3.34 \(\int (f+g x)^3 \sqrt {d+c^2 d x^2} (a+b \sinh ^{-1}(c x)) \, dx\)

Optimal. Leaf size=640 \[ \frac {1}{2} f^3 x \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )+\frac {f^3 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c \sqrt {c^2 x^2+1}}+\frac {f^2 g \left (c^2 x^2+1\right ) \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{c^2}+\frac {3 f g^2 x \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2}+\frac {3}{4} f g^2 x^3 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )+\frac {g^3 \left (c^2 x^2+1\right )^2 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^4}-\frac {g^3 \left (c^2 x^2+1\right ) \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4}-\frac {3 f g^2 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c^3 \sqrt {c^2 x^2+1}}-\frac {b c f^3 x^2 \sqrt {c^2 d x^2+d}}{4 \sqrt {c^2 x^2+1}}-\frac {b f^2 g x \sqrt {c^2 d x^2+d}}{c \sqrt {c^2 x^2+1}}-\frac {b c f^2 g x^3 \sqrt {c^2 d x^2+d}}{3 \sqrt {c^2 x^2+1}}-\frac {3 b f g^2 x^2 \sqrt {c^2 d x^2+d}}{16 c \sqrt {c^2 x^2+1}}-\frac {3 b c f g^2 x^4 \sqrt {c^2 d x^2+d}}{16 \sqrt {c^2 x^2+1}}-\frac {b c g^3 x^5 \sqrt {c^2 d x^2+d}}{25 \sqrt {c^2 x^2+1}}-\frac {b g^3 x^3 \sqrt {c^2 d x^2+d}}{45 c \sqrt {c^2 x^2+1}}+\frac {2 b g^3 x \sqrt {c^2 d x^2+d}}{15 c^3 \sqrt {c^2 x^2+1}} \]

[Out]

1/2*f^3*x*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)+3/8*f*g^2*x*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)/c^2+3/4*f*
g^2*x^3*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)+f^2*g*(c^2*x^2+1)*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)/c^2-1/
3*g^3*(c^2*x^2+1)*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)/c^4+1/5*g^3*(c^2*x^2+1)^2*(a+b*arcsinh(c*x))*(c^2*d*x
^2+d)^(1/2)/c^4-b*f^2*g*x*(c^2*d*x^2+d)^(1/2)/c/(c^2*x^2+1)^(1/2)+2/15*b*g^3*x*(c^2*d*x^2+d)^(1/2)/c^3/(c^2*x^
2+1)^(1/2)-1/4*b*c*f^3*x^2*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-3/16*b*f*g^2*x^2*(c^2*d*x^2+d)^(1/2)/c/(c^2*x
^2+1)^(1/2)-1/3*b*c*f^2*g*x^3*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-1/45*b*g^3*x^3*(c^2*d*x^2+d)^(1/2)/c/(c^2*
x^2+1)^(1/2)-3/16*b*c*f*g^2*x^4*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-1/25*b*c*g^3*x^5*(c^2*d*x^2+d)^(1/2)/(c^
2*x^2+1)^(1/2)+1/4*f^3*(a+b*arcsinh(c*x))^2*(c^2*d*x^2+d)^(1/2)/b/c/(c^2*x^2+1)^(1/2)-3/16*f*g^2*(a+b*arcsinh(
c*x))^2*(c^2*d*x^2+d)^(1/2)/b/c^3/(c^2*x^2+1)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.69, antiderivative size = 640, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5835, 5821, 5682, 5675, 30, 5717, 5742, 5758, 266, 43, 5732, 12} \[ \frac {f^2 g \left (c^2 x^2+1\right ) \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{c^2}+\frac {1}{2} f^3 x \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )+\frac {f^3 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c \sqrt {c^2 x^2+1}}+\frac {3}{4} f g^2 x^3 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )+\frac {3 f g^2 x \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2}-\frac {3 f g^2 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c^3 \sqrt {c^2 x^2+1}}+\frac {g^3 \left (c^2 x^2+1\right )^2 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^4}-\frac {g^3 \left (c^2 x^2+1\right ) \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4}-\frac {b c f^2 g x^3 \sqrt {c^2 d x^2+d}}{3 \sqrt {c^2 x^2+1}}-\frac {b f^2 g x \sqrt {c^2 d x^2+d}}{c \sqrt {c^2 x^2+1}}-\frac {b c f^3 x^2 \sqrt {c^2 d x^2+d}}{4 \sqrt {c^2 x^2+1}}-\frac {3 b c f g^2 x^4 \sqrt {c^2 d x^2+d}}{16 \sqrt {c^2 x^2+1}}-\frac {3 b f g^2 x^2 \sqrt {c^2 d x^2+d}}{16 c \sqrt {c^2 x^2+1}}-\frac {b c g^3 x^5 \sqrt {c^2 d x^2+d}}{25 \sqrt {c^2 x^2+1}}-\frac {b g^3 x^3 \sqrt {c^2 d x^2+d}}{45 c \sqrt {c^2 x^2+1}}+\frac {2 b g^3 x \sqrt {c^2 d x^2+d}}{15 c^3 \sqrt {c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

