3.3.9 \(\int x^2 (d+c^2 d x^2)^2 (a+b \sinh ^{-1}(c x))^2 \, dx\) [209]

Optimal. Leaf size=303 \[ -\frac {1636 b^2 d^2 x}{11025 c^2}+\frac {818 b^2 d^2 x^3}{33075}+\frac {136 b^2 c^2 d^2 x^5}{6125}+\frac {2}{343} b^2 c^4 d^2 x^7+\frac {32 b d^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{315 c^3}-\frac {16 b d^2 x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{315 c}+\frac {8 b d^2 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{105 c^3}+\frac {2 b d^2 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{175 c^3}-\frac {2 b d^2 \left (1+c^2 x^2\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{49 c^3}+\frac {8}{105} d^2 x^3 \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {4}{35} d^2 x^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{7} d^2 x^3 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2 \]

[Out]

-1636/11025*b^2*d^2*x/c^2+818/33075*b^2*d^2*x^3+136/6125*b^2*c^2*d^2*x^5+2/343*b^2*c^4*d^2*x^7+8/105*b*d^2*(c^
2*x^2+1)^(3/2)*(a+b*arcsinh(c*x))/c^3+2/175*b*d^2*(c^2*x^2+1)^(5/2)*(a+b*arcsinh(c*x))/c^3-2/49*b*d^2*(c^2*x^2
+1)^(7/2)*(a+b*arcsinh(c*x))/c^3+8/105*d^2*x^3*(a+b*arcsinh(c*x))^2+4/35*d^2*x^3*(c^2*x^2+1)*(a+b*arcsinh(c*x)
)^2+1/7*d^2*x^3*(c^2*x^2+1)^2*(a+b*arcsinh(c*x))^2+32/315*b*d^2*(a+b*arcsinh(c*x))*(c^2*x^2+1)^(1/2)/c^3-16/31
5*b*d^2*x^2*(a+b*arcsinh(c*x))*(c^2*x^2+1)^(1/2)/c

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Rubi [A]
time = 0.41, antiderivative size = 303, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {5808, 5776, 5812, 5798, 8, 30, 272, 45, 5804, 12, 380} \begin {gather*} -\frac {16 b d^2 x^2 \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{315 c}+\frac {1}{7} d^2 x^3 \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {4}{35} d^2 x^3 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {2 b d^2 \left (c^2 x^2+1\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{49 c^3}+\frac {2 b d^2 \left (c^2 x^2+1\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{175 c^3}+\frac {8 b d^2 \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{105 c^3}+\frac {32 b d^2 \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{315 c^3}+\frac {8}{105} d^2 x^3 \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {2}{343} b^2 c^4 d^2 x^7+\frac {136 b^2 c^2 d^2 x^5}{6125}-\frac {1636 b^2 d^2 x}{11025 c^2}+\frac {818 b^2 d^2 x^3}{33075} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-1636*b^2*d^2*x)/(11025*c^2) + (818*b^2*d^2*x^3)/33075 + (136*b^2*c^2*d^2*x^5)/6125 + (2*b^2*c^4*d^2*x^7)/343
+ (32*b*d^2*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(315*c^3) - (16*b*d^2*x^2*Sqrt[1 + c^2*x^2]*(a + b*ArcSin
h[c*x]))/(315*c) + (8*b*d^2*(1 + c^2*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/(105*c^3) + (2*b*d^2*(1 + c^2*x^2)^(5/2)
*(a + b*ArcSinh[c*x]))/(175*c^3) - (2*b*d^2*(1 + c^2*x^2)^(7/2)*(a + b*ArcSinh[c*x]))/(49*c^3) + (8*d^2*x^3*(a
+ b*ArcSinh[c*x])^2)/105 + (4*d^2*x^3*(1 + c^2*x^2)*(a + b*ArcSinh[c*x])^2)/35 + (d^2*x^3*(1 + c^2*x^2)^2*(a
+ b*ArcSinh[c*x])^2)/7

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 12

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 45

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 272

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 380

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

Rule 5776

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcS
inh[c*x])^n/(d*(m + 1))), x] - Dist[b*c*(n/(d*(m + 1))), Int[(d*x)^(m + 1)*((a + b*ArcSinh[c*x])^(n - 1)/Sqrt[
1 + c^2*x^2]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 5798

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/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)
^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 5804

