3.148 \(\int \frac {(a+b \log (c (d+e x)^n))^2}{(f+g x)^{3/2}} \, dx\)

Optimal. Leaf size=312 \[ -\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{g \sqrt {f+g x}}-\frac {8 b \sqrt {e} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g \sqrt {e f-d g}}-\frac {8 b^2 \sqrt {e} n^2 \text {Li}_2\left (1-\frac {2}{1-\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}}\right )}{g \sqrt {e f-d g}}+\frac {8 b^2 \sqrt {e} n^2 \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )^2}{g \sqrt {e f-d g}}-\frac {16 b^2 \sqrt {e} n^2 \log \left (\frac {2}{1-\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}}\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{g \sqrt {e f-d g}} \]

[Out]

8*b^2*n^2*arctanh(e^(1/2)*(g*x+f)^(1/2)/(-d*g+e*f)^(1/2))^2*e^(1/2)/g/(-d*g+e*f)^(1/2)-8*b*n*arctanh(e^(1/2)*(
g*x+f)^(1/2)/(-d*g+e*f)^(1/2))*(a+b*ln(c*(e*x+d)^n))*e^(1/2)/g/(-d*g+e*f)^(1/2)-16*b^2*n^2*arctanh(e^(1/2)*(g*
x+f)^(1/2)/(-d*g+e*f)^(1/2))*ln(2/(1-e^(1/2)*(g*x+f)^(1/2)/(-d*g+e*f)^(1/2)))*e^(1/2)/g/(-d*g+e*f)^(1/2)-8*b^2
*n^2*polylog(2,1-2/(1-e^(1/2)*(g*x+f)^(1/2)/(-d*g+e*f)^(1/2)))*e^(1/2)/g/(-d*g+e*f)^(1/2)-2*(a+b*ln(c*(e*x+d)^
n))^2/g/(g*x+f)^(1/2)

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Rubi [A]  time = 0.77, antiderivative size = 312, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 12, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2398, 2411, 63, 208, 2348, 12, 1587, 6741, 5984, 5918, 2402, 2315} \[ -\frac {8 b^2 \sqrt {e} n^2 \text {PolyLog}\left (2,1-\frac {2}{1-\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}}\right )}{g \sqrt {e f-d g}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{g \sqrt {f+g x}}-\frac {8 b \sqrt {e} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g \sqrt {e f-d g}}+\frac {8 b^2 \sqrt {e} n^2 \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )^2}{g \sqrt {e f-d g}}-\frac {16 b^2 \sqrt {e} n^2 \log \left (\frac {2}{1-\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}}\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{g \sqrt {e f-d g}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(d + e*x)^n])^2/(f + g*x)^(3/2),x]

[Out]

(8*b^2*Sqrt[e]*n^2*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]]^2)/(g*Sqrt[e*f - d*g]) - (8*b*Sqrt[e]*n*Ar
cTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]]*(a + b*Log[c*(d + e*x)^n]))/(g*Sqrt[e*f - d*g]) - (2*(a + b*Log
[c*(d + e*x)^n])^2)/(g*Sqrt[f + g*x]) - (16*b^2*Sqrt[e]*n^2*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]]*L
og[2/(1 - (Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g])])/(g*Sqrt[e*f - d*g]) - (8*b^2*Sqrt[e]*n^2*PolyLog[2, 1 - 2
/(1 - (Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g])])/(g*Sqrt[e*f - d*g])

Rule 12

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 1587

Int[(Pp_)/(Qq_), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[(Coeff[Pp, x, p]*Log[RemoveConte
nt[Qq, x]])/(q*Coeff[Qq, x, q]), x] /; EqQ[p, q - 1] && EqQ[Pp, Simplify[(Coeff[Pp, x, p]*D[Qq, x])/(q*Coeff[Q
q, x, q])]]] /; PolyQ[Pp, x] && PolyQ[Qq, x]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2348

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.))/(x_), x_Symbol] :> With[{u = IntHi
de[(d + e*x^r)^q/x, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[Dist[1/x, u, x], x], x]] /; FreeQ[{a, b
, c, d, e, n, r}, x] && IntegerQ[q - 1/2]

Rule 2398

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

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

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 5918

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

Rule 5984

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c
*x])^(p + 1)/(b*e*(p + 1)), x] + Dist[1/(c*d), Int[(a + b*ArcTanh[c*x])^p/(1 - c*x), x], x] /; FreeQ[{a, b, c,
d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps

