3.291 \(\int \frac{x^4 (a+b \sinh ^{-1}(c x))^2}{\sqrt{d+c^2 d x^2}} \, dx\)

Optimal. Leaf size=323 \[ -\frac{b x^4 \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{8 c \sqrt{c^2 d x^2+d}}+\frac{x^3 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2 d}+\frac{3 b x^2 \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{8 c^3 \sqrt{c^2 d x^2+d}}-\frac{3 x \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{8 c^4 d}+\frac{\sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^3}{8 b c^5 \sqrt{c^2 d x^2+d}}+\frac{b^2 x^3 \left (c^2 x^2+1\right )}{32 c^2 \sqrt{c^2 d x^2+d}}-\frac{15 b^2 x \left (c^2 x^2+1\right )}{64 c^4 \sqrt{c^2 d x^2+d}}+\frac{15 b^2 \sqrt{c^2 x^2+1} \sinh ^{-1}(c x)}{64 c^5 \sqrt{c^2 d x^2+d}} \]

[Out]

(-15*b^2*x*(1 + c^2*x^2))/(64*c^4*Sqrt[d + c^2*d*x^2]) + (b^2*x^3*(1 + c^2*x^2))/(32*c^2*Sqrt[d + c^2*d*x^2])
+ (15*b^2*Sqrt[1 + c^2*x^2]*ArcSinh[c*x])/(64*c^5*Sqrt[d + c^2*d*x^2]) + (3*b*x^2*Sqrt[1 + c^2*x^2]*(a + b*Arc
Sinh[c*x]))/(8*c^3*Sqrt[d + c^2*d*x^2]) - (b*x^4*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(8*c*Sqrt[d + c^2*d*x
^2]) - (3*x*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^2)/(8*c^4*d) + (x^3*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*
x])^2)/(4*c^2*d) + (Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^3)/(8*b*c^5*Sqrt[d + c^2*d*x^2])

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Rubi [A]  time = 0.478891, antiderivative size = 323, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {5758, 5677, 5675, 5661, 321, 215} \[ -\frac{b x^4 \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{8 c \sqrt{c^2 d x^2+d}}+\frac{x^3 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2 d}+\frac{3 b x^2 \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{8 c^3 \sqrt{c^2 d x^2+d}}-\frac{3 x \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{8 c^4 d}+\frac{\sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^3}{8 b c^5 \sqrt{c^2 d x^2+d}}+\frac{b^2 x^3 \left (c^2 x^2+1\right )}{32 c^2 \sqrt{c^2 d x^2+d}}-\frac{15 b^2 x \left (c^2 x^2+1\right )}{64 c^4 \sqrt{c^2 d x^2+d}}+\frac{15 b^2 \sqrt{c^2 x^2+1} \sinh ^{-1}(c x)}{64 c^5 \sqrt{c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*ArcSinh[c*x])^2)/Sqrt[d + c^2*d*x^2],x]

[Out]

(-15*b^2*x*(1 + c^2*x^2))/(64*c^4*Sqrt[d + c^2*d*x^2]) + (b^2*x^3*(1 + c^2*x^2))/(32*c^2*Sqrt[d + c^2*d*x^2])
+ (15*b^2*Sqrt[1 + c^2*x^2]*ArcSinh[c*x])/(64*c^5*Sqrt[d + c^2*d*x^2]) + (3*b*x^2*Sqrt[1 + c^2*x^2]*(a + b*Arc
Sinh[c*x]))/(8*c^3*Sqrt[d + c^2*d*x^2]) - (b*x^4*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(8*c*Sqrt[d + c^2*d*x
^2]) - (3*x*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^2)/(8*c^4*d) + (x^3*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*
x])^2)/(4*c^2*d) + (Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^3)/(8*b*c^5*Sqrt[d + c^2*d*x^2])

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 5677

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[1 + c^2*x^2]/S
qrt[d + e*x^2], Int[(a + b*ArcSinh[c*x])^n/Sqrt[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e,
c^2*d] &&  !GtQ[d, 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 5661

