3.12 \(\int (d+e x)^3 (a+b \sinh ^{-1}(c x))^2 \, dx\)

Optimal. Leaf size=368 \[ -\frac{3 b d^2 e x \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{2 c}+\frac{3 d^2 e \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2}-\frac{2 b d^3 \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{c}+\frac{4 b d e^2 \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^3}-\frac{2 b d e^2 x^2 \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{3 c}-\frac{b e^3 x^3 \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{8 c}+\frac{3 b e^3 x \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{16 c^3}-\frac{3 e^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{32 c^4}-\frac{d^4 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 e}+\frac{(d+e x)^4 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 e}-\frac{4 b^2 d e^2 x}{3 c^2}-\frac{3 b^2 e^3 x^2}{32 c^2}+\frac{3}{4} b^2 d^2 e x^2+2 b^2 d^3 x+\frac{2}{9} b^2 d e^2 x^3+\frac{1}{32} b^2 e^3 x^4 \]

[Out]

2*b^2*d^3*x - (4*b^2*d*e^2*x)/(3*c^2) + (3*b^2*d^2*e*x^2)/4 - (3*b^2*e^3*x^2)/(32*c^2) + (2*b^2*d*e^2*x^3)/9 +
(b^2*e^3*x^4)/32 - (2*b*d^3*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/c + (4*b*d*e^2*Sqrt[1 + c^2*x^2]*(a + b*A
rcSinh[c*x]))/(3*c^3) - (3*b*d^2*e*x*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(2*c) + (3*b*e^3*x*Sqrt[1 + c^2*x
^2]*(a + b*ArcSinh[c*x]))/(16*c^3) - (2*b*d*e^2*x^2*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(3*c) - (b*e^3*x^3
*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(8*c) - (d^4*(a + b*ArcSinh[c*x])^2)/(4*e) + (3*d^2*e*(a + b*ArcSinh[
c*x])^2)/(4*c^2) - (3*e^3*(a + b*ArcSinh[c*x])^2)/(32*c^4) + ((d + e*x)^4*(a + b*ArcSinh[c*x])^2)/(4*e)

________________________________________________________________________________________

Rubi [A]  time = 0.759384, antiderivative size = 368, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {5801, 5821, 5675, 5717, 8, 5758, 30} \[ -\frac{3 b d^2 e x \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{2 c}+\frac{3 d^2 e \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2}-\frac{2 b d^3 \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{c}+\frac{4 b d e^2 \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^3}-\frac{2 b d e^2 x^2 \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{3 c}-\frac{b e^3 x^3 \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{8 c}+\frac{3 b e^3 x \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{16 c^3}-\frac{3 e^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{32 c^4}-\frac{d^4 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 e}+\frac{(d+e x)^4 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 e}-\frac{4 b^2 d e^2 x}{3 c^2}-\frac{3 b^2 e^3 x^2}{32 c^2}+\frac{3}{4} b^2 d^2 e x^2+2 b^2 d^3 x+\frac{2}{9} b^2 d e^2 x^3+\frac{1}{32} b^2 e^3 x^4 \]

Antiderivative was successfully verified.

[In]

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

[Out]

2*b^2*d^3*x - (4*b^2*d*e^2*x)/(3*c^2) + (3*b^2*d^2*e*x^2)/4 - (3*b^2*e^3*x^2)/(32*c^2) + (2*b^2*d*e^2*x^3)/9 +
(b^2*e^3*x^4)/32 - (2*b*d^3*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/c + (4*b*d*e^2*Sqrt[1 + c^2*x^2]*(a + b*A
rcSinh[c*x]))/(3*c^3) - (3*b*d^2*e*x*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(2*c) + (3*b*e^3*x*Sqrt[1 + c^2*x
^2]*(a + b*ArcSinh[c*x]))/(16*c^3) - (2*b*d*e^2*x^2*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(3*c) - (b*e^3*x^3
*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(8*c) - (d^4*(a + b*ArcSinh[c*x])^2)/(4*e) + (3*d^2*e*(a + b*ArcSinh[
c*x])^2)/(4*c^2) - (3*e^3*(a + b*ArcSinh[c*x])^2)/(32*c^4) + ((d + e*x)^4*(a + b*ArcSinh[c*x])^2)/(4*e)

