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

Optimal. Leaf size=145 \[ \frac{d^3 \left (c^2 x^2+1\right )^4 \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2}-\frac{b d^3 x \left (c^2 x^2+1\right )^{7/2}}{64 c}-\frac{7 b d^3 x \left (c^2 x^2+1\right )^{5/2}}{384 c}-\frac{35 b d^3 x \left (c^2 x^2+1\right )^{3/2}}{1536 c}-\frac{35 b d^3 x \sqrt{c^2 x^2+1}}{1024 c}-\frac{35 b d^3 \sinh ^{-1}(c x)}{1024 c^2} \]

[Out]

(-35*b*d^3*x*Sqrt[1 + c^2*x^2])/(1024*c) - (35*b*d^3*x*(1 + c^2*x^2)^(3/2))/(1536*c) - (7*b*d^3*x*(1 + c^2*x^2
)^(5/2))/(384*c) - (b*d^3*x*(1 + c^2*x^2)^(7/2))/(64*c) - (35*b*d^3*ArcSinh[c*x])/(1024*c^2) + (d^3*(1 + c^2*x
^2)^4*(a + b*ArcSinh[c*x]))/(8*c^2)

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Rubi [A]  time = 0.0697191, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {5717, 195, 215} \[ \frac{d^3 \left (c^2 x^2+1\right )^4 \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2}-\frac{b d^3 x \left (c^2 x^2+1\right )^{7/2}}{64 c}-\frac{7 b d^3 x \left (c^2 x^2+1\right )^{5/2}}{384 c}-\frac{35 b d^3 x \left (c^2 x^2+1\right )^{3/2}}{1536 c}-\frac{35 b d^3 x \sqrt{c^2 x^2+1}}{1024 c}-\frac{35 b d^3 \sinh ^{-1}(c x)}{1024 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-35*b*d^3*x*Sqrt[1 + c^2*x^2])/(1024*c) - (35*b*d^3*x*(1 + c^2*x^2)^(3/2))/(1536*c) - (7*b*d^3*x*(1 + c^2*x^2
)^(5/2))/(384*c) - (b*d^3*x*(1 + c^2*x^2)^(7/2))/(64*c) - (35*b*d^3*ArcSinh[c*x])/(1024*c^2) + (d^3*(1 + c^2*x
^2)^4*(a + b*ArcSinh[c*x]))/(8*c^2)

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 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 x \left (d+c^2 d x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac{d^3 \left (1+c^2 x^2\right )^4 \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2}-\frac{\left (b d^3\right ) \int \left (1+c^2 x^2\right )^{7/2} \, dx}{8 c}\\ &=-\frac{b d^3 x \left (1+c^2 x^2\right )^{7/2}}{64 c}+\frac{d^3 \left (1+c^2 x^2\right )^4 \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2}-\frac{\left (7 b d^3\right ) \int \left (1+c^2 x^2\right )^{5/2} \, dx}{64 c}\\ &=-\frac{7 b d^3 x \left (1+c^2 x^2\right )^{5/2}}{384 c}-\frac{b d^3 x \left (1+c^2 x^2\right )^{7/2}}{64 c}+\frac{d^3 \left (1+c^2 x^2\right )^4 \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2}-\frac{\left (35 b d^3\right ) \int \left (1+c^2 x^2\right )^{3/2} \, dx}{384 c}\\ &=-\frac{35 b d^3 x \left (1+c^2 x^2\right )^{3/2}}{1536 c}-\frac{7 b d^3 x \left (1+c^2 x^2\right )^{5/2}}{384 c}-\frac{b d^3 x \left (1+c^2 x^2\right )^{7/2}}{64 c}+\frac{d^3 \left (1+c^2 x^2\right )^4 \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2}-\frac{\left (35 b d^3\right ) \int \sqrt{1+c^2 x^2} \, dx}{512 c}\\ &=-\frac{35 b d^3 x \sqrt{1+c^2 x^2}}{1024 c}-\frac{35 b d^3 x \left (1+c^2 x^2\right )^{3/2}}{1536 c}-\frac{7 b d^3 x \left (1+c^2 x^2\right )^{5/2}}{384 c}-\frac{b d^3 x \left (1+c^2 x^2\right )^{7/2}}{64 c}+\frac{d^3 \left (1+c^2 x^2\right )^4 \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2}-\frac{\left (35 b d^3\right ) \int \frac{1}{\sqrt{1+c^2 x^2}} \, dx}{1024 c}\\ &=-\frac{35 b d^3 x \sqrt{1+c^2 x^2}}{1024 c}-\frac{35 b d^3 x \left (1+c^2 x^2\right )^{3/2}}{1536 c}-\frac{7 b d^3 x \left (1+c^2 x^2\right )^{5/2}}{384 c}-\frac{b d^3 x \left (1+c^2 x^2\right )^{7/2}}{64 c}-\frac{35 b d^3 \sinh ^{-1}(c x)}{1024 c^2}+\frac{d^3 \left (1+c^2 x^2\right )^4 \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2}\\ \end{align*}

