3.182 \(\int \frac {a+b \log (c (e+f x))}{(d e+d f x) (h+i x)^3} \, dx\)

Optimal. Leaf size=250 \[ -\frac {f^2 \log \left (\frac {f h-e i}{i (e+f x)}+1\right ) (a+b \log (c (e+f x)))}{d (f h-e i)^3}-\frac {f i (e+f x) (a+b \log (c (e+f x)))}{d (h+i x) (f h-e i)^3}+\frac {a+b \log (c (e+f x))}{2 d (h+i x)^2 (f h-e i)}+\frac {b f^2 \text {Li}_2\left (-\frac {f h-e i}{i (e+f x)}\right )}{d (f h-e i)^3}-\frac {b f^2 \log (e+f x)}{2 d (f h-e i)^3}+\frac {3 b f^2 \log (h+i x)}{2 d (f h-e i)^3}-\frac {b f}{2 d (h+i x) (f h-e i)^2} \]

[Out]

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

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Rubi [A]  time = 0.57, antiderivative size = 282, normalized size of antiderivative = 1.13, number of steps used = 13, number of rules used = 11, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.367, Rules used = {2411, 12, 2347, 2344, 2301, 2317, 2391, 2314, 31, 2319, 44} \[ -\frac {b f^2 \text {PolyLog}\left (2,-\frac {i (e+f x)}{f h-e i}\right )}{d (f h-e i)^3}+\frac {f^2 (a+b \log (c (e+f x)))^2}{2 b d (f h-e i)^3}-\frac {f^2 \log \left (\frac {f (h+i x)}{f h-e i}\right ) (a+b \log (c (e+f x)))}{d (f h-e i)^3}-\frac {f i (e+f x) (a+b \log (c (e+f x)))}{d (h+i x) (f h-e i)^3}+\frac {a+b \log (c (e+f x))}{2 d (h+i x)^2 (f h-e i)}-\frac {b f^2 \log (e+f x)}{2 d (f h-e i)^3}+\frac {3 b f^2 \log (h+i x)}{2 d (f h-e i)^3}-\frac {b f}{2 d (h+i x) (f h-e i)^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(e + f*x)])/((d*e + d*f*x)*(h + i*x)^3),x]

[Out]