\begin {align*} \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{(f+g x)^{3/2}} \, dx &=-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{g \sqrt {f+g x}}+\frac {(4 b e n) \int \frac {a+b \log \left (c (d+e x)^n\right )}{(d+e x) \sqrt {f+g x}} \, dx}{g}\\ &=-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{g \sqrt {f+g x}}+\frac {(4 b n) \operatorname {Subst}\left (\int \frac {a+b \log \left (c x^n\right )}{x \sqrt {\frac {e f-d g}{e}+\frac {g x}{e}}} \, dx,x,d+e x\right )}{g}\\ &=-\frac {8 b \sqrt {e} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g \sqrt {e f-d g}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{g \sqrt {f+g x}}-\frac {\left (4 b^2 n^2\right ) \operatorname {Subst}\left (\int -\frac {2 \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f-\frac {d g}{e}+\frac {g x}{e}}}{\sqrt {e f-d g}}\right )}{\sqrt {e f-d g} x} \, dx,x,d+e x\right )}{g}\\ &=-\frac {8 b \sqrt {e} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g \sqrt {e f-d g}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{g \sqrt {f+g x}}+\frac {\left (8 b^2 \sqrt {e} n^2\right ) \operatorname {Subst}\left (\int \frac {\tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f-\frac {d g}{e}+\frac {g x}{e}}}{\sqrt {e f-d g}}\right )}{x} \, dx,x,d+e x\right )}{g \sqrt {e f-d g}}\\ &=-\frac {8 b \sqrt {e} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g \sqrt {e f-d g}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{g \sqrt {f+g x}}+\frac {\left (16 b^2 e^{3/2} n^2\right ) \operatorname {Subst}\left (\int \frac {x \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {e f-d g}}\right )}{d g+e \left (-f+x^2\right )} \, dx,x,\sqrt {f+g x}\right )}{g \sqrt {e f-d g}}\\ &=-\frac {8 b \sqrt {e} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g \sqrt {e f-d g}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{g \sqrt {f+g x}}+\frac {\left (16 b^2 e^{3/2} n^2\right ) \operatorname {Subst}\left (\int \frac {x \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {e f-d g}}\right )}{-e f+d g+e x^2} \, dx,x,\sqrt {f+g x}\right )}{g \sqrt {e f-d g}}\\ &=\frac {8 b^2 \sqrt {e} n^2 \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )^2}{g \sqrt {e f-d g}}-\frac {8 b \sqrt {e} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g \sqrt {e f-d g}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{g \sqrt {f+g x}}-\frac {\left (16 b^2 e n^2\right ) \operatorname {Subst}\left (\int \frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {e f-d g}}\right )}{1-\frac {\sqrt {e} x}{\sqrt {e f-d g}}} \, dx,x,\sqrt {f+g x}\right )}{g (e f-d g)}\\ &=\frac {8 b^2 \sqrt {e} n^2 \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )^2}{g \sqrt {e f-d g}}-\frac {8 b \sqrt {e} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g \sqrt {e f-d g}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{g \sqrt {f+g x}}-\frac {16 b^2 \sqrt {e} n^2 \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \log \left (\frac {2}{1-\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}}\right )}{g \sqrt {e f-d g}}+\frac {\left (16 b^2 e n^2\right ) \operatorname {Subst}\left (\int \frac {\log \left (\frac {2}{1-\frac {\sqrt {e} x}{\sqrt {e f-d g}}}\right )}{1-\frac {e x^2}{e f-d g}} \, dx,x,\sqrt {f+g x}\right )}{g (e f-d g)}\\ &=\frac {8 b^2 \sqrt {e} n^2 \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )^2}{g \sqrt {e f-d g}}-\frac {8 b \sqrt {e} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g \sqrt {e f-d g}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{g \sqrt {f+g x}}-\frac {16 b^2 \sqrt {e} n^2 \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \log \left (\frac {2}{1-\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}}\right )}{g \sqrt {e f-d g}}-\frac {\left (16 b^2 \sqrt {e} n^2\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}}\right )}{g \sqrt {e f-d g}}\\ &=\frac {8 b^2 \sqrt {e} n^2 \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )^2}{g \sqrt {e f-d g}}-\frac {8 b \sqrt {e} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g \sqrt {e f-d g}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{g \sqrt {f+g x}}-\frac {16 b^2 \sqrt {e} n^2 \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \log \left (\frac {2}{1-\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}}\right )}{g \sqrt {e f-d g}}-\frac {8 b^2 \sqrt {e} n^2 \text {Li}_2\left (1-\frac {2}{1-\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}}\right )}{g \sqrt {e f-d g}}\\ \end {align*}

