3.9 \(\int (a+b x)^2 \log (e (f (a+b x)^p (c+d x)^q)^r) \, dx\)

Optimal. Leaf size=143 \[ -\frac{q r x (b c-a d)^2}{3 d^2}+\frac{q r (b c-a d)^3 \log (c+d x)}{3 b d^3}+\frac{(a+b x)^3 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{3 b}+\frac{q r (a+b x)^2 (b c-a d)}{6 b d}-\frac{p r (a+b x)^3}{9 b}-\frac{q r (a+b x)^3}{9 b} \]

[Out]

-((b*c - a*d)^2*q*r*x)/(3*d^2) + ((b*c - a*d)*q*r*(a + b*x)^2)/(6*b*d) - (p*r*(a + b*x)^3)/(9*b) - (q*r*(a + b
*x)^3)/(9*b) + ((b*c - a*d)^3*q*r*Log[c + d*x])/(3*b*d^3) + ((a + b*x)^3*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r])
/(3*b)

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Rubi [A]  time = 0.0608917, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {2495, 32, 43} \[ -\frac{q r x (b c-a d)^2}{3 d^2}+\frac{q r (b c-a d)^3 \log (c+d x)}{3 b d^3}+\frac{(a+b x)^3 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{3 b}+\frac{q r (a+b x)^2 (b c-a d)}{6 b d}-\frac{p r (a+b x)^3}{9 b}-\frac{q r (a+b x)^3}{9 b} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r],x]

[Out]

-((b*c - a*d)^2*q*r*x)/(3*d^2) + ((b*c - a*d)*q*r*(a + b*x)^2)/(6*b*d) - (p*r*(a + b*x)^3)/(9*b) - (q*r*(a + b
*x)^3)/(9*b) + ((b*c - a*d)^3*q*r*Log[c + d*x])/(3*b*d^3) + ((a + b*x)^3*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r])
/(3*b)

Rule 2495

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]*((g_.) + (h_.)*(x_))^(m_.),
x_Symbol] :> Simp[((g + h*x)^(m + 1)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r])/(h*(m + 1)), x] + (-Dist[(b*p*r)/(
h*(m + 1)), Int[(g + h*x)^(m + 1)/(a + b*x), x], x] - Dist[(d*q*r)/(h*(m + 1)), Int[(g + h*x)^(m + 1)/(c + d*x
), x], x]) /; FreeQ[{a, b, c, d, e, f, g, h, m, p, q, r}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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])

Rubi steps

\begin{align*} \int (a+b x)^2 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right ) \, dx &=\frac{(a+b x)^3 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{3 b}-\frac{1}{3} (p r) \int (a+b x)^2 \, dx-\frac{(d q r) \int \frac{(a+b x)^3}{c+d x} \, dx}{3 b}\\ &=-\frac{p r (a+b x)^3}{9 b}+\frac{(a+b x)^3 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{3 b}-\frac{(d q r) \int \left (\frac{b (b c-a d)^2}{d^3}-\frac{b (b c-a d) (a+b x)}{d^2}+\frac{b (a+b x)^2}{d}+\frac{(-b c+a d)^3}{d^3 (c+d x)}\right ) \, dx}{3 b}\\ &=-\frac{(b c-a d)^2 q r x}{3 d^2}+\frac{(b c-a d) q r (a+b x)^2}{6 b d}-\frac{p r (a+b x)^3}{9 b}-\frac{q r (a+b x)^3}{9 b}+\frac{(b c-a d)^3 q r \log (c+d x)}{3 b d^3}+\frac{(a+b x)^3 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{3 b}\\ \end{align*}

Mathematica [A]  time = 0.134685, size = 127, normalized size = 0.89 \[ \frac{(a+b x)^3 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )-\frac{r \left (-3 b^2 (2 p+3 q) (c+d x)^2 (b c-a d)+6 b d x (p+3 q) (b c-a d)^2-6 q (b c-a d)^3 \log (c+d x)+2 b^3 (p+q) (c+d x)^3\right )}{6 d^3}}{3 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^2*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r],x]

[Out]

(-(r*(6*b*d*(b*c - a*d)^2*(p + 3*q)*x - 3*b^2*(b*c - a*d)*(2*p + 3*q)*(c + d*x)^2 + 2*b^3*(p + q)*(c + d*x)^3
- 6*(b*c - a*d)^3*q*Log[c + d*x]))/(6*d^3) + (a + b*x)^3*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r])/(3*b)