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

Rule 12

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

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 2314

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

Rule 2317

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

Rule 2319

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

Rule 2344

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

Rule 2347

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

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

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 {a+b \log (c (e+f x))}{(h+182 x)^3 (d e+d f x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {a+b \log (c x)}{d x \left (\frac {-182 e+f h}{f}+\frac {182 x}{f}\right )^3} \, dx,x,e+f x\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \frac {a+b \log (c x)}{x \left (\frac {-182 e+f h}{f}+\frac {182 x}{f}\right )^3} \, dx,x,e+f x\right )}{d f}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {a+b \log (c x)}{x \left (\frac {-182 e+f h}{f}+\frac {182 x}{f}\right )^2} \, dx,x,e+f x\right )}{d (182 e-f h)}+\frac {182 \operatorname {Subst}\left (\int \frac {a+b \log (c x)}{\left (\frac {-182 e+f h}{f}+\frac {182 x}{f}\right )^3} \, dx,x,e+f x\right )}{d f (182 e-f h)}\\ &=-\frac {a+b \log (c (e+f x))}{2 d (182 e-f h) (h+182 x)^2}-\frac {182 \operatorname {Subst}\left (\int \frac {a+b \log (c x)}{\left (\frac {-182 e+f h}{f}+\frac {182 x}{f}\right )^2} \, dx,x,e+f x\right )}{d (182 e-f h)^2}+\frac {f \operatorname {Subst}\left (\int \frac {a+b \log (c x)}{x \left (\frac {-182 e+f h}{f}+\frac {182 x}{f}\right )} \, dx,x,e+f x\right )}{d (182 e-f h)^2}+\frac {b \operatorname {Subst}\left (\int \frac {1}{x \left (\frac {-182 e+f h}{f}+\frac {182 x}{f}\right )^2} \, dx,x,e+f x\right )}{2 d (182 e-f h)}\\ &=-\frac {a+b \log (c (e+f x))}{2 d (182 e-f h) (h+182 x)^2}+\frac {182 f (e+f x) (a+b \log (c (e+f x)))}{d (182 e-f h)^3 (h+182 x)}+\frac {(182 f) \operatorname {Subst}\left (\int \frac {a+b \log (c x)}{\frac {-182 e+f h}{f}+\frac {182 x}{f}} \, dx,x,e+f x\right )}{d (182 e-f h)^3}-\frac {(182 b f) \operatorname {Subst}\left (\int \frac {1}{\frac {-182 e+f h}{f}+\frac {182 x}{f}} \, dx,x,e+f x\right )}{d (182 e-f h)^3}-\frac {f^2 \operatorname {Subst}\left (\int \frac {a+b \log (c x)}{x} \, dx,x,e+f x\right )}{d (182 e-f h)^3}+\frac {b \operatorname {Subst}\left (\int \left (\frac {182 f^2}{(182 e-f h) (182 e-f h-182 x)^2}+\frac {182 f^2}{(182 e-f h)^2 (182 e-f h-182 x)}+\frac {f^2}{(182 e-f h)^2 x}\right ) \, dx,x,e+f x\right )}{2 d (182 e-f h)}\\ &=-\frac {b f}{2 d (182 e-f h)^2 (h+182 x)}-\frac {3 b f^2 \log (h+182 x)}{2 d (182 e-f h)^3}+\frac {b f^2 \log (e+f x)}{2 d (182 e-f h)^3}-\frac {a+b \log (c (e+f x))}{2 d (182 e-f h) (h+182 x)^2}+\frac {182 f (e+f x) (a+b \log (c (e+f x)))}{d (182 e-f h)^3 (h+182 x)}+\frac {f^2 \log \left (-\frac {f (h+182 x)}{182 e-f h}\right ) (a+b \log (c (e+f x)))}{d (182 e-f h)^3}-\frac {f^2 (a+b \log (c (e+f x)))^2}{2 b d (182 e-f h)^3}-\frac {\left (b f^2\right ) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {182 x}{-182 e+f h}\right )}{x} \, dx,x,e+f x\right )}{d (182 e-f h)^3}\\ &=-\frac {b f}{2 d (182 e-f h)^2 (h+182 x)}-\frac {3 b f^2 \log (h+182 x)}{2 d (182 e-f h)^3}+\frac {b f^2 \log (e+f x)}{2 d (182 e-f h)^3}-\frac {a+b \log (c (e+f x))}{2 d (182 e-f h) (h+182 x)^2}+\frac {182 f (e+f x) (a+b \log (c (e+f x)))}{d (182 e-f h)^3 (h+182 x)}+\frac {f^2 \log \left (-\frac {f (h+182 x)}{182 e-f h}\right ) (a+b \log (c (e+f x)))}{d (182 e-f h)^3}-\frac {f^2 (a+b \log (c (e+f x)))^2}{2 b d (182 e-f h)^3}+\frac {b f^2 \text {Li}_2\left (\frac {182 (e+f x)}{182 e-f h}\right )}{d (182 e-f h)^3}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 226, normalized size = 0.90 \[ \frac {-2 f^2 \log \left (\frac {f (h+i x)}{f h-e i}\right ) (a+b \log (c (e+f x)))+\frac {f^2 (a+b \log (c (e+f x)))^2}{b}+\frac {2 f (f h-e i) (a+b \log (c (e+f x)))}{h+i x}+\frac {(f h-e i)^2 (a+b \log (c (e+f x)))}{(h+i x)^2}-2 b f^2 \text {Li}_2\left (\frac {i (e+f x)}{e i-f h}\right )-2 b f^2 (\log (e+f x)-\log (h+i x))-\frac {b f (f (h+i x) \log (e+f x)-e i-f (h+i x) \log (h+i x)+f h)}{h+i x}}{2 d (f h-e i)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*(e + f*x)])/((d*e + d*f*x)*(h + i*x)^3),x]

[Out]

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

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \log \left (c f x + c e\right ) + a}{d f i^{3} x^{4} + d e h^{3} + {\left (3 \, d f h i^{2} + d e i^{3}\right )} x^{3} + 3 \, {\left (d f h^{2} i + d e h i^{2}\right )} x^{2} + {\left (d f h^{3} + 3 \, d e h^{2} i\right )} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(f*x+e)))/(d*f*x+d*e)/(i*x+h)^3,x, algorithm="fricas")

