3.178 \(\int \frac{(h+i x) (a+b \log (c (e+f x)))}{d e+d f x} \, dx\)

Optimal. Leaf size=79 \[ \frac{(f h-e i) (a+b \log (c (e+f x)))^2}{2 b d f^2}+\frac{a i x}{d f}+\frac{b i (e+f x) \log (c (e+f x))}{d f^2}-\frac{b i x}{d f} \]

[Out]

(a*i*x)/(d*f) - (b*i*x)/(d*f) + (b*i*(e + f*x)*Log[c*(e + f*x)])/(d*f^2) + ((f*h - e*i)*(a + b*Log[c*(e + f*x)
])^2)/(2*b*d*f^2)

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Rubi [A]  time = 0.129091, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {2411, 12, 2346, 2301, 2295} \[ \frac{(f h-e i) (a+b \log (c (e+f x)))^2}{2 b d f^2}+\frac{a i x}{d f}+\frac{b i (e+f x) \log (c (e+f x))}{d f^2}-\frac{b i x}{d f} \]

Antiderivative was successfully verified.

[In]

Int[((h + i*x)*(a + b*Log[c*(e + f*x)]))/(d*e + d*f*x),x]

[Out]

(a*i*x)/(d*f) - (b*i*x)/(d*f) + (b*i*(e + f*x)*Log[c*(e + f*x)])/(d*f^2) + ((f*h - e*i)*(a + b*Log[c*(e + f*x)
])^2)/(2*b*d*f^2)

Rule 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2346

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.))/(x_), x_Symbol] :> Dist[d, Int[((d
+ e*x)^(q - 1)*(a + b*Log[c*x^n])^p)/x, x], x] + Dist[e, Int[(d + e*x)^(q - 1)*(a + b*Log[c*x^n])^p, x], x] /
; FreeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && GtQ[q, 0] && IntegerQ[2*q]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps

\begin{align*} \int \frac{(h+178 x) (a+b \log (c (e+f x)))}{d e+d f x} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (\frac{-178 e+f h}{f}+\frac{178 x}{f}\right ) (a+b \log (c x))}{d x} \, dx,x,e+f x\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (\frac{-178 e+f h}{f}+\frac{178 x}{f}\right ) (a+b \log (c x))}{x} \, dx,x,e+f x\right )}{d f}\\ &=\frac{178 \operatorname{Subst}(\int (a+b \log (c x)) \, dx,x,e+f x)}{d f^2}-\frac{(178 e-f h) \operatorname{Subst}\left (\int \frac{a+b \log (c x)}{x} \, dx,x,e+f x\right )}{d f^2}\\ &=\frac{178 a x}{d f}-\frac{(178 e-f h) (a+b \log (c (e+f x)))^2}{2 b d f^2}+\frac{(178 b) \operatorname{Subst}(\int \log (c x) \, dx,x,e+f x)}{d f^2}\\ &=\frac{178 a x}{d f}-\frac{178 b x}{d f}+\frac{178 b (e+f x) \log (c (e+f x))}{d f^2}-\frac{(178 e-f h) (a+b \log (c (e+f x)))^2}{2 b d f^2}\\ \end{align*}

Mathematica [A]  time = 0.0523261, size = 66, normalized size = 0.84 \[ \frac{\frac{(f h-e i) (a+b \log (c (e+f x)))^2}{b}+2 a f i x+2 b i (e+f x) \log (c (e+f x))-2 b f i x}{2 d f^2} \]

Antiderivative was successfully verified.

[In]

Integrate[((h + i*x)*(a + b*Log[c*(e + f*x)]))/(d*e + d*f*x),x]

[Out]

(2*a*f*i*x - 2*b*f*i*x + 2*b*i*(e + f*x)*Log[c*(e + f*x)] + ((f*h - e*i)*(a + b*Log[c*(e + f*x)])^2)/b)/(2*d*f
^2)

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Maple [B]  time = 0.06, size = 163, normalized size = 2.1 \begin{align*} -{\frac{aei\ln \left ( cfx+ce \right ) }{d{f}^{2}}}+{\frac{ah\ln \left ( cfx+ce \right ) }{df}}+{\frac{aix}{df}}+{\frac{aei}{d{f}^{2}}}-{\frac{bei \left ( \ln \left ( cfx+ce \right ) \right ) ^{2}}{2\,d{f}^{2}}}+{\frac{bh \left ( \ln \left ( cfx+ce \right ) \right ) ^{2}}{2\,df}}+{\frac{bi\ln \left ( cfx+ce \right ) x}{df}}+{\frac{bi\ln \left ( cfx+ce \right ) e}{d{f}^{2}}}-{\frac{bix}{df}}-{\frac{bei}{d{f}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((i*x+h)*(a+b*ln(c*(f*x+e)))/(d*f*x+d*e),x)

[Out]

