I*c*x^n)^2*(x^r)^2-2*I*Pi*b*d*e*csgn(I*c*x^n)^3*x^r+8*I*Pi*b*d*e*r^3*csgn(I*c*x^n)^3*x^r+10*I*Pi*b*d*e*r
*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)*x^r-4*I*Pi*b*d^2*r^4*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-I*Pi*b*e^2*csgn(
I*c)*csgn(I*x^n)*csgn(I*c*x^n)*(x^r)^2+5*I*Pi*b*e^2*r^2*csgn(I*x^n)*csgn(I*c*x^n)^2*(x^r)^2+2*I*Pi*b*e^2*r^3*c
sgn(I*c*x^n)^3*(x^r)^2+2*I*Pi*b*d*e*csgn(I*c)*csgn(I*c*x^n)^2*x^r+2*I*Pi*b*d*e*csgn(I*x^n)*csgn(I*c*x^n)^2*x^r
+2*I*Pi*b*e^2*r^3*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)*(x^r)^2-8*I*Pi*b*d*e*r^3*csgn(I*c)*csgn(I*c*x^n)^2*x^r-4
*I*Pi*b*e^2*r*csgn(I*c)*csgn(I*c*x^n)^2*(x^r)^2-20*ln(c)*b*d*e*r*x^r-4*I*Pi*b*d^2*r^4*csgn(I*c*x^n)^3-13*I*Pi*
b*d^2*r^2*csgn(I*c*x^n)^3-I*Pi*b*e^2*csgn(I*c*x^n)^3*(x^r)^2+I*Pi*b*d^2*csgn(I*x^n)*csgn(I*c*x^n)^2+I*Pi*b*d^2
*csgn(I*c)*csgn(I*c*x^n)^2-16*ln(c)*b*d*e*r^3*x^r+16*b*d*e*n*r^2*x^r-16*I*Pi*b*d*e*r^2*csgn(I*c)*csgn(I*x^n)*c
sgn(I*c*x^n)*x^r+10*I*Pi*b*d*e*r*csgn(I*c*x^n)^3*x^r-2*I*Pi*b*e^2*r^3*csgn(I*x^n)*csgn(I*c*x^n)^2*(x^r)^2+12*I
*Pi*b*d^2*r^3*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n))/(-1+2*r)^2/x/(-1+r)^2

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(r-2>0)', see `assume?` for mor
e details)Is

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 411 vs. \(2 (118) = 236\).
time = 0.37, size = 411, normalized size = 3.34 \begin {gather*} -\frac {4 \, {\left (b d^{2} n + a d^{2}\right )} r^{4} + b d^{2} n - 12 \, {\left (b d^{2} n + a d^{2}\right )} r^{3} + a d^{2} + 13 \, {\left (b d^{2} n + a d^{2}\right )} r^{2} - 6 \, {\left (b d^{2} n + a d^{2}\right )} r - {\left ({\left (2 \, b r^{3} - 5 \, b r^{2} + 4 \, b r - b\right )} e^{2} \log \left (c\right ) + {\left (2 \, b n r^{3} - 5 \, b n r^{2} + 4 \, b n r - b n\right )} e^{2} \log \left (x\right ) + {\left (2 \, a r^{3} - {\left (b n + 5 \, a\right )} r^{2} - b n + 2 \, {\left (b n + 2 \, a\right )} r - a\right )} e^{2}\right )} x^{2 \, r} - 2 \, {\left ({\left (4 \, b d r^{3} - 8 \, b d r^{2} + 5 \, b d r - b d\right )} e \log \left (c\right ) + {\left (4 \, b d n r^{3} - 8 \, b d n r^{2} + 5 \, b d n r - b d n\right )} e \log \left (x\right ) + {\left (4 \, a d r^{3} - b d n - 4 \, {\left (b d n + 2 \, a d\right )} r^{2} - a d + {\left (4 \, b d n + 5 \, a d\right )} r\right )} e\right )} x^{r} + {\left (4 \, b d^{2} r^{4} - 12 \, b d^{2} r^{3} + 13 \, b d^{2} r^{2} - 6 \, b d^{2} r + b d^{2}\right )} \log \left (c\right ) + {\left (4 \, b d^{2} n r^{4} - 12 \, b d^{2} n r^{3} + 13 \, b d^{2} n r^{2} - 6 \, b d^{2} n r + b d^{2} n\right )} \log \left (x\right )}{{\left (4 \, r^{4} - 12 \, r^{3} + 13 \, r^{2} - 6 \, r + 1\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*(b*d^2*n + a*d^2)*r^4 + b*d^2*n - 12*(b*d^2*n + a*d^2)*r^3 + a*d^2 + 13*(b*d^2*n + a*d^2)*r^2 - 6*(b*d^2*n
+ a*d^2)*r - ((2*b*r^3 - 5*b*r^2 + 4*b*r - b)*e^2*log(c) + (2*b*n*r^3 - 5*b*n*r^2 + 4*b*n*r - b*n)*e^2*log(x)
+ (2*a*r^3 - (b*n + 5*a)*r^2 - b*n + 2*(b*n + 2*a)*r - a)*e^2)*x^(2*r) - 2*((4*b*d*r^3 - 8*b*d*r^2 + 5*b*d*r
- b*d)*e*log(c) + (4*b*d*n*r^3 - 8*b*d*n*r^2 + 5*b*d*n*r - b*d*n)*e*log(x) + (4*a*d*r^3 - b*d*n - 4*(b*d*n + 2
*a*d)*r^2 - a*d + (4*b*d*n + 5*a*d)*r)*e)*x^r + (4*b*d^2*r^4 - 12*b*d^2*r^3 + 13*b*d^2*r^2 - 6*b*d^2*r + b*d^2
)*log(c) + (4*b*d^2*n*r^4 - 12*b*d^2*n*r^3 + 13*b*d^2*n*r^2 - 6*b*d^2*n*r + b*d^2*n)*log(x))/((4*r^4 - 12*r^3
+ 13*r^2 - 6*r + 1)*x)

