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

Optimal. Leaf size=120 \[ \frac{1}{6} c^2 d x^6 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{4} d x^4 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{36} b c d x^5 \sqrt{c^2 x^2+1}-\frac{b d x^3 \sqrt{c^2 x^2+1}}{36 c}+\frac{b d x \sqrt{c^2 x^2+1}}{24 c^3}-\frac{b d \sinh ^{-1}(c x)}{24 c^4} \]

[Out]

(b*d*x*Sqrt[1 + c^2*x^2])/(24*c^3) - (b*d*x^3*Sqrt[1 + c^2*x^2])/(36*c) - (b*c*d*x^5*Sqrt[1 + c^2*x^2])/36 - (
b*d*ArcSinh[c*x])/(24*c^4) + (d*x^4*(a + b*ArcSinh[c*x]))/4 + (c^2*d*x^6*(a + b*ArcSinh[c*x]))/6

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Rubi [A]  time = 0.0981831, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {14, 5730, 12, 459, 321, 215} \[ \frac{1}{6} c^2 d x^6 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{4} d x^4 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{36} b c d x^5 \sqrt{c^2 x^2+1}-\frac{b d x^3 \sqrt{c^2 x^2+1}}{36 c}+\frac{b d x \sqrt{c^2 x^2+1}}{24 c^3}-\frac{b d \sinh ^{-1}(c x)}{24 c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d*x*Sqrt[1 + c^2*x^2])/(24*c^3) - (b*d*x^3*Sqrt[1 + c^2*x^2])/(36*c) - (b*c*d*x^5*Sqrt[1 + c^2*x^2])/36 - (
b*d*ArcSinh[c*x])/(24*c^4) + (d*x^4*(a + b*ArcSinh[c*x]))/4 + (c^2*d*x^6*(a + b*ArcSinh[c*x]))/6

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 5730

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSinh[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1
+ c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 459

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

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 x^3 \left (d+c^2 d x^2\right ) \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac{1}{4} d x^4 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{6} c^2 d x^6 \left (a+b \sinh ^{-1}(c x)\right )-(b c) \int \frac{d x^4 \left (3+2 c^2 x^2\right )}{12 \sqrt{1+c^2 x^2}} \, dx\\ &=\frac{1}{4} d x^4 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{6} c^2 d x^6 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{12} (b c d) \int \frac{x^4 \left (3+2 c^2 x^2\right )}{\sqrt{1+c^2 x^2}} \, dx\\ &=-\frac{1}{36} b c d x^5 \sqrt{1+c^2 x^2}+\frac{1}{4} d x^4 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{6} c^2 d x^6 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{9} (b c d) \int \frac{x^4}{\sqrt{1+c^2 x^2}} \, dx\\ &=-\frac{b d x^3 \sqrt{1+c^2 x^2}}{36 c}-\frac{1}{36} b c d x^5 \sqrt{1+c^2 x^2}+\frac{1}{4} d x^4 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{6} c^2 d x^6 \left (a+b \sinh ^{-1}(c x)\right )+\frac{(b d) \int \frac{x^2}{\sqrt{1+c^2 x^2}} \, dx}{12 c}\\ &=\frac{b d x \sqrt{1+c^2 x^2}}{24 c^3}-\frac{b d x^3 \sqrt{1+c^2 x^2}}{36 c}-\frac{1}{36} b c d x^5 \sqrt{1+c^2 x^2}+\frac{1}{4} d x^4 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{6} c^2 d x^6 \left (a+b \sinh ^{-1}(c x)\right )-\frac{(b d) \int \frac{1}{\sqrt{1+c^2 x^2}} \, dx}{24 c^3}\\ &=\frac{b d x \sqrt{1+c^2 x^2}}{24 c^3}-\frac{b d x^3 \sqrt{1+c^2 x^2}}{36 c}-\frac{1}{36} b c d x^5 \sqrt{1+c^2 x^2}-\frac{b d \sinh ^{-1}(c x)}{24 c^4}+\frac{1}{4} d x^4 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{6} c^2 d x^6 \left (a+b \sinh ^{-1}(c x)\right )\\ \end{align*}

Mathematica [A]  time = 0.0547256, size = 88, normalized size = 0.73 \[ \frac{d \left (6 a c^4 x^4 \left (2 c^2 x^2+3\right )+b c x \sqrt{c^2 x^2+1} \left (-2 c^4 x^4-2 c^2 x^2+3\right )+3 b \left (4 c^6 x^6+6 c^4 x^4-1\right ) \sinh ^{-1}(c x)\right )}{72 c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(6*a*c^4*x^4*(3 + 2*c^2*x^2) + b*c*x*Sqrt[1 + c^2*x^2]*(3 - 2*c^2*x^2 - 2*c^4*x^4) + 3*b*(-1 + 6*c^4*x^4 +
4*c^6*x^6)*ArcSinh[c*x]))/(72*c^4)

