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

Optimal. Leaf size=283 \[ \frac {1}{7} d x^5 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {4 b d x^4 \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{175 c}-\frac {2 b d \left (c^2 x^2+1\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{49 c^5}+\frac {4 b d \left (c^2 x^2+1\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{35 c^5}-\frac {2 b d \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{21 c^5}-\frac {32 b d \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{525 c^5}+\frac {16 b d x^2 \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{525 c^3}+\frac {2}{35} d x^5 \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {304 b^2 d x}{3675 c^4}+\frac {2}{343} b^2 c^2 d x^7-\frac {152 b^2 d x^3}{11025 c^2}+\frac {38 b^2 d x^5}{6125} \]

[Out]

304/3675*b^2*d*x/c^4-152/11025*b^2*d*x^3/c^2+38/6125*b^2*d*x^5+2/343*b^2*c^2*d*x^7-2/21*b*d*(c^2*x^2+1)^(3/2)*
(a+b*arcsinh(c*x))/c^5+4/35*b*d*(c^2*x^2+1)^(5/2)*(a+b*arcsinh(c*x))/c^5-2/49*b*d*(c^2*x^2+1)^(7/2)*(a+b*arcsi
nh(c*x))/c^5+2/35*d*x^5*(a+b*arcsinh(c*x))^2+1/7*d*x^5*(c^2*x^2+1)*(a+b*arcsinh(c*x))^2-32/525*b*d*(a+b*arcsin
h(c*x))*(c^2*x^2+1)^(1/2)/c^5+16/525*b*d*x^2*(a+b*arcsinh(c*x))*(c^2*x^2+1)^(1/2)/c^3-4/175*b*d*x^4*(a+b*arcsi
nh(c*x))*(c^2*x^2+1)^(1/2)/c

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Rubi [A]  time = 0.48, antiderivative size = 283, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {5744, 5661, 5758, 5717, 8, 30, 266, 43, 5732, 12} \[ \frac {1}{7} d x^5 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {4 b d x^4 \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{175 c}+\frac {16 b d x^2 \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{525 c^3}-\frac {2 b d \left (c^2 x^2+1\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{49 c^5}+\frac {4 b d \left (c^2 x^2+1\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{35 c^5}-\frac {2 b d \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{21 c^5}-\frac {32 b d \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{525 c^5}+\frac {2}{35} d x^5 \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {2}{343} b^2 c^2 d x^7-\frac {152 b^2 d x^3}{11025 c^2}+\frac {304 b^2 d x}{3675 c^4}+\frac {38 b^2 d x^5}{6125} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(304*b^2*d*x)/(3675*c^4) - (152*b^2*d*x^3)/(11025*c^2) + (38*b^2*d*x^5)/6125 + (2*b^2*c^2*d*x^7)/343 - (32*b*d
*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(525*c^5) + (16*b*d*x^2*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(525*
c^3) - (4*b*d*x^4*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(175*c) - (2*b*d*(1 + c^2*x^2)^(3/2)*(a + b*ArcSinh[
c*x]))/(21*c^5) + (4*b*d*(1 + c^2*x^2)^(5/2)*(a + b*ArcSinh[c*x]))/(35*c^5) - (2*b*d*(1 + c^2*x^2)^(7/2)*(a +
b*ArcSinh[c*x]))/(49*c^5) + (2*d*x^5*(a + b*ArcSinh[c*x])^2)/35 + (d*x^5*(1 + c^2*x^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 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 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 5661

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 5717

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*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p +
1)*(1 + c^2*x^2)^FracPart[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 5732

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

Rule 5744

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*d^IntPart[p]*(d + e*x^2)^FracPart[p]
)/(f*(m + 2*p + 1)*(1 + c^2*x^2)^FracPart[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]
&& (RationalQ[m] || EqQ[n, 1])