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Sympy [A]
time = 10.13, size = 197, normalized size = 1.60 \begin {gather*} - \frac {a d^{2}}{x} + 2 a d e \left (\begin {cases} \frac {x^{r}}{r x - x} & \text {for}\: r \neq 1 \\\log {\left (x \right )} & \text {otherwise} \end {cases}\right ) + a e^{2} \left (\begin {cases} \frac {x^{2 r}}{2 r x - x} & \text {for}\: r \neq \frac {1}{2} \\\log {\left (x \right )} & \text {otherwise} \end {cases}\right ) - \frac {b d^{2} n}{x} - \frac {b d^{2} \log {\left (c x^{n} \right )}}{x} - 2 b d e n \left (\begin {cases} \frac {\begin {cases} \frac {x^{r}}{r x - x} & \text {for}\: r \neq 1 \\\log {\left (x \right )} & \text {otherwise} \end {cases}}{r - 1} & \text {for}\: r > -\infty \wedge r < \infty \wedge r \neq 1 \\\frac {\log {\left (x \right )}^{2}}{2} & \text {otherwise} \end {cases}\right ) + 2 b d e \left (\begin {cases} \frac {x^{r - 1}}{r - 1} & \text {for}\: r \neq 1 \\\log {\left (x \right )} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} - b e^{2} n \left (\begin {cases} \frac {\begin {cases} \frac {x^{2 r}}{2 r x - x} & \text {for}\: r \neq \frac {1}{2} \\\log {\left (x \right )} & \text {otherwise} \end {cases}}{2 r - 1} & \text {for}\: r > -\infty \wedge r < \infty \wedge r \neq \frac {1}{2} \\\frac {\log {\left (x \right )}^{2}}{2} & \text {otherwise} \end {cases}\right ) + b e^{2} \left (\begin {cases} \frac {x^{2 r - 1}}{2 r - 1} & \text {for}\: r \neq \frac {1}{2} \\\log {\left (x \right )} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+e*x**r)**2*(a+b*ln(c*x**n))/x**2,x)

[Out]

-a*d**2/x + 2*a*d*e*Piecewise((x**r/(r*x - x), Ne(r, 1)), (log(x), True)) + a*e**2*Piecewise((x**(2*r)/(2*r*x
- x), Ne(r, 1/2)), (log(x), True)) - b*d**2*n/x - b*d**2*log(c*x**n)/x - 2*b*d*e*n*Piecewise((Piecewise((x**r/
(r*x - x), Ne(r, 1)), (log(x), True))/(r - 1), (r > -oo) & (r < oo) & Ne(r, 1)), (log(x)**2/2, True)) + 2*b*d*
e*Piecewise((x**(r - 1)/(r - 1), Ne(r, 1)), (log(x), True))*log(c*x**n) - b*e**2*n*Piecewise((Piecewise((x**(2
*r)/(2*r*x - x), Ne(r, 1/2)), (log(x), True))/(2*r - 1), (r > -oo) & (r < oo) & Ne(r, 1/2)), (log(x)**2/2, Tru
e)) + b*e**2*Piecewise((x**(2*r - 1)/(2*r - 1), Ne(r, 1/2)), (log(x), True))*log(c*x**n)

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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((d+e*x^r)^2*(a+b*log(c*x^n))/x^2,x, algorithm="giac")

[Out]

integrate((x^r*e + d)^2*(b*log(c*x^n) + a)/x^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d+e\,x^r\right )}^2\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((d + e*x^r)^2*(a + b*log(c*x^n)))/x^2,x)

[Out]

int(((d + e*x^r)^2*(a + b*log(c*x^n)))/x^2, x)

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