Rule 5801

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

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 (d+e x)^3 \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx &=\frac{(d+e x)^4 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 e}-\frac{(b c) \int \frac{(d+e x)^4 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}} \, dx}{2 e}\\ &=\frac{(d+e x)^4 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 e}-\frac{(b c) \int \left (\frac{d^4 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}}+\frac{4 d^3 e x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}}+\frac{6 d^2 e^2 x^2 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}}+\frac{4 d e^3 x^3 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}}+\frac{e^4 x^4 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}}\right ) \, dx}{2 e}\\ &=\frac{(d+e x)^4 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 e}-\left (2 b c d^3\right ) \int \frac{x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}} \, dx-\frac{\left (b c d^4\right ) \int \frac{a+b \sinh ^{-1}(c x)}{\sqrt{1+c^2 x^2}} \, dx}{2 e}-\left (3 b c d^2 e\right ) \int \frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}} \, dx-\left (2 b c d e^2\right ) \int \frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}} \, dx-\frac{1}{2} \left (b c e^3\right ) \int \frac{x^4 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}} \, dx\\ &=-\frac{2 b d^3 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c}-\frac{3 b d^2 e x \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 c}-\frac{2 b d e^2 x^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c}-\frac{b e^3 x^3 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c}-\frac{d^4 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 e}+\frac{(d+e x)^4 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 e}+\left (2 b^2 d^3\right ) \int 1 \, dx+\frac{1}{2} \left (3 b^2 d^2 e\right ) \int x \, dx+\frac{\left (3 b d^2 e\right ) \int \frac{a+b \sinh ^{-1}(c x)}{\sqrt{1+c^2 x^2}} \, dx}{2 c}+\frac{1}{3} \left (2 b^2 d e^2\right ) \int x^2 \, dx+\frac{\left (4 b d e^2\right ) \int \frac{x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}} \, dx}{3 c}+\frac{1}{8} \left (b^2 e^3\right ) \int x^3 \, dx+\frac{\left (3 b e^3\right ) \int \frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}} \, dx}{8 c}\\ &=2 b^2 d^3 x+\frac{3}{4} b^2 d^2 e x^2+\frac{2}{9} b^2 d e^2 x^3+\frac{1}{32} b^2 e^3 x^4-\frac{2 b d^3 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c}+\frac{4 b d e^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^3}-\frac{3 b d^2 e x \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 c}+\frac{3 b e^3 x \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{16 c^3}-\frac{2 b d e^2 x^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c}-\frac{b e^3 x^3 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c}-\frac{d^4 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 e}+\frac{3 d^2 e \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2}+\frac{(d+e x)^4 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 e}-\frac{\left (4 b^2 d e^2\right ) \int 1 \, dx}{3 c^2}-\frac{\left (3 b e^3\right ) \int \frac{a+b \sinh ^{-1}(c x)}{\sqrt{1+c^2 x^2}} \, dx}{16 c^3}-\frac{\left (3 b^2 e^3\right ) \int x \, dx}{16 c^2}\\ &=2 b^2 d^3 x-\frac{4 b^2 d e^2 x}{3 c^2}+\frac{3}{4} b^2 d^2 e x^2-\frac{3 b^2 e^3 x^2}{32 c^2}+\frac{2}{9} b^2 d e^2 x^3+\frac{1}{32} b^2 e^3 x^4-\frac{2 b d^3 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c}+\frac{4 b d e^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^3}-\frac{3 b d^2 e x \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 c}+\frac{3 b e^3 x \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{16 c^3}-\frac{2 b d e^2 x^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c}-\frac{b e^3 x^3 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c}-\frac{d^4 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 e}+\frac{3 d^2 e \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2}-\frac{3 e^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{32 c^4}+\frac{(d+e x)^4 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 e}\\ \end{align*}