Rule 5758

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

Rubi steps

\begin {align*} \int x^4 \left (d+c^2 d x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx &=\frac {1}{7} d x^5 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{7} (2 d) \int x^4 \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx-\frac {1}{7} (2 b c d) \int x^5 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx\\ &=-\frac {2 b d \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{21 c^5}+\frac {4 b d \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{35 c^5}-\frac {2 b d \left (1+c^2 x^2\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{49 c^5}+\frac {2}{35} d x^5 \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{7} d x^5 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {1}{35} (4 b c d) \int \frac {x^5 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}} \, dx+\frac {1}{7} \left (2 b^2 c^2 d\right ) \int \frac {8-4 c^2 x^2+3 c^4 x^4+15 c^6 x^6}{105 c^6} \, dx\\ &=-\frac {4 b d x^4 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{175 c}-\frac {2 b d \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{21 c^5}+\frac {4 b d \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{35 c^5}-\frac {2 b d \left (1+c^2 x^2\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{49 c^5}+\frac {2}{35} d x^5 \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{7} d x^5 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{175} \left (4 b^2 d\right ) \int x^4 \, dx+\frac {\left (2 b^2 d\right ) \int \left (8-4 c^2 x^2+3 c^4 x^4+15 c^6 x^6\right ) \, dx}{735 c^4}+\frac {(16 b d) \int \frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}} \, dx}{175 c}\\ &=\frac {16 b^2 d x}{735 c^4}-\frac {8 b^2 d x^3}{2205 c^2}+\frac {38 b^2 d x^5}{6125}+\frac {2}{343} b^2 c^2 d x^7+\frac {16 b d x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{525 c^3}-\frac {4 b d x^4 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{175 c}-\frac {2 b d \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{21 c^5}+\frac {4 b d \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{35 c^5}-\frac {2 b d \left (1+c^2 x^2\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{49 c^5}+\frac {2}{35} d x^5 \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{7} d x^5 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {(32 b d) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}} \, dx}{525 c^3}-\frac {\left (16 b^2 d\right ) \int x^2 \, dx}{525 c^2}\\ &=\frac {16 b^2 d x}{735 c^4}-\frac {152 b^2 d x^3}{11025 c^2}+\frac {38 b^2 d x^5}{6125}+\frac {2}{343} b^2 c^2 d x^7-\frac {32 b d \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{525 c^5}+\frac {16 b d x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{525 c^3}-\frac {4 b d x^4 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{175 c}-\frac {2 b d \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{21 c^5}+\frac {4 b d \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{35 c^5}-\frac {2 b d \left (1+c^2 x^2\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{49 c^5}+\frac {2}{35} d x^5 \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{7} d x^5 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {\left (32 b^2 d\right ) \int 1 \, dx}{525 c^4}\\ &=\frac {304 b^2 d x}{3675 c^4}-\frac {152 b^2 d x^3}{11025 c^2}+\frac {38 b^2 d x^5}{6125}+\frac {2}{343} b^2 c^2 d x^7-\frac {32 b d \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{525 c^5}+\frac {16 b d x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{525 c^3}-\frac {4 b d x^4 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{175 c}-\frac {2 b d \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{21 c^5}+\frac {4 b d \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{35 c^5}-\frac {2 b d \left (1+c^2 x^2\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{49 c^5}+\frac {2}{35} d x^5 \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{7} d x^5 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 201, normalized size = 0.71 \[ \frac {d \left (11025 a^2 c^5 x^5 \left (5 c^2 x^2+7\right )-210 a b \sqrt {c^2 x^2+1} \left (75 c^6 x^6+57 c^4 x^4-76 c^2 x^2+152\right )-210 b \sinh ^{-1}(c x) \left (b \sqrt {c^2 x^2+1} \left (75 c^6 x^6+57 c^4 x^4-76 c^2 x^2+152\right )-105 a c^5 x^5 \left (5 c^2 x^2+7\right )\right )+11025 b^2 c^5 x^5 \left (5 c^2 x^2+7\right ) \sinh ^{-1}(c x)^2+b^2 \left (2250 c^7 x^7+2394 c^5 x^5-5320 c^3 x^3+31920 c x\right )\right )}{385875 c^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(11025*a^2*c^5*x^5*(7 + 5*c^2*x^2) - 210*a*b*Sqrt[1 + c^2*x^2]*(152 - 76*c^2*x^2 + 57*c^4*x^4 + 75*c^6*x^6)
+ b^2*(31920*c*x - 5320*c^3*x^3 + 2394*c^5*x^5 + 2250*c^7*x^7) - 210*b*(-105*a*c^5*x^5*(7 + 5*c^2*x^2) + b*Sq
rt[1 + c^2*x^2]*(152 - 76*c^2*x^2 + 57*c^4*x^4 + 75*c^6*x^6))*ArcSinh[c*x] + 11025*b^2*c^5*x^5*(7 + 5*c^2*x^2)
*ArcSinh[c*x]^2))/(385875*c^5)

