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

Optimal. Leaf size=157 \[ \frac{1}{7} c^4 d^2 x^7 \left (a+b \sinh ^{-1}(c x)\right )+\frac{2}{5} c^2 d^2 x^5 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} d^2 x^3 \left (a+b \sinh ^{-1}(c x)\right )-\frac{b d^2 \left (c^2 x^2+1\right )^{7/2}}{49 c^3}+\frac{b d^2 \left (c^2 x^2+1\right )^{5/2}}{175 c^3}+\frac{4 b d^2 \left (c^2 x^2+1\right )^{3/2}}{315 c^3}+\frac{8 b d^2 \sqrt{c^2 x^2+1}}{105 c^3} \]

[Out]

(8*b*d^2*Sqrt[1 + c^2*x^2])/(105*c^3) + (4*b*d^2*(1 + c^2*x^2)^(3/2))/(315*c^3) + (b*d^2*(1 + c^2*x^2)^(5/2))/
(175*c^3) - (b*d^2*(1 + c^2*x^2)^(7/2))/(49*c^3) + (d^2*x^3*(a + b*ArcSinh[c*x]))/3 + (2*c^2*d^2*x^5*(a + b*Ar
cSinh[c*x]))/5 + (c^4*d^2*x^7*(a + b*ArcSinh[c*x]))/7

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Rubi [A]  time = 0.168962, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {270, 5730, 12, 1251, 771} \[ \frac{1}{7} c^4 d^2 x^7 \left (a+b \sinh ^{-1}(c x)\right )+\frac{2}{5} c^2 d^2 x^5 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} d^2 x^3 \left (a+b \sinh ^{-1}(c x)\right )-\frac{b d^2 \left (c^2 x^2+1\right )^{7/2}}{49 c^3}+\frac{b d^2 \left (c^2 x^2+1\right )^{5/2}}{175 c^3}+\frac{4 b d^2 \left (c^2 x^2+1\right )^{3/2}}{315 c^3}+\frac{8 b d^2 \sqrt{c^2 x^2+1}}{105 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*b*d^2*Sqrt[1 + c^2*x^2])/(105*c^3) + (4*b*d^2*(1 + c^2*x^2)^(3/2))/(315*c^3) + (b*d^2*(1 + c^2*x^2)^(5/2))/
(175*c^3) - (b*d^2*(1 + c^2*x^2)^(7/2))/(49*c^3) + (d^2*x^3*(a + b*ArcSinh[c*x]))/3 + (2*c^2*d^2*x^5*(a + b*Ar
cSinh[c*x]))/5 + (c^4*d^2*x^7*(a + b*ArcSinh[c*x]))/7

Rule 270

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

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 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rule 771

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(d + e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && N
eQ[b^2 - 4*a*c, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

\begin{align*} \int x^2 \left (d+c^2 d x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac{1}{3} d^2 x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac{2}{5} c^2 d^2 x^5 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{7} c^4 d^2 x^7 \left (a+b \sinh ^{-1}(c x)\right )-(b c) \int \frac{d^2 x^3 \left (35+42 c^2 x^2+15 c^4 x^4\right )}{105 \sqrt{1+c^2 x^2}} \, dx\\ &=\frac{1}{3} d^2 x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac{2}{5} c^2 d^2 x^5 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{7} c^4 d^2 x^7 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{105} \left (b c d^2\right ) \int \frac{x^3 \left (35+42 c^2 x^2+15 c^4 x^4\right )}{\sqrt{1+c^2 x^2}} \, dx\\ &=\frac{1}{3} d^2 x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac{2}{5} c^2 d^2 x^5 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{7} c^4 d^2 x^7 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{210} \left (b c d^2\right ) \operatorname{Subst}\left (\int \frac{x \left (35+42 c^2 x+15 c^4 x^2\right )}{\sqrt{1+c^2 x}} \, dx,x,x^2\right )\\ &=\frac{1}{3} d^2 x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac{2}{5} c^2 d^2 x^5 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{7} c^4 d^2 x^7 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{210} \left (b c d^2\right ) \operatorname{Subst}\left (\int \left (-\frac{8}{c^2 \sqrt{1+c^2 x}}-\frac{4 \sqrt{1+c^2 x}}{c^2}-\frac{3 \left (1+c^2 x\right )^{3/2}}{c^2}+\frac{15 \left (1+c^2 x\right )^{5/2}}{c^2}\right ) \, dx,x,x^2\right )\\ &=\frac{8 b d^2 \sqrt{1+c^2 x^2}}{105 c^3}+\frac{4 b d^2 \left (1+c^2 x^2\right )^{3/2}}{315 c^3}+\frac{b d^2 \left (1+c^2 x^2\right )^{5/2}}{175 c^3}-\frac{b d^2 \left (1+c^2 x^2\right )^{7/2}}{49 c^3}+\frac{1}{3} d^2 x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac{2}{5} c^2 d^2 x^5 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{7} c^4 d^2 x^7 \left (a+b \sinh ^{-1}(c x)\right )\\ \end{align*}