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> With[{u = IntHide[x
^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSinh[c*x], u, x] - Dist[b*c*Simp[Sqrt[d + e*x^2]/Sqrt[1 + c^2*x^2]], Int[
SimplifyIntegrand[u/Sqrt[d + e*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IntegerQ[p -
1/2] && NeQ[p, -2^(-1)] && (IGtQ[(m + 1)/2, 0] || ILtQ[(m + 2*p + 3)/2, 0])

Rule 5808

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp
[(f*x)^(m + 1)*(d + e*x^2)^p*((a + b*ArcSinh[c*x])^n/(f*(m + 2*p + 1))), x] + (Dist[2*d*(p/(m + 2*p + 1)), Int
[(f*x)^m*(d + e*x^2)^(p - 1)*(a + b*ArcSinh[c*x])^n, x], x] - Dist[b*c*(n/(f*(m + 2*p + 1)))*Simp[(d + e*x^2)^
p/(1 + c^2*x^2)^p], Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p - 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{
a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && GtQ[p, 0] &&  !LtQ[m, -1]

Rule 5812

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

Rubi steps

\begin {align*} \int x^2 \left (d+c^2 d x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx &=\frac {1}{7} d^2 x^3 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{7} (4 d) \int x^2 \left (d+c^2 d x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx-\frac {1}{7} \left (2 b c d^2\right ) \int x^3 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx\\ &=\frac {2 b d^2 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{35 c^3}-\frac {2 b d^2 \left (1+c^2 x^2\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{49 c^3}+\frac {4}{35} d^2 x^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{7} d^2 x^3 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{35} \left (8 d^2\right ) \int x^2 \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx-\frac {1}{35} \left (8 b c d^2\right ) \int x^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx+\frac {1}{7} \left (2 b^2 c^2 d^2\right ) \int \frac {\left (1+c^2 x^2\right )^2 \left (-2+5 c^2 x^2\right )}{35 c^4} \, dx\\ &=\frac {8 b d^2 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{105 c^3}+\frac {2 b d^2 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{175 c^3}-\frac {2 b d^2 \left (1+c^2 x^2\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{49 c^3}+\frac {8}{105} d^2 x^3 \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {4}{35} d^2 x^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{7} d^2 x^3 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {\left (2 b^2 d^2\right ) \int \left (1+c^2 x^2\right )^2 \left (-2+5 c^2 x^2\right ) \, dx}{245 c^2}-\frac {1}{105} \left (16 b c d^2\right ) \int \frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}} \, dx+\frac {1}{35} \left (8 b^2 c^2 d^2\right ) \int \frac {-2+c^2 x^2+3 c^4 x^4}{15 c^4} \, dx\\ &=-\frac {16 b d^2 x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{315 c}+\frac {8 b d^2 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{105 c^3}+\frac {2 b d^2 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{175 c^3}-\frac {2 b d^2 \left (1+c^2 x^2\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{49 c^3}+\frac {8}{105} d^2 x^3 \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {4}{35} d^2 x^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{7} d^2 x^3 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{315} \left (16 b^2 d^2\right ) \int x^2 \, dx+\frac {\left (2 b^2 d^2\right ) \int \left (-2+c^2 x^2+8 c^4 x^4+5 c^6 x^6\right ) \, dx}{245 c^2}+\frac {\left (8 b^2 d^2\right ) \int \left (-2+c^2 x^2+3 c^4 x^4\right ) \, dx}{525 c^2}+\frac {\left (32 b d^2\right ) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}} \, dx}{315 c}\\ &=-\frac {172 b^2 d^2 x}{3675 c^2}+\frac {818 b^2 d^2 x^3}{33075}+\frac {136 b^2 c^2 d^2 x^5}{6125}+\frac {2}{343} b^2 c^4 d^2 x^7+\frac {32 b d^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{315 c^3}-\frac {16 b d^2 x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{315 c}+\frac {8 b d^2 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{105 c^3}+\frac {2 b d^2 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{175 c^3}-\frac {2 b d^2 \left (1+c^2 x^2\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{49 c^3}+\frac {8}{105} d^2 x^3 \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {4}{35} d^2 x^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{7} d^2 x^3 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {\left (32 b^2 d^2\right ) \int 1 \, dx}{315 c^2}\\ &=-\frac {1636 b^2 d^2 x}{11025 c^2}+\frac {818 b^2 d^2 x^3}{33075}+\frac {136 b^2 c^2 d^2 x^5}{6125}+\frac {2}{343} b^2 c^4 d^2 x^7+\frac {32 b d^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{315 c^3}-\frac {16 b d^2 x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{315 c}+\frac {8 b d^2 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{105 c^3}+\frac {2 b d^2 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{175 c^3}-\frac {2 b d^2 \left (1+c^2 x^2\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{49 c^3}+\frac {8}{105} d^2 x^3 \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {4}{35} d^2 x^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{7} d^2 x^3 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2\\ \end {align*}

