3.54 \(\int \frac {(a+b \sinh ^{-1}(c x))^2 \log (h (f+g x)^m)}{\sqrt {1+c^2 x^2}} \, dx\)

Optimal. Leaf size=438 \[ \frac {m \left (a+b \sinh ^{-1}(c x)\right )^4}{12 b^2 c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^2 \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^2 \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{c}+\frac {2 b m \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_3\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}+\frac {2 b m \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_3\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^3 \log \left (\frac {g e^{\sinh ^{-1}(c x)}}{c f-\sqrt {c^2 f^2+g^2}}+1\right )}{3 b c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^3 \log \left (\frac {g e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 f^2+g^2}+c f}+1\right )}{3 b c}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}-\frac {2 b^2 m \text {Li}_4\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}-\frac {2 b^2 m \text {Li}_4\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{c} \]

[Out]

1/12*m*(a+b*arcsinh(c*x))^4/b^2/c+1/3*(a+b*arcsinh(c*x))^3*ln(h*(g*x+f)^m)/b/c-1/3*m*(a+b*arcsinh(c*x))^3*ln(1
+(c*x+(c^2*x^2+1)^(1/2))*g/(c*f-(c^2*f^2+g^2)^(1/2)))/b/c-1/3*m*(a+b*arcsinh(c*x))^3*ln(1+(c*x+(c^2*x^2+1)^(1/
2))*g/(c*f+(c^2*f^2+g^2)^(1/2)))/b/c-m*(a+b*arcsinh(c*x))^2*polylog(2,-(c*x+(c^2*x^2+1)^(1/2))*g/(c*f-(c^2*f^2
+g^2)^(1/2)))/c-m*(a+b*arcsinh(c*x))^2*polylog(2,-(c*x+(c^2*x^2+1)^(1/2))*g/(c*f+(c^2*f^2+g^2)^(1/2)))/c+2*b*m
*(a+b*arcsinh(c*x))*polylog(3,-(c*x+(c^2*x^2+1)^(1/2))*g/(c*f-(c^2*f^2+g^2)^(1/2)))/c+2*b*m*(a+b*arcsinh(c*x))
*polylog(3,-(c*x+(c^2*x^2+1)^(1/2))*g/(c*f+(c^2*f^2+g^2)^(1/2)))/c-2*b^2*m*polylog(4,-(c*x+(c^2*x^2+1)^(1/2))*
g/(c*f-(c^2*f^2+g^2)^(1/2)))/c-2*b^2*m*polylog(4,-(c*x+(c^2*x^2+1)^(1/2))*g/(c*f+(c^2*f^2+g^2)^(1/2)))/c

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Rubi [A]  time = 0.73, antiderivative size = 438, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.265, Rules used = {5675, 5838, 5799, 5561, 2190, 2531, 6609, 2282, 6589} \[ -\frac {m \left (a+b \sinh ^{-1}(c x)\right )^2 \text {PolyLog}\left (2,-\frac {g e^{\sinh ^{-1}(c x)}}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^2 \text {PolyLog}\left (2,-\frac {g e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 f^2+g^2}+c f}\right )}{c}+\frac {2 b m \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (3,-\frac {g e^{\sinh ^{-1}(c x)}}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}+\frac {2 b m \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (3,-\frac {g e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 f^2+g^2}+c f}\right )}{c}-\frac {2 b^2 m \text {PolyLog}\left (4,-\frac {g e^{\sinh ^{-1}(c x)}}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}-\frac {2 b^2 m \text {PolyLog}\left (4,-\frac {g e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 f^2+g^2}+c f}\right )}{c}+\frac {m \left (a+b \sinh ^{-1}(c x)\right )^4}{12 b^2 c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^3 \log \left (\frac {g e^{\sinh ^{-1}(c x)}}{c f-\sqrt {c^2 f^2+g^2}}+1\right )}{3 b c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^3 \log \left (\frac {g e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 f^2+g^2}+c f}+1\right )}{3 b c}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*ArcSinh[c*x])^2*Log[h*(f + g*x)^m])/Sqrt[1 + c^2*x^2],x]

[Out]