Mathematica [A]  time = 0.173852, size = 128, normalized size = 0.88 \[ \frac{d^3 \left (c x \left (384 a c x \left (c^6 x^6+4 c^4 x^4+6 c^2 x^2+4\right )-b \sqrt{c^2 x^2+1} \left (48 c^6 x^6+200 c^4 x^4+326 c^2 x^2+279\right )\right )+3 b \left (128 c^8 x^8+512 c^6 x^6+768 c^4 x^4+512 c^2 x^2+93\right ) \sinh ^{-1}(c x)\right )}{3072 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^3*(c*x*(384*a*c*x*(4 + 6*c^2*x^2 + 4*c^4*x^4 + c^6*x^6) - b*Sqrt[1 + c^2*x^2]*(279 + 326*c^2*x^2 + 200*c^4*
x^4 + 48*c^6*x^6)) + 3*b*(93 + 512*c^2*x^2 + 768*c^4*x^4 + 512*c^6*x^6 + 128*c^8*x^8)*ArcSinh[c*x]))/(3072*c^2
)

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Maple [A]  time = 0.004, size = 176, normalized size = 1.2 \begin{align*}{\frac{1}{{c}^{2}} \left ({d}^{3}a \left ({\frac{{c}^{8}{x}^{8}}{8}}+{\frac{{c}^{6}{x}^{6}}{2}}+{\frac{3\,{c}^{4}{x}^{4}}{4}}+{\frac{{c}^{2}{x}^{2}}{2}} \right ) +{d}^{3}b \left ({\frac{{\it Arcsinh} \left ( cx \right ){c}^{8}{x}^{8}}{8}}+{\frac{{\it Arcsinh} \left ( cx \right ){c}^{6}{x}^{6}}{2}}+{\frac{3\,{\it Arcsinh} \left ( cx \right ){c}^{4}{x}^{4}}{4}}+{\frac{{\it Arcsinh} \left ( cx \right ){c}^{2}{x}^{2}}{2}}-{\frac{{c}^{7}{x}^{7}}{64}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{25\,{c}^{5}{x}^{5}}{384}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{163\,{c}^{3}{x}^{3}}{1536}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{93\,cx}{1024}\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{93\,{\it Arcsinh} \left ( cx \right ) }{1024}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^2*(d^3*a*(1/8*c^8*x^8+1/2*c^6*x^6+3/4*c^4*x^4+1/2*c^2*x^2)+d^3*b*(1/8*arcsinh(c*x)*c^8*x^8+1/2*arcsinh(c*x
)*c^6*x^6+3/4*arcsinh(c*x)*c^4*x^4+1/2*arcsinh(c*x)*c^2*x^2-1/64*c^7*x^7*(c^2*x^2+1)^(1/2)-25/384*c^5*x^5*(c^2
*x^2+1)^(1/2)-163/1536*c^3*x^3*(c^2*x^2+1)^(1/2)-93/1024*c*x*(c^2*x^2+1)^(1/2)+93/1024*arcsinh(c*x)))