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Maple [F]  time = 0.386, size = 0, normalized size = 0. \begin{align*} \int \left ( bx+a \right ) ^{2}\ln \left ( e \left ( f \left ( bx+a \right ) ^{p} \left ( dx+c \right ) ^{q} \right ) ^{r} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^2*ln(e*(f*(b*x+a)^p*(d*x+c)^q)^r),x)

[Out]

int((b*x+a)^2*ln(e*(f*(b*x+a)^p*(d*x+c)^q)^r),x)

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Maxima [A]  time = 1.28831, size = 262, normalized size = 1.83 \begin{align*} \frac{1}{3} \,{\left (b^{2} x^{3} + 3 \, a b x^{2} + 3 \, a^{2} x\right )} \log \left (\left ({\left (b x + a\right )}^{p}{\left (d x + c\right )}^{q} f\right )^{r} e\right ) + \frac{{\left (\frac{6 \, a^{3} f p \log \left (b x + a\right )}{b} - \frac{2 \, b^{2} d^{2} f{\left (p + q\right )} x^{3} + 3 \,{\left (a b d^{2} f{\left (2 \, p + 3 \, q\right )} - b^{2} c d f q\right )} x^{2} + 6 \,{\left (a^{2} d^{2} f{\left (p + 3 \, q\right )} + b^{2} c^{2} f q - 3 \, a b c d f q\right )} x}{d^{2}} + \frac{6 \,{\left (b^{2} c^{3} f q - 3 \, a b c^{2} d f q + 3 \, a^{2} c d^{2} f q\right )} \log \left (d x + c\right )}{d^{3}}\right )} r}{18 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b^2*x^3 + 3*a*b*x^2 + 3*a^2*x)*log(((b*x + a)^p*(d*x + c)^q*f)^r*e) + 1/18*(6*a^3*f*p*log(b*x + a)/b - (2
*b^2*d^2*f*(p + q)*x^3 + 3*(a*b*d^2*f*(2*p + 3*q) - b^2*c*d*f*q)*x^2 + 6*(a^2*d^2*f*(p + 3*q) + b^2*c^2*f*q -
3*a*b*c*d*f*q)*x)/d^2 + 6*(b^2*c^3*f*q - 3*a*b*c^2*d*f*q + 3*a^2*c*d^2*f*q)*log(d*x + c)/d^3)*r/f

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Fricas [B]  time = 0.880294, size = 693, normalized size = 4.85 \begin{align*} -\frac{2 \,{\left (b^{3} d^{3} p + b^{3} d^{3} q\right )} r x^{3} + 3 \,{\left (2 \, a b^{2} d^{3} p -{\left (b^{3} c d^{2} - 3 \, a b^{2} d^{3}\right )} q\right )} r x^{2} + 6 \,{\left (a^{2} b d^{3} p +{\left (b^{3} c^{2} d - 3 \, a b^{2} c d^{2} + 3 \, a^{2} b d^{3}\right )} q\right )} r x - 6 \,{\left (b^{3} d^{3} p r x^{3} + 3 \, a b^{2} d^{3} p r x^{2} + 3 \, a^{2} b d^{3} p r x + a^{3} d^{3} p r\right )} \log \left (b x + a\right ) - 6 \,{\left (b^{3} d^{3} q r x^{3} + 3 \, a b^{2} d^{3} q r x^{2} + 3 \, a^{2} b d^{3} q r x +{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2}\right )} q r\right )} \log \left (d x + c\right ) - 6 \,{\left (b^{3} d^{3} x^{3} + 3 \, a b^{2} d^{3} x^{2} + 3 \, a^{2} b d^{3} x\right )} \log \left (e\right ) - 6 \,{\left (b^{3} d^{3} r x^{3} + 3 \, a b^{2} d^{3} r x^{2} + 3 \, a^{2} b d^{3} r x\right )} \log \left (f\right )}{18 \, b d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/18*(2*(b^3*d^3*p + b^3*d^3*q)*r*x^3 + 3*(2*a*b^2*d^3*p - (b^3*c*d^2 - 3*a*b^2*d^3)*q)*r*x^2 + 6*(a^2*b*d^3*
p + (b^3*c^2*d - 3*a*b^2*c*d^2 + 3*a^2*b*d^3)*q)*r*x - 6*(b^3*d^3*p*r*x^3 + 3*a*b^2*d^3*p*r*x^2 + 3*a^2*b*d^3*
p*r*x + a^3*d^3*p*r)*log(b*x + a) - 6*(b^3*d^3*q*r*x^3 + 3*a*b^2*d^3*q*r*x^2 + 3*a^2*b*d^3*q*r*x + (b^3*c^3 -
3*a*b^2*c^2*d + 3*a^2*b*c*d^2)*q*r)*log(d*x + c) - 6*(b^3*d^3*x^3 + 3*a*b^2*d^3*x^2 + 3*a^2*b*d^3*x)*log(e) -
6*(b^3*d^3*r*x^3 + 3*a*b^2*d^3*r*x^2 + 3*a^2*b*d^3*r*x)*log(f))/(b*d^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**2*ln(e*(f*(b*x+a)**p*(d*x+c)**q)**r),x)