Mathematica [A]  time = 0.524257, size = 354, normalized size = 0.96 \[ \frac{c \left (72 a^2 c^3 x \left (6 d^2 e x+4 d^3+4 d e^2 x^2+e^3 x^3\right )-6 a b \sqrt{c^2 x^2+1} \left (c^2 \left (72 d^2 e x+96 d^3+32 d e^2 x^2+6 e^3 x^3\right )-e^2 (64 d+9 e x)\right )+b^2 c x \left (c^2 \left (216 d^2 e x+576 d^3+64 d e^2 x^2+9 e^3 x^3\right )-3 e^2 (128 d+9 e x)\right )\right )-6 b \sinh ^{-1}(c x) \left (b c \sqrt{c^2 x^2+1} \left (c^2 \left (72 d^2 e x+96 d^3+32 d e^2 x^2+6 e^3 x^3\right )-e^2 (64 d+9 e x)\right )-3 a \left (8 c^4 x \left (6 d^2 e x+4 d^3+4 d e^2 x^2+e^3 x^3\right )+24 c^2 d^2 e-3 e^3\right )\right )+9 b^2 \sinh ^{-1}(c x)^2 \left (8 c^4 x \left (6 d^2 e x+4 d^3+4 d e^2 x^2+e^3 x^3\right )+24 c^2 d^2 e-3 e^3\right )}{288 c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(72*a^2*c^3*x*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3) - 6*a*b*Sqrt[1 + c^2*x^2]*(-(e^2*(64*d + 9*e*x))
+ c^2*(96*d^3 + 72*d^2*e*x + 32*d*e^2*x^2 + 6*e^3*x^3)) + b^2*c*x*(-3*e^2*(128*d + 9*e*x) + c^2*(576*d^3 + 216
*d^2*e*x + 64*d*e^2*x^2 + 9*e^3*x^3))) - 6*b*(-3*a*(24*c^2*d^2*e - 3*e^3 + 8*c^4*x*(4*d^3 + 6*d^2*e*x + 4*d*e^
2*x^2 + e^3*x^3)) + b*c*Sqrt[1 + c^2*x^2]*(-(e^2*(64*d + 9*e*x)) + c^2*(96*d^3 + 72*d^2*e*x + 32*d*e^2*x^2 + 6
*e^3*x^3)))*ArcSinh[c*x] + 9*b^2*(24*c^2*d^2*e - 3*e^3 + 8*c^4*x*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3))*
ArcSinh[c*x]^2)/(288*c^4)

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Maple [A]  time = 0.085, size = 641, normalized size = 1.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3*(a+b*arcsinh(c*x))^2,x)

[Out]

1/c*(1/4*(c*e*x+c*d)^4*a^2/c^3/e+b^2/c^3*(c^3*d^3*(arcsinh(c*x)^2*c*x-2*arcsinh(c*x)*(c^2*x^2+1)^(1/2)+2*c*x)+
3/4*c^2*d^2*e*(2*arcsinh(c*x)^2*c^2*x^2-2*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*c*x+arcsinh(c*x)^2+c^2*x^2+1)+1/9*c*d
*e^2*(9*arcsinh(c*x)^2*c^3*x^3-6*arcsinh(c*x)*c^2*x^2*(c^2*x^2+1)^(1/2)+27*arcsinh(c*x)^2*c*x+2*c^3*x^3-42*arc
sinh(c*x)*(c^2*x^2+1)^(1/2)+42*c*x)+1/32*e^3*(8*arcsinh(c*x)^2*c^4*x^4-4*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*c^3*x^
3+16*arcsinh(c*x)^2*c^2*x^2+c^4*x^4-10*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*c*x+5*arcsinh(c*x)^2+5*c^2*x^2+4)-3*c*d*
e^2*(arcsinh(c*x)^2*c*x-2*arcsinh(c*x)*(c^2*x^2+1)^(1/2)+2*c*x)-1/4*e^3*(2*arcsinh(c*x)^2*c^2*x^2-2*arcsinh(c*
x)*(c^2*x^2+1)^(1/2)*c*x+arcsinh(c*x)^2+c^2*x^2+1))+2*a*b/c^3*(1/4*e^3*arcsinh(c*x)*c^4*x^4+e^2*arcsinh(c*x)*c
^4*x^3*d+3/2*e*arcsinh(c*x)*c^4*x^2*d^2+arcsinh(c*x)*c^4*x*d^3+1/4/e*arcsinh(c*x)*c^4*d^4-1/4/e*(e^4*(1/4*c^3*
x^3*(c^2*x^2+1)^(1/2)-3/8*c*x*(c^2*x^2+1)^(1/2)+3/8*arcsinh(c*x))+4*c*d*e^3*(1/3*c^2*x^2*(c^2*x^2+1)^(1/2)-2/3
*(c^2*x^2+1)^(1/2))+6*c^2*d^2*e^2*(1/2*c*x*(c^2*x^2+1)^(1/2)-1/2*arcsinh(c*x))+4*c^3*d^3*e*(c^2*x^2+1)^(1/2)+c
^4*d^4*arcsinh(c*x))))

