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

Optimal. Leaf size=124 \[ \frac{1}{7} c^2 d x^7 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} d x^5 \left (a+b \sinh ^{-1}(c x)\right )-\frac{b d \left (c^2 x^2+1\right )^{7/2}}{49 c^5}+\frac{8 b d \left (c^2 x^2+1\right )^{5/2}}{175 c^5}-\frac{b d \left (c^2 x^2+1\right )^{3/2}}{105 c^5}-\frac{2 b d \sqrt{c^2 x^2+1}}{35 c^5} \]

[Out]

(-2*b*d*Sqrt[1 + c^2*x^2])/(35*c^5) - (b*d*(1 + c^2*x^2)^(3/2))/(105*c^5) + (8*b*d*(1 + c^2*x^2)^(5/2))/(175*c
^5) - (b*d*(1 + c^2*x^2)^(7/2))/(49*c^5) + (d*x^5*(a + b*ArcSinh[c*x]))/5 + (c^2*d*x^7*(a + b*ArcSinh[c*x]))/7

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Rubi [A]  time = 0.118849, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {14, 5730, 12, 446, 77} \[ \frac{1}{7} c^2 d x^7 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} d x^5 \left (a+b \sinh ^{-1}(c x)\right )-\frac{b d \left (c^2 x^2+1\right )^{7/2}}{49 c^5}+\frac{8 b d \left (c^2 x^2+1\right )^{5/2}}{175 c^5}-\frac{b d \left (c^2 x^2+1\right )^{3/2}}{105 c^5}-\frac{2 b d \sqrt{c^2 x^2+1}}{35 c^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b*d*Sqrt[1 + c^2*x^2])/(35*c^5) - (b*d*(1 + c^2*x^2)^(3/2))/(105*c^5) + (8*b*d*(1 + c^2*x^2)^(5/2))/(175*c
^5) - (b*d*(1 + c^2*x^2)^(7/2))/(49*c^5) + (d*x^5*(a + b*ArcSinh[c*x]))/5 + (c^2*d*x^7*(a + b*ArcSinh[c*x]))/7

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 446

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int x^4 \left (d+c^2 d x^2\right ) \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac{1}{5} d x^5 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{7} c^2 d x^7 \left (a+b \sinh ^{-1}(c x)\right )-(b c) \int \frac{d x^5 \left (7+5 c^2 x^2\right )}{35 \sqrt{1+c^2 x^2}} \, dx\\ &=\frac{1}{5} d x^5 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{7} c^2 d x^7 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{35} (b c d) \int \frac{x^5 \left (7+5 c^2 x^2\right )}{\sqrt{1+c^2 x^2}} \, dx\\ &=\frac{1}{5} d x^5 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{7} c^2 d x^7 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{70} (b c d) \operatorname{Subst}\left (\int \frac{x^2 \left (7+5 c^2 x\right )}{\sqrt{1+c^2 x}} \, dx,x,x^2\right )\\ &=\frac{1}{5} d x^5 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{7} c^2 d x^7 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{70} (b c d) \operatorname{Subst}\left (\int \left (\frac{2}{c^4 \sqrt{1+c^2 x}}+\frac{\sqrt{1+c^2 x}}{c^4}-\frac{8 \left (1+c^2 x\right )^{3/2}}{c^4}+\frac{5 \left (1+c^2 x\right )^{5/2}}{c^4}\right ) \, dx,x,x^2\right )\\ &=-\frac{2 b d \sqrt{1+c^2 x^2}}{35 c^5}-\frac{b d \left (1+c^2 x^2\right )^{3/2}}{105 c^5}+\frac{8 b d \left (1+c^2 x^2\right )^{5/2}}{175 c^5}-\frac{b d \left (1+c^2 x^2\right )^{7/2}}{49 c^5}+\frac{1}{5} d x^5 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{7} c^2 d x^7 \left (a+b \sinh ^{-1}(c x)\right )\\ \end{align*}

