3.186 \(\int \frac {(h+i x) (a+b \log (c (e+f x)))^2}{d e+d f x} \, dx\)

Optimal. Leaf size=113 \[ \frac {(f h-e i) (a+b \log (c (e+f x)))^3}{3 b d f^2}+\frac {i (e+f x) (a+b \log (c (e+f x)))^2}{d f^2}-\frac {2 a b i x}{d f}-\frac {2 b^2 i (e+f x) \log (c (e+f x))}{d f^2}+\frac {2 b^2 i x}{d f} \]

[Out]

-2*a*b*i*x/d/f+2*b^2*i*x/d/f-2*b^2*i*(f*x+e)*ln(c*(f*x+e))/d/f^2+i*(f*x+e)*(a+b*ln(c*(f*x+e)))^2/d/f^2+1/3*(-e
*i+f*h)*(a+b*ln(c*(f*x+e)))^3/b/d/f^2

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Rubi [A]  time = 0.20, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {2411, 12, 2346, 2302, 30, 2296, 2295} \[ \frac {(f h-e i) (a+b \log (c (e+f x)))^3}{3 b d f^2}+\frac {i (e+f x) (a+b \log (c (e+f x)))^2}{d f^2}-\frac {2 a b i x}{d f}-\frac {2 b^2 i (e+f x) \log (c (e+f x))}{d f^2}+\frac {2 b^2 i x}{d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*b*i*x)/(d*f) + (2*b^2*i*x)/(d*f) - (2*b^2*i*(e + f*x)*Log[c*(e + f*x)])/(d*f^2) + (i*(e + f*x)*(a + b*Lo
g[c*(e + f*x)])^2)/(d*f^2) + ((f*h - e*i)*(a + b*Log[c*(e + f*x)])^3)/(3*b*d*f^2)

Rule 12

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2295

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

Rule 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, 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 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]

Rubi steps

\begin {align*} \int \frac {(h+186 x) (a+b \log (c (e+f x)))^2}{d e+d f x} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (\frac {-186 e+f h}{f}+\frac {186 x}{f}\right ) (a+b \log (c x))^2}{d x} \, dx,x,e+f x\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (\frac {-186 e+f h}{f}+\frac {186 x}{f}\right ) (a+b \log (c x))^2}{x} \, dx,x,e+f x\right )}{d f}\\ &=\frac {186 \operatorname {Subst}\left (\int (a+b \log (c x))^2 \, dx,x,e+f x\right )}{d f^2}-\frac {(186 e-f h) \operatorname {Subst}\left (\int \frac {(a+b \log (c x))^2}{x} \, dx,x,e+f x\right )}{d f^2}\\ &=\frac {186 (e+f x) (a+b \log (c (e+f x)))^2}{d f^2}-\frac {(372 b) \operatorname {Subst}(\int (a+b \log (c x)) \, dx,x,e+f x)}{d f^2}-\frac {(186 e-f h) \operatorname {Subst}\left (\int x^2 \, dx,x,a+b \log (c (e+f x))\right )}{b d f^2}\\ &=-\frac {372 a b x}{d f}+\frac {186 (e+f x) (a+b \log (c (e+f x)))^2}{d f^2}-\frac {(186 e-f h) (a+b \log (c (e+f x)))^3}{3 b d f^2}-\frac {\left (372 b^2\right ) \operatorname {Subst}(\int \log (c x) \, dx,x,e+f x)}{d f^2}\\ &=-\frac {372 a b x}{d f}+\frac {372 b^2 x}{d f}-\frac {372 b^2 (e+f x) \log (c (e+f x))}{d f^2}+\frac {186 (e+f x) (a+b \log (c (e+f x)))^2}{d f^2}-\frac {(186 e-f h) (a+b \log (c (e+f x)))^3}{3 b d f^2}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 89, normalized size = 0.79 \[ \frac {\frac {(f h-e i) (a+b \log (c (e+f x)))^3}{b}+3 i (e+f x) (a+b \log (c (e+f x)))^2-6 b f i x (a-b)-6 b^2 i (e+f x) \log (c (e+f x))}{3 d f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 141, normalized size = 1.25 \[ \frac {3 \, {\left (a^{2} - 2 \, a b + 2 \, b^{2}\right )} f i x + {\left (b^{2} f h - b^{2} e i\right )} \log \left (c f x + c e\right )^{3} + 3 \, {\left (b^{2} f i x + a b f h - {\left (a b - b^{2}\right )} e i\right )} \log \left (c f x + c e\right )^{2} + 3 \, {\left (a^{2} f h + 2 \, {\left (a b - b^{2}\right )} f i x - {\left (a^{2} - 2 \, a b + 2 \, b^{2}\right )} e i\right )} \log \left (c f x + c e\right )}{3 \, d f^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*(a^2 - 2*a*b + 2*b^2)*f*i*x + (b^2*f*h - b^2*e*i)*log(c*f*x + c*e)^3 + 3*(b^2*f*i*x + a*b*f*h - (a*b -
b^2)*e*i)*log(c*f*x + c*e)^2 + 3*(a^2*f*h + 2*(a*b - b^2)*f*i*x - (a^2 - 2*a*b + 2*b^2)*e*i)*log(c*f*x + c*e))
/(d*f^2)