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Maxima [B]  time = 1.12706, size = 540, normalized size = 3.72 \begin{align*} \frac{1}{8} \, a c^{6} d^{3} x^{8} + \frac{1}{2} \, a c^{4} d^{3} x^{6} + \frac{1}{3072} \,{\left (384 \, x^{8} \operatorname{arsinh}\left (c x\right ) -{\left (\frac{48 \, \sqrt{c^{2} x^{2} + 1} x^{7}}{c^{2}} - \frac{56 \, \sqrt{c^{2} x^{2} + 1} x^{5}}{c^{4}} + \frac{70 \, \sqrt{c^{2} x^{2} + 1} x^{3}}{c^{6}} - \frac{105 \, \sqrt{c^{2} x^{2} + 1} x}{c^{8}} + \frac{105 \, \operatorname{arsinh}\left (\frac{c^{2} x}{\sqrt{c^{2}}}\right )}{\sqrt{c^{2}} c^{8}}\right )} c\right )} b c^{6} d^{3} + \frac{3}{4} \, a c^{2} d^{3} x^{4} + \frac{1}{96} \,{\left (48 \, x^{6} \operatorname{arsinh}\left (c x\right ) -{\left (\frac{8 \, \sqrt{c^{2} x^{2} + 1} x^{5}}{c^{2}} - \frac{10 \, \sqrt{c^{2} x^{2} + 1} x^{3}}{c^{4}} + \frac{15 \, \sqrt{c^{2} x^{2} + 1} x}{c^{6}} - \frac{15 \, \operatorname{arsinh}\left (\frac{c^{2} x}{\sqrt{c^{2}}}\right )}{\sqrt{c^{2}} c^{6}}\right )} c\right )} b c^{4} d^{3} + \frac{3}{32} \,{\left (8 \, x^{4} \operatorname{arsinh}\left (c x\right ) -{\left (\frac{2 \, \sqrt{c^{2} x^{2} + 1} x^{3}}{c^{2}} - \frac{3 \, \sqrt{c^{2} x^{2} + 1} x}{c^{4}} + \frac{3 \, \operatorname{arsinh}\left (\frac{c^{2} x}{\sqrt{c^{2}}}\right )}{\sqrt{c^{2}} c^{4}}\right )} c\right )} b c^{2} d^{3} + \frac{1}{2} \, a d^{3} x^{2} + \frac{1}{4} \,{\left (2 \, x^{2} \operatorname{arsinh}\left (c x\right ) - c{\left (\frac{\sqrt{c^{2} x^{2} + 1} x}{c^{2}} - \frac{\operatorname{arsinh}\left (\frac{c^{2} x}{\sqrt{c^{2}}}\right )}{\sqrt{c^{2}} c^{2}}\right )}\right )} b d^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*a*c^6*d^3*x^8 + 1/2*a*c^4*d^3*x^6 + 1/3072*(384*x^8*arcsinh(c*x) - (48*sqrt(c^2*x^2 + 1)*x^7/c^2 - 56*sqrt
(c^2*x^2 + 1)*x^5/c^4 + 70*sqrt(c^2*x^2 + 1)*x^3/c^6 - 105*sqrt(c^2*x^2 + 1)*x/c^8 + 105*arcsinh(c^2*x/sqrt(c^
2))/(sqrt(c^2)*c^8))*c)*b*c^6*d^3 + 3/4*a*c^2*d^3*x^4 + 1/96*(48*x^6*arcsinh(c*x) - (8*sqrt(c^2*x^2 + 1)*x^5/c
^2 - 10*sqrt(c^2*x^2 + 1)*x^3/c^4 + 15*sqrt(c^2*x^2 + 1)*x/c^6 - 15*arcsinh(c^2*x/sqrt(c^2))/(sqrt(c^2)*c^6))*
c)*b*c^4*d^3 + 3/32*(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/sqrt(c^2))/(sqrt(c^2)*c^4))*c)*b*c^2*d^3 + 1/2*a*d^3*x^2 + 1/4*(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)))*b*d^3

