3.8 \(\int (a+b x)^3 \log (e (f (a+b x)^p (c+d x)^q)^r) \, dx\)

Optimal. Leaf size=172 \[ -\frac {q r (b c-a d)^4 \log (c+d x)}{4 b d^4}+\frac {q r x (b c-a d)^3}{4 d^3}-\frac {q r (a+b x)^2 (b c-a d)^2}{8 b d^2}+\frac {(a+b x)^4 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{4 b}+\frac {q r (a+b x)^3 (b c-a d)}{12 b d}-\frac {p r (a+b x)^4}{16 b}-\frac {q r (a+b x)^4}{16 b} \]

[Out]

1/4*(-a*d+b*c)^3*q*r*x/d^3-1/8*(-a*d+b*c)^2*q*r*(b*x+a)^2/b/d^2+1/12*(-a*d+b*c)*q*r*(b*x+a)^3/b/d-1/16*p*r*(b*
x+a)^4/b-1/16*q*r*(b*x+a)^4/b-1/4*(-a*d+b*c)^4*q*r*ln(d*x+c)/b/d^4+1/4*(b*x+a)^4*ln(e*(f*(b*x+a)^p*(d*x+c)^q)^
r)/b

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Rubi [A]  time = 0.07, antiderivative size = 172, normalized size of antiderivative = 1.00, 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} \[ \frac {q r x (b c-a d)^3}{4 d^3}-\frac {q r (a+b x)^2 (b c-a d)^2}{8 b d^2}-\frac {q r (b c-a d)^4 \log (c+d x)}{4 b d^4}+\frac {(a+b x)^4 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{4 b}+\frac {q r (a+b x)^3 (b c-a d)}{12 b d}-\frac {p r (a+b x)^4}{16 b}-\frac {q r (a+b x)^4}{16 b} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^3*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r],x]

[Out]

((b*c - a*d)^3*q*r*x)/(4*d^3) - ((b*c - a*d)^2*q*r*(a + b*x)^2)/(8*b*d^2) + ((b*c - a*d)*q*r*(a + b*x)^3)/(12*
b*d) - (p*r*(a + b*x)^4)/(16*b) - (q*r*(a + b*x)^4)/(16*b) - ((b*c - a*d)^4*q*r*Log[c + d*x])/(4*b*d^4) + ((a
+ b*x)^4*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r])/(4*b)

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])

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]

Rubi steps

\begin {align*} \int (a+b x)^3 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right ) \, dx &=\frac {(a+b x)^4 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{4 b}-\frac {1}{4} (p r) \int (a+b x)^3 \, dx-\frac {(d q r) \int \frac {(a+b x)^4}{c+d x} \, dx}{4 b}\\ &=-\frac {p r (a+b x)^4}{16 b}+\frac {(a+b x)^4 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{4 b}-\frac {(d q r) \int \left (-\frac {b (b c-a d)^3}{d^4}+\frac {b (b c-a d)^2 (a+b x)}{d^3}-\frac {b (b c-a d) (a+b x)^2}{d^2}+\frac {b (a+b x)^3}{d}+\frac {(-b c+a d)^4}{d^4 (c+d x)}\right ) \, dx}{4 b}\\ &=\frac {(b c-a d)^3 q r x}{4 d^3}-\frac {(b c-a d)^2 q r (a+b x)^2}{8 b d^2}+\frac {(b c-a d) q r (a+b x)^3}{12 b d}-\frac {p r (a+b x)^4}{16 b}-\frac {q r (a+b x)^4}{16 b}-\frac {(b c-a d)^4 q r \log (c+d x)}{4 b d^4}+\frac {(a+b x)^4 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{4 b}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 154, normalized size = 0.90 \[ \frac {\frac {r \left (4 b^3 (3 p+4 q) (c+d x)^3 (b c-a d)-18 b^2 (p+2 q) (c+d x)^2 (b c-a d)^2+12 b d x (p+4 q) (b c-a d)^3-12 q (b c-a d)^4 \log (c+d x)-3 b^4 (p+q) (c+d x)^4\right )}{12 d^4}+(a+b x)^4 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^3*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r],x]