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

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rubi steps

\begin{align*} \int \frac{x^4 \left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt{d+c^2 d x^2}} \, dx &=\frac{x^3 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2 d}-\frac{3 \int \frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt{d+c^2 d x^2}} \, dx}{4 c^2}-\frac{\left (b \sqrt{1+c^2 x^2}\right ) \int x^3 \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{2 c \sqrt{d+c^2 d x^2}}\\ &=-\frac{b x^4 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c \sqrt{d+c^2 d x^2}}-\frac{3 x \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{8 c^4 d}+\frac{x^3 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2 d}+\frac{3 \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt{d+c^2 d x^2}} \, dx}{8 c^4}+\frac{\left (b^2 \sqrt{1+c^2 x^2}\right ) \int \frac{x^4}{\sqrt{1+c^2 x^2}} \, dx}{8 \sqrt{d+c^2 d x^2}}+\frac{\left (3 b \sqrt{1+c^2 x^2}\right ) \int x \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{4 c^3 \sqrt{d+c^2 d x^2}}\\ &=\frac{b^2 x^3 \left (1+c^2 x^2\right )}{32 c^2 \sqrt{d+c^2 d x^2}}+\frac{3 b x^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c^3 \sqrt{d+c^2 d x^2}}-\frac{b x^4 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c \sqrt{d+c^2 d x^2}}-\frac{3 x \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{8 c^4 d}+\frac{x^3 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2 d}+\frac{\left (3 \sqrt{1+c^2 x^2}\right ) \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt{1+c^2 x^2}} \, dx}{8 c^4 \sqrt{d+c^2 d x^2}}-\frac{\left (3 b^2 \sqrt{1+c^2 x^2}\right ) \int \frac{x^2}{\sqrt{1+c^2 x^2}} \, dx}{32 c^2 \sqrt{d+c^2 d x^2}}-\frac{\left (3 b^2 \sqrt{1+c^2 x^2}\right ) \int \frac{x^2}{\sqrt{1+c^2 x^2}} \, dx}{8 c^2 \sqrt{d+c^2 d x^2}}\\ &=-\frac{15 b^2 x \left (1+c^2 x^2\right )}{64 c^4 \sqrt{d+c^2 d x^2}}+\frac{b^2 x^3 \left (1+c^2 x^2\right )}{32 c^2 \sqrt{d+c^2 d x^2}}+\frac{3 b x^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c^3 \sqrt{d+c^2 d x^2}}-\frac{b x^4 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c \sqrt{d+c^2 d x^2}}-\frac{3 x \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{8 c^4 d}+\frac{x^3 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2 d}+\frac{\sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^3}{8 b c^5 \sqrt{d+c^2 d x^2}}+\frac{\left (3 b^2 \sqrt{1+c^2 x^2}\right ) \int \frac{1}{\sqrt{1+c^2 x^2}} \, dx}{64 c^4 \sqrt{d+c^2 d x^2}}+\frac{\left (3 b^2 \sqrt{1+c^2 x^2}\right ) \int \frac{1}{\sqrt{1+c^2 x^2}} \, dx}{16 c^4 \sqrt{d+c^2 d x^2}}\\ &=-\frac{15 b^2 x \left (1+c^2 x^2\right )}{64 c^4 \sqrt{d+c^2 d x^2}}+\frac{b^2 x^3 \left (1+c^2 x^2\right )}{32 c^2 \sqrt{d+c^2 d x^2}}+\frac{15 b^2 \sqrt{1+c^2 x^2} \sinh ^{-1}(c x)}{64 c^5 \sqrt{d+c^2 d x^2}}+\frac{3 b x^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c^3 \sqrt{d+c^2 d x^2}}-\frac{b x^4 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c \sqrt{d+c^2 d x^2}}-\frac{3 x \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{8 c^4 d}+\frac{x^3 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2 d}+\frac{\sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^3}{8 b c^5 \sqrt{d+c^2 d x^2}}\\ \end{align*}

Mathematica [A]  time = 0.834756, size = 268, normalized size = 0.83 \[ \frac{32 a^2 c \sqrt{d} x \left (c^2 x^2+1\right ) \left (2 c^2 x^2-3\right )+96 a^2 \sqrt{c^2 d x^2+d} \log \left (\sqrt{d} \sqrt{c^2 d x^2+d}+c d x\right )+4 a b \sqrt{d} \sqrt{c^2 x^2+1} \left (4 \sinh ^{-1}(c x) \left (6 \sinh ^{-1}(c x)-8 \sinh \left (2 \sinh ^{-1}(c x)\right )+\sinh \left (4 \sinh ^{-1}(c x)\right )\right )+16 \cosh \left (2 \sinh ^{-1}(c x)\right )-\cosh \left (4 \sinh ^{-1}(c x)\right )\right )+b^2 \sqrt{d} \sqrt{c^2 x^2+1} \left (32 \sinh ^{-1}(c x)^3+8 \left (\sinh \left (4 \sinh ^{-1}(c x)\right )-8 \sinh \left (2 \sinh ^{-1}(c x)\right )\right ) \sinh ^{-1}(c x)^2-32 \sinh \left (2 \sinh ^{-1}(c x)\right )+\sinh \left (4 \sinh ^{-1}(c x)\right )-4 \sinh ^{-1}(c x) \left (\cosh \left (4 \sinh ^{-1}(c x)\right )-16 \cosh \left (2 \sinh ^{-1}(c x)\right )\right )\right )}{256 c^5 \sqrt{d} \sqrt{c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(a + b*ArcSinh[c*x])^2)/Sqrt[d + c^2*d*x^2],x]

[Out]

