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

Optimal. Leaf size=116 \[ -\frac{q r (b c-a d)^2 \log (c+d x)}{2 b d^2}+\frac{(a+b x)^2 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{2 b}+\frac{q r x (b c-a d)}{2 d}-\frac{q r (a+b x)^2}{4 b}-\frac{1}{2} a p r x-\frac{1}{4} b p r x^2 \]

[Out]

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

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Rubi [A]  time = 0.0432106, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.074, Rules used = {2495, 43} \[ -\frac{q r (b c-a d)^2 \log (c+d x)}{2 b d^2}+\frac{(a+b x)^2 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{2 b}+\frac{q r x (b c-a d)}{2 d}-\frac{q r (a+b x)^2}{4 b}-\frac{1}{2} a p r x-\frac{1}{4} b p r x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*p*r*x)/2 + ((b*c - a*d)*q*r*x)/(2*d) - (b*p*r*x^2)/4 - (q*r*(a + b*x)^2)/(4*b) - ((b*c - a*d)^2*q*r*Log[c
+ d*x])/(2*b*d^2) + ((a + b*x)^2*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r])/(2*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 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) \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right ) \, dx &=\frac{(a+b x)^2 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{2 b}-\frac{1}{2} (p r) \int (a+b x) \, dx-\frac{(d q r) \int \frac{(a+b x)^2}{c+d x} \, dx}{2 b}\\ &=-\frac{1}{2} a p r x-\frac{1}{4} b p r x^2+\frac{(a+b x)^2 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{2 b}-\frac{(d q r) \int \left (-\frac{b (b c-a d)}{d^2}+\frac{b (a+b x)}{d}+\frac{(-b c+a d)^2}{d^2 (c+d x)}\right ) \, dx}{2 b}\\ &=-\frac{1}{2} a p r x+\frac{(b c-a d) q r x}{2 d}-\frac{1}{4} b p r x^2-\frac{q r (a+b x)^2}{4 b}-\frac{(b c-a d)^2 q r \log (c+d x)}{2 b d^2}+\frac{(a+b x)^2 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.19288, size = 105, normalized size = 0.91 \[ \frac{a^2 p r \log (a+b x)}{2 b}-\frac{d x \left (r (2 a d (p+2 q)-2 b c q+b d x (p+q))-2 d (2 a+b x) \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )\right )+2 c q r (b c-2 a d) \log (c+d x)}{4 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.174, size = 0, normalized size = 0. \begin{align*} \int \left ( bx+a \right ) \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)*ln(e*(f*(b*x+a)^p*(d*x+c)^q)^r),x)

[Out]

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

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Maxima [A]  time = 1.35446, size = 159, normalized size = 1.37 \begin{align*} \frac{1}{2} \,{\left (b x^{2} + 2 \, a 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{2 \, a^{2} f p \log \left (b x + a\right )}{b} - \frac{b d f{\left (p + q\right )} x^{2} + 2 \,{\left (a d f{\left (p + 2 \, q\right )} - b c f q\right )} x}{d} - \frac{2 \,{\left (b c^{2} f q - 2 \, a c d f q\right )} \log \left (d x + c\right )}{d^{2}}\right )} r}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.820055, size = 435, normalized size = 3.75 \begin{align*} -\frac{{\left (b^{2} d^{2} p + b^{2} d^{2} q\right )} r x^{2} + 2 \,{\left (a b d^{2} p -{\left (b^{2} c d - 2 \, a b d^{2}\right )} q\right )} r x - 2 \,{\left (b^{2} d^{2} p r x^{2} + 2 \, a b d^{2} p r x + a^{2} d^{2} p r\right )} \log \left (b x + a\right ) - 2 \,{\left (b^{2} d^{2} q r x^{2} + 2 \, a b d^{2} q r x -{\left (b^{2} c^{2} - 2 \, a b c d\right )} q r\right )} \log \left (d x + c\right ) - 2 \,{\left (b^{2} d^{2} x^{2} + 2 \, a b d^{2} x\right )} \log \left (e\right ) - 2 \,{\left (b^{2} d^{2} r x^{2} + 2 \, a b d^{2} r x\right )} \log \left (f\right )}{4 \, b d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [B]  time = 1.35044, size = 381, normalized size = 3.28 \begin{align*} -\frac{1}{4} \,{\left (b p r + b q r - 2 \, b r \log \left (f\right ) - 2 \, b\right )} x^{2} + \frac{1}{2} \,{\left (b p r x^{2} + 2 \, a p r x\right )} \log \left (b x + a\right ) + \frac{1}{2} \,{\left (b q r x^{2} + 2 \, a q r x\right )} \log \left (d x + c\right ) - \frac{{\left (a d p r - b c q r + 2 \, a d q r - 2 \, a d r \log \left (f\right ) - 2 \, a d\right )} x}{2 \, d} + \frac{{\left (a^{2} d^{2} p r - b^{2} c^{2} q r + 2 \, a b c d q r\right )} \log \left ({\left | b d x^{2} + b c x + a d x + a c \right |}\right )}{4 \, b d^{2}} + \frac{{\left (a^{2} b c d^{2} p r - a^{3} d^{3} p r + b^{3} c^{3} q r - 3 \, a b^{2} c^{2} d q r + 2 \, a^{2} b c d^{2} 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 )}{4 \, b d^{2}{\left | b c - a d \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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