[Out]

((r*(12*b*d*(b*c - a*d)^3*(p + 4*q)*x - 18*b^2*(b*c - a*d)^2*(p + 2*q)*(c + d*x)^2 + 4*b^3*(b*c - a*d)*(3*p +
4*q)*(c + d*x)^3 - 3*b^4*(p + q)*(c + d*x)^4 - 12*(b*c - a*d)^4*q*Log[c + d*x]))/(12*d^4) + (a + b*x)^4*Log[e*
(f*(a + b*x)^p*(c + d*x)^q)^r])/(4*b)

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fricas [B]  time = 0.44, size = 469, normalized size = 2.73 \[ -\frac {3 \, {\left (b^{4} d^{4} p + b^{4} d^{4} q\right )} r x^{4} + 4 \, {\left (3 \, a b^{3} d^{4} p - {\left (b^{4} c d^{3} - 4 \, a b^{3} d^{4}\right )} q\right )} r x^{3} + 6 \, {\left (3 \, a^{2} b^{2} d^{4} p + {\left (b^{4} c^{2} d^{2} - 4 \, a b^{3} c d^{3} + 6 \, a^{2} b^{2} d^{4}\right )} q\right )} r x^{2} + 12 \, {\left (a^{3} b d^{4} p - {\left (b^{4} c^{3} d - 4 \, a b^{3} c^{2} d^{2} + 6 \, a^{2} b^{2} c d^{3} - 4 \, a^{3} b d^{4}\right )} q\right )} r x - 12 \, {\left (b^{4} d^{4} p r x^{4} + 4 \, a b^{3} d^{4} p r x^{3} + 6 \, a^{2} b^{2} d^{4} p r x^{2} + 4 \, a^{3} b d^{4} p r x + a^{4} d^{4} p r\right )} \log \left (b x + a\right ) - 12 \, {\left (b^{4} d^{4} q r x^{4} + 4 \, a b^{3} d^{4} q r x^{3} + 6 \, a^{2} b^{2} d^{4} q r x^{2} + 4 \, a^{3} b d^{4} q r x - {\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3}\right )} q r\right )} \log \left (d x + c\right ) - 12 \, {\left (b^{4} d^{4} x^{4} + 4 \, a b^{3} d^{4} x^{3} + 6 \, a^{2} b^{2} d^{4} x^{2} + 4 \, a^{3} b d^{4} x\right )} \log \relax (e) - 12 \, {\left (b^{4} d^{4} r x^{4} + 4 \, a b^{3} d^{4} r x^{3} + 6 \, a^{2} b^{2} d^{4} r x^{2} + 4 \, a^{3} b d^{4} r x\right )} \log \relax (f)}{48 \, b d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(3*(b^4*d^4*p + b^4*d^4*q)*r*x^4 + 4*(3*a*b^3*d^4*p - (b^4*c*d^3 - 4*a*b^3*d^4)*q)*r*x^3 + 6*(3*a^2*b^2*
d^4*p + (b^4*c^2*d^2 - 4*a*b^3*c*d^3 + 6*a^2*b^2*d^4)*q)*r*x^2 + 12*(a^3*b*d^4*p - (b^4*c^3*d - 4*a*b^3*c^2*d^
2 + 6*a^2*b^2*c*d^3 - 4*a^3*b*d^4)*q)*r*x - 12*(b^4*d^4*p*r*x^4 + 4*a*b^3*d^4*p*r*x^3 + 6*a^2*b^2*d^4*p*r*x^2
+ 4*a^3*b*d^4*p*r*x + a^4*d^4*p*r)*log(b*x + a) - 12*(b^4*d^4*q*r*x^4 + 4*a*b^3*d^4*q*r*x^3 + 6*a^2*b^2*d^4*q*
r*x^2 + 4*a^3*b*d^4*q*r*x - (b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3)*q*r)*log(d*x + c) -
12*(b^4*d^4*x^4 + 4*a*b^3*d^4*x^3 + 6*a^2*b^2*d^4*x^2 + 4*a^3*b*d^4*x)*log(e) - 12*(b^4*d^4*r*x^4 + 4*a*b^3*d^
4*r*x^3 + 6*a^2*b^2*d^4*r*x^2 + 4*a^3*b*d^4*r*x)*log(f))/(b*d^4)

