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

Optimal. Leaf size=180 \[ \frac{1}{8} c^4 d^2 x^8 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} c^2 d^2 x^6 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{4} d^2 x^4 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{64} b c^3 d^2 x^7 \sqrt{c^2 x^2+1}-\frac{43 b c d^2 x^5 \sqrt{c^2 x^2+1}}{1152}-\frac{73 b d^2 x^3 \sqrt{c^2 x^2+1}}{4608 c}+\frac{73 b d^2 x \sqrt{c^2 x^2+1}}{3072 c^3}-\frac{73 b d^2 \sinh ^{-1}(c x)}{3072 c^4} \]

[Out]

(73*b*d^2*x*Sqrt[1 + c^2*x^2])/(3072*c^3) - (73*b*d^2*x^3*Sqrt[1 + c^2*x^2])/(4608*c) - (43*b*c*d^2*x^5*Sqrt[1
+ c^2*x^2])/1152 - (b*c^3*d^2*x^7*Sqrt[1 + c^2*x^2])/64 - (73*b*d^2*ArcSinh[c*x])/(3072*c^4) + (d^2*x^4*(a +
b*ArcSinh[c*x]))/4 + (c^2*d^2*x^6*(a + b*ArcSinh[c*x]))/3 + (c^4*d^2*x^8*(a + b*ArcSinh[c*x]))/8

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Rubi [A]  time = 0.17493, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 8, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {266, 43, 5730, 12, 1267, 459, 321, 215} \[ \frac{1}{8} c^4 d^2 x^8 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} c^2 d^2 x^6 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{4} d^2 x^4 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{64} b c^3 d^2 x^7 \sqrt{c^2 x^2+1}-\frac{43 b c d^2 x^5 \sqrt{c^2 x^2+1}}{1152}-\frac{73 b d^2 x^3 \sqrt{c^2 x^2+1}}{4608 c}+\frac{73 b d^2 x \sqrt{c^2 x^2+1}}{3072 c^3}-\frac{73 b d^2 \sinh ^{-1}(c x)}{3072 c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(73*b*d^2*x*Sqrt[1 + c^2*x^2])/(3072*c^3) - (73*b*d^2*x^3*Sqrt[1 + c^2*x^2])/(4608*c) - (43*b*c*d^2*x^5*Sqrt[1
+ c^2*x^2])/1152 - (b*c^3*d^2*x^7*Sqrt[1 + c^2*x^2])/64 - (73*b*d^2*ArcSinh[c*x])/(3072*c^4) + (d^2*x^4*(a +
b*ArcSinh[c*x]))/4 + (c^2*d^2*x^6*(a + b*ArcSinh[c*x]))/3 + (c^4*d^2*x^8*(a + b*ArcSinh[c*x]))/8

Rule 266

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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 1267

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Si
mp[(c^p*(f*x)^(m + 4*p - 1)*(d + e*x^2)^(q + 1))/(e*f^(4*p - 1)*(m + 4*p + 2*q + 1)), x] + Dist[1/(e*(m + 4*p
+ 2*q + 1)), Int[(f*x)^m*(d + e*x^2)^q*ExpandToSum[e*(m + 4*p + 2*q + 1)*((a + b*x^2 + c*x^4)^p - c^p*x^(4*p))
- d*c^p*(m + 4*p - 1)*x^(4*p - 2), x], x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[b^2 - 4*a*c, 0] &&
IGtQ[p, 0] &&  !IntegerQ[q] && NeQ[m + 4*p + 2*q + 1, 0]