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Mathematica [A]  time = 0.67, size = 424, normalized size = 1.36 \[ \frac {2 \left (\frac {b \sqrt {e} n \left (2 \log \left (\sqrt {e f-d g}-\sqrt {e} \sqrt {f+g x}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-2 \log \left (\sqrt {e f-d g}+\sqrt {e} \sqrt {f+g x}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-b n \left (2 \text {Li}_2\left (\frac {1}{2}-\frac {\sqrt {e} \sqrt {f+g x}}{2 \sqrt {e f-d g}}\right )+\log \left (\sqrt {e f-d g}-\sqrt {e} \sqrt {f+g x}\right ) \left (\log \left (\sqrt {e f-d g}-\sqrt {e} \sqrt {f+g x}\right )+2 \log \left (\frac {1}{2} \left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}+1\right )\right )\right )\right )+b n \left (2 \text {Li}_2\left (\frac {1}{2} \left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}+1\right )\right )+\log \left (\sqrt {e f-d g}+\sqrt {e} \sqrt {f+g x}\right ) \left (\log \left (\sqrt {e f-d g}+\sqrt {e} \sqrt {f+g x}\right )+2 \log \left (\frac {1}{2}-\frac {\sqrt {e} \sqrt {f+g x}}{2 \sqrt {e f-d g}}\right )\right )\right )\right )}{\sqrt {e f-d g}}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{\sqrt {f+g x}}\right )}{g} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*(d + e*x)^n])^2/(f + g*x)^(3/2),x]

[Out]

(2*(-((a + b*Log[c*(d + e*x)^n])^2/Sqrt[f + g*x]) + (b*Sqrt[e]*n*(2*(a + b*Log[c*(d + e*x)^n])*Log[Sqrt[e*f -
d*g] - Sqrt[e]*Sqrt[f + g*x]] - 2*(a + b*Log[c*(d + e*x)^n])*Log[Sqrt[e*f - d*g] + Sqrt[e]*Sqrt[f + g*x]] - b*
n*(Log[Sqrt[e*f - d*g] - Sqrt[e]*Sqrt[f + g*x]]*(Log[Sqrt[e*f - d*g] - Sqrt[e]*Sqrt[f + g*x]] + 2*Log[(1 + (Sq
rt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g])/2]) + 2*PolyLog[2, 1/2 - (Sqrt[e]*Sqrt[f + g*x])/(2*Sqrt[e*f - d*g])]) +
b*n*(Log[Sqrt[e*f - d*g] + Sqrt[e]*Sqrt[f + g*x]]*(Log[Sqrt[e*f - d*g] + Sqrt[e]*Sqrt[f + g*x]] + 2*Log[1/2 -
(Sqrt[e]*Sqrt[f + g*x])/(2*Sqrt[e*f - d*g])]) + 2*PolyLog[2, (1 + (Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g])/2]
)))/Sqrt[e*f - d*g]))/g

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {g x + f} b^{2} \log \left ({\left (e x + d\right )}^{n} c\right )^{2} + 2 \, \sqrt {g x + f} a b \log \left ({\left (e x + d\right )}^{n} c\right ) + \sqrt {g x + f} a^{2}}{g^{2} x^{2} + 2 \, f g x + f^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))^2/(g*x+f)^(3/2),x, algorithm="fricas")

[Out]

integral((sqrt(g*x + f)*b^2*log((e*x + d)^n*c)^2 + 2*sqrt(g*x + f)*a*b*log((e*x + d)^n*c) + sqrt(g*x + f)*a^2)
/(g^2*x^2 + 2*f*g*x + f^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{2}}{{\left (g x + f\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))^2/(g*x+f)^(3/2),x, algorithm="giac")

[Out]

integrate((b*log((e*x + d)^n*c) + a)^2/(g*x + f)^(3/2), x)

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maple [F]  time = 0.49, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \left (e x +d \right )^{n}\right )+a \right )^{2}}{\left (g x +f \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*ln(c*(e*x+d)^n)+a)^2/(g*x+f)^(3/2),x)

[Out]

int((b*ln(c*(e*x+d)^n)+a)^2/(g*x+f)^(3/2),x)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))^2/(g*x+f)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(d*g-e*f>0)', see `assume?` for
more details)Is d*g-e*f positive or negative?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}^2}{{\left (f+g\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*log(c*(d + e*x)^n))^2/(f + g*x)^(3/2),x)

[Out]

int((a + b*log(c*(d + e*x)^n))^2/(f + g*x)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*(e*x+d)**n))**2/(g*x+f)**(3/2),x)

[Out]

Integral((a + b*log(c*(d + e*x)**n))**2/(f + g*x)**(3/2), x)

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