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Fricas [A]  time = 2.39912, size = 424, normalized size = 2.92 \begin{align*} \frac{384 \, a c^{8} d^{3} x^{8} + 1536 \, a c^{6} d^{3} x^{6} + 2304 \, a c^{4} d^{3} x^{4} + 1536 \, a c^{2} d^{3} x^{2} + 3 \,{\left (128 \, b c^{8} d^{3} x^{8} + 512 \, b c^{6} d^{3} x^{6} + 768 \, b c^{4} d^{3} x^{4} + 512 \, b c^{2} d^{3} x^{2} + 93 \, b d^{3}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) -{\left (48 \, b c^{7} d^{3} x^{7} + 200 \, b c^{5} d^{3} x^{5} + 326 \, b c^{3} d^{3} x^{3} + 279 \, b c d^{3} x\right )} \sqrt{c^{2} x^{2} + 1}}{3072 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3072*(384*a*c^8*d^3*x^8 + 1536*a*c^6*d^3*x^6 + 2304*a*c^4*d^3*x^4 + 1536*a*c^2*d^3*x^2 + 3*(128*b*c^8*d^3*x^
8 + 512*b*c^6*d^3*x^6 + 768*b*c^4*d^3*x^4 + 512*b*c^2*d^3*x^2 + 93*b*d^3)*log(c*x + sqrt(c^2*x^2 + 1)) - (48*b
*c^7*d^3*x^7 + 200*b*c^5*d^3*x^5 + 326*b*c^3*d^3*x^3 + 279*b*c*d^3*x)*sqrt(c^2*x^2 + 1))/c^2

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Sympy [A]  time = 16.0277, size = 253, normalized size = 1.74 \begin{align*} \begin{cases} \frac{a c^{6} d^{3} x^{8}}{8} + \frac{a c^{4} d^{3} x^{6}}{2} + \frac{3 a c^{2} d^{3} x^{4}}{4} + \frac{a d^{3} x^{2}}{2} + \frac{b c^{6} d^{3} x^{8} \operatorname{asinh}{\left (c x \right )}}{8} - \frac{b c^{5} d^{3} x^{7} \sqrt{c^{2} x^{2} + 1}}{64} + \frac{b c^{4} d^{3} x^{6} \operatorname{asinh}{\left (c x \right )}}{2} - \frac{25 b c^{3} d^{3} x^{5} \sqrt{c^{2} x^{2} + 1}}{384} + \frac{3 b c^{2} d^{3} x^{4} \operatorname{asinh}{\left (c x \right )}}{4} - \frac{163 b c d^{3} x^{3} \sqrt{c^{2} x^{2} + 1}}{1536} + \frac{b d^{3} x^{2} \operatorname{asinh}{\left (c x \right )}}{2} - \frac{93 b d^{3} x \sqrt{c^{2} x^{2} + 1}}{1024 c} + \frac{93 b d^{3} \operatorname{asinh}{\left (c x \right )}}{1024 c^{2}} & \text{for}\: c \neq 0 \\\frac{a d^{3} x^{2}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*c**6*d**3*x**8/8 + a*c**4*d**3*x**6/2 + 3*a*c**2*d**3*x**4/4 + a*d**3*x**2/2 + b*c**6*d**3*x**8*a
sinh(c*x)/8 - b*c**5*d**3*x**7*sqrt(c**2*x**2 + 1)/64 + b*c**4*d**3*x**6*asinh(c*x)/2 - 25*b*c**3*d**3*x**5*sq
rt(c**2*x**2 + 1)/384 + 3*b*c**2*d**3*x**4*asinh(c*x)/4 - 163*b*c*d**3*x**3*sqrt(c**2*x**2 + 1)/1536 + b*d**3*
x**2*asinh(c*x)/2 - 93*b*d**3*x*sqrt(c**2*x**2 + 1)/(1024*c) + 93*b*d**3*asinh(c*x)/(1024*c**2), Ne(c, 0)), (a
*d**3*x**2/2, True))