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giac [B]  time = 8.96, size = 407, normalized size = 2.37 \[ \frac {a^{4} p r \log \left (b x + a\right )}{4 \, b} - \frac {1}{16} \, {\left (b^{3} p r + b^{3} q r - 4 \, b^{3} r \log \relax (f) - 4 \, b^{3}\right )} x^{4} - \frac {{\left (3 \, a b^{2} d p r - b^{3} c q r + 4 \, a b^{2} d q r - 12 \, a b^{2} d r \log \relax (f) - 12 \, a b^{2} d\right )} x^{3}}{12 \, d} + \frac {1}{4} \, {\left (b^{3} p r x^{4} + 4 \, a b^{2} p r x^{3} + 6 \, a^{2} b p r x^{2} + 4 \, a^{3} p r x\right )} \log \left (b x + a\right ) + \frac {1}{4} \, {\left (b^{3} q r x^{4} + 4 \, a b^{2} q r x^{3} + 6 \, a^{2} b q r x^{2} + 4 \, a^{3} q r x\right )} \log \left (d x + c\right ) - \frac {{\left (3 \, a^{2} b d^{2} p r + b^{3} c^{2} q r - 4 \, a b^{2} c d q r + 6 \, a^{2} b d^{2} q r - 12 \, a^{2} b d^{2} r \log \relax (f) - 12 \, a^{2} b d^{2}\right )} x^{2}}{8 \, d^{2}} - \frac {{\left (a^{3} d^{3} p r - b^{3} c^{3} q r + 4 \, a b^{2} c^{2} d q r - 6 \, a^{2} b c d^{2} q r + 4 \, a^{3} d^{3} q r - 4 \, a^{3} d^{3} r \log \relax (f) - 4 \, a^{3} d^{3}\right )} x}{4 \, d^{3}} - \frac {{\left (b^{3} c^{4} q r - 4 \, a b^{2} c^{3} d q r + 6 \, a^{2} b c^{2} d^{2} q r - 4 \, a^{3} c d^{3} q r\right )} \log \left (-d x - c\right )}{4 \, d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^4*p*r*log(b*x + a)/b - 1/16*(b^3*p*r + b^3*q*r - 4*b^3*r*log(f) - 4*b^3)*x^4 - 1/12*(3*a*b^2*d*p*r - b^3
*c*q*r + 4*a*b^2*d*q*r - 12*a*b^2*d*r*log(f) - 12*a*b^2*d)*x^3/d + 1/4*(b^3*p*r*x^4 + 4*a*b^2*p*r*x^3 + 6*a^2*
b*p*r*x^2 + 4*a^3*p*r*x)*log(b*x + a) + 1/4*(b^3*q*r*x^4 + 4*a*b^2*q*r*x^3 + 6*a^2*b*q*r*x^2 + 4*a^3*q*r*x)*lo
g(d*x + c) - 1/8*(3*a^2*b*d^2*p*r + b^3*c^2*q*r - 4*a*b^2*c*d*q*r + 6*a^2*b*d^2*q*r - 12*a^2*b*d^2*r*log(f) -
12*a^2*b*d^2)*x^2/d^2 - 1/4*(a^3*d^3*p*r - b^3*c^3*q*r + 4*a*b^2*c^2*d*q*r - 6*a^2*b*c*d^2*q*r + 4*a^3*d^3*q*r
- 4*a^3*d^3*r*log(f) - 4*a^3*d^3)*x/d^3 - 1/4*(b^3*c^4*q*r - 4*a*b^2*c^3*d*q*r + 6*a^2*b*c^2*d^2*q*r - 4*a^3*
c*d^3*q*r)*log(-d*x - c)/d^4