Mathematica [A]  time = 0.0851899, size = 111, normalized size = 0.71 \[ \frac{d^2 \left (105 a c^3 x^3 \left (15 c^4 x^4+42 c^2 x^2+35\right )-b \sqrt{c^2 x^2+1} \left (225 c^6 x^6+612 c^4 x^4+409 c^2 x^2-818\right )+105 b c^3 x^3 \left (15 c^4 x^4+42 c^2 x^2+35\right ) \sinh ^{-1}(c x)\right )}{11025 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(105*a*c^3*x^3*(35 + 42*c^2*x^2 + 15*c^4*x^4) - b*Sqrt[1 + c^2*x^2]*(-818 + 409*c^2*x^2 + 612*c^4*x^4 + 2
25*c^6*x^6) + 105*b*c^3*x^3*(35 + 42*c^2*x^2 + 15*c^4*x^4)*ArcSinh[c*x]))/(11025*c^3)

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Maple [A]  time = 0.003, size = 148, normalized size = 0.9 \begin{align*}{\frac{1}{{c}^{3}} \left ({d}^{2}a \left ({\frac{{c}^{7}{x}^{7}}{7}}+{\frac{2\,{c}^{5}{x}^{5}}{5}}+{\frac{{c}^{3}{x}^{3}}{3}} \right ) +{d}^{2}b \left ({\frac{{\it Arcsinh} \left ( cx \right ){c}^{7}{x}^{7}}{7}}+{\frac{2\,{\it Arcsinh} \left ( cx \right ){c}^{5}{x}^{5}}{5}}+{\frac{{\it Arcsinh} \left ( cx \right ){c}^{3}{x}^{3}}{3}}-{\frac{{c}^{6}{x}^{6}}{49}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{68\,{c}^{4}{x}^{4}}{1225}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{409\,{c}^{2}{x}^{2}}{11025}\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{818}{11025}\sqrt{{c}^{2}{x}^{2}+1}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^3*(d^2*a*(1/7*c^7*x^7+2/5*c^5*x^5+1/3*c^3*x^3)+d^2*b*(1/7*arcsinh(c*x)*c^7*x^7+2/5*arcsinh(c*x)*c^5*x^5+1/
3*arcsinh(c*x)*c^3*x^3-1/49*c^6*x^6*(c^2*x^2+1)^(1/2)-68/1225*c^4*x^4*(c^2*x^2+1)^(1/2)-409/11025*c^2*x^2*(c^2
*x^2+1)^(1/2)+818/11025*(c^2*x^2+1)^(1/2)))

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Maxima [A]  time = 1.18923, size = 352, normalized size = 2.24 \begin{align*} \frac{1}{7} \, a c^{4} d^{2} x^{7} + \frac{2}{5} \, a c^{2} d^{2} 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^{4} d^{2} + \frac{2}{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 c^{2} d^{2} + \frac{1}{3} \, a d^{2} x^{3} + \frac{1}{9} \,{\left (3 \, x^{3} \operatorname{arsinh}\left (c x\right ) - c{\left (\frac{\sqrt{c^{2} x^{2} + 1} x^{2}}{c^{2}} - \frac{2 \, \sqrt{c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} b d^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*a*c^4*d^2*x^7 + 2/5*a*c^2*d^2*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^4*d^2 + 2/75*(15*x^5*arcsin
h(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*c^2*d^2 +
1/3*a*d^2*x^3 + 1/9*(3*x^3*arcsinh(c*x) - c*(sqrt(c^2*x^2 + 1)*x^2/c^2 - 2*sqrt(c^2*x^2 + 1)/c^4))*b*d^2