Int[(f + g*x)^3*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]),x]

[Out]

-((b*f^2*g*x*Sqrt[d + c^2*d*x^2])/(c*Sqrt[1 + c^2*x^2])) + (2*b*g^3*x*Sqrt[d + c^2*d*x^2])/(15*c^3*Sqrt[1 + c^
2*x^2]) - (b*c*f^3*x^2*Sqrt[d + c^2*d*x^2])/(4*Sqrt[1 + c^2*x^2]) - (3*b*f*g^2*x^2*Sqrt[d + c^2*d*x^2])/(16*c*
Sqrt[1 + c^2*x^2]) - (b*c*f^2*g*x^3*Sqrt[d + c^2*d*x^2])/(3*Sqrt[1 + c^2*x^2]) - (b*g^3*x^3*Sqrt[d + c^2*d*x^2
])/(45*c*Sqrt[1 + c^2*x^2]) - (3*b*c*f*g^2*x^4*Sqrt[d + c^2*d*x^2])/(16*Sqrt[1 + c^2*x^2]) - (b*c*g^3*x^5*Sqrt
[d + c^2*d*x^2])/(25*Sqrt[1 + c^2*x^2]) + (f^3*x*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/2 + (3*f*g^2*x*Sqrt
[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(8*c^2) + (3*f*g^2*x^3*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/4 + (f^
2*g*(1 + c^2*x^2)*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/c^2 - (g^3*(1 + c^2*x^2)*Sqrt[d + c^2*d*x^2]*(a +
b*ArcSinh[c*x]))/(3*c^4) + (g^3*(1 + c^2*x^2)^2*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(5*c^4) + (f^3*Sqrt[
d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^2)/(4*b*c*Sqrt[1 + c^2*x^2]) - (3*f*g^2*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh
[c*x])^2)/(16*b*c^3*Sqrt[1 + c^2*x^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 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 5682

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

Rule 5717

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

Rule 5732

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> With[{u = IntHide[x
^m*(1 + c^2*x^2)^p, x]}, Dist[d^p*(a + b*ArcSinh[c*x]), u, x] - Dist[b*c*d^p, Int[SimplifyIntegrand[u/Sqrt[1 +
c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IntegerQ[p - 1/2] && (IGtQ[(m + 1)/2,
0] || ILtQ[(m + 2*p + 3)/2, 0]) && NeQ[p, -2^(-1)] && GtQ[d, 0]

Rule 5742

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

Rule 5758

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

Rule 5821

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol]
:> Int[ExpandIntegrand[(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g
}, x] && EqQ[e, c^2*d] && IGtQ[m, 0] && IntegerQ[p + 1/2] && GtQ[d, 0] && IGtQ[n, 0] && ((EqQ[n, 1] && GtQ[p,
-1]) || GtQ[p, 0] || EqQ[m, 1] || (EqQ[m, 2] && LtQ[p, -2]))