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Mathematica [A]
time = 0.22, size = 227, normalized size = 0.75 \begin {gather*} \frac {d^2 \left (11025 a^2 c^3 x^3 \left (35+42 c^2 x^2+15 c^4 x^4\right )-210 a b \sqrt {1+c^2 x^2} \left (-818+409 c^2 x^2+612 c^4 x^4+225 c^6 x^6\right )+2 b^2 c x \left (-85890+14315 c^2 x^2+12852 c^4 x^4+3375 c^6 x^6\right )-210 b \left (-105 a c^3 x^3 \left (35+42 c^2 x^2+15 c^4 x^4\right )+b \sqrt {1+c^2 x^2} \left (-818+409 c^2 x^2+612 c^4 x^4+225 c^6 x^6\right )\right ) \sinh ^{-1}(c x)+11025 b^2 c^3 x^3 \left (35+42 c^2 x^2+15 c^4 x^4\right ) \sinh ^{-1}(c x)^2\right )}{1157625 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(11025*a^2*c^3*x^3*(35 + 42*c^2*x^2 + 15*c^4*x^4) - 210*a*b*Sqrt[1 + c^2*x^2]*(-818 + 409*c^2*x^2 + 612*c
^4*x^4 + 225*c^6*x^6) + 2*b^2*c*x*(-85890 + 14315*c^2*x^2 + 12852*c^4*x^4 + 3375*c^6*x^6) - 210*b*(-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 + 225*c^6*x^6))*Arc
Sinh[c*x] + 11025*b^2*c^3*x^3*(35 + 42*c^2*x^2 + 15*c^4*x^4)*ArcSinh[c*x]^2))/(1157625*c^3)

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int x^{2} \left (c^{2} d \,x^{2}+d \right )^{2} \left (a +b \arcsinh \left (c x \right )\right )^{2}\, dx\]

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))^2,x)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 619 vs. \(2 (269) = 538\).
time = 0.32, size = 619, normalized size = 2.04 \begin {gather*} \frac {1}{7} \, b^{2} c^{4} d^{2} x^{7} \operatorname {arsinh}\left (c x\right )^{2} + \frac {1}{7} \, a^{2} c^{4} d^{2} x^{7} + \frac {2}{5} \, b^{2} c^{2} d^{2} x^{5} \operatorname {arsinh}\left (c x\right )^{2} + \frac {2}{5} \, a^{2} c^{2} d^{2} x^{5} + \frac {2}{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 )} a b c^{4} d^{2} - \frac {2}{25725} \, {\left (105 \, {\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 \operatorname {arsinh}\left (c x\right ) - \frac {75 \, c^{6} x^{7} - 126 \, c^{4} x^{5} + 280 \, c^{2} x^{3} - 1680 \, x}{c^{6}}\right )} b^{2} c^{4} d^{2} + \frac {1}{3} \, b^{2} d^{2} x^{3} \operatorname {arsinh}\left (c x\right )^{2} + \frac {4}{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 )} a b c^{2} d^{2} - \frac {4}{1125} \, {\left (15 \, {\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 \operatorname {arsinh}\left (c x\right ) - \frac {9 \, c^{4} x^{5} - 20 \, c^{2} x^{3} + 120 \, x}{c^{4}}\right )} b^{2} c^{2} d^{2} + \frac {1}{3} \, a^{2} d^{2} x^{3} + \frac {2}{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 )} a b d^{2} - \frac {2}{27} \, {\left (3 \, c {\left (\frac {\sqrt {c^{2} x^{2} + 1} x^{2}}{c^{2}} - \frac {2 \, \sqrt {c^{2} x^{2} + 1}}{c^{4}}\right )} \operatorname {arsinh}\left (c x\right ) - \frac {c^{2} x^{3} - 6 \, x}{c^{2}}\right )} b^{2} d^{2} \end {gather*}