(m*(a + b*ArcSinh[c*x])^4)/(12*b^2*c) - (m*(a + b*ArcSinh[c*x])^3*Log[1 + (E^ArcSinh[c*x]*g)/(c*f - Sqrt[c^2*f
^2 + g^2])])/(3*b*c) - (m*(a + b*ArcSinh[c*x])^3*Log[1 + (E^ArcSinh[c*x]*g)/(c*f + Sqrt[c^2*f^2 + g^2])])/(3*b
*c) + ((a + b*ArcSinh[c*x])^3*Log[h*(f + g*x)^m])/(3*b*c) - (m*(a + b*ArcSinh[c*x])^2*PolyLog[2, -((E^ArcSinh[
c*x]*g)/(c*f - Sqrt[c^2*f^2 + g^2]))])/c - (m*(a + b*ArcSinh[c*x])^2*PolyLog[2, -((E^ArcSinh[c*x]*g)/(c*f + Sq
rt[c^2*f^2 + g^2]))])/c + (2*b*m*(a + b*ArcSinh[c*x])*PolyLog[3, -((E^ArcSinh[c*x]*g)/(c*f - Sqrt[c^2*f^2 + g^
2]))])/c + (2*b*m*(a + b*ArcSinh[c*x])*PolyLog[3, -((E^ArcSinh[c*x]*g)/(c*f + Sqrt[c^2*f^2 + g^2]))])/c - (2*b
^2*m*PolyLog[4, -((E^ArcSinh[c*x]*g)/(c*f - Sqrt[c^2*f^2 + g^2]))])/c - (2*b^2*m*PolyLog[4, -((E^ArcSinh[c*x]*
g)/(c*f + Sqrt[c^2*f^2 + g^2]))])/c

Rule 2190

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

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 5561

Int[(Cosh[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Symbol] :
> -Simp[(e + f*x)^(m + 1)/(b*f*(m + 1)), x] + (Int[((e + f*x)^m*E^(c + d*x))/(a - Rt[a^2 + b^2, 2] + b*E^(c +
d*x)), x] + Int[((e + f*x)^m*E^(c + d*x))/(a + Rt[a^2 + b^2, 2] + b*E^(c + d*x)), x]) /; FreeQ[{a, b, c, d, e,
f}, x] && IGtQ[m, 0] && NeQ[a^2 + b^2, 0]

Rule 5675

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(a + b*ArcSinh[c*x]
)^(n + 1)/(b*c*Sqrt[d]*(n + 1)), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && GtQ[d, 0] && NeQ[n, -1
]

Rule 5799

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((d_.) + (e_.)*(x_)), x_Symbol] :> Subst[Int[((a + b*x)^n*Cosh[x
])/(c*d + e*Sinh[x]), x], x, ArcSinh[c*x]] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[n, 0]

Rule 5838

Int[(Log[(h_.)*((f_.) + (g_.)*(x_))^(m_.)]*((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.))/Sqrt[(d_) + (e_.)*(x_)^2
], x_Symbol] :> Simp[(Log[h*(f + g*x)^m]*(a + b*ArcSinh[c*x])^(n + 1))/(b*c*Sqrt[d]*(n + 1)), x] - Dist[(g*m)/
(b*c*Sqrt[d]*(n + 1)), Int[(a + b*ArcSinh[c*x])^(n + 1)/(f + g*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m}
, x] && EqQ[e, c^2*d] && GtQ[d, 0] && IGtQ[n, 0]