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maple [F]  time = 0.30, size = 0, normalized size = 0.00 \[ \int \left (b x +a \right )^{3} \ln \left (e \left (f \left (b x +a \right )^{p} \left (d x +c \right )^{q}\right )^{r}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^3*ln(e*(f*(b*x+a)^p*(d*x+c)^q)^r),x)

[Out]

int((b*x+a)^3*ln(e*(f*(b*x+a)^p*(d*x+c)^q)^r),x)

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maxima [A]  time = 0.91, size = 285, normalized size = 1.66 \[ \frac {1}{4} \, {\left (b^{3} x^{4} + 4 \, a b^{2} x^{3} + 6 \, a^{2} b x^{2} + 4 \, a^{3} x\right )} \log \left (\left ({\left (b x + a\right )}^{p} {\left (d x + c\right )}^{q} f\right )^{r} e\right ) + \frac {{\left (\frac {12 \, a^{4} f p \log \left (b x + a\right )}{b} - \frac {3 \, b^{3} d^{3} f {\left (p + q\right )} x^{4} + 4 \, {\left (a b^{2} d^{3} f {\left (3 \, p + 4 \, q\right )} - b^{3} c d^{2} f q\right )} x^{3} + 6 \, {\left (3 \, a^{2} b d^{3} f {\left (p + 2 \, q\right )} + b^{3} c^{2} d f q - 4 \, a b^{2} c d^{2} f q\right )} x^{2} + 12 \, {\left (a^{3} d^{3} f {\left (p + 4 \, q\right )} - b^{3} c^{3} f q + 4 \, a b^{2} c^{2} d f q - 6 \, a^{2} b c d^{2} f q\right )} x}{d^{3}} - \frac {12 \, {\left (b^{3} c^{4} f q - 4 \, a b^{2} c^{3} d f q + 6 \, a^{2} b c^{2} d^{2} f q - 4 \, a^{3} c d^{3} f q\right )} \log \left (d x + c\right )}{d^{4}}\right )} r}{48 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b^3*x^4 + 4*a*b^2*x^3 + 6*a^2*b*x^2 + 4*a^3*x)*log(((b*x + a)^p*(d*x + c)^q*f)^r*e) + 1/48*(12*a^4*f*p*lo
g(b*x + a)/b - (3*b^3*d^3*f*(p + q)*x^4 + 4*(a*b^2*d^3*f*(3*p + 4*q) - b^3*c*d^2*f*q)*x^3 + 6*(3*a^2*b*d^3*f*(
p + 2*q) + b^3*c^2*d*f*q - 4*a*b^2*c*d^2*f*q)*x^2 + 12*(a^3*d^3*f*(p + 4*q) - b^3*c^3*f*q + 4*a*b^2*c^2*d*f*q
- 6*a^2*b*c*d^2*f*q)*x)/d^3 - 12*(b^3*c^4*f*q - 4*a*b^2*c^3*d*f*q + 6*a^2*b*c^2*d^2*f*q - 4*a^3*c*d^3*f*q)*log
(d*x + c)/d^4)*r/f