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Giac [B]  time = 2.01012, size = 572, normalized size = 3.94 \begin{align*} \frac{1}{8} \, a c^{6} d^{3} x^{8} + \frac{1}{2} \, a c^{4} d^{3} x^{6} + \frac{1}{3072} \,{\left (384 \, x^{8} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) -{\left (\sqrt{c^{2} x^{2} + 1}{\left (2 \,{\left (4 \, x^{2}{\left (\frac{6 \, x^{2}}{c^{2}} - \frac{7}{c^{4}}\right )} + \frac{35}{c^{6}}\right )} x^{2} - \frac{105}{c^{8}}\right )} x - \frac{105 \, \log \left ({\left | -x{\left | c \right |} + \sqrt{c^{2} x^{2} + 1} \right |}\right )}{c^{8}{\left | c \right |}}\right )} c\right )} b c^{6} d^{3} + \frac{3}{4} \, a c^{2} d^{3} x^{4} + \frac{1}{96} \,{\left (48 \, x^{6} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) -{\left (\sqrt{c^{2} x^{2} + 1}{\left (2 \, x^{2}{\left (\frac{4 \, x^{2}}{c^{2}} - \frac{5}{c^{4}}\right )} + \frac{15}{c^{6}}\right )} x + \frac{15 \, \log \left ({\left | -x{\left | c \right |} + \sqrt{c^{2} x^{2} + 1} \right |}\right )}{c^{6}{\left | c \right |}}\right )} c\right )} b c^{4} d^{3} + \frac{3}{32} \,{\left (8 \, x^{4} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) -{\left (\sqrt{c^{2} x^{2} + 1} x{\left (\frac{2 \, x^{2}}{c^{2}} - \frac{3}{c^{4}}\right )} - \frac{3 \, \log \left ({\left | -x{\left | c \right |} + \sqrt{c^{2} x^{2} + 1} \right |}\right )}{c^{4}{\left | c \right |}}\right )} c\right )} b c^{2} d^{3} + \frac{1}{2} \, a d^{3} x^{2} + \frac{1}{4} \,{\left (2 \, x^{2} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) - c{\left (\frac{\sqrt{c^{2} x^{2} + 1} x}{c^{2}} + \frac{\log \left ({\left | -x{\left | c \right |} + \sqrt{c^{2} x^{2} + 1} \right |}\right )}{c^{2}{\left | c \right |}}\right )}\right )} b d^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*a*c^6*d^3*x^8 + 1/2*a*c^4*d^3*x^6 + 1/3072*(384*x^8*log(c*x + sqrt(c^2*x^2 + 1)) - (sqrt(c^2*x^2 + 1)*(2*(
4*x^2*(6*x^2/c^2 - 7/c^4) + 35/c^6)*x^2 - 105/c^8)*x - 105*log(abs(-x*abs(c) + sqrt(c^2*x^2 + 1)))/(c^8*abs(c)
))*c)*b*c^6*d^3 + 3/4*a*c^2*d^3*x^4 + 1/96*(48*x^6*log(c*x + sqrt(c^2*x^2 + 1)) - (sqrt(c^2*x^2 + 1)*(2*x^2*(4
*x^2/c^2 - 5/c^4) + 15/c^6)*x + 15*log(abs(-x*abs(c) + sqrt(c^2*x^2 + 1)))/(c^6*abs(c)))*c)*b*c^4*d^3 + 3/32*(
8*x^4*log(c*x + sqrt(c^2*x^2 + 1)) - (sqrt(c^2*x^2 + 1)*x*(2*x^2/c^2 - 3/c^4) - 3*log(abs(-x*abs(c) + sqrt(c^2
*x^2 + 1)))/(c^4*abs(c)))*c)*b*c^2*d^3 + 1/2*a*d^3*x^2 + 1/4*(2*x^2*log(c*x + sqrt(c^2*x^2 + 1)) - c*(sqrt(c^2
*x^2 + 1)*x/c^2 + log(abs(-x*abs(c) + sqrt(c^2*x^2 + 1)))/(c^2*abs(c))))*b*d^3