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 )^2 \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac{1}{4} d^2 x^4 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} c^2 d^2 x^6 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{8} c^4 d^2 x^8 \left (a+b \sinh ^{-1}(c x)\right )-(b c) \int \frac{d^2 x^4 \left (6+8 c^2 x^2+3 c^4 x^4\right )}{24 \sqrt{1+c^2 x^2}} \, dx\\ &=\frac{1}{4} d^2 x^4 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} c^2 d^2 x^6 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{8} c^4 d^2 x^8 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{24} \left (b c d^2\right ) \int \frac{x^4 \left (6+8 c^2 x^2+3 c^4 x^4\right )}{\sqrt{1+c^2 x^2}} \, dx\\ &=-\frac{1}{64} b c^3 d^2 x^7 \sqrt{1+c^2 x^2}+\frac{1}{4} d^2 x^4 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} c^2 d^2 x^6 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{8} c^4 d^2 x^8 \left (a+b \sinh ^{-1}(c x)\right )-\frac{\left (b d^2\right ) \int \frac{x^4 \left (48 c^2+43 c^4 x^2\right )}{\sqrt{1+c^2 x^2}} \, dx}{192 c}\\ &=-\frac{43 b c d^2 x^5 \sqrt{1+c^2 x^2}}{1152}-\frac{1}{64} b c^3 d^2 x^7 \sqrt{1+c^2 x^2}+\frac{1}{4} d^2 x^4 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} c^2 d^2 x^6 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{8} c^4 d^2 x^8 \left (a+b \sinh ^{-1}(c x)\right )-\frac{\left (73 b c d^2\right ) \int \frac{x^4}{\sqrt{1+c^2 x^2}} \, dx}{1152}\\ &=-\frac{73 b d^2 x^3 \sqrt{1+c^2 x^2}}{4608 c}-\frac{43 b c d^2 x^5 \sqrt{1+c^2 x^2}}{1152}-\frac{1}{64} b c^3 d^2 x^7 \sqrt{1+c^2 x^2}+\frac{1}{4} d^2 x^4 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} c^2 d^2 x^6 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{8} c^4 d^2 x^8 \left (a+b \sinh ^{-1}(c x)\right )+\frac{\left (73 b d^2\right ) \int \frac{x^2}{\sqrt{1+c^2 x^2}} \, dx}{1536 c}\\ &=\frac{73 b d^2 x \sqrt{1+c^2 x^2}}{3072 c^3}-\frac{73 b d^2 x^3 \sqrt{1+c^2 x^2}}{4608 c}-\frac{43 b c d^2 x^5 \sqrt{1+c^2 x^2}}{1152}-\frac{1}{64} b c^3 d^2 x^7 \sqrt{1+c^2 x^2}+\frac{1}{4} d^2 x^4 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} c^2 d^2 x^6 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{8} c^4 d^2 x^8 \left (a+b \sinh ^{-1}(c x)\right )-\frac{\left (73 b d^2\right ) \int \frac{1}{\sqrt{1+c^2 x^2}} \, dx}{3072 c^3}\\ &=\frac{73 b d^2 x \sqrt{1+c^2 x^2}}{3072 c^3}-\frac{73 b d^2 x^3 \sqrt{1+c^2 x^2}}{4608 c}-\frac{43 b c d^2 x^5 \sqrt{1+c^2 x^2}}{1152}-\frac{1}{64} b c^3 d^2 x^7 \sqrt{1+c^2 x^2}-\frac{73 b d^2 \sinh ^{-1}(c x)}{3072 c^4}+\frac{1}{4} d^2 x^4 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} c^2 d^2 x^6 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{8} c^4 d^2 x^8 \left (a+b \sinh ^{-1}(c x)\right )\\ \end{align*}

Mathematica [A]  time = 0.0821231, size = 115, normalized size = 0.64 \[ \frac{d^2 \left (384 a c^4 x^4 \left (3 c^4 x^4+8 c^2 x^2+6\right )-b c x \sqrt{c^2 x^2+1} \left (144 c^6 x^6+344 c^4 x^4+146 c^2 x^2-219\right )+3 b \left (384 c^8 x^8+1024 c^6 x^6+768 c^4 x^4-73\right ) \sinh ^{-1}(c x)\right )}{9216 c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(384*a*c^4*x^4*(6 + 8*c^2*x^2 + 3*c^4*x^4) - b*c*x*Sqrt[1 + c^2*x^2]*(-219 + 146*c^2*x^2 + 344*c^4*x^4 +
144*c^6*x^6) + 3*b*(-73 + 768*c^4*x^4 + 1024*c^6*x^6 + 384*c^8*x^8)*ArcSinh[c*x]))/(9216*c^4)

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Maple [A]  time = 0.009, size = 156, normalized size = 0.9 \begin{align*}{\frac{1}{{c}^{4}} \left ({d}^{2}a \left ({\frac{{c}^{8}{x}^{8}}{8}}+{\frac{{c}^{6}{x}^{6}}{3}}+{\frac{{c}^{4}{x}^{4}}{4}} \right ) +{d}^{2}b \left ({\frac{{\it Arcsinh} \left ( cx \right ){c}^{8}{x}^{8}}{8}}+{\frac{{\it Arcsinh} \left ( cx \right ){c}^{6}{x}^{6}}{3}}+{\frac{{\it Arcsinh} \left ( cx \right ){c}^{4}{x}^{4}}{4}}-{\frac{{c}^{7}{x}^{7}}{64}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{43\,{c}^{5}{x}^{5}}{1152}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{73\,{c}^{3}{x}^{3}}{4608}\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{73\,cx}{3072}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{73\,{\it Arcsinh} \left ( cx \right ) }{3072}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^4*(d^2*a*(1/8*c^8*x^8+1/3*c^6*x^6+1/4*c^4*x^4)+d^2*b*(1/8*arcsinh(c*x)*c^8*x^8+1/3*arcsinh(c*x)*c^6*x^6+1/
4*arcsinh(c*x)*c^4*x^4-1/64*c^7*x^7*(c^2*x^2+1)^(1/2)-43/1152*c^5*x^5*(c^2*x^2+1)^(1/2)-73/4608*c^3*x^3*(c^2*x
^2+1)^(1/2)+73/3072*c*x*(c^2*x^2+1)^(1/2)-73/3072*arcsinh(c*x)))