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mupad [B]  time = 0.53, size = 501, normalized size = 2.91 \[ x^2\,\left (\frac {\left (\frac {b^2\,r\,\left (4\,a\,d\,p+b\,c\,p+5\,a\,d\,q\right )}{4\,d}-\frac {b^2\,r\,\left (p+q\right )\,\left (4\,a\,d+4\,b\,c\right )}{16\,d}\right )\,\left (4\,a\,d+4\,b\,c\right )}{8\,b\,d}-\frac {a\,b\,r\,\left (3\,a\,d\,p+2\,b\,c\,p+5\,a\,d\,q\right )}{4\,d}+\frac {a\,b^2\,c\,r\,\left (p+q\right )}{8\,d}\right )-x^3\,\left (\frac {b^2\,r\,\left (4\,a\,d\,p+b\,c\,p+5\,a\,d\,q\right )}{12\,d}-\frac {b^2\,r\,\left (p+q\right )\,\left (4\,a\,d+4\,b\,c\right )}{48\,d}\right )+\ln \left (e\,{\left (f\,{\left (a+b\,x\right )}^p\,{\left (c+d\,x\right )}^q\right )}^r\right )\,\left (a^3\,x+\frac {3\,a^2\,b\,x^2}{2}+a\,b^2\,x^3+\frac {b^3\,x^4}{4}\right )-x\,\left (\frac {\left (4\,a\,d+4\,b\,c\right )\,\left (\frac {\left (\frac {b^2\,r\,\left (4\,a\,d\,p+b\,c\,p+5\,a\,d\,q\right )}{4\,d}-\frac {b^2\,r\,\left (p+q\right )\,\left (4\,a\,d+4\,b\,c\right )}{16\,d}\right )\,\left (4\,a\,d+4\,b\,c\right )}{4\,b\,d}-\frac {a\,b\,r\,\left (3\,a\,d\,p+2\,b\,c\,p+5\,a\,d\,q\right )}{2\,d}+\frac {a\,b^2\,c\,r\,\left (p+q\right )}{4\,d}\right )}{4\,b\,d}+\frac {a^2\,r\,\left (2\,a\,d\,p+3\,b\,c\,p+5\,a\,d\,q\right )}{2\,d}-\frac {a\,c\,\left (\frac {b^2\,r\,\left (4\,a\,d\,p+b\,c\,p+5\,a\,d\,q\right )}{4\,d}-\frac {b^2\,r\,\left (p+q\right )\,\left (4\,a\,d+4\,b\,c\right )}{16\,d}\right )}{b\,d}\right )-\frac {\ln \left (c+d\,x\right )\,\left (-4\,q\,r\,a^3\,c\,d^3+6\,q\,r\,a^2\,b\,c^2\,d^2-4\,q\,r\,a\,b^2\,c^3\,d+q\,r\,b^3\,c^4\right )}{4\,d^4}-\frac {b^3\,r\,x^4\,\left (p+q\right )}{16}+\frac {a^4\,p\,r\,\ln \left (a+b\,x\right )}{4\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(e*(f*(a + b*x)^p*(c + d*x)^q)^r)*(a + b*x)^3,x)

[Out]

x^2*((((b^2*r*(4*a*d*p + b*c*p + 5*a*d*q))/(4*d) - (b^2*r*(p + q)*(4*a*d + 4*b*c))/(16*d))*(4*a*d + 4*b*c))/(8
*b*d) - (a*b*r*(3*a*d*p + 2*b*c*p + 5*a*d*q))/(4*d) + (a*b^2*c*r*(p + q))/(8*d)) - x^3*((b^2*r*(4*a*d*p + b*c*
p + 5*a*d*q))/(12*d) - (b^2*r*(p + q)*(4*a*d + 4*b*c))/(48*d)) + log(e*(f*(a + b*x)^p*(c + d*x)^q)^r)*(a^3*x +
(b^3*x^4)/4 + (3*a^2*b*x^2)/2 + a*b^2*x^3) - x*(((4*a*d + 4*b*c)*((((b^2*r*(4*a*d*p + b*c*p + 5*a*d*q))/(4*d)
- (b^2*r*(p + q)*(4*a*d + 4*b*c))/(16*d))*(4*a*d + 4*b*c))/(4*b*d) - (a*b*r*(3*a*d*p + 2*b*c*p + 5*a*d*q))/(2
*d) + (a*b^2*c*r*(p + q))/(4*d)))/(4*b*d) + (a^2*r*(2*a*d*p + 3*b*c*p + 5*a*d*q))/(2*d) - (a*c*((b^2*r*(4*a*d*
p + b*c*p + 5*a*d*q))/(4*d) - (b^2*r*(p + q)*(4*a*d + 4*b*c))/(16*d)))/(b*d)) - (log(c + d*x)*(b^3*c^4*q*r - 4
*a^3*c*d^3*q*r - 4*a*b^2*c^3*d*q*r + 6*a^2*b*c^2*d^2*q*r))/(4*d^4) - (b^3*r*x^4*(p + q))/16 + (a^4*p*r*log(a +
b*x))/(4*b)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**3*ln(e*(f*(b*x+a)**p*(d*x+c)**q)**r),x)

[Out]

Timed out

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