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))^2,x, algorithm="maxima")

[Out]

1/7*b^2*c^4*d^2*x^7*arcsinh(c*x)^2 + 1/7*a^2*c^4*d^2*x^7 + 2/5*b^2*c^2*d^2*x^5*arcsinh(c*x)^2 + 2/5*a^2*c^2*d^
2*x^5 + 2/245*(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)*a*b*c^4*d^2 - 2/25725*(105*(5*sqrt(c^2*x^2 + 1)*x^6/c^2 - 6*sqr
t(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*arcsinh(c*x) - (75*c^6*x^7
- 126*c^4*x^5 + 280*c^2*x^3 - 1680*x)/c^6)*b^2*c^4*d^2 + 1/3*b^2*d^2*x^3*arcsinh(c*x)^2 + 4/75*(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)*a*b*c^2*d^2 -
4/1125*(15*(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*arcsinh(c*
x) - (9*c^4*x^5 - 20*c^2*x^3 + 120*x)/c^4)*b^2*c^2*d^2 + 1/3*a^2*d^2*x^3 + 2/9*(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))*a*b*d^2 - 2/27*(3*c*(sqrt(c^2*x^2 + 1)*x^2/c^2 - 2*sqrt(c^2*x^
2 + 1)/c^4)*arcsinh(c*x) - (c^2*x^3 - 6*x)/c^2)*b^2*d^2

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Fricas [A]
time = 0.35, size = 327, normalized size = 1.08 \begin {gather*} \frac {3375 \, {\left (49 \, a^{2} + 2 \, b^{2}\right )} c^{7} d^{2} x^{7} + 378 \, {\left (1225 \, a^{2} + 68 \, b^{2}\right )} c^{5} d^{2} x^{5} + 35 \, {\left (11025 \, a^{2} + 818 \, b^{2}\right )} c^{3} d^{2} x^{3} - 171780 \, b^{2} c d^{2} x + 11025 \, {\left (15 \, b^{2} c^{7} d^{2} x^{7} + 42 \, b^{2} c^{5} d^{2} x^{5} + 35 \, b^{2} c^{3} d^{2} x^{3}\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2} + 210 \, {\left (1575 \, a b c^{7} d^{2} x^{7} + 4410 \, a b c^{5} d^{2} x^{5} + 3675 \, a b c^{3} d^{2} x^{3} - {\left (225 \, b^{2} c^{6} d^{2} x^{6} + 612 \, b^{2} c^{4} d^{2} x^{4} + 409 \, b^{2} c^{2} d^{2} x^{2} - 818 \, b^{2} d^{2}\right )} \sqrt {c^{2} x^{2} + 1}\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - 210 \, {\left (225 \, a b c^{6} d^{2} x^{6} + 612 \, a b c^{4} d^{2} x^{4} + 409 \, a b c^{2} d^{2} x^{2} - 818 \, a b d^{2}\right )} \sqrt {c^{2} x^{2} + 1}}{1157625 \, c^{3}} \end {gather*}

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))^2,x, algorithm="fricas")

[Out]

1/1157625*(3375*(49*a^2 + 2*b^2)*c^7*d^2*x^7 + 378*(1225*a^2 + 68*b^2)*c^5*d^2*x^5 + 35*(11025*a^2 + 818*b^2)*
c^3*d^2*x^3 - 171780*b^2*c*d^2*x + 11025*(15*b^2*c^7*d^2*x^7 + 42*b^2*c^5*d^2*x^5 + 35*b^2*c^3*d^2*x^3)*log(c*
x + sqrt(c^2*x^2 + 1))^2 + 210*(1575*a*b*c^7*d^2*x^7 + 4410*a*b*c^5*d^2*x^5 + 3675*a*b*c^3*d^2*x^3 - (225*b^2*
c^6*d^2*x^6 + 612*b^2*c^4*d^2*x^4 + 409*b^2*c^2*d^2*x^2 - 818*b^2*d^2)*sqrt(c^2*x^2 + 1))*log(c*x + sqrt(c^2*x
^2 + 1)) - 210*(225*a*b*c^6*d^2*x^6 + 612*a*b*c^4*d^2*x^4 + 409*a*b*c^2*d^2*x^2 - 818*a*b*d^2)*sqrt(c^2*x^2 +
1))/c^3