Mathematica [A]  time = 0.0937715, size = 87, normalized size = 0.7 \[ \frac{d \left (105 a x^5 \left (5 c^2 x^2+7\right )-\frac{b \sqrt{c^2 x^2+1} \left (75 c^6 x^6+57 c^4 x^4-76 c^2 x^2+152\right )}{c^5}+105 b x^5 \left (5 c^2 x^2+7\right ) \sinh ^{-1}(c x)\right )}{3675} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(105*a*x^5*(7 + 5*c^2*x^2) - (b*Sqrt[1 + c^2*x^2]*(152 - 76*c^2*x^2 + 57*c^4*x^4 + 75*c^6*x^6))/c^5 + 105*b
*x^5*(7 + 5*c^2*x^2)*ArcSinh[c*x]))/3675

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Maple [A]  time = 0.009, size = 124, normalized size = 1. \begin{align*}{\frac{1}{{c}^{5}} \left ( da \left ({\frac{{c}^{7}{x}^{7}}{7}}+{\frac{{c}^{5}{x}^{5}}{5}} \right ) +db \left ({\frac{{\it Arcsinh} \left ( cx \right ){c}^{7}{x}^{7}}{7}}+{\frac{{\it Arcsinh} \left ( cx \right ){c}^{5}{x}^{5}}{5}}-{\frac{{c}^{6}{x}^{6}}{49}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{19\,{c}^{4}{x}^{4}}{1225}\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{76\,{c}^{2}{x}^{2}}{3675}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{152}{3675}\sqrt{{c}^{2}{x}^{2}+1}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^5*(d*a*(1/7*c^7*x^7+1/5*c^5*x^5)+d*b*(1/7*arcsinh(c*x)*c^7*x^7+1/5*arcsinh(c*x)*c^5*x^5-1/49*c^6*x^6*(c^2*
x^2+1)^(1/2)-19/1225*c^4*x^4*(c^2*x^2+1)^(1/2)+76/3675*c^2*x^2*(c^2*x^2+1)^(1/2)-152/3675*(c^2*x^2+1)^(1/2)))

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Maxima [A]  time = 1.20355, size = 248, normalized size = 2. \begin{align*} \frac{1}{7} \, a c^{2} d x^{7} + \frac{1}{5} \, a d x^{5} + \frac{1}{245} \,{\left (35 \, x^{7} \operatorname{arsinh}\left (c x\right ) -{\left (\frac{5 \, \sqrt{c^{2} x^{2} + 1} x^{6}}{c^{2}} - \frac{6 \, \sqrt{c^{2} x^{2} + 1} x^{4}}{c^{4}} + \frac{8 \, \sqrt{c^{2} x^{2} + 1} x^{2}}{c^{6}} - \frac{16 \, \sqrt{c^{2} x^{2} + 1}}{c^{8}}\right )} c\right )} b c^{2} d + \frac{1}{75} \,{\left (15 \, x^{5} \operatorname{arsinh}\left (c x\right ) -{\left (\frac{3 \, \sqrt{c^{2} x^{2} + 1} x^{4}}{c^{2}} - \frac{4 \, \sqrt{c^{2} x^{2} + 1} x^{2}}{c^{4}} + \frac{8 \, \sqrt{c^{2} x^{2} + 1}}{c^{6}}\right )} c\right )} b d \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*a*c^2*d*x^7 + 1/5*a*d*x^5 + 1/245*(35*x^7*arcsinh(c*x) - (5*sqrt(c^2*x^2 + 1)*x^6/c^2 - 6*sqrt(c^2*x^2 + 1
)*x^4/c^4 + 8*sqrt(c^2*x^2 + 1)*x^2/c^6 - 16*sqrt(c^2*x^2 + 1)/c^8)*c)*b*c^2*d + 1/75*(15*x^5*arcsinh(c*x) - (
3*sqrt(c^2*x^2 + 1)*x^4/c^2 - 4*sqrt(c^2*x^2 + 1)*x^2/c^4 + 8*sqrt(c^2*x^2 + 1)/c^6)*c)*b*d