(32*a^2*c*Sqrt[d]*x*(1 + c^2*x^2)*(-3 + 2*c^2*x^2) + 96*a^2*Sqrt[d + c^2*d*x^2]*Log[c*d*x + Sqrt[d]*Sqrt[d + c
^2*d*x^2]] + b^2*Sqrt[d]*Sqrt[1 + c^2*x^2]*(32*ArcSinh[c*x]^3 - 4*ArcSinh[c*x]*(-16*Cosh[2*ArcSinh[c*x]] + Cos
h[4*ArcSinh[c*x]]) - 32*Sinh[2*ArcSinh[c*x]] + Sinh[4*ArcSinh[c*x]] + 8*ArcSinh[c*x]^2*(-8*Sinh[2*ArcSinh[c*x]
] + Sinh[4*ArcSinh[c*x]])) + 4*a*b*Sqrt[d]*Sqrt[1 + c^2*x^2]*(16*Cosh[2*ArcSinh[c*x]] - Cosh[4*ArcSinh[c*x]] +
4*ArcSinh[c*x]*(6*ArcSinh[c*x] - 8*Sinh[2*ArcSinh[c*x]] + Sinh[4*ArcSinh[c*x]])))/(256*c^5*Sqrt[d]*Sqrt[d + c
^2*d*x^2])

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Maple [B]  time = 0.38, size = 760, normalized size = 2.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d)^(1/2),x)

[Out]

1/4*a^2*x^3/c^2/d*(c^2*d*x^2+d)^(1/2)-3/8*a^2/c^4*x/d*(c^2*d*x^2+d)^(1/2)+3/8*a^2/c^4*ln(x*c^2*d/(c^2*d)^(1/2)
+(c^2*d*x^2+d)^(1/2))/(c^2*d)^(1/2)+1/8*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^5/d*arcsinh(c*x)^3+3/8*b
^2*(d*(c^2*x^2+1))^(1/2)/c^3/d/(c^2*x^2+1)^(1/2)*arcsinh(c*x)*x^2-3/8*b^2*(d*(c^2*x^2+1))^(1/2)/c^4/d/(c^2*x^2
+1)*arcsinh(c*x)^2*x+15/64*b^2*(d*(c^2*x^2+1))^(1/2)/c^5/d/(c^2*x^2+1)^(1/2)*arcsinh(c*x)+1/4*b^2*(d*(c^2*x^2+
1))^(1/2)/d/(c^2*x^2+1)*arcsinh(c*x)^2*x^5-1/8*b^2*(d*(c^2*x^2+1))^(1/2)/c^2/d/(c^2*x^2+1)*arcsinh(c*x)^2*x^3-
1/8*b^2*(d*(c^2*x^2+1))^(1/2)/c/d/(c^2*x^2+1)^(1/2)*arcsinh(c*x)*x^4+1/32*b^2*(d*(c^2*x^2+1))^(1/2)/d/(c^2*x^2
+1)*x^5-13/64*b^2*(d*(c^2*x^2+1))^(1/2)/c^2/d/(c^2*x^2+1)*x^3-15/64*b^2*(d*(c^2*x^2+1))^(1/2)/c^4/d/(c^2*x^2+1
)*x+3/8*a*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^5/d*arcsinh(c*x)^2+3/8*a*b*(d*(c^2*x^2+1))^(1/2)/c^3/d/(
c^2*x^2+1)^(1/2)*x^2-3/4*a*b*(d*(c^2*x^2+1))^(1/2)/c^4/d/(c^2*x^2+1)*arcsinh(c*x)*x+1/2*a*b*(d*(c^2*x^2+1))^(1
/2)/d/(c^2*x^2+1)*arcsinh(c*x)*x^5-1/8*a*b*(d*(c^2*x^2+1))^(1/2)/c/d/(c^2*x^2+1)^(1/2)*x^4-1/4*a*b*(d*(c^2*x^2
+1))^(1/2)/c^2/d/(c^2*x^2+1)*arcsinh(c*x)*x^3+15/64*a*b*(d*(c^2*x^2+1))^(1/2)/c^5/d/(c^2*x^2+1)^(1/2)

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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(x^4*(a+b*arcsinh(c*x))^2/(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{b^{2} x^{4} \operatorname{arsinh}\left (c x\right )^{2} + 2 \, a b x^{4} \operatorname{arsinh}\left (c x\right ) + a^{2} x^{4}}{\sqrt{c^{2} d x^{2} + d}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

integral((b^2*x^4*arcsinh(c*x)^2 + 2*a*b*x^4*arcsinh(c*x) + a^2*x^4)/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{x^{4} \left (a + b \operatorname{asinh}{\left (c x \right )}\right )^{2}}{\sqrt{d \left (c^{2} x^{2} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(a+b*asinh(c*x))**2/(c**2*d*x**2+d)**(1/2),x)

[Out]

Integral(x**4*(a + b*asinh(c*x))**2/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 (b \operatorname{arsinh}\left (c x\right ) + a\right )}^{2} x^{4}}{\sqrt{c^{2} d x^{2} + d}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d)^(1/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)^2*x^4/sqrt(c^2*d*x^2 + d), x)