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

Optimal. Leaf size=430 \[ \frac{3 f^2 g \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^2 \sqrt{c^2 d x^2+d}}+\frac{f^3 \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c \sqrt{c^2 d x^2+d}}-\frac{3 f g^2 \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c^3 \sqrt{c^2 d x^2+d}}+\frac{3 f g^2 x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 \sqrt{c^2 d x^2+d}}-\frac{2 g^3 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 \sqrt{c^2 d x^2+d}}+\frac{g^3 x^2 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 \sqrt{c^2 d x^2+d}}-\frac{3 b f^2 g x \sqrt{c^2 x^2+1}}{c \sqrt{c^2 d x^2+d}}-\frac{3 b f g^2 x^2 \sqrt{c^2 x^2+1}}{4 c \sqrt{c^2 d x^2+d}}-\frac{b g^3 x^3 \sqrt{c^2 x^2+1}}{9 c \sqrt{c^2 d x^2+d}}+\frac{2 b g^3 x \sqrt{c^2 x^2+1}}{3 c^3 \sqrt{c^2 d x^2+d}} \]

[Out]

(-3*b*f^2*g*x*Sqrt[1 + c^2*x^2])/(c*Sqrt[d + c^2*d*x^2]) + (2*b*g^3*x*Sqrt[1 + c^2*x^2])/(3*c^3*Sqrt[d + c^2*d
*x^2]) - (3*b*f*g^2*x^2*Sqrt[1 + c^2*x^2])/(4*c*Sqrt[d + c^2*d*x^2]) - (b*g^3*x^3*Sqrt[1 + c^2*x^2])/(9*c*Sqrt
[d + c^2*d*x^2]) + (3*f^2*g*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]))/(c^2*Sqrt[d + c^2*d*x^2]) - (2*g^3*(1 + c^2*x^
2)*(a + b*ArcSinh[c*x]))/(3*c^4*Sqrt[d + c^2*d*x^2]) + (3*f*g^2*x*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]))/(2*c^2*S
qrt[d + c^2*d*x^2]) + (g^3*x^2*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]))/(3*c^2*Sqrt[d + c^2*d*x^2]) + (f^3*Sqrt[1 +
c^2*x^2]*(a + b*ArcSinh[c*x])^2)/(2*b*c*Sqrt[d + c^2*d*x^2]) - (3*f*g^2*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]
)^2)/(4*b*c^3*Sqrt[d + c^2*d*x^2])

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Rubi [A]  time = 0.575307, antiderivative size = 430, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 7, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.233, Rules used = {5835, 5821, 5675, 5717, 8, 5758, 30} \[ \frac{3 f^2 g \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^2 \sqrt{c^2 d x^2+d}}+\frac{f^3 \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c \sqrt{c^2 d x^2+d}}-\frac{3 f g^2 \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c^3 \sqrt{c^2 d x^2+d}}+\frac{3 f g^2 x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 \sqrt{c^2 d x^2+d}}-\frac{2 g^3 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 \sqrt{c^2 d x^2+d}}+\frac{g^3 x^2 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 \sqrt{c^2 d x^2+d}}-\frac{3 b f^2 g x \sqrt{c^2 x^2+1}}{c \sqrt{c^2 d x^2+d}}-\frac{3 b f g^2 x^2 \sqrt{c^2 x^2+1}}{4 c \sqrt{c^2 d x^2+d}}-\frac{b g^3 x^3 \sqrt{c^2 x^2+1}}{9 c \sqrt{c^2 d x^2+d}}+\frac{2 b g^3 x \sqrt{c^2 x^2+1}}{3 c^3 \sqrt{c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*f^2*g*x*Sqrt[1 + c^2*x^2])/(c*Sqrt[d + c^2*d*x^2]) + (2*b*g^3*x*Sqrt[1 + c^2*x^2])/(3*c^3*Sqrt[d + c^2*d
*x^2]) - (3*b*f*g^2*x^2*Sqrt[1 + c^2*x^2])/(4*c*Sqrt[d + c^2*d*x^2]) - (b*g^3*x^3*Sqrt[1 + c^2*x^2])/(9*c*Sqrt
[d + c^2*d*x^2]) + (3*f^2*g*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]))/(c^2*Sqrt[d + c^2*d*x^2]) - (2*g^3*(1 + c^2*x^
2)*(a + b*ArcSinh[c*x]))/(3*c^4*Sqrt[d + c^2*d*x^2]) + (3*f*g^2*x*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]))/(2*c^2*S
qrt[d + c^2*d*x^2]) + (g^3*x^2*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]))/(3*c^2*Sqrt[d + c^2*d*x^2]) + (f^3*Sqrt[1 +
c^2*x^2]*(a + b*ArcSinh[c*x])^2)/(2*b*c*Sqrt[d + c^2*d*x^2]) - (3*f*g^2*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]
)^2)/(4*b*c^3*Sqrt[d + c^2*d*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 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 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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 30