-1/f^2/d*a*e*i*ln(c*f*x+c*e)+1/f/d*a*h*ln(c*f*x+c*e)+a*i*x/d/f+1/f^2/d*a*e*i-1/2/f^2/d*b*e*i*ln(c*f*x+c*e)^2+1
/2/f/d*b*h*ln(c*f*x+c*e)^2+1/f/d*b*i*ln(c*f*x+c*e)*x+1/f^2/d*b*i*ln(c*f*x+c*e)*e-b*i*x/d/f-1/f^2/d*b*e*i

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Maxima [B]  time = 1.1177, size = 271, normalized size = 3.43 \begin{align*} b i{\left (\frac{x}{d f} - \frac{e \log \left (f x + e\right )}{d f^{2}}\right )} \log \left (c f x + c e\right ) - \frac{1}{2} \, b h{\left (\frac{2 \, \log \left (c f x + c e\right ) \log \left (d f x + d e\right )}{d f} - \frac{\log \left (f x + e\right )^{2} + 2 \, \log \left (f x + e\right ) \log \left (c\right )}{d f}\right )} + a i{\left (\frac{x}{d f} - \frac{e \log \left (f x + e\right )}{d f^{2}}\right )} + \frac{b h \log \left (c f x + c e\right ) \log \left (d f x + d e\right )}{d f} + \frac{a h \log \left (d f x + d e\right )}{d f} + \frac{{\left (e \log \left (f x + e\right )^{2} - 2 \, f x + 2 \, e \log \left (f x + e\right )\right )} b i}{2 \, d f^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x+h)*(a+b*log(c*(f*x+e)))/(d*f*x+d*e),x, algorithm="maxima")

[Out]

b*i*(x/(d*f) - e*log(f*x + e)/(d*f^2))*log(c*f*x + c*e) - 1/2*b*h*(2*log(c*f*x + c*e)*log(d*f*x + d*e)/(d*f) -
(log(f*x + e)^2 + 2*log(f*x + e)*log(c))/(d*f)) + a*i*(x/(d*f) - e*log(f*x + e)/(d*f^2)) + b*h*log(c*f*x + c*
e)*log(d*f*x + d*e)/(d*f) + a*h*log(d*f*x + d*e)/(d*f) + 1/2*(e*log(f*x + e)^2 - 2*f*x + 2*e*log(f*x + e))*b*i
/(d*f^2)

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Fricas [A]  time = 1.62039, size = 163, normalized size = 2.06 \begin{align*} \frac{2 \,{\left (a - b\right )} f i x +{\left (b f h - b e i\right )} \log \left (c f x + c e\right )^{2} + 2 \,{\left (b f i x + a f h -{\left (a - b\right )} e i\right )} \log \left (c f x + c e\right )}{2 \, d f^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x+h)*(a+b*log(c*(f*x+e)))/(d*f*x+d*e),x, algorithm="fricas")

[Out]

1/2*(2*(a - b)*f*i*x + (b*f*h - b*e*i)*log(c*f*x + c*e)^2 + 2*(b*f*i*x + a*f*h - (a - b)*e*i)*log(c*f*x + c*e)
)/(d*f^2)

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Sympy [A]  time = 0.89032, size = 82, normalized size = 1.04 \begin{align*} \frac{b i x \log{\left (c \left (e + f x\right ) \right )}}{d f} + \frac{x \left (a i - b i\right )}{d f} + \frac{\left (- b e i + b f h\right ) \log{\left (c \left (e + f x\right ) \right )}^{2}}{2 d f^{2}} - \frac{\left (a e i - a f h - b e i\right ) \log{\left (e + f x \right )}}{d f^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x+h)*(a+b*ln(c*(f*x+e)))/(d*f*x+d*e),x)

[Out]

b*i*x*log(c*(e + f*x))/(d*f) + x*(a*i - b*i)/(d*f) + (-b*e*i + b*f*h)*log(c*(e + f*x))**2/(2*d*f**2) - (a*e*i
- a*f*h - b*e*i)*log(e + f*x)/(d*f**2)

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Giac [A]  time = 1.17167, size = 147, normalized size = 1.86 \begin{align*} \frac{2 \, b f i x \log \left (c f x + c e\right ) + b f h \log \left (c f x + c e\right )^{2} - b i e \log \left (c f x + c e\right )^{2} + 2 \, a f i x - 2 \, b f i x + 2 \, a f h \log \left (f x + e\right ) - 2 \, a i e \log \left (f x + e\right ) + 2 \, b i e \log \left (f x + e\right )}{2 \, d f^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x+h)*(a+b*log(c*(f*x+e)))/(d*f*x+d*e),x, algorithm="giac")

[Out]

1/2*(2*b*f*i*x*log(c*f*x + c*e) + b*f*h*log(c*f*x + c*e)^2 - b*i*e*log(c*f*x + c*e)^2 + 2*a*f*i*x - 2*b*f*i*x
+ 2*a*f*h*log(f*x + e) - 2*a*i*e*log(f*x + e) + 2*b*i*e*log(f*x + e))/(d*f^2)