[Out]

Timed out

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Giac [B]  time = 1.30847, size = 566, normalized size = 3.96 \begin{align*} -\frac{1}{9} \,{\left (b^{2} p r + b^{2} q r - 3 \, b^{2} r \log \left (f\right ) - 3 \, b^{2}\right )} x^{3} - \frac{{\left (2 \, a b d p r - b^{2} c q r + 3 \, a b d q r - 6 \, a b d r \log \left (f\right ) - 6 \, a b d\right )} x^{2}}{6 \, d} + \frac{1}{3} \,{\left (b^{2} p r x^{3} + 3 \, a b p r x^{2} + 3 \, a^{2} p r x\right )} \log \left (b x + a\right ) + \frac{1}{3} \,{\left (b^{2} q r x^{3} + 3 \, a b q r x^{2} + 3 \, a^{2} q r x\right )} \log \left (d x + c\right ) - \frac{{\left (a^{2} d^{2} p r + b^{2} c^{2} q r - 3 \, a b c d q r + 3 \, a^{2} d^{2} q r - 3 \, a^{2} d^{2} r \log \left (f\right ) - 3 \, a^{2} d^{2}\right )} x}{3 \, d^{2}} + \frac{{\left (a^{3} d^{3} p r + b^{3} c^{3} q r - 3 \, a b^{2} c^{2} d q r + 3 \, a^{2} b c d^{2} q r\right )} \log \left ({\left | b d x^{2} + b c x + a d x + a c \right |}\right )}{6 \, b d^{3}} + \frac{{\left (a^{3} b c d^{3} p r - a^{4} d^{4} p r - b^{4} c^{4} q r + 4 \, a b^{3} c^{3} d q r - 6 \, a^{2} b^{2} c^{2} d^{2} q r + 3 \, a^{3} b c d^{3} q r\right )} \log \left ({\left | \frac{2 \, b d x + b c + a d -{\left | b c - a d \right |}}{2 \, b d x + b c + a d +{\left | b c - a d \right |}} \right |}\right )}{6 \, b d^{3}{\left | b c - a d \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*(b^2*p*r + b^2*q*r - 3*b^2*r*log(f) - 3*b^2)*x^3 - 1/6*(2*a*b*d*p*r - b^2*c*q*r + 3*a*b*d*q*r - 6*a*b*d*r
*log(f) - 6*a*b*d)*x^2/d + 1/3*(b^2*p*r*x^3 + 3*a*b*p*r*x^2 + 3*a^2*p*r*x)*log(b*x + a) + 1/3*(b^2*q*r*x^3 + 3
*a*b*q*r*x^2 + 3*a^2*q*r*x)*log(d*x + c) - 1/3*(a^2*d^2*p*r + b^2*c^2*q*r - 3*a*b*c*d*q*r + 3*a^2*d^2*q*r - 3*
a^2*d^2*r*log(f) - 3*a^2*d^2)*x/d^2 + 1/6*(a^3*d^3*p*r + b^3*c^3*q*r - 3*a*b^2*c^2*d*q*r + 3*a^2*b*c*d^2*q*r)*
log(abs(b*d*x^2 + b*c*x + a*d*x + a*c))/(b*d^3) + 1/6*(a^3*b*c*d^3*p*r - a^4*d^4*p*r - b^4*c^4*q*r + 4*a*b^3*c
^3*d*q*r - 6*a^2*b^2*c^2*d^2*q*r + 3*a^3*b*c*d^3*q*r)*log(abs((2*b*d*x + b*c + a*d - abs(b*c - a*d))/(2*b*d*x
+ b*c + a*d + abs(b*c - a*d))))/(b*d^3*abs(b*c - a*d))