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giac [B]  time = 0.22, size = 240, normalized size = 2.12 \[ \frac {3 \, b^{2} f i x \log \left (c f x + c e\right )^{2} + b^{2} f h \log \left (c f x + c e\right )^{3} - b^{2} i e \log \left (c f x + c e\right )^{3} + 6 \, a b f i x \log \left (c f x + c e\right ) - 6 \, b^{2} f i x \log \left (c f x + c e\right ) + 3 \, a b f h \log \left (c f x + c e\right )^{2} - 3 \, a b i e \log \left (c f x + c e\right )^{2} + 3 \, b^{2} i e \log \left (c f x + c e\right )^{2} + 3 \, a^{2} f i x - 6 \, a b f i x + 6 \, b^{2} f i x + 3 \, a^{2} f h \log \left (f x + e\right ) - 3 \, a^{2} i e \log \left (f x + e\right ) + 6 \, a b i e \log \left (f x + e\right ) - 6 \, b^{2} i e \log \left (f x + e\right )}{3 \, d f^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x+h)*(a+b*log(c*(f*x+e)))^2/(d*f*x+d*e),x, algorithm="giac")

[Out]

1/3*(3*b^2*f*i*x*log(c*f*x + c*e)^2 + b^2*f*h*log(c*f*x + c*e)^3 - b^2*i*e*log(c*f*x + c*e)^3 + 6*a*b*f*i*x*lo
g(c*f*x + c*e) - 6*b^2*f*i*x*log(c*f*x + c*e) + 3*a*b*f*h*log(c*f*x + c*e)^2 - 3*a*b*i*e*log(c*f*x + c*e)^2 +
3*b^2*i*e*log(c*f*x + c*e)^2 + 3*a^2*f*i*x - 6*a*b*f*i*x + 6*b^2*f*i*x + 3*a^2*f*h*log(f*x + e) - 3*a^2*i*e*lo
g(f*x + e) + 6*a*b*i*e*log(f*x + e) - 6*b^2*i*e*log(f*x + e))/(d*f^2)

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maple [B]  time = 0.05, size = 341, normalized size = 3.02 \[ -\frac {b^{2} e i \ln \left (c f x +c e \right )^{3}}{3 d \,f^{2}}+\frac {b^{2} h \ln \left (c f x +c e \right )^{3}}{3 d f}+\frac {b^{2} i x \ln \left (c f x +c e \right )^{2}}{d f}-\frac {a b e i \ln \left (c f x +c e \right )^{2}}{d \,f^{2}}+\frac {a b h \ln \left (c f x +c e \right )^{2}}{d f}+\frac {2 a b i x \ln \left (c f x +c e \right )}{d f}+\frac {b^{2} e i \ln \left (c f x +c e \right )^{2}}{d \,f^{2}}-\frac {2 b^{2} i x \ln \left (c f x +c e \right )}{d f}-\frac {a^{2} e i \ln \left (c f x +c e \right )}{d \,f^{2}}+\frac {a^{2} h \ln \left (c f x +c e \right )}{d f}+\frac {a^{2} i x}{d f}+\frac {2 a b e i \ln \left (c f x +c e \right )}{d \,f^{2}}-\frac {2 a b i x}{d f}-\frac {2 b^{2} e i \ln \left (c f x +c e \right )}{d \,f^{2}}+\frac {2 b^{2} i x}{d f}+\frac {a^{2} e i}{d \,f^{2}}-\frac {2 a b e i}{d \,f^{2}}+\frac {2 b^{2} e i}{d \,f^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/f^2/d*a^2*e*i*ln(c*f*x+c*e)+1/f/d*a^2*h*ln(c*f*x+c*e)+1/f/d*a^2*i*x+1/f^2/d*a^2*e*i-1/f^2/d*a*b*e*i*ln(c*f*
x+c*e)^2+1/f/d*a*b*h*ln(c*f*x+c*e)^2+2/f/d*a*b*i*ln(c*f*x+c*e)*x+2/f^2/d*a*b*i*ln(c*f*x+c*e)*e-2*a*b*i*x/d/f-2
/f^2/d*a*b*e*i-1/3/f^2/d*b^2*e*i*ln(c*f*x+c*e)^3+1/3/f/d*b^2*h*ln(c*f*x+c*e)^3+1/f/d*b^2*i*ln(c*f*x+c*e)^2*x+1
/f^2/d*b^2*i*ln(c*f*x+c*e)^2*e-2/f/d*b^2*i*ln(c*f*x+c*e)*x-2/f^2/d*b^2*i*ln(c*f*x+c*e)*e+2*b^2*i*x/d/f+2/f^2/d
*b^2*e*i