[Out]

integral((b*log(c*f*x + c*e) + a)/(d*f*i^3*x^4 + d*e*h^3 + (3*d*f*h*i^2 + d*e*i^3)*x^3 + 3*(d*f*h^2*i + d*e*h*
i^2)*x^2 + (d*f*h^3 + 3*d*e*h^2*i)*x), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \log \left ({\left (f x + e\right )} c\right ) + a}{{\left (d f x + d e\right )} {\left (i x + h\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(f*x+e)))/(d*f*x+d*e)/(i*x+h)^3,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.17, size = 656, normalized size = 2.62 \[ \frac {b \,c^{2} f^{4} i^{2} x^{2} \ln \left (c f x +c e \right )}{2 \left (e i -f h \right )^{3} \left (c f i x +c f h \right )^{2} d}+\frac {b \,c^{2} f^{4} h i x \ln \left (c f x +c e \right )}{\left (e i -f h \right )^{3} \left (c f i x +c f h \right )^{2} d}-\frac {b \,c^{2} e^{2} f^{2} i^{2} \ln \left (c f x +c e \right )}{2 \left (e i -f h \right )^{3} \left (c f i x +c f h \right )^{2} d}+\frac {b \,c^{2} e \,f^{3} h i \ln \left (c f x +c e \right )}{\left (e i -f h \right )^{3} \left (c f i x +c f h \right )^{2} d}+\frac {b c \,f^{3} i x \ln \left (c f x +c e \right )}{\left (e i -f h \right )^{3} \left (c f i x +c f h \right ) d}+\frac {b c e \,f^{2} i \ln \left (c f x +c e \right )}{\left (e i -f h \right )^{3} \left (c f i x +c f h \right ) d}-\frac {b c e \,f^{2} i}{2 \left (e i -f h \right )^{3} \left (c f i x +c f h \right ) d}+\frac {b c \,f^{3} h}{2 \left (e i -f h \right )^{3} \left (c f i x +c f h \right ) d}-\frac {a \,c^{2} f^{2}}{2 \left (e i -f h \right ) \left (c f i x +c f h \right )^{2} d}+\frac {b \,f^{2} \ln \left (\frac {-c e i +c f h +\left (c f x +c e \right ) i}{-c e i +c f h}\right ) \ln \left (c f x +c e \right )}{\left (e i -f h \right )^{3} d}-\frac {b \,f^{2} \ln \left (c f x +c e \right )^{2}}{2 \left (e i -f h \right )^{3} d}+\frac {a c \,f^{2}}{\left (e i -f h \right )^{2} \left (c f i x +c f h \right ) d}+\frac {a \,f^{2} \ln \left (-c e i +c f h +\left (c f x +c e \right ) i \right )}{\left (e i -f h \right )^{3} d}-\frac {a \,f^{2} \ln \left (c f x +c e \right )}{\left (e i -f h \right )^{3} d}+\frac {b \,f^{2} \dilog \left (\frac {-c e i +c f h +\left (c f x +c e \right ) i}{-c e i +c f h}\right )}{\left (e i -f h \right )^{3} d}-\frac {3 b \,f^{2} \ln \left (-c e i +c f h +\left (c f x +c e \right ) i \right )}{2 \left (e i -f h \right )^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*(f*x+e)))/(d*f*x+d*e)/(i*x+h)^3,x)

[Out]