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Sympy [A]
time = 1.13, size = 483, normalized size = 1.59 \begin {gather*} \begin {cases} \frac {a^{2} c^{4} d^{2} x^{7}}{7} + \frac {2 a^{2} c^{2} d^{2} x^{5}}{5} + \frac {a^{2} d^{2} x^{3}}{3} + \frac {2 a b c^{4} d^{2} x^{7} \operatorname {asinh}{\left (c x \right )}}{7} - \frac {2 a b c^{3} d^{2} x^{6} \sqrt {c^{2} x^{2} + 1}}{49} + \frac {4 a b c^{2} d^{2} x^{5} \operatorname {asinh}{\left (c x \right )}}{5} - \frac {136 a b c d^{2} x^{4} \sqrt {c^{2} x^{2} + 1}}{1225} + \frac {2 a b d^{2} x^{3} \operatorname {asinh}{\left (c x \right )}}{3} - \frac {818 a b d^{2} x^{2} \sqrt {c^{2} x^{2} + 1}}{11025 c} + \frac {1636 a b d^{2} \sqrt {c^{2} x^{2} + 1}}{11025 c^{3}} + \frac {b^{2} c^{4} d^{2} x^{7} \operatorname {asinh}^{2}{\left (c x \right )}}{7} + \frac {2 b^{2} c^{4} d^{2} x^{7}}{343} - \frac {2 b^{2} c^{3} d^{2} x^{6} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{49} + \frac {2 b^{2} c^{2} d^{2} x^{5} \operatorname {asinh}^{2}{\left (c x \right )}}{5} + \frac {136 b^{2} c^{2} d^{2} x^{5}}{6125} - \frac {136 b^{2} c d^{2} x^{4} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{1225} + \frac {b^{2} d^{2} x^{3} \operatorname {asinh}^{2}{\left (c x \right )}}{3} + \frac {818 b^{2} d^{2} x^{3}}{33075} - \frac {818 b^{2} d^{2} x^{2} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{11025 c} - \frac {1636 b^{2} d^{2} x}{11025 c^{2}} + \frac {1636 b^{2} d^{2} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{11025 c^{3}} & \text {for}\: c \neq 0 \\\frac {a^{2} d^{2} x^{3}}{3} & \text {otherwise} \end {cases} \end {gather*}

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))**2,x)

[Out]

Piecewise((a**2*c**4*d**2*x**7/7 + 2*a**2*c**2*d**2*x**5/5 + a**2*d**2*x**3/3 + 2*a*b*c**4*d**2*x**7*asinh(c*x
)/7 - 2*a*b*c**3*d**2*x**6*sqrt(c**2*x**2 + 1)/49 + 4*a*b*c**2*d**2*x**5*asinh(c*x)/5 - 136*a*b*c*d**2*x**4*sq
rt(c**2*x**2 + 1)/1225 + 2*a*b*d**2*x**3*asinh(c*x)/3 - 818*a*b*d**2*x**2*sqrt(c**2*x**2 + 1)/(11025*c) + 1636
*a*b*d**2*sqrt(c**2*x**2 + 1)/(11025*c**3) + b**2*c**4*d**2*x**7*asinh(c*x)**2/7 + 2*b**2*c**4*d**2*x**7/343 -
2*b**2*c**3*d**2*x**6*sqrt(c**2*x**2 + 1)*asinh(c*x)/49 + 2*b**2*c**2*d**2*x**5*asinh(c*x)**2/5 + 136*b**2*c*
*2*d**2*x**5/6125 - 136*b**2*c*d**2*x**4*sqrt(c**2*x**2 + 1)*asinh(c*x)/1225 + b**2*d**2*x**3*asinh(c*x)**2/3
+ 818*b**2*d**2*x**3/33075 - 818*b**2*d**2*x**2*sqrt(c**2*x**2 + 1)*asinh(c*x)/(11025*c) - 1636*b**2*d**2*x/(1
1025*c**2) + 1636*b**2*d**2*sqrt(c**2*x**2 + 1)*asinh(c*x)/(11025*c**3), Ne(c, 0)), (a**2*d**2*x**3/3, True))

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

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))^2,x, algorithm="giac")

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:sym2poly/r2sym(co
nst gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^2\,{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2\,{\left (d\,c^2\,x^2+d\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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