Rule 6589

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

Rule 6609

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

Rubi steps

\begin {align*} \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{\sqrt {1+c^2 x^2}} \, dx &=\frac {\left (a+b \sinh ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}-\frac {(g m) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^3}{f+g x} \, dx}{3 b c}\\ &=\frac {\left (a+b \sinh ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}-\frac {(g m) \operatorname {Subst}\left (\int \frac {(a+b x)^3 \cosh (x)}{c f+g \sinh (x)} \, dx,x,\sinh ^{-1}(c x)\right )}{3 b c}\\ &=\frac {m \left (a+b \sinh ^{-1}(c x)\right )^4}{12 b^2 c}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}-\frac {(g m) \operatorname {Subst}\left (\int \frac {e^x (a+b x)^3}{c f+e^x g-\sqrt {c^2 f^2+g^2}} \, dx,x,\sinh ^{-1}(c x)\right )}{3 b c}-\frac {(g m) \operatorname {Subst}\left (\int \frac {e^x (a+b x)^3}{c f+e^x g+\sqrt {c^2 f^2+g^2}} \, dx,x,\sinh ^{-1}(c x)\right )}{3 b c}\\ &=\frac {m \left (a+b \sinh ^{-1}(c x)\right )^4}{12 b^2 c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^3 \log \left (1+\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{3 b c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^3 \log \left (1+\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{3 b c}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}+\frac {m \operatorname {Subst}\left (\int (a+b x)^2 \log \left (1+\frac {e^x g}{c f-\sqrt {c^2 f^2+g^2}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c}+\frac {m \operatorname {Subst}\left (\int (a+b x)^2 \log \left (1+\frac {e^x g}{c f+\sqrt {c^2 f^2+g^2}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c}\\ &=\frac {m \left (a+b \sinh ^{-1}(c x)\right )^4}{12 b^2 c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^3 \log \left (1+\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{3 b c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^3 \log \left (1+\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{3 b c}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^2 \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^2 \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{c}+\frac {(2 b m) \operatorname {Subst}\left (\int (a+b x) \text {Li}_2\left (-\frac {e^x g}{c f-\sqrt {c^2 f^2+g^2}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c}+\frac {(2 b m) \operatorname {Subst}\left (\int (a+b x) \text {Li}_2\left (-\frac {e^x g}{c f+\sqrt {c^2 f^2+g^2}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c}\\ &=\frac {m \left (a+b \sinh ^{-1}(c x)\right )^4}{12 b^2 c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^3 \log \left (1+\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{3 b c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^3 \log \left (1+\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{3 b c}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^2 \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^2 \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{c}+\frac {2 b m \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_3\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}+\frac {2 b m \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_3\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{c}-\frac {\left (2 b^2 m\right ) \operatorname {Subst}\left (\int \text {Li}_3\left (-\frac {e^x g}{c f-\sqrt {c^2 f^2+g^2}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c}-\frac {\left (2 b^2 m\right ) \operatorname {Subst}\left (\int \text {Li}_3\left (-\frac {e^x g}{c f+\sqrt {c^2 f^2+g^2}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c}\\ &=\frac {m \left (a+b \sinh ^{-1}(c x)\right )^4}{12 b^2 c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^3 \log \left (1+\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{3 b c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^3 \log \left (1+\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{3 b c}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^2 \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^2 \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{c}+\frac {2 b m \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_3\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}+\frac {2 b m \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_3\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{c}-\frac {\left (2 b^2 m\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (\frac {g x}{-c f+\sqrt {c^2 f^2+g^2}}\right )}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{c}-\frac {\left (2 b^2 m\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (-\frac {g x}{c f+\sqrt {c^2 f^2+g^2}}\right )}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{c}\\ &=\frac {m \left (a+b \sinh ^{-1}(c x)\right )^4}{12 b^2 c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^3 \log \left (1+\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{3 b c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^3 \log \left (1+\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{3 b c}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^2 \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^2 \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{c}+\frac {2 b m \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_3\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}+\frac {2 b m \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_3\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{c}-\frac {2 b^2 m \text {Li}_4\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}-\frac {2 b^2 m \text {Li}_4\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{c}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 397, normalized size = 0.91 \[ -\frac {3 b m \left (-2 b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_3\left (\frac {e^{\sinh ^{-1}(c x)} g}{\sqrt {c^2 f^2+g^2}-c f}\right )+\left (a+b \sinh ^{-1}(c x)\right )^2 \text {Li}_2\left (\frac {e^{\sinh ^{-1}(c x)} g}{\sqrt {c^2 f^2+g^2}-c f}\right )+2 b^2 \text {Li}_4\left (\frac {e^{\sinh ^{-1}(c x)} g}{\sqrt {c^2 f^2+g^2}-c f}\right )\right )+3 b m \left (-2 b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_3\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )+\left (a+b \sinh ^{-1}(c x)\right )^2 \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )+2 b^2 \text {Li}_4\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )\right )+m \left (a+b \sinh ^{-1}(c x)\right )^3 \log \left (\frac {g e^{\sinh ^{-1}(c x)}}{c f-\sqrt {c^2 f^2+g^2}}+1\right )+m \left (a+b \sinh ^{-1}(c x)\right )^3 \log \left (\frac {g e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 f^2+g^2}+c f}+1\right )-\left (a+b \sinh ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^4}{4 b}}{3 b c} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*ArcSinh[c*x])^2*Log[h*(f + g*x)^m])/Sqrt[1 + c^2*x^2],x]

[Out]