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Fricas [A]  time = 2.255, size = 265, normalized size = 2.14 \begin{align*} \frac{525 \, a c^{7} d x^{7} + 735 \, a c^{5} d x^{5} + 105 \,{\left (5 \, b c^{7} d x^{7} + 7 \, b c^{5} d x^{5}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) -{\left (75 \, b c^{6} d x^{6} + 57 \, b c^{4} d x^{4} - 76 \, b c^{2} d x^{2} + 152 \, b d\right )} \sqrt{c^{2} x^{2} + 1}}{3675 \, c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3675*(525*a*c^7*d*x^7 + 735*a*c^5*d*x^5 + 105*(5*b*c^7*d*x^7 + 7*b*c^5*d*x^5)*log(c*x + sqrt(c^2*x^2 + 1)) -
(75*b*c^6*d*x^6 + 57*b*c^4*d*x^4 - 76*b*c^2*d*x^2 + 152*b*d)*sqrt(c^2*x^2 + 1))/c^5

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Sympy [A]  time = 7.9766, size = 151, normalized size = 1.22 \begin{align*} \begin{cases} \frac{a c^{2} d x^{7}}{7} + \frac{a d x^{5}}{5} + \frac{b c^{2} d x^{7} \operatorname{asinh}{\left (c x \right )}}{7} - \frac{b c d x^{6} \sqrt{c^{2} x^{2} + 1}}{49} + \frac{b d x^{5} \operatorname{asinh}{\left (c x \right )}}{5} - \frac{19 b d x^{4} \sqrt{c^{2} x^{2} + 1}}{1225 c} + \frac{76 b d x^{2} \sqrt{c^{2} x^{2} + 1}}{3675 c^{3}} - \frac{152 b d \sqrt{c^{2} x^{2} + 1}}{3675 c^{5}} & \text{for}\: c \neq 0 \\\frac{a d x^{5}}{5} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*c**2*d*x**7/7 + a*d*x**5/5 + b*c**2*d*x**7*asinh(c*x)/7 - b*c*d*x**6*sqrt(c**2*x**2 + 1)/49 + b*d
*x**5*asinh(c*x)/5 - 19*b*d*x**4*sqrt(c**2*x**2 + 1)/(1225*c) + 76*b*d*x**2*sqrt(c**2*x**2 + 1)/(3675*c**3) -
152*b*d*sqrt(c**2*x**2 + 1)/(3675*c**5), Ne(c, 0)), (a*d*x**5/5, True))

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Giac [A]  time = 1.49798, size = 238, normalized size = 1.92 \begin{align*} \frac{1}{7} \, a c^{2} d x^{7} + \frac{1}{5} \, a d x^{5} + \frac{1}{245} \,{\left (35 \, x^{7} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) - \frac{5 \,{\left (c^{2} x^{2} + 1\right )}^{\frac{7}{2}} - 21 \,{\left (c^{2} x^{2} + 1\right )}^{\frac{5}{2}} + 35 \,{\left (c^{2} x^{2} + 1\right )}^{\frac{3}{2}} - 35 \, \sqrt{c^{2} x^{2} + 1}}{c^{7}}\right )} b c^{2} d + \frac{1}{75} \,{\left (15 \, x^{5} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) - \frac{3 \,{\left (c^{2} x^{2} + 1\right )}^{\frac{5}{2}} - 10 \,{\left (c^{2} x^{2} + 1\right )}^{\frac{3}{2}} + 15 \, \sqrt{c^{2} x^{2} + 1}}{c^{5}}\right )} b d \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*a*c^2*d*x^7 + 1/5*a*d*x^5 + 1/245*(35*x^7*log(c*x + sqrt(c^2*x^2 + 1)) - (5*(c^2*x^2 + 1)^(7/2) - 21*(c^2*
x^2 + 1)^(5/2) + 35*(c^2*x^2 + 1)^(3/2) - 35*sqrt(c^2*x^2 + 1))/c^7)*b*c^2*d + 1/75*(15*x^5*log(c*x + sqrt(c^2
*x^2 + 1)) - (3*(c^2*x^2 + 1)^(5/2) - 10*(c^2*x^2 + 1)^(3/2) + 15*sqrt(c^2*x^2 + 1))/c^5)*b*d