Rule 5835

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol]
:> Dist[(d^IntPart[p]*(d + e*x^2)^FracPart[p])/(1 + c^2*x^2)^FracPart[p], Int[(f + g*x)^m*(1 + c^2*x^2)^p*(a +
b*ArcSinh[c*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && EqQ[e, c^2*d] && IntegerQ[m] && IntegerQ[p
- 1/2] &&  !GtQ[d, 0]

Rubi steps

\begin {align*} \int (f+g x)^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac {\sqrt {d+c^2 d x^2} \int (f+g x)^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{\sqrt {1+c^2 x^2}}\\ &=\frac {\sqrt {d+c^2 d x^2} \int \left (f^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )+3 f^2 g x \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )+3 f g^2 x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )+g^3 x^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )\right ) \, dx}{\sqrt {1+c^2 x^2}}\\ &=\frac {\left (f^3 \sqrt {d+c^2 d x^2}\right ) \int \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{\sqrt {1+c^2 x^2}}+\frac {\left (3 f^2 g \sqrt {d+c^2 d x^2}\right ) \int x \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{\sqrt {1+c^2 x^2}}+\frac {\left (3 f g^2 \sqrt {d+c^2 d x^2}\right ) \int x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{\sqrt {1+c^2 x^2}}+\frac {\left (g^3 \sqrt {d+c^2 d x^2}\right ) \int x^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{\sqrt {1+c^2 x^2}}\\ &=\frac {1}{2} f^3 x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {3}{4} f g^2 x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {f^2 g \left (1+c^2 x^2\right ) \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c^2}-\frac {g^3 \left (1+c^2 x^2\right ) \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4}+\frac {g^3 \left (1+c^2 x^2\right )^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^4}+\frac {\left (f^3 \sqrt {d+c^2 d x^2}\right ) \int \frac {a+b \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}} \, dx}{2 \sqrt {1+c^2 x^2}}-\frac {\left (b c f^3 \sqrt {d+c^2 d x^2}\right ) \int x \, dx}{2 \sqrt {1+c^2 x^2}}-\frac {\left (b f^2 g \sqrt {d+c^2 d x^2}\right ) \int \left (1+c^2 x^2\right ) \, dx}{c \sqrt {1+c^2 x^2}}+\frac {\left (3 f g^2 \sqrt {d+c^2 d x^2}\right ) \int \frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}} \, dx}{4 \sqrt {1+c^2 x^2}}-\frac {\left (3 b c f g^2 \sqrt {d+c^2 d x^2}\right ) \int x^3 \, dx}{4 \sqrt {1+c^2 x^2}}-\frac {\left (b c g^3 \sqrt {d+c^2 d x^2}\right ) \int \frac {-2+c^2 x^2+3 c^4 x^4}{15 c^4} \, dx}{\sqrt {1+c^2 x^2}}\\ &=-\frac {b f^2 g x \sqrt {d+c^2 d x^2}}{c \sqrt {1+c^2 x^2}}-\frac {b c f^3 x^2 \sqrt {d+c^2 d x^2}}{4 \sqrt {1+c^2 x^2}}-\frac {b c f^2 g x^3 \sqrt {d+c^2 d x^2}}{3 \sqrt {1+c^2 x^2}}-\frac {3 b c f g^2 x^4 \sqrt {d+c^2 d x^2}}{16 \sqrt {1+c^2 x^2}}+\frac {1}{2} f^3 x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {3 f g^2 x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2}+\frac {3}{4} f g^2 x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {f^2 g \left (1+c^2 x^2\right ) \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c^2}-\frac {g^3 \left (1+c^2 x^2\right ) \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4}+\frac {g^3 \left (1+c^2 x^2\right )^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^4}+\frac {f^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c \sqrt {1+c^2 x^2}}-\frac {\left (3 f g^2 \sqrt {d+c^2 d x^2}\right ) \int \frac {a+b \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}} \, dx}{8 c^2 \sqrt {1+c^2 x^2}}-\frac {\left (3 b f g^2 \sqrt {d+c^2 d x^2}\right ) \int x \, dx}{8 c \sqrt {1+c^2 x^2}}-\frac {\left (b g^3 \sqrt {d+c^2 d x^2}\right ) \int \left (-2+c^2 x^2+3 c^4 x^4\right ) \, dx}{15 c^3 \sqrt {1+c^2 x^2}}\\ &=-\frac {b f^2 g x \sqrt {d+c^2 d x^2}}{c \sqrt {1+c^2 x^2}}+\frac {2 b g^3 x \sqrt {d+c^2 d x^2}}{15 c^3 \sqrt {1+c^2 x^2}}-\frac {b c f^3 x^2 \sqrt {d+c^2 d x^2}}{4 \sqrt {1+c^2 x^2}}-\frac {3 b f g^2 x^2 \sqrt {d+c^2 d x^2}}{16 c \sqrt {1+c^2 x^2}}-\frac {b c f^2 g x^3 \sqrt {d+c^2 d x^2}}{3 \sqrt {1+c^2 x^2}}-\frac {b g^3 x^3 \sqrt {d+c^2 d x^2}}{45 c \sqrt {1+c^2 x^2}}-\frac {3 b c f g^2 x^4 \sqrt {d+c^2 d x^2}}{16 \sqrt {1+c^2 x^2}}-\frac {b c g^3 x^5 \sqrt {d+c^2 d x^2}}{25 \sqrt {1+c^2 x^2}}+\frac {1}{2} f^3 x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {3 f g^2 x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2}+\frac {3}{4} f g^2 x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {f^2 g \left (1+c^2 x^2\right ) \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c^2}-\frac {g^3 \left (1+c^2 x^2\right ) \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4}+\frac {g^3 \left (1+c^2 x^2\right )^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^4}+\frac {f^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c \sqrt {1+c^2 x^2}}-\frac {3 f g^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c^3 \sqrt {1+c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 1.49, size = 413, normalized size = 0.65 \[ \frac {3600 a c \sqrt {d} f \sqrt {c^2 x^2+1} \left (4 c^2 f^2-3 g^2\right ) \log \left (\sqrt {d} \sqrt {c^2 d x^2+d}+c d x\right )+240 a \sqrt {c^2 x^2+1} \sqrt {c^2 d x^2+d} \left (6 c^4 x \left (10 f^3+20 f^2 g x+15 f g^2 x^2+4 g^3 x^3\right )+c^2 g \left (120 f^2+45 f g x+8 g^2 x^2\right )-16 g^3\right )-675 b c f g^2 \sqrt {c^2 d x^2+d} \left (8 \sinh ^{-1}(c x)^2-4 \sinh \left (4 \sinh ^{-1}(c x)\right ) \sinh ^{-1}(c x)+\cosh \left (4 \sinh ^{-1}(c x)\right )\right )-128 b g^3 \sqrt {c^2 d x^2+d} \left (c x \left (9 c^4 x^4+5 c^2 x^2-30\right )-15 \sqrt {c^2 x^2+1} \left (3 c^4 x^4+c^2 x^2-2\right ) \sinh ^{-1}(c x)\right )-3600 b c^3 f^3 \sqrt {c^2 d x^2+d} \left (\cosh \left (2 \sinh ^{-1}(c x)\right )-2 \sinh ^{-1}(c x) \left (\sinh ^{-1}(c x)+\sinh \left (2 \sinh ^{-1}(c x)\right )\right )\right )-9600 b c^2 f^2 g \sqrt {c^2 d x^2+d} \left (c^3 x^3-3 \left (c^2 x^2+1\right )^{3/2} \sinh ^{-1}(c x)+3 c x\right )}{28800 c^4 \sqrt {c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(f + g*x)^3*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]),x]