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Maxima [B]  time = 1.21305, size = 443, normalized size = 2.46 \begin{align*} \frac{1}{8} \, a c^{4} d^{2} x^{8} + \frac{1}{3} \, a c^{2} d^{2} 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^{4} d^{2} + \frac{1}{4} \, a d^{2} x^{4} + \frac{1}{144} \,{\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^{2} + \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^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*a*c^4*d^2*x^8 + 1/3*a*c^2*d^2*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^4*d^2 + 1/4*a*d^2*x^4 + 1/144*(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^2 + 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^2

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Fricas [A]  time = 2.15863, size = 373, normalized size = 2.07 \begin{align*} \frac{1152 \, a c^{8} d^{2} x^{8} + 3072 \, a c^{6} d^{2} x^{6} + 2304 \, a c^{4} d^{2} x^{4} + 3 \,{\left (384 \, b c^{8} d^{2} x^{8} + 1024 \, b c^{6} d^{2} x^{6} + 768 \, b c^{4} d^{2} x^{4} - 73 \, b d^{2}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) -{\left (144 \, b c^{7} d^{2} x^{7} + 344 \, b c^{5} d^{2} x^{5} + 146 \, b c^{3} d^{2} x^{3} - 219 \, b c d^{2} x\right )} \sqrt{c^{2} x^{2} + 1}}{9216 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9216*(1152*a*c^8*d^2*x^8 + 3072*a*c^6*d^2*x^6 + 2304*a*c^4*d^2*x^4 + 3*(384*b*c^8*d^2*x^8 + 1024*b*c^6*d^2*x
^6 + 768*b*c^4*d^2*x^4 - 73*b*d^2)*log(c*x + sqrt(c^2*x^2 + 1)) - (144*b*c^7*d^2*x^7 + 344*b*c^5*d^2*x^5 + 146
*b*c^3*d^2*x^3 - 219*b*c*d^2*x)*sqrt(c^2*x^2 + 1))/c^4

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Sympy [A]  time = 16.4422, size = 218, normalized size = 1.21 \begin{align*} \begin{cases} \frac{a c^{4} d^{2} x^{8}}{8} + \frac{a c^{2} d^{2} x^{6}}{3} + \frac{a d^{2} x^{4}}{4} + \frac{b c^{4} d^{2} x^{8} \operatorname{asinh}{\left (c x \right )}}{8} - \frac{b c^{3} d^{2} x^{7} \sqrt{c^{2} x^{2} + 1}}{64} + \frac{b c^{2} d^{2} x^{6} \operatorname{asinh}{\left (c x \right )}}{3} - \frac{43 b c d^{2} x^{5} \sqrt{c^{2} x^{2} + 1}}{1152} + \frac{b d^{2} x^{4} \operatorname{asinh}{\left (c x \right )}}{4} - \frac{73 b d^{2} x^{3} \sqrt{c^{2} x^{2} + 1}}{4608 c} + \frac{73 b d^{2} x \sqrt{c^{2} x^{2} + 1}}{3072 c^{3}} - \frac{73 b d^{2} \operatorname{asinh}{\left (c x \right )}}{3072 c^{4}} & \text{for}\: c \neq 0 \\\frac{a d^{2} 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)**2*(a+b*asinh(c*x)),x)

[Out]

Piecewise((a*c**4*d**2*x**8/8 + a*c**2*d**2*x**6/3 + a*d**2*x**4/4 + b*c**4*d**2*x**8*asinh(c*x)/8 - b*c**3*d*
*2*x**7*sqrt(c**2*x**2 + 1)/64 + b*c**2*d**2*x**6*asinh(c*x)/3 - 43*b*c*d**2*x**5*sqrt(c**2*x**2 + 1)/1152 + b
*d**2*x**4*asinh(c*x)/4 - 73*b*d**2*x**3*sqrt(c**2*x**2 + 1)/(4608*c) + 73*b*d**2*x*sqrt(c**2*x**2 + 1)/(3072*
c**3) - 73*b*d**2*asinh(c*x)/(3072*c**4), Ne(c, 0)), (a*d**2*x**4/4, True))

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Giac [B]  time = 1.79885, size = 451, normalized size = 2.51 \begin{align*} \frac{1}{8} \, a c^{4} d^{2} x^{8} + \frac{1}{3} \, a c^{2} d^{2} 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^{4} d^{2} + \frac{1}{4} \, a d^{2} x^{4} + \frac{1}{144} \,{\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^{2} + \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^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*a*c^4*d^2*x^8 + 1/3*a*c^2*d^2*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^4*d^2 + 1/4*a*d^2*x^4 + 1/144*(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^2 + 1/32*(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*d^2