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

Optimal. Leaf size=353 \[ \frac{3}{8} d f x \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{4} d f x \left (c^2 x^2+1\right ) \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )+\frac{3 d f \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c \sqrt{c^2 x^2+1}}+\frac{d g \left (c^2 x^2+1\right )^2 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^2}-\frac{b c^3 d f x^4 \sqrt{c^2 d x^2+d}}{16 \sqrt{c^2 x^2+1}}-\frac{5 b c d f x^2 \sqrt{c^2 d x^2+d}}{16 \sqrt{c^2 x^2+1}}-\frac{b c^3 d g x^5 \sqrt{c^2 d x^2+d}}{25 \sqrt{c^2 x^2+1}}-\frac{2 b c d g x^3 \sqrt{c^2 d x^2+d}}{15 \sqrt{c^2 x^2+1}}-\frac{b d g x \sqrt{c^2 d x^2+d}}{5 c \sqrt{c^2 x^2+1}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.335457, antiderivative size = 353, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.321, Rules used = {5835, 5821, 5684, 5682, 5675, 30, 14, 5717, 194} \[ \frac{3}{8} d f x \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{4} d f x \left (c^2 x^2+1\right ) \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )+\frac{3 d f \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c \sqrt{c^2 x^2+1}}+\frac{d g \left (c^2 x^2+1\right )^2 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^2}-\frac{b c^3 d f x^4 \sqrt{c^2 d x^2+d}}{16 \sqrt{c^2 x^2+1}}-\frac{5 b c d f x^2 \sqrt{c^2 d x^2+d}}{16 \sqrt{c^2 x^2+1}}-\frac{b c^3 d g x^5 \sqrt{c^2 d x^2+d}}{25 \sqrt{c^2 x^2+1}}-\frac{2 b c d g x^3 \sqrt{c^2 d x^2+d}}{15 \sqrt{c^2 x^2+1}}-\frac{b d g x \sqrt{c^2 d x^2+d}}{5 c \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 5684

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

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 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 30

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rubi steps

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

Mathematica [A]  time = 1.20627, size = 392, normalized size = 1.11 \[ \frac{3600 a c d^{3/2} f \sqrt{c^2 x^2+1} \log \left (\sqrt{d} \sqrt{c^2 d x^2+d}+c d x\right )+240 a d \sqrt{c^2 x^2+1} \sqrt{c^2 d x^2+d} \left (5 c^2 f x \left (2 c^2 x^2+5\right )+8 g \left (c^2 x^2+1\right )^2\right )+2400 b c d f \sqrt{c^2 d x^2+d} \sinh ^{-1}(c x) \left (\sinh ^{-1}(c x)+\sinh \left (2 \sinh ^{-1}(c x)\right )\right )-1200 b c d f \sqrt{c^2 d x^2+d} \cosh \left (2 \sinh ^{-1}(c x)\right )-75 b c d f \sqrt{c^2 d x^2+d} \left (8 \sinh ^{-1}(c x)^2-4 \sinh \left (4 \sinh ^{-1}(c x)\right ) \sinh ^{-1}(c x)+\cosh \left (4 \sinh ^{-1}(c x)\right )\right )-128 b c^3 d g x^3 \left (3 c^2 x^2+5\right ) \sqrt{c^2 d x^2+d}-640 b c d g x \left (c^2 x^2+3\right ) \sqrt{c^2 d x^2+d}+3200 b d g \left (c^2 x^2+1\right )^{3/2} \sqrt{c^2 d x^2+d} \sinh ^{-1}(c x)+640 b d g \left (c^2 x^2+1\right )^{3/2} \left (3 c^2 x^2-2\right ) \sqrt{c^2 d x^2+d} \sinh ^{-1}(c x)}{9600 c^2 \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-640*b*c*d*g*x*(3 + c^2*x^2)*Sqrt[d + c^2*d*x^2] - 128*b*c^3*d*g*x^3*(5 + 3*c^2*x^2)*Sqrt[d + c^2*d*x^2] + 24
0*a*d*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x^2]*(8*g*(1 + c^2*x^2)^2 + 5*c^2*f*x*(5 + 2*c^2*x^2)) + 3200*b*d*g*(1
+ c^2*x^2)^(3/2)*Sqrt[d + c^2*d*x^2]*ArcSinh[c*x] + 640*b*d*g*(1 + c^2*x^2)^(3/2)*(-2 + 3*c^2*x^2)*Sqrt[d + c^
2*d*x^2]*ArcSinh[c*x] - 1200*b*c*d*f*Sqrt[d + c^2*d*x^2]*Cosh[2*ArcSinh[c*x]] + 3600*a*c*d^(3/2)*f*Sqrt[1 + c^
2*x^2]*Log[c*d*x + Sqrt[d]*Sqrt[d + c^2*d*x^2]] + 2400*b*c*d*f*Sqrt[d + c^2*d*x^2]*ArcSinh[c*x]*(ArcSinh[c*x]
+ Sinh[2*ArcSinh[c*x]]) - 75*b*c*d*f*Sqrt[d + c^2*d*x^2]*(8*ArcSinh[c*x]^2 + Cosh[4*ArcSinh[c*x]] - 4*ArcSinh[
c*x]*Sinh[4*ArcSinh[c*x]]))/(9600*c^2*Sqrt[1 + c^2*x^2])