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fricas [A]  time = 0.45, size = 260, normalized size = 0.92 \[ \frac {1125 \, {\left (49 \, a^{2} + 2 \, b^{2}\right )} c^{7} d x^{7} + 63 \, {\left (1225 \, a^{2} + 38 \, b^{2}\right )} c^{5} d x^{5} - 5320 \, b^{2} c^{3} d x^{3} + 31920 \, b^{2} c d x + 11025 \, {\left (5 \, b^{2} c^{7} d x^{7} + 7 \, b^{2} c^{5} d x^{5}\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2} + 210 \, {\left (525 \, a b c^{7} d x^{7} + 735 \, a b c^{5} d x^{5} - {\left (75 \, b^{2} c^{6} d x^{6} + 57 \, b^{2} c^{4} d x^{4} - 76 \, b^{2} c^{2} d x^{2} + 152 \, b^{2} d\right )} \sqrt {c^{2} x^{2} + 1}\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - 210 \, {\left (75 \, a b c^{6} d x^{6} + 57 \, a b c^{4} d x^{4} - 76 \, a b c^{2} d x^{2} + 152 \, a b d\right )} \sqrt {c^{2} x^{2} + 1}}{385875 \, c^{5}} \]

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))^2,x, algorithm="fricas")

[Out]

1/385875*(1125*(49*a^2 + 2*b^2)*c^7*d*x^7 + 63*(1225*a^2 + 38*b^2)*c^5*d*x^5 - 5320*b^2*c^3*d*x^3 + 31920*b^2*
c*d*x + 11025*(5*b^2*c^7*d*x^7 + 7*b^2*c^5*d*x^5)*log(c*x + sqrt(c^2*x^2 + 1))^2 + 210*(525*a*b*c^7*d*x^7 + 73
5*a*b*c^5*d*x^5 - (75*b^2*c^6*d*x^6 + 57*b^2*c^4*d*x^4 - 76*b^2*c^2*d*x^2 + 152*b^2*d)*sqrt(c^2*x^2 + 1))*log(
c*x + sqrt(c^2*x^2 + 1)) - 210*(75*a*b*c^6*d*x^6 + 57*a*b*c^4*d*x^4 - 76*a*b*c^2*d*x^2 + 152*a*b*d)*sqrt(c^2*x
^2 + 1))/c^5

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

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))^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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maple [A]  time = 0.06, size = 330, normalized size = 1.17 \[ \frac {d \,a^{2} \left (\frac {1}{7} c^{7} x^{7}+\frac {1}{5} c^{5} x^{5}\right )+d \,b^{2} \left (\frac {\arcsinh \left (c x \right )^{2} c^{3} x^{3} \left (c^{2} x^{2}+1\right )^{2}}{7}-\frac {3 \arcsinh \left (c x \right )^{2} c x \left (c^{2} x^{2}+1\right )^{2}}{35}+\frac {2 \arcsinh \left (c x \right )^{2} c x}{35}+\frac {\arcsinh \left (c x \right )^{2} c x \left (c^{2} x^{2}+1\right )}{35}-\frac {2 \arcsinh \left (c x \right ) c^{2} x^{2} \left (c^{2} x^{2}+1\right )^{\frac {5}{2}}}{49}+\frac {62 \arcsinh \left (c x \right ) \left (c^{2} x^{2}+1\right )^{\frac {5}{2}}}{1225}+\frac {2 c x \left (c^{2} x^{2}+1\right )^{3}}{343}+\frac {37384 c x}{385875}-\frac {484 c x \left (c^{2} x^{2}+1\right )^{2}}{42875}-\frac {3358 c x \left (c^{2} x^{2}+1\right )}{385875}-\frac {4 \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}}{35}-\frac {2 \arcsinh \left (c x \right ) \left (c^{2} x^{2}+1\right )^{\frac {3}{2}}}{105}\right )+2 d a b \left (\frac {\arcsinh \left (c x \right ) c^{7} x^{7}}{7}+\frac {\arcsinh \left (c x \right ) c^{5} x^{5}}{5}-\frac {c^{6} x^{6} \sqrt {c^{2} x^{2}+1}}{49}-\frac {19 c^{4} x^{4} \sqrt {c^{2} x^{2}+1}}{1225}+\frac {76 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}}{3675}-\frac {152 \sqrt {c^{2} x^{2}+1}}{3675}\right )}{c^{5}} \]