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

Rubi steps

\begin{align*} \int \frac{(f+g x)^3 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{d+c^2 d x^2}} \, dx &=\frac{\sqrt{1+c^2 x^2} \int \frac{(f+g x)^3 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}} \, dx}{\sqrt{d+c^2 d x^2}}\\ &=\frac{\sqrt{1+c^2 x^2} \int \left (\frac{f^3 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}}+\frac{3 f^2 g x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}}+\frac{3 f g^2 x^2 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}}+\frac{g^3 x^3 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}}\right ) \, dx}{\sqrt{d+c^2 d x^2}}\\ &=\frac{\left (f^3 \sqrt{1+c^2 x^2}\right ) \int \frac{a+b \sinh ^{-1}(c x)}{\sqrt{1+c^2 x^2}} \, dx}{\sqrt{d+c^2 d x^2}}+\frac{\left (3 f^2 g \sqrt{1+c^2 x^2}\right ) \int \frac{x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}} \, dx}{\sqrt{d+c^2 d x^2}}+\frac{\left (3 f g^2 \sqrt{1+c^2 x^2}\right ) \int \frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}} \, dx}{\sqrt{d+c^2 d x^2}}+\frac{\left (g^3 \sqrt{1+c^2 x^2}\right ) \int \frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}} \, dx}{\sqrt{d+c^2 d x^2}}\\ &=\frac{3 f^2 g \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^2 \sqrt{d+c^2 d x^2}}+\frac{3 f g^2 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 \sqrt{d+c^2 d x^2}}+\frac{g^3 x^2 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 \sqrt{d+c^2 d x^2}}+\frac{f^3 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c \sqrt{d+c^2 d x^2}}-\frac{\left (3 b f^2 g \sqrt{1+c^2 x^2}\right ) \int 1 \, dx}{c \sqrt{d+c^2 d x^2}}-\frac{\left (3 f g^2 \sqrt{1+c^2 x^2}\right ) \int \frac{a+b \sinh ^{-1}(c x)}{\sqrt{1+c^2 x^2}} \, dx}{2 c^2 \sqrt{d+c^2 d x^2}}-\frac{\left (3 b f g^2 \sqrt{1+c^2 x^2}\right ) \int x \, dx}{2 c \sqrt{d+c^2 d x^2}}-\frac{\left (2 g^3 \sqrt{1+c^2 x^2}\right ) \int \frac{x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}} \, dx}{3 c^2 \sqrt{d+c^2 d x^2}}-\frac{\left (b g^3 \sqrt{1+c^2 x^2}\right ) \int x^2 \, dx}{3 c \sqrt{d+c^2 d x^2}}\\ &=-\frac{3 b f^2 g x \sqrt{1+c^2 x^2}}{c \sqrt{d+c^2 d x^2}}-\frac{3 b f g^2 x^2 \sqrt{1+c^2 x^2}}{4 c \sqrt{d+c^2 d x^2}}-\frac{b g^3 x^3 \sqrt{1+c^2 x^2}}{9 c \sqrt{d+c^2 d x^2}}+\frac{3 f^2 g \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^2 \sqrt{d+c^2 d x^2}}-\frac{2 g^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 \sqrt{d+c^2 d x^2}}+\frac{3 f g^2 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 \sqrt{d+c^2 d x^2}}+\frac{g^3 x^2 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 \sqrt{d+c^2 d x^2}}+\frac{f^3 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c \sqrt{d+c^2 d x^2}}-\frac{3 f g^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c^3 \sqrt{d+c^2 d x^2}}+\frac{\left (2 b g^3 \sqrt{1+c^2 x^2}\right ) \int 1 \, dx}{3 c^3 \sqrt{d+c^2 d x^2}}\\ &=-\frac{3 b f^2 g x \sqrt{1+c^2 x^2}}{c \sqrt{d+c^2 d x^2}}+\frac{2 b g^3 x \sqrt{1+c^2 x^2}}{3 c^3 \sqrt{d+c^2 d x^2}}-\frac{3 b f g^2 x^2 \sqrt{1+c^2 x^2}}{4 c \sqrt{d+c^2 d x^2}}-\frac{b g^3 x^3 \sqrt{1+c^2 x^2}}{9 c \sqrt{d+c^2 d x^2}}+\frac{3 f^2 g \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^2 \sqrt{d+c^2 d x^2}}-\frac{2 g^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 \sqrt{d+c^2 d x^2}}+\frac{3 f g^2 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 \sqrt{d+c^2 d x^2}}+\frac{g^3 x^2 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 \sqrt{d+c^2 d x^2}}+\frac{f^3 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c \sqrt{d+c^2 d x^2}}-\frac{3 f g^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c^3 \sqrt{d+c^2 d x^2}}\\ \end{align*}