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Fricas [A]  time = 2.43742, size = 351, normalized size = 2.24 \begin{align*} \frac{1575 \, a c^{7} d^{2} x^{7} + 4410 \, a c^{5} d^{2} x^{5} + 3675 \, a c^{3} d^{2} x^{3} + 105 \,{\left (15 \, b c^{7} d^{2} x^{7} + 42 \, b c^{5} d^{2} x^{5} + 35 \, b c^{3} d^{2} x^{3}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) -{\left (225 \, b c^{6} d^{2} x^{6} + 612 \, b c^{4} d^{2} x^{4} + 409 \, b c^{2} d^{2} x^{2} - 818 \, b d^{2}\right )} \sqrt{c^{2} x^{2} + 1}}{11025 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/11025*(1575*a*c^7*d^2*x^7 + 4410*a*c^5*d^2*x^5 + 3675*a*c^3*d^2*x^3 + 105*(15*b*c^7*d^2*x^7 + 42*b*c^5*d^2*x
^5 + 35*b*c^3*d^2*x^3)*log(c*x + sqrt(c^2*x^2 + 1)) - (225*b*c^6*d^2*x^6 + 612*b*c^4*d^2*x^4 + 409*b*c^2*d^2*x
^2 - 818*b*d^2)*sqrt(c^2*x^2 + 1))/c^3

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Sympy [A]  time = 9.19659, size = 202, normalized size = 1.29 \begin{align*} \begin{cases} \frac{a c^{4} d^{2} x^{7}}{7} + \frac{2 a c^{2} d^{2} x^{5}}{5} + \frac{a d^{2} x^{3}}{3} + \frac{b c^{4} d^{2} x^{7} \operatorname{asinh}{\left (c x \right )}}{7} - \frac{b c^{3} d^{2} x^{6} \sqrt{c^{2} x^{2} + 1}}{49} + \frac{2 b c^{2} d^{2} x^{5} \operatorname{asinh}{\left (c x \right )}}{5} - \frac{68 b c d^{2} x^{4} \sqrt{c^{2} x^{2} + 1}}{1225} + \frac{b d^{2} x^{3} \operatorname{asinh}{\left (c x \right )}}{3} - \frac{409 b d^{2} x^{2} \sqrt{c^{2} x^{2} + 1}}{11025 c} + \frac{818 b d^{2} \sqrt{c^{2} x^{2} + 1}}{11025 c^{3}} & \text{for}\: c \neq 0 \\\frac{a d^{2} x^{3}}{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*c**4*d**2*x**7/7 + 2*a*c**2*d**2*x**5/5 + a*d**2*x**3/3 + b*c**4*d**2*x**7*asinh(c*x)/7 - b*c**3*
d**2*x**6*sqrt(c**2*x**2 + 1)/49 + 2*b*c**2*d**2*x**5*asinh(c*x)/5 - 68*b*c*d**2*x**4*sqrt(c**2*x**2 + 1)/1225
+ b*d**2*x**3*asinh(c*x)/3 - 409*b*d**2*x**2*sqrt(c**2*x**2 + 1)/(11025*c) + 818*b*d**2*sqrt(c**2*x**2 + 1)/(
11025*c**3), Ne(c, 0)), (a*d**2*x**3/3, True))

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Giac [A]  time = 1.71359, size = 347, normalized size = 2.21 \begin{align*} \frac{1}{7} \, a c^{4} d^{2} x^{7} + \frac{2}{5} \, a c^{2} d^{2} 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^{4} d^{2} + \frac{2}{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 c^{2} d^{2} + \frac{1}{3} \, a d^{2} x^{3} + \frac{1}{9} \,{\left (3 \, x^{3} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) - \frac{{\left (c^{2} x^{2} + 1\right )}^{\frac{3}{2}} - 3 \, \sqrt{c^{2} x^{2} + 1}}{c^{3}}\right )} b d^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*a*c^4*d^2*x^7 + 2/5*a*c^2*d^2*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^4*d^2 + 2/75*(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*c^2*d^2
+ 1/3*a*d^2*x^3 + 1/9*(3*x^3*log(c*x + sqrt(c^2*x^2 + 1)) - ((c^2*x^2 + 1)^(3/2) - 3*sqrt(c^2*x^2 + 1))/c^3)*b
*d^2