[Out]

(240*a*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x^2]*(-16*g^3 + c^2*g*(120*f^2 + 45*f*g*x + 8*g^2*x^2) + 6*c^4*x*(10*f
^3 + 20*f^2*g*x + 15*f*g^2*x^2 + 4*g^3*x^3)) - 9600*b*c^2*f^2*g*Sqrt[d + c^2*d*x^2]*(3*c*x + c^3*x^3 - 3*(1 +
c^2*x^2)^(3/2)*ArcSinh[c*x]) - 128*b*g^3*Sqrt[d + c^2*d*x^2]*(c*x*(-30 + 5*c^2*x^2 + 9*c^4*x^4) - 15*Sqrt[1 +
c^2*x^2]*(-2 + c^2*x^2 + 3*c^4*x^4)*ArcSinh[c*x]) + 3600*a*c*Sqrt[d]*f*(4*c^2*f^2 - 3*g^2)*Sqrt[1 + c^2*x^2]*L
og[c*d*x + Sqrt[d]*Sqrt[d + c^2*d*x^2]] - 3600*b*c^3*f^3*Sqrt[d + c^2*d*x^2]*(Cosh[2*ArcSinh[c*x]] - 2*ArcSinh
[c*x]*(ArcSinh[c*x] + Sinh[2*ArcSinh[c*x]])) - 675*b*c*f*g^2*Sqrt[d + c^2*d*x^2]*(8*ArcSinh[c*x]^2 + Cosh[4*Ar
cSinh[c*x]] - 4*ArcSinh[c*x]*Sinh[4*ArcSinh[c*x]]))/(28800*c^4*Sqrt[1 + c^2*x^2])

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a g^{3} x^{3} + 3 \, a f g^{2} x^{2} + 3 \, a f^{2} g x + a f^{3} + {\left (b g^{3} x^{3} + 3 \, b f g^{2} x^{2} + 3 \, b f^{2} g x + b f^{3}\right )} \operatorname {arsinh}\left (c x\right )\right )} \sqrt {c^{2} d x^{2} + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^3*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

integral((a*g^3*x^3 + 3*a*f*g^2*x^2 + 3*a*f^2*g*x + a*f^3 + (b*g^3*x^3 + 3*b*f*g^2*x^2 + 3*b*f^2*g*x + b*f^3)*
arcsinh(c*x))*sqrt(c^2*d*x^2 + d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^3*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2),x, algorithm="giac")

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:sym2poly/r2sym(co
nst gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.80, size = 1119, normalized size = 1.75 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x+f)^3*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2),x)

[Out]

