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

Optimal. Leaf size=192 \[ \frac{x^3 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{4 c^2 d}-\frac{3 x \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{8 c^4 d}+\frac{3 \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c^5 \sqrt{c^2 d x^2+d}}-\frac{b x^4 \sqrt{c^2 x^2+1}}{16 c \sqrt{c^2 d x^2+d}}+\frac{3 b x^2 \sqrt{c^2 x^2+1}}{16 c^3 \sqrt{c^2 d x^2+d}} \]

[Out]

(3*b*x^2*Sqrt[1 + c^2*x^2])/(16*c^3*Sqrt[d + c^2*d*x^2]) - (b*x^4*Sqrt[1 + c^2*x^2])/(16*c*Sqrt[d + c^2*d*x^2]
) - (3*x*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(8*c^4*d) + (x^3*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/
(4*c^2*d) + (3*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^2)/(16*b*c^5*Sqrt[d + c^2*d*x^2])

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Rubi [A]  time = 0.252467, antiderivative size = 192, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {5758, 5677, 5675, 30} \[ \frac{x^3 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{4 c^2 d}-\frac{3 x \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{8 c^4 d}+\frac{3 \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c^5 \sqrt{c^2 d x^2+d}}-\frac{b x^4 \sqrt{c^2 x^2+1}}{16 c \sqrt{c^2 d x^2+d}}+\frac{3 b x^2 \sqrt{c^2 x^2+1}}{16 c^3 \sqrt{c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*x^2*Sqrt[1 + c^2*x^2])/(16*c^3*Sqrt[d + c^2*d*x^2]) - (b*x^4*Sqrt[1 + c^2*x^2])/(16*c*Sqrt[d + c^2*d*x^2]
) - (3*x*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(8*c^4*d) + (x^3*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/
(4*c^2*d) + (3*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^2)/(16*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 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{x^4 \left (a+b \sinh ^{-1}(c x)\right )}{\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 )}{4 c^2 d}-\frac{3 \int \frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{d+c^2 d x^2}} \, dx}{4 c^2}-\frac{\left (b \sqrt{1+c^2 x^2}\right ) \int x^3 \, dx}{4 c \sqrt{d+c^2 d x^2}}\\ &=-\frac{b x^4 \sqrt{1+c^2 x^2}}{16 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 )}{8 c^4 d}+\frac{x^3 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{4 c^2 d}+\frac{3 \int \frac{a+b \sinh ^{-1}(c x)}{\sqrt{d+c^2 d x^2}} \, dx}{8 c^4}+\frac{\left (3 b \sqrt{1+c^2 x^2}\right ) \int x \, dx}{8 c^3 \sqrt{d+c^2 d x^2}}\\ &=\frac{3 b x^2 \sqrt{1+c^2 x^2}}{16 c^3 \sqrt{d+c^2 d x^2}}-\frac{b x^4 \sqrt{1+c^2 x^2}}{16 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 )}{8 c^4 d}+\frac{x^3 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{4 c^2 d}+\frac{\left (3 \sqrt{1+c^2 x^2}\right ) \int \frac{a+b \sinh ^{-1}(c x)}{\sqrt{1+c^2 x^2}} \, dx}{8 c^4 \sqrt{d+c^2 d x^2}}\\ &=\frac{3 b x^2 \sqrt{1+c^2 x^2}}{16 c^3 \sqrt{d+c^2 d x^2}}-\frac{b x^4 \sqrt{1+c^2 x^2}}{16 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 )}{8 c^4 d}+\frac{x^3 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{4 c^2 d}+\frac{3 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c^5 \sqrt{d+c^2 d x^2}}\\ \end{align*}

Mathematica [A]  time = 0.6033, size = 151, normalized size = 0.79 \[ \frac{\frac{16 a c x \left (2 c^2 x^2-3\right ) \sqrt{c^2 d x^2+d}}{d}+\frac{48 a \log \left (\sqrt{d} \sqrt{c^2 d x^2+d}+c d x\right )}{\sqrt{d}}+\frac{b \sqrt{c^2 x^2+1} \left (4 \sinh ^{-1}(c x) \left (6 \sinh ^{-1}(c x)-8 \sinh \left (2 \sinh ^{-1}(c x)\right )+\sinh \left (4 \sinh ^{-1}(c x)\right )\right )+16 \cosh \left (2 \sinh ^{-1}(c x)\right )-\cosh \left (4 \sinh ^{-1}(c x)\right )\right )}{\sqrt{c^2 d x^2+d}}}{128 c^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((16*a*c*x*(-3 + 2*c^2*x^2)*Sqrt[d + c^2*d*x^2])/d + (48*a*Log[c*d*x + Sqrt[d]*Sqrt[d + c^2*d*x^2]])/Sqrt[d] +
(b*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*Sin
h[2*ArcSinh[c*x]] + Sinh[4*ArcSinh[c*x]])))/Sqrt[d + c^2*d*x^2])/(128*c^5)

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Maple [B]  time = 0.253, size = 347, normalized size = 1.8 \begin{align*}{\frac{a{x}^{3}}{4\,{c}^{2}d}\sqrt{{c}^{2}d{x}^{2}+d}}-{\frac{3\,ax}{8\,{c}^{4}d}\sqrt{{c}^{2}d{x}^{2}+d}}+{\frac{3\,a}{8\,{c}^{4}}\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{b{\it Arcsinh} \left ( cx \right ){x}^{5}}{4\,d \left ({c}^{2}{x}^{2}+1 \right ) }\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{b{x}^{4}}{16\,cd}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}-{\frac{b{\it Arcsinh} \left ( cx \right ){x}^{3}}{8\,{c}^{2}d \left ({c}^{2}{x}^{2}+1 \right ) }\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}+{\frac{3\,b{x}^{2}}{16\,{c}^{3}d}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}-{\frac{3\,b{\it Arcsinh} \left ( cx \right ) x}{8\,{c}^{4}d \left ({c}^{2}{x}^{2}+1 \right ) }\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}+{\frac{3\,b \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{16\,{c}^{5}d}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{15\,b}{128\,{c}^{5}d}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*x^3/c^2/d*(c^2*d*x^2+d)^(1/2)-3/8*a/c^4*x/d*(c^2*d*x^2+d)^(1/2)+3/8*a/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)+1/4*b*(d*(c^2*x^2+1))^(1/2)/d/(c^2*x^2+1)*arcsinh(c*x)*x^5-1/16*b*(d*(c^2*x^2+1)
)^(1/2)/c/d/(c^2*x^2+1)^(1/2)*x^4-1/8*b*(d*(c^2*x^2+1))^(1/2)/c^2/d/(c^2*x^2+1)*arcsinh(c*x)*x^3+3/16*b*(d*(c^
2*x^2+1))^(1/2)/c^3/d/(c^2*x^2+1)^(1/2)*x^2-3/8*b*(d*(c^2*x^2+1))^(1/2)/c^4/d/(c^2*x^2+1)*arcsinh(c*x)*x+3/16*
b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^5/d*arcsinh(c*x)^2+15/128*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))/(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 x^{4} \operatorname{arsinh}\left (c x\right ) + a 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))/(c^2*d*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

integral((b*x^4*arcsinh(c*x) + a*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 )}{\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))/(c**2*d*x**2+d)**(1/2),x)

[Out]

Integral(x**4*(a + b*asinh(c*x))/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 )} 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))/(c^2*d*x^2+d)^(1/2),x, algorithm="giac")

[Out]

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