Mathematica [A]  time = 0.920952, size = 304, normalized size = 0.71 \[ \frac{4 \sqrt{d} g \left (3 a \left (c^2 x^2+1\right ) \left (c^2 \left (18 f^2+9 f g x+2 g^2 x^2\right )-4 g^2\right )-2 b c x \sqrt{c^2 x^2+1} \left (c^2 \left (27 f^2+g^2 x^2\right )-6 g^2\right )\right )+36 a c f \sqrt{c^2 d x^2+d} \left (2 c^2 f^2-3 g^2\right ) \log \left (\sqrt{d} \sqrt{c^2 d x^2+d}+c d x\right )+12 b \sqrt{d} g \left (c^2 x^2+1\right ) \sinh ^{-1}(c x) \left (c^2 \left (18 f^2+9 f g x+2 g^2 x^2\right )-4 g^2\right )+18 b c \sqrt{d} f \sqrt{c^2 x^2+1} \left (2 c^2 f^2-3 g^2\right ) \sinh ^{-1}(c x)^2-27 b c \sqrt{d} f g^2 \sqrt{c^2 x^2+1} \cosh \left (2 \sinh ^{-1}(c x)\right )}{72 c^4 \sqrt{d} \sqrt{c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[d]*g*(-2*b*c*x*Sqrt[1 + c^2*x^2]*(-6*g^2 + c^2*(27*f^2 + g^2*x^2)) + 3*a*(1 + c^2*x^2)*(-4*g^2 + c^2*(
18*f^2 + 9*f*g*x + 2*g^2*x^2))) + 12*b*Sqrt[d]*g*(1 + c^2*x^2)*(-4*g^2 + c^2*(18*f^2 + 9*f*g*x + 2*g^2*x^2))*A
rcSinh[c*x] + 18*b*c*Sqrt[d]*f*(2*c^2*f^2 - 3*g^2)*Sqrt[1 + c^2*x^2]*ArcSinh[c*x]^2 - 27*b*c*Sqrt[d]*f*g^2*Sqr
t[1 + c^2*x^2]*Cosh[2*ArcSinh[c*x]] + 36*a*c*f*(2*c^2*f^2 - 3*g^2)*Sqrt[d + c^2*d*x^2]*Log[c*d*x + Sqrt[d]*Sqr
t[d + c^2*d*x^2]])/(72*c^4*Sqrt[d]*Sqrt[d + c^2*d*x^2])

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Maple [A]  time = 0.348, size = 751, normalized size = 1.8 \begin{align*} \text{result too large to display} \end{align*}

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]

1/3*a*g^3*x^2/c^2/d*(c^2*d*x^2+d)^(1/2)-2/3*a*g^3/d/c^4*(c^2*d*x^2+d)^(1/2)+3/2*a*f*g^2*x/c^2/d*(c^2*d*x^2+d)^
(1/2)-3/2*a*f*g^2/c^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*a*f^2*g/c^2/d*(c^2*d*x^2+d
)^(1/2)+a*f^3*ln(x*c^2*d/(c^2*d)^(1/2)+(c^2*d*x^2+d)^(1/2))/(c^2*d)^(1/2)+3/2*b*(d*(c^2*x^2+1))^(1/2)*f*g^2/d/
(c^2*x^2+1)*arcsinh(c*x)*x^3-3/4*b*(d*(c^2*x^2+1))^(1/2)*f*g^2/c/d/(c^2*x^2+1)^(1/2)*x^2+3/2*b*(d*(c^2*x^2+1))
^(1/2)*f*g^2/c^2/d/(c^2*x^2+1)*arcsinh(c*x)*x+3*b*(d*(c^2*x^2+1))^(1/2)*g/d/(c^2*x^2+1)*arcsinh(c*x)*x^2*f^2-3
*b*(d*(c^2*x^2+1))^(1/2)*g/c/d/(c^2*x^2+1)^(1/2)*x*f^2-1/9*b*(d*(c^2*x^2+1))^(1/2)*g^3/c/d/(c^2*x^2+1)^(1/2)*x
^3-1/3*b*(d*(c^2*x^2+1))^(1/2)*g^3/c^2/d/(c^2*x^2+1)*arcsinh(c*x)*x^2+2/3*b*(d*(c^2*x^2+1))^(1/2)*g^3/c^3/d/(c
^2*x^2+1)^(1/2)*x-3/8*b*(d*(c^2*x^2+1))^(1/2)*f*g^2/c^3/d/(c^2*x^2+1)^(1/2)+3*b*(d*(c^2*x^2+1))^(1/2)*g/c^2/d/
(c^2*x^2+1)*arcsinh(c*x)*f^2+1/2*b*(d*(c^2*x^2+1))^(1/2)*f^3*arcsinh(c*x)^2/(c^2*x^2+1)^(1/2)/c/d-2/3*b*(d*(c^
2*x^2+1))^(1/2)*g^3/c^4/d/(c^2*x^2+1)*arcsinh(c*x)-3/4*b*(d*(c^2*x^2+1))^(1/2)*f*arcsinh(c*x)^2/(c^2*x^2+1)^(1
/2)/c^3/d*g^2+1/3*b*(d*(c^2*x^2+1))^(1/2)*g^3/d/(c^2*x^2+1)*arcsinh(c*x)*x^4

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

[Out]

Exception raised: ValueError

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

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

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

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]

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