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Maxima [A]  time = 1.18135, size = 880, normalized size = 2.39 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^2*e^3*x^4*arcsinh(c*x)^2 + b^2*d*e^2*x^3*arcsinh(c*x)^2 + 1/4*a^2*e^3*x^4 + 3/2*b^2*d^2*e*x^2*arcsinh(c*
x)^2 + a^2*d*e^2*x^3 + b^2*d^3*x*arcsinh(c*x)^2 + 3/2*a^2*d^2*e*x^2 + 3/2*(2*x^2*arcsinh(c*x) - c*(sqrt(c^2*x^
2 + 1)*x/c^2 - arcsinh(c^2*x/sqrt(c^2))/(sqrt(c^2)*c^2)))*a*b*d^2*e + 3/4*(c^2*(x^2/c^2 - log(c^2*x/sqrt(c^2)
+ sqrt(c^2*x^2 + 1))^2/c^4) - 2*c*(sqrt(c^2*x^2 + 1)*x/c^2 - arcsinh(c^2*x/sqrt(c^2))/(sqrt(c^2)*c^2))*arcsinh
(c*x))*b^2*d^2*e + 2/3*(3*x^3*arcsinh(c*x) - c*(sqrt(c^2*x^2 + 1)*x^2/c^2 - 2*sqrt(c^2*x^2 + 1)/c^4))*a*b*d*e^
2 - 2/9*(3*c*(sqrt(c^2*x^2 + 1)*x^2/c^2 - 2*sqrt(c^2*x^2 + 1)/c^4)*arcsinh(c*x) - (c^2*x^3 - 6*x)/c^2)*b^2*d*e
^2 + 1/16*(8*x^4*arcsinh(c*x) - (2*sqrt(c^2*x^2 + 1)*x^3/c^2 - 3*sqrt(c^2*x^2 + 1)*x/c^4 + 3*arcsinh(c^2*x/sqr
t(c^2))/(sqrt(c^2)*c^4))*c)*a*b*e^3 + 1/32*((x^4/c^2 - 3*x^2/c^4 + 3*log(c^2*x/sqrt(c^2) + sqrt(c^2*x^2 + 1))^
2/c^6)*c^2 - 2*(2*sqrt(c^2*x^2 + 1)*x^3/c^2 - 3*sqrt(c^2*x^2 + 1)*x/c^4 + 3*arcsinh(c^2*x/sqrt(c^2))/(sqrt(c^2
)*c^4))*c*arcsinh(c*x))*b^2*e^3 + 2*b^2*d^3*(x - sqrt(c^2*x^2 + 1)*arcsinh(c*x)/c) + a^2*d^3*x + 2*(c*x*arcsin
h(c*x) - sqrt(c^2*x^2 + 1))*a*b*d^3/c

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Fricas [A]  time = 2.59181, size = 1010, normalized size = 2.74 \begin{align*} \frac{9 \,{\left (8 \, a^{2} + b^{2}\right )} c^{4} e^{3} x^{4} + 32 \,{\left (9 \, a^{2} + 2 \, b^{2}\right )} c^{4} d e^{2} x^{3} + 27 \,{\left (8 \,{\left (2 \, a^{2} + b^{2}\right )} c^{4} d^{2} e - b^{2} c^{2} e^{3}\right )} x^{2} + 9 \,{\left (8 \, b^{2} c^{4} e^{3} x^{4} + 32 \, b^{2} c^{4} d e^{2} x^{3} + 48 \, b^{2} c^{4} d^{2} e x^{2} + 32 \, b^{2} c^{4} d^{3} x + 24 \, b^{2} c^{2} d^{2} e - 3 \, b^{2} e^{3}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )^{2} + 96 \,{\left (3 \,{\left (a^{2} + 2 \, b^{2}\right )} c^{4} d^{3} - 4 \, b^{2} c^{2} d e^{2}\right )} x + 6 \,{\left (24 \, a b c^{4} e^{3} x^{4} + 96 \, a b c^{4} d e^{2} x^{3} + 144 \, a b c^{4} d^{2} e x^{2} + 96 \, a b c^{4} d^{3} x + 72 \, a b c^{2} d^{2} e - 9 \, a b e^{3} -{\left (6 \, b^{2} c^{3} e^{3} x^{3} + 32 \, b^{2} c^{3} d e^{2} x^{2} + 96 \, b^{2} c^{3} d^{3} - 64 \, b^{2} c d e^{2} + 9 \,{\left (8 \, b^{2} c^{3} d^{2} e - b^{2} c e^{3}\right )} x\right )} \sqrt{c^{2} x^{2} + 1}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) - 6 \,{\left (6 \, a b c^{3} e^{3} x^{3} + 32 \, a b c^{3} d e^{2} x^{2} + 96 \, a b c^{3} d^{3} - 64 \, a b c d e^{2} + 9 \,{\left (8 \, a b c^{3} d^{2} e - a b c e^{3}\right )} x\right )} \sqrt{c^{2} x^{2} + 1}}{288 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/288*(9*(8*a^2 + b^2)*c^4*e^3*x^4 + 32*(9*a^2 + 2*b^2)*c^4*d*e^2*x^3 + 27*(8*(2*a^2 + b^2)*c^4*d^2*e - b^2*c^
2*e^3)*x^2 + 9*(8*b^2*c^4*e^3*x^4 + 32*b^2*c^4*d*e^2*x^3 + 48*b^2*c^4*d^2*e*x^2 + 32*b^2*c^4*d^3*x + 24*b^2*c^
2*d^2*e - 3*b^2*e^3)*log(c*x + sqrt(c^2*x^2 + 1))^2 + 96*(3*(a^2 + 2*b^2)*c^4*d^3 - 4*b^2*c^2*d*e^2)*x + 6*(24
*a*b*c^4*e^3*x^4 + 96*a*b*c^4*d*e^2*x^3 + 144*a*b*c^4*d^2*e*x^2 + 96*a*b*c^4*d^3*x + 72*a*b*c^2*d^2*e - 9*a*b*
e^3 - (6*b^2*c^3*e^3*x^3 + 32*b^2*c^3*d*e^2*x^2 + 96*b^2*c^3*d^3 - 64*b^2*c*d*e^2 + 9*(8*b^2*c^3*d^2*e - b^2*c
*e^3)*x)*sqrt(c^2*x^2 + 1))*log(c*x + sqrt(c^2*x^2 + 1)) - 6*(6*a*b*c^3*e^3*x^3 + 32*a*b*c^3*d*e^2*x^2 + 96*a*
b*c^3*d^3 - 64*a*b*c*d*e^2 + 9*(8*a*b*c^3*d^2*e - a*b*c*e^3)*x)*sqrt(c^2*x^2 + 1))/c^4