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))^2,x)

[Out]

1/c^5*(d*a^2*(1/7*c^7*x^7+1/5*c^5*x^5)+d*b^2*(1/7*arcsinh(c*x)^2*c^3*x^3*(c^2*x^2+1)^2-3/35*arcsinh(c*x)^2*c*x
*(c^2*x^2+1)^2+2/35*arcsinh(c*x)^2*c*x+1/35*arcsinh(c*x)^2*c*x*(c^2*x^2+1)-2/49*arcsinh(c*x)*c^2*x^2*(c^2*x^2+
1)^(5/2)+62/1225*arcsinh(c*x)*(c^2*x^2+1)^(5/2)+2/343*c*x*(c^2*x^2+1)^3+37384/385875*c*x-484/42875*c*x*(c^2*x^
2+1)^2-3358/385875*c*x*(c^2*x^2+1)-4/35*arcsinh(c*x)*(c^2*x^2+1)^(1/2)-2/105*arcsinh(c*x)*(c^2*x^2+1)^(3/2))+2
*d*a*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 = 0.54, size = 441, normalized size = 1.56 \[ \frac {1}{7} \, b^{2} c^{2} d x^{7} \operatorname {arsinh}\left (c x\right )^{2} + \frac {1}{7} \, a^{2} c^{2} d x^{7} + \frac {1}{5} \, b^{2} d x^{5} \operatorname {arsinh}\left (c x\right )^{2} + \frac {1}{5} \, a^{2} d x^{5} + \frac {2}{245} \, {\left (35 \, x^{7} \operatorname {arsinh}\left (c x\right ) - {\left (\frac {5 \, \sqrt {c^{2} x^{2} + 1} x^{6}}{c^{2}} - \frac {6 \, \sqrt {c^{2} x^{2} + 1} x^{4}}{c^{4}} + \frac {8 \, \sqrt {c^{2} x^{2} + 1} x^{2}}{c^{6}} - \frac {16 \, \sqrt {c^{2} x^{2} + 1}}{c^{8}}\right )} c\right )} a b c^{2} d - \frac {2}{25725} \, {\left (105 \, {\left (\frac {5 \, \sqrt {c^{2} x^{2} + 1} x^{6}}{c^{2}} - \frac {6 \, \sqrt {c^{2} x^{2} + 1} x^{4}}{c^{4}} + \frac {8 \, \sqrt {c^{2} x^{2} + 1} x^{2}}{c^{6}} - \frac {16 \, \sqrt {c^{2} x^{2} + 1}}{c^{8}}\right )} c \operatorname {arsinh}\left (c x\right ) - \frac {75 \, c^{6} x^{7} - 126 \, c^{4} x^{5} + 280 \, c^{2} x^{3} - 1680 \, x}{c^{6}}\right )} b^{2} c^{2} d + \frac {2}{75} \, {\left (15 \, x^{5} \operatorname {arsinh}\left (c x\right ) - {\left (\frac {3 \, \sqrt {c^{2} x^{2} + 1} x^{4}}{c^{2}} - \frac {4 \, \sqrt {c^{2} x^{2} + 1} x^{2}}{c^{4}} + \frac {8 \, \sqrt {c^{2} x^{2} + 1}}{c^{6}}\right )} c\right )} a b d - \frac {2}{1125} \, {\left (15 \, {\left (\frac {3 \, \sqrt {c^{2} x^{2} + 1} x^{4}}{c^{2}} - \frac {4 \, \sqrt {c^{2} x^{2} + 1} x^{2}}{c^{4}} + \frac {8 \, \sqrt {c^{2} x^{2} + 1}}{c^{6}}\right )} c \operatorname {arsinh}\left (c x\right ) - \frac {9 \, c^{4} x^{5} - 20 \, c^{2} x^{3} + 120 \, x}{c^{4}}\right )} b^{2} d \]