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maxima [B]  time = 0.52, size = 304, normalized size = 2.69 \[ 2 \, a b i {\left (\frac {x}{d f} - \frac {e \log \left (f x + e\right )}{d f^{2}}\right )} \log \left (c f x + c e\right ) - a b h {\left (\frac {2 \, \log \left (c f x + c e\right ) \log \left (d f x + d e\right )}{d f} - \frac {\log \left (f x + e\right )^{2} + 2 \, \log \left (f x + e\right ) \log \relax (c)}{d f}\right )} + a^{2} i {\left (\frac {x}{d f} - \frac {e \log \left (f x + e\right )}{d f^{2}}\right )} + \frac {b^{2} h \log \left (c f x + c e\right )^{3}}{3 \, d f} + \frac {2 \, a b h \log \left (c f x + c e\right ) \log \left (d f x + d e\right )}{d f} + \frac {a^{2} h \log \left (d f x + d e\right )}{d f} + \frac {{\left (e \log \left (f x + e\right )^{2} - 2 \, f x + 2 \, e \log \left (f x + e\right )\right )} a b i}{d f^{2}} - \frac {{\left (c^{2} e \log \left (c f x + c e\right )^{3} - 3 \, {\left (c f x + c e\right )} {\left (c \log \left (c f x + c e\right )^{2} - 2 \, c \log \left (c f x + c e\right ) + 2 \, c\right )}\right )} b^{2} i}{3 \, c^{2} d f^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a*b*i*(x/(d*f) - e*log(f*x + e)/(d*f^2))*log(c*f*x + c*e) - a*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^2*i*(x/(d*f) - e*log(f*x + e)/(d*f^2)) + 1/3*b^2*h*log(
c*f*x + c*e)^3/(d*f) + 2*a*b*h*log(c*f*x + c*e)*log(d*f*x + d*e)/(d*f) + a^2*h*log(d*f*x + d*e)/(d*f) + (e*log
(f*x + e)^2 - 2*f*x + 2*e*log(f*x + e))*a*b*i/(d*f^2) - 1/3*(c^2*e*log(c*f*x + c*e)^3 - 3*(c*f*x + c*e)*(c*log
(c*f*x + c*e)^2 - 2*c*log(c*f*x + c*e) + 2*c))*b^2*i/(c^2*d*f^2)

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mupad [B]  time = 0.34, size = 163, normalized size = 1.44 \[ {\ln \left (c\,\left (e+f\,x\right )\right )}^2\,\left (\frac {b\,\left (a\,f\,h-a\,e\,i+b\,e\,i\right )}{d\,f^2}+\frac {b^2\,i\,x}{d\,f}\right )-\frac {\ln \left (e+f\,x\right )\,\left (a^2\,e\,i-a^2\,f\,h+2\,b^2\,e\,i-2\,a\,b\,e\,i\right )}{d\,f^2}+\frac {i\,x\,\left (a^2-2\,a\,b+2\,b^2\right )}{d\,f}-\frac {b^2\,{\ln \left (c\,\left (e+f\,x\right )\right )}^3\,\left (e\,i-f\,h\right )}{3\,d\,f^2}+\frac {2\,b\,i\,x\,\ln \left (c\,\left (e+f\,x\right )\right )\,\left (a-b\right )}{d\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(c*(e + f*x))^2*((b*(a*f*h - a*e*i + b*e*i))/(d*f^2) + (b^2*i*x)/(d*f)) - (log(e + f*x)*(a^2*e*i - a^2*f*h
+ 2*b^2*e*i - 2*a*b*e*i))/(d*f^2) + (i*x*(a^2 - 2*a*b + 2*b^2))/(d*f) - (b^2*log(c*(e + f*x))^3*(e*i - f*h))/(
3*d*f^2) + (2*b*i*x*log(c*(e + f*x))*(a - b))/(d*f)

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sympy [A]  time = 2.08, size = 175, normalized size = 1.55 \[ x \left (\frac {a^{2} i}{d f} - \frac {2 a b i}{d f} + \frac {2 b^{2} i}{d f}\right ) + \frac {\left (2 a b i x - 2 b^{2} i x\right ) \log {\left (c \left (e + f x\right ) \right )}}{d f} + \frac {\left (- b^{2} e i + b^{2} f h\right ) \log {\left (c \left (e + f x\right ) \right )}^{3}}{3 d f^{2}} - \frac {\left (a^{2} e i - a^{2} f h - 2 a b e i + 2 b^{2} e i\right ) \log {\left (e + f x \right )}}{d f^{2}} + \frac {\left (- a b e i + a b f h + b^{2} e i + b^{2} f i x\right ) \log {\left (c \left (e + f x\right ) \right )}^{2}}{d f^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x+h)*(a+b*ln(c*(f*x+e)))**2/(d*f*x+d*e),x)

[Out]

x*(a**2*i/(d*f) - 2*a*b*i/(d*f) + 2*b**2*i/(d*f)) + (2*a*b*i*x - 2*b**2*i*x)*log(c*(e + f*x))/(d*f) + (-b**2*e
*i + b**2*f*h)*log(c*(e + f*x))**3/(3*d*f**2) - (a**2*e*i - a**2*f*h - 2*a*b*e*i + 2*b**2*e*i)*log(e + f*x)/(d
*f**2) + (-a*b*e*i + a*b*f*h + b**2*e*i + b**2*f*i*x)*log(c*(e + f*x))**2/(d*f**2)

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