-1/3*(-1/4*(m*(a + b*ArcSinh[c*x])^4)/b + m*(a + b*ArcSinh[c*x])^3*Log[1 + (E^ArcSinh[c*x]*g)/(c*f - Sqrt[c^2*
f^2 + g^2])] + m*(a + b*ArcSinh[c*x])^3*Log[1 + (E^ArcSinh[c*x]*g)/(c*f + Sqrt[c^2*f^2 + g^2])] - (a + b*ArcSi
nh[c*x])^3*Log[h*(f + g*x)^m] + 3*b*m*((a + b*ArcSinh[c*x])^2*PolyLog[2, (E^ArcSinh[c*x]*g)/(-(c*f) + Sqrt[c^2
*f^2 + g^2])] - 2*b*(a + b*ArcSinh[c*x])*PolyLog[3, (E^ArcSinh[c*x]*g)/(-(c*f) + Sqrt[c^2*f^2 + g^2])] + 2*b^2
*PolyLog[4, (E^ArcSinh[c*x]*g)/(-(c*f) + Sqrt[c^2*f^2 + g^2])]) + 3*b*m*((a + b*ArcSinh[c*x])^2*PolyLog[2, -((
E^ArcSinh[c*x]*g)/(c*f + Sqrt[c^2*f^2 + g^2]))] - 2*b*(a + b*ArcSinh[c*x])*PolyLog[3, -((E^ArcSinh[c*x]*g)/(c*
f + Sqrt[c^2*f^2 + g^2]))] + 2*b^2*PolyLog[4, -((E^ArcSinh[c*x]*g)/(c*f + Sqrt[c^2*f^2 + g^2]))]))/(b*c)

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} \operatorname {arsinh}\left (c x\right )^{2} + 2 \, a b \operatorname {arsinh}\left (c x\right ) + a^{2}\right )} \log \left ({\left (g x + f\right )}^{m} h\right )}{\sqrt {c^{2} x^{2} + 1}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))^2*log(h*(g*x+f)^m)/(c^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

integral((b^2*arcsinh(c*x)^2 + 2*a*b*arcsinh(c*x) + a^2)*log((g*x + f)^m*h)/sqrt(c^2*x^2 + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))^2*log(h*(g*x+f)^m)/(c^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Eval
uation time: 7.14Unable to divide, perhaps due to rounding error%%%{720,[0,4,1,1,1,1,4,0]%%%}+%%%{-1260,[0,4,1
,1,1,1,3,1]%%%}+%%%{360,[0,4,1,1,1,1,2,2]%%%}+%%%{180,[0,4,1,1,1,1,1,3]%%%}+%%%{-144,[0,4,1,1,1,0,4,0]%%%}+%%%
{387,[0,4,1,1,1,0,3,1]%%%}+%%%{-282,[0,4,1,1,1,0,2,2]%%%}+%%%{-51,[0,4,1,1,1,0,1,3]%%%}+%%%{180,[0,4,1,1,1,0,0
,4]%%%}+%%%{480,[0,2,3,1,1,1,2,0]%%%}+%%%{420,[0,2,3,1,1,1,1,1]%%%}+%%%{-256,[0,2,3,1,1,0,2,0]%%%}+%%%{1,[0,2,
3,1,1,0,1,1]%%%}+%%%{540,[0,2,3,1,1,0,0,2]%%%}+%%%{-480,[0,0,5,1,1,1,0,0]%%%}+%%%{256,[0,0,5,1,1,0,0,0]%%%} /
%%%{1800,[0,1,0,0,0,0,0,0]%%%} Error: Bad Argument Value

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maple [F]  time = 2.64, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arcsinh \left (c x \right )\right )^{2} \ln \left (h \left (g x +f \right )^{m}\right )}{\sqrt {c^{2} x^{2}+1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))^2*ln(h*(g*x+f)^m)/(c^2*x^2+1)^(1/2),x)

[Out]

int((a+b*arcsinh(c*x))^2*ln(h*(g*x+f)^m)/(c^2*x^2+1)^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2} \log \left ({\left (g x + f\right )}^{m} h\right )}{\sqrt {c^{2} x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))^2*log(h*(g*x+f)^m)/(c^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

integrate((b*arcsinh(c*x) + a)^2*log((g*x + f)^m*h)/sqrt(c^2*x^2 + 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\ln \left (h\,{\left (f+g\,x\right )}^m\right )\,{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2}{\sqrt {c^2\,x^2+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(h*(f + g*x)^m)*(a + b*asinh(c*x))^2)/(c^2*x^2 + 1)^(1/2),x)

[Out]

int((log(h*(f + g*x)^m)*(a + b*asinh(c*x))^2)/(c^2*x^2 + 1)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2} \log {\left (h \left (f + g x\right )^{m} \right )}}{\sqrt {c^{2} x^{2} + 1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(c*x))**2*ln(h*(g*x+f)**m)/(c**2*x**2+1)**(1/2),x)

[Out]

Integral((a + b*asinh(c*x))**2*log(h*(f + g*x)**m)/sqrt(c**2*x**2 + 1), x)

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