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Sympy [A]  time = 5.12166, size = 743, normalized size = 2.02 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3*(a+b*asinh(c*x))**2,x)

[Out]

Piecewise((a**2*d**3*x + 3*a**2*d**2*e*x**2/2 + a**2*d*e**2*x**3 + a**2*e**3*x**4/4 + 2*a*b*d**3*x*asinh(c*x)
+ 3*a*b*d**2*e*x**2*asinh(c*x) + 2*a*b*d*e**2*x**3*asinh(c*x) + a*b*e**3*x**4*asinh(c*x)/2 - 2*a*b*d**3*sqrt(c
**2*x**2 + 1)/c - 3*a*b*d**2*e*x*sqrt(c**2*x**2 + 1)/(2*c) - 2*a*b*d*e**2*x**2*sqrt(c**2*x**2 + 1)/(3*c) - a*b
*e**3*x**3*sqrt(c**2*x**2 + 1)/(8*c) + 3*a*b*d**2*e*asinh(c*x)/(2*c**2) + 4*a*b*d*e**2*sqrt(c**2*x**2 + 1)/(3*
c**3) + 3*a*b*e**3*x*sqrt(c**2*x**2 + 1)/(16*c**3) - 3*a*b*e**3*asinh(c*x)/(16*c**4) + b**2*d**3*x*asinh(c*x)*
*2 + 2*b**2*d**3*x + 3*b**2*d**2*e*x**2*asinh(c*x)**2/2 + 3*b**2*d**2*e*x**2/4 + b**2*d*e**2*x**3*asinh(c*x)**
2 + 2*b**2*d*e**2*x**3/9 + b**2*e**3*x**4*asinh(c*x)**2/4 + b**2*e**3*x**4/32 - 2*b**2*d**3*sqrt(c**2*x**2 + 1
)*asinh(c*x)/c - 3*b**2*d**2*e*x*sqrt(c**2*x**2 + 1)*asinh(c*x)/(2*c) - 2*b**2*d*e**2*x**2*sqrt(c**2*x**2 + 1)
*asinh(c*x)/(3*c) - b**2*e**3*x**3*sqrt(c**2*x**2 + 1)*asinh(c*x)/(8*c) + 3*b**2*d**2*e*asinh(c*x)**2/(4*c**2)
- 4*b**2*d*e**2*x/(3*c**2) - 3*b**2*e**3*x**2/(32*c**2) + 4*b**2*d*e**2*sqrt(c**2*x**2 + 1)*asinh(c*x)/(3*c**
3) + 3*b**2*e**3*x*sqrt(c**2*x**2 + 1)*asinh(c*x)/(16*c**3) - 3*b**2*e**3*asinh(c*x)**2/(32*c**4), Ne(c, 0)),
(a**2*(d**3*x + 3*d**2*e*x**2/2 + d*e**2*x**3 + e**3*x**4/4), True))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x + d)^3*(b*arcsinh(c*x) + a)^2, x)