-2/15*a*g^3/d/c^4*(c^2*d*x^2+d)^(3/2)+1/2*a*f^3*x*(c^2*d*x^2+d)^(1/2)+3/4*b*(d*(c^2*x^2+1))^(1/2)*f*g^2*c^2/(c
^2*x^2+1)*arcsinh(c*x)*x^5+3/8*b*(d*(c^2*x^2+1))^(1/2)*f*g^2/c^2/(c^2*x^2+1)*arcsinh(c*x)*x+b*(d*(c^2*x^2+1))^
(1/2)*g*c^2/(c^2*x^2+1)*arcsinh(c*x)*x^4*f^2+1/5*a*g^3*x^2*(c^2*d*x^2+d)^(3/2)/c^2/d-3/8*a*f*g^2/c^2*x*(c^2*d*
x^2+d)^(1/2)+a*f^2*g/c^2/d*(c^2*d*x^2+d)^(3/2)+1/2*b*(d*(c^2*x^2+1))^(1/2)*f^3/(c^2*x^2+1)*arcsinh(c*x)*x+1/4*
b*(d*(c^2*x^2+1))^(1/2)*f^3*arcsinh(c*x)^2/(c^2*x^2+1)^(1/2)/c-2/15*b*(d*(c^2*x^2+1))^(1/2)*g^3/c^4/(c^2*x^2+1
)*arcsinh(c*x)+4/15*b*(d*(c^2*x^2+1))^(1/2)*g^3/(c^2*x^2+1)*arcsinh(c*x)*x^4-1/25*b*(d*(c^2*x^2+1))^(1/2)*g^3*
c/(c^2*x^2+1)^(1/2)*x^5-1/45*b*(d*(c^2*x^2+1))^(1/2)*g^3/c/(c^2*x^2+1)^(1/2)*x^3+2/15*b*(d*(c^2*x^2+1))^(1/2)*
g^3/c^3/(c^2*x^2+1)^(1/2)*x-3/128*b*(d*(c^2*x^2+1))^(1/2)*f*g^2/c^3/(c^2*x^2+1)^(1/2)-1/4*b*(d*(c^2*x^2+1))^(1
/2)*f^3*c/(c^2*x^2+1)^(1/2)*x^2-1/8*b*(d*(c^2*x^2+1))^(1/2)*f^3/c/(c^2*x^2+1)^(1/2)+1/2*a*f^3*d*ln(x*c^2*d/(c^
2*d)^(1/2)+(c^2*d*x^2+d)^(1/2))/(c^2*d)^(1/2)-3/8*a*f*g^2/c^2*d*ln(x*c^2*d/(c^2*d)^(1/2)+(c^2*d*x^2+d)^(1/2))/
(c^2*d)^(1/2)-3/16*b*(d*(c^2*x^2+1))^(1/2)*f*g^2*c/(c^2*x^2+1)^(1/2)*x^4-3/16*b*(d*(c^2*x^2+1))^(1/2)*f*g^2/c/
(c^2*x^2+1)^(1/2)*x^2-1/3*b*(d*(c^2*x^2+1))^(1/2)*g*c/(c^2*x^2+1)^(1/2)*x^3*f^2-b*(d*(c^2*x^2+1))^(1/2)*g/c/(c
^2*x^2+1)^(1/2)*x*f^2-3/16*b*(d*(c^2*x^2+1))^(1/2)*f*arcsinh(c*x)^2/(c^2*x^2+1)^(1/2)/c^3*g^2+1/5*b*(d*(c^2*x^
2+1))^(1/2)*g^3*c^2/(c^2*x^2+1)*arcsinh(c*x)*x^6-1/15*b*(d*(c^2*x^2+1))^(1/2)*g^3/c^2/(c^2*x^2+1)*arcsinh(c*x)
*x^2+b*(d*(c^2*x^2+1))^(1/2)*g/c^2/(c^2*x^2+1)*arcsinh(c*x)*f^2+1/2*b*(d*(c^2*x^2+1))^(1/2)*f^3*c^2/(c^2*x^2+1
)*arcsinh(c*x)*x^3+2*b*(d*(c^2*x^2+1))^(1/2)*g/(c^2*x^2+1)*arcsinh(c*x)*x^2*f^2+9/8*b*(d*(c^2*x^2+1))^(1/2)*f*
g^2/(c^2*x^2+1)*arcsinh(c*x)*x^3+3/4*a*f*g^2*x*(c^2*d*x^2+d)^(3/2)/c^2/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^3*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (f+g\,x\right )}^3\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,\sqrt {d\,c^2\,x^2+d} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f + g*x)^3*(a + b*asinh(c*x))*(d + c^2*d*x^2)^(1/2),x)

[Out]

int((f + g*x)^3*(a + b*asinh(c*x))*(d + c^2*d*x^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {d \left (c^{2} x^{2} + 1\right )} \left (a + b \operatorname {asinh}{\left (c x \right )}\right ) \left (f + g x\right )^{3}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)**3*(a+b*asinh(c*x))*(c**2*d*x**2+d)**(1/2),x)

[Out]

Integral(sqrt(d*(c**2*x**2 + 1))*(a + b*asinh(c*x))*(f + g*x)**3, x)

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