-f^2/d*a/(e*i-f*h)^3*ln(c*f*x+c*e)+f^2/d*a/(e*i-f*h)^3*ln(-c*e*i+c*f*h+(c*f*x+c*e)*i)-1/2*c^2*f^2/d*a/(e*i-f*h
)/(c*f*i*x+c*f*h)^2+c*f^2/d*a/(e*i-f*h)^2/(c*f*i*x+c*f*h)-1/2*c*f^2/d*b/(e*i-f*h)^3*i/(c*f*i*x+c*f*h)*e+1/2*c*
f^3/d*b/(e*i-f*h)^3/(c*f*i*x+c*f*h)*h-3/2*f^2/d*b/(e*i-f*h)^3*ln(-c*e*i+c*f*h+(c*f*x+c*e)*i)-1/2*c^2*f^2/d*b/(
e*i-f*h)^3*i^2*ln(c*f*x+c*e)/(c*f*i*x+c*f*h)^2*e^2+c^2*f^4/d*b/(e*i-f*h)^3*i*ln(c*f*x+c*e)/(c*f*i*x+c*f*h)^2*h
*x+c^2*f^3/d*b/(e*i-f*h)^3*i*ln(c*f*x+c*e)/(c*f*i*x+c*f*h)^2*h*e+1/2*c^2*f^4/d*b/(e*i-f*h)^3*i^2*ln(c*f*x+c*e)
/(c*f*i*x+c*f*h)^2*x^2+c*f^3/d*b/(e*i-f*h)^3*i*ln(c*f*x+c*e)/(c*f*i*x+c*f*h)*x+c*f^2/d*b/(e*i-f*h)^3*i*ln(c*f*
x+c*e)/(c*f*i*x+c*f*h)*e-1/2*f^2/d*b*ln(c*f*x+c*e)^2/(e*i-f*h)^3+f^2/d*b/(e*i-f*h)^3*dilog((-c*e*i+c*f*h+(c*f*
x+c*e)*i)/(-c*e*i+c*f*h))+f^2/d*b/(e*i-f*h)^3*ln(c*f*x+c*e)*ln((-c*e*i+c*f*h+(c*f*x+c*e)*i)/(-c*e*i+c*f*h))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, {\left (\frac {2 \, f^{2} \log \left (f x + e\right )}{d f^{3} h^{3} - 3 \, d e f^{2} h^{2} i + 3 \, d e^{2} f h i^{2} - d e^{3} i^{3}} - \frac {2 \, f^{2} \log \left (i x + h\right )}{d f^{3} h^{3} - 3 \, d e f^{2} h^{2} i + 3 \, d e^{2} f h i^{2} - d e^{3} i^{3}} + \frac {2 \, f i x + 3 \, f h - e i}{d f^{2} h^{4} - 2 \, d e f h^{3} i + d e^{2} h^{2} i^{2} + {\left (d f^{2} h^{2} i^{2} - 2 \, d e f h i^{3} + d e^{2} i^{4}\right )} x^{2} + 2 \, {\left (d f^{2} h^{3} i - 2 \, d e f h^{2} i^{2} + d e^{2} h i^{3}\right )} x}\right )} a + b \int \frac {\log \left (f x + e\right ) + \log \relax (c)}{d f i^{3} x^{4} + d e h^{3} + {\left (3 \, f h i^{2} + e i^{3}\right )} d x^{3} + 3 \, {\left (f h^{2} i + e h i^{2}\right )} d x^{2} + {\left (f h^{3} + 3 \, e h^{2} i\right )} d x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(f*x+e)))/(d*f*x+d*e)/(i*x+h)^3,x, algorithm="maxima")

[Out]

1/2*(2*f^2*log(f*x + e)/(d*f^3*h^3 - 3*d*e*f^2*h^2*i + 3*d*e^2*f*h*i^2 - d*e^3*i^3) - 2*f^2*log(i*x + h)/(d*f^
3*h^3 - 3*d*e*f^2*h^2*i + 3*d*e^2*f*h*i^2 - d*e^3*i^3) + (2*f*i*x + 3*f*h - e*i)/(d*f^2*h^4 - 2*d*e*f*h^3*i +
d*e^2*h^2*i^2 + (d*f^2*h^2*i^2 - 2*d*e*f*h*i^3 + d*e^2*i^4)*x^2 + 2*(d*f^2*h^3*i - 2*d*e*f*h^2*i^2 + d*e^2*h*i
^3)*x))*a + b*integrate((log(f*x + e) + log(c))/(d*f*i^3*x^4 + d*e*h^3 + (3*f*h*i^2 + e*i^3)*d*x^3 + 3*(f*h^2*
i + e*h*i^2)*d*x^2 + (f*h^3 + 3*e*h^2*i)*d*x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {a+b\,\ln \left (c\,\left (e+f\,x\right )\right )}{{\left (h+i\,x\right )}^3\,\left (d\,e+d\,f\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a}{e h^{3} + 3 e h^{2} i x + 3 e h i^{2} x^{2} + e i^{3} x^{3} + f h^{3} x + 3 f h^{2} i x^{2} + 3 f h i^{2} x^{3} + f i^{3} x^{4}}\, dx + \int \frac {b \log {\left (c e + c f x \right )}}{e h^{3} + 3 e h^{2} i x + 3 e h i^{2} x^{2} + e i^{3} x^{3} + f h^{3} x + 3 f h^{2} i x^{2} + 3 f h i^{2} x^{3} + f i^{3} x^{4}}\, dx}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*(f*x+e)))/(d*f*x+d*e)/(i*x+h)**3,x)

[Out]

(Integral(a/(e*h**3 + 3*e*h**2*i*x + 3*e*h*i**2*x**2 + e*i**3*x**3 + f*h**3*x + 3*f*h**2*i*x**2 + 3*f*h*i**2*x
**3 + f*i**3*x**4), x) + Integral(b*log(c*e + c*f*x)/(e*h**3 + 3*e*h**2*i*x + 3*e*h*i**2*x**2 + e*i**3*x**3 +
f*h**3*x + 3*f*h**2*i*x**2 + 3*f*h*i**2*x**3 + f*i**3*x**4), x))/d

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