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Maple [A]  time = 0.009, size = 113, normalized size = 0.9 \begin{align*}{\frac{1}{{c}^{4}} \left ( da \left ({\frac{{c}^{6}{x}^{6}}{6}}+{\frac{{c}^{4}{x}^{4}}{4}} \right ) +db \left ({\frac{{\it Arcsinh} \left ( cx \right ){c}^{6}{x}^{6}}{6}}+{\frac{{\it Arcsinh} \left ( cx \right ){c}^{4}{x}^{4}}{4}}-{\frac{{c}^{5}{x}^{5}}{36}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{{c}^{3}{x}^{3}}{36}\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{cx}{24}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{{\it Arcsinh} \left ( cx \right ) }{24}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^4*(d*a*(1/6*c^6*x^6+1/4*c^4*x^4)+d*b*(1/6*arcsinh(c*x)*c^6*x^6+1/4*arcsinh(c*x)*c^4*x^4-1/36*c^5*x^5*(c^2*
x^2+1)^(1/2)-1/36*c^3*x^3*(c^2*x^2+1)^(1/2)+1/24*c*x*(c^2*x^2+1)^(1/2)-1/24*arcsinh(c*x)))

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Maxima [A]  time = 1.15503, size = 257, normalized size = 2.14 \begin{align*} \frac{1}{6} \, a c^{2} d x^{6} + \frac{1}{4} \, a d x^{4} + \frac{1}{288} \,{\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^{2} d + \frac{1}{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 d \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*a*c^2*d*x^6 + 1/4*a*d*x^4 + 1/288*(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^2*d + 1/32*(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*d

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Fricas [A]  time = 2.50227, size = 242, normalized size = 2.02 \begin{align*} \frac{12 \, a c^{6} d x^{6} + 18 \, a c^{4} d x^{4} + 3 \,{\left (4 \, b c^{6} d x^{6} + 6 \, b c^{4} d x^{4} - b d\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) -{\left (2 \, b c^{5} d x^{5} + 2 \, b c^{3} d x^{3} - 3 \, b c d x\right )} \sqrt{c^{2} x^{2} + 1}}{72 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/72*(12*a*c^6*d*x^6 + 18*a*c^4*d*x^4 + 3*(4*b*c^6*d*x^6 + 6*b*c^4*d*x^4 - b*d)*log(c*x + sqrt(c^2*x^2 + 1)) -
(2*b*c^5*d*x^5 + 2*b*c^3*d*x^3 - 3*b*c*d*x)*sqrt(c^2*x^2 + 1))/c^4

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Sympy [A]  time = 5.18128, size = 138, normalized size = 1.15 \begin{align*} \begin{cases} \frac{a c^{2} d x^{6}}{6} + \frac{a d x^{4}}{4} + \frac{b c^{2} d x^{6} \operatorname{asinh}{\left (c x \right )}}{6} - \frac{b c d x^{5} \sqrt{c^{2} x^{2} + 1}}{36} + \frac{b d x^{4} \operatorname{asinh}{\left (c x \right )}}{4} - \frac{b d x^{3} \sqrt{c^{2} x^{2} + 1}}{36 c} + \frac{b d x \sqrt{c^{2} x^{2} + 1}}{24 c^{3}} - \frac{b d \operatorname{asinh}{\left (c x \right )}}{24 c^{4}} & \text{for}\: c \neq 0 \\\frac{a d x^{4}}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*c**2*d*x**6/6 + a*d*x**4/4 + b*c**2*d*x**6*asinh(c*x)/6 - b*c*d*x**5*sqrt(c**2*x**2 + 1)/36 + b*d
*x**4*asinh(c*x)/4 - b*d*x**3*sqrt(c**2*x**2 + 1)/(36*c) + b*d*x*sqrt(c**2*x**2 + 1)/(24*c**3) - b*d*asinh(c*x
)/(24*c**4), Ne(c, 0)), (a*d*x**4/4, True))

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Giac [A]  time = 1.58097, size = 273, normalized size = 2.28 \begin{align*} \frac{1}{6} \, a c^{2} d x^{6} + \frac{1}{4} \, a d x^{4} + \frac{1}{288} \,{\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^{2} d + \frac{1}{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 d \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*a*c^2*d*x^6 + 1/4*a*d*x^4 + 1/288*(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^2*d + 1/32*(8*x^4*l
og(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*d