________________________________________________________________________________________

Maple [A]  time = 0.284, size = 601, normalized size = 1.7 \begin{align*}{\frac{ag}{5\,{c}^{2}d} \left ({c}^{2}d{x}^{2}+d \right ) ^{{\frac{5}{2}}}}+{\frac{afx}{4} \left ({c}^{2}d{x}^{2}+d \right ) ^{{\frac{3}{2}}}}+{\frac{3\,afdx}{8}\sqrt{{c}^{2}d{x}^{2}+d}}+{\frac{3\,af{d}^{2}}{8}\ln \left ({{c}^{2}dx{\frac{1}{\sqrt{{c}^{2}d}}}}+\sqrt{{c}^{2}d{x}^{2}+d} \right ){\frac{1}{\sqrt{{c}^{2}d}}}}+{\frac{3\,bf \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}d}{16\,c}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{bgd{\it Arcsinh} \left ( cx \right ) }{5\,{c}^{2} \left ({c}^{2}{x}^{2}+1 \right ) }\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{17\,bfd}{128\,c}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}-{\frac{bgdx}{5\,c}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{bfd{c}^{4}{\it Arcsinh} \left ( cx \right ){x}^{5}}{4\,{c}^{2}{x}^{2}+4}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{bfd{c}^{3}{x}^{4}}{16}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{7\,bfd{c}^{2}{\it Arcsinh} \left ( cx \right ){x}^{3}}{8\,{c}^{2}{x}^{2}+8}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{5\,bfdc{x}^{2}}{16}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{5\,bfd{\it Arcsinh} \left ( cx \right ) x}{8\,{c}^{2}{x}^{2}+8}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}+{\frac{bgd{c}^{4}{\it Arcsinh} \left ( cx \right ){x}^{6}}{5\,{c}^{2}{x}^{2}+5}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{bgd{c}^{3}{x}^{5}}{25}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{3\,bgd{c}^{2}{\it Arcsinh} \left ( cx \right ){x}^{4}}{5\,{c}^{2}{x}^{2}+5}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{2\,bgdc{x}^{3}}{15}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{3\,bgd{\it Arcsinh} \left ( cx \right ){x}^{2}}{5\,{c}^{2}{x}^{2}+5}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a*g/c^2/d*(c^2*d*x^2+d)^(5/2)+1/4*a*f*x*(c^2*d*x^2+d)^(3/2)+3/8*a*f*d*x*(c^2*d*x^2+d)^(1/2)+3/8*a*f*d^2*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)/(c^2*x^2+1)^(1/2)/c*f*a
rcsinh(c*x)^2*d+1/5*b*(d*(c^2*x^2+1))^(1/2)*g*d/c^2/(c^2*x^2+1)*arcsinh(c*x)-17/128*b*(d*(c^2*x^2+1))^(1/2)*f*
d/c/(c^2*x^2+1)^(1/2)-1/5*b*(d*(c^2*x^2+1))^(1/2)*g*d/c/(c^2*x^2+1)^(1/2)*x+1/4*b*(d*(c^2*x^2+1))^(1/2)*f*d*c^
4/(c^2*x^2+1)*arcsinh(c*x)*x^5-1/16*b*(d*(c^2*x^2+1))^(1/2)*f*d*c^3/(c^2*x^2+1)^(1/2)*x^4+7/8*b*(d*(c^2*x^2+1)
)^(1/2)*f*d*c^2/(c^2*x^2+1)*arcsinh(c*x)*x^3-5/16*b*(d*(c^2*x^2+1))^(1/2)*f*d*c/(c^2*x^2+1)^(1/2)*x^2+5/8*b*(d
*(c^2*x^2+1))^(1/2)*f*d/(c^2*x^2+1)*arcsinh(c*x)*x+1/5*b*(d*(c^2*x^2+1))^(1/2)*g*d*c^4/(c^2*x^2+1)*arcsinh(c*x
)*x^6-1/25*b*(d*(c^2*x^2+1))^(1/2)*g*d*c^3/(c^2*x^2+1)^(1/2)*x^5+3/5*b*(d*(c^2*x^2+1))^(1/2)*g*d*c^2/(c^2*x^2+
1)*arcsinh(c*x)*x^4-2/15*b*(d*(c^2*x^2+1))^(1/2)*g*d*c/(c^2*x^2+1)^(1/2)*x^3+3/5*b*(d*(c^2*x^2+1))^(1/2)*g*d/(
c^2*x^2+1)*arcsinh(c*x)*x^2

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a c^{2} d g x^{3} + a c^{2} d f x^{2} + a d g x + a d f +{\left (b c^{2} d g x^{3} + b c^{2} d f x^{2} + b d g x + b d f\right )} \operatorname{arsinh}\left (c x\right )\right )} \sqrt{c^{2} d x^{2} + d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac{3}{2}} \left (a + b \operatorname{asinh}{\left (c x \right )}\right ) \left (f + g x\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c^{2} d x^{2} + d\right )}^{\frac{3}{2}}{\left (g x + f\right )}{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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