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))^2,x, algorithm="maxima")

[Out]

1/7*b^2*c^2*d*x^7*arcsinh(c*x)^2 + 1/7*a^2*c^2*d*x^7 + 1/5*b^2*d*x^5*arcsinh(c*x)^2 + 1/5*a^2*d*x^5 + 2/245*(3
5*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^2*d - 2/25725*(105*(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*arcsinh(c*x) - (75*c^6*x^7 - 126*c^4*x^5 + 28
0*c^2*x^3 - 1680*x)/c^6)*b^2*c^2*d + 2/75*(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*d - 2/1125*(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*d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^4\,{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2\,\left (d\,c^2\,x^2+d\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 11.40, size = 388, normalized size = 1.37 \[ \begin {cases} \frac {a^{2} c^{2} d x^{7}}{7} + \frac {a^{2} d x^{5}}{5} + \frac {2 a b c^{2} d x^{7} \operatorname {asinh}{\left (c x \right )}}{7} - \frac {2 a b c d x^{6} \sqrt {c^{2} x^{2} + 1}}{49} + \frac {2 a b d x^{5} \operatorname {asinh}{\left (c x \right )}}{5} - \frac {38 a b d x^{4} \sqrt {c^{2} x^{2} + 1}}{1225 c} + \frac {152 a b d x^{2} \sqrt {c^{2} x^{2} + 1}}{3675 c^{3}} - \frac {304 a b d \sqrt {c^{2} x^{2} + 1}}{3675 c^{5}} + \frac {b^{2} c^{2} d x^{7} \operatorname {asinh}^{2}{\left (c x \right )}}{7} + \frac {2 b^{2} c^{2} d x^{7}}{343} - \frac {2 b^{2} c d x^{6} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{49} + \frac {b^{2} d x^{5} \operatorname {asinh}^{2}{\left (c x \right )}}{5} + \frac {38 b^{2} d x^{5}}{6125} - \frac {38 b^{2} d x^{4} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{1225 c} - \frac {152 b^{2} d x^{3}}{11025 c^{2}} + \frac {152 b^{2} d x^{2} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{3675 c^{3}} + \frac {304 b^{2} d x}{3675 c^{4}} - \frac {304 b^{2} d \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{3675 c^{5}} & \text {for}\: c \neq 0 \\\frac {a^{2} d x^{5}}{5} & \text {otherwise} \end {cases} \]

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))**2,x)

[Out]

Piecewise((a**2*c**2*d*x**7/7 + a**2*d*x**5/5 + 2*a*b*c**2*d*x**7*asinh(c*x)/7 - 2*a*b*c*d*x**6*sqrt(c**2*x**2
+ 1)/49 + 2*a*b*d*x**5*asinh(c*x)/5 - 38*a*b*d*x**4*sqrt(c**2*x**2 + 1)/(1225*c) + 152*a*b*d*x**2*sqrt(c**2*x
**2 + 1)/(3675*c**3) - 304*a*b*d*sqrt(c**2*x**2 + 1)/(3675*c**5) + b**2*c**2*d*x**7*asinh(c*x)**2/7 + 2*b**2*c
**2*d*x**7/343 - 2*b**2*c*d*x**6*sqrt(c**2*x**2 + 1)*asinh(c*x)/49 + b**2*d*x**5*asinh(c*x)**2/5 + 38*b**2*d*x
**5/6125 - 38*b**2*d*x**4*sqrt(c**2*x**2 + 1)*asinh(c*x)/(1225*c) - 152*b**2*d*x**3/(11025*c**2) + 152*b**2*d*
x**2*sqrt(c**2*x**2 + 1)*asinh(c*x)/(3675*c**3) + 304*b**2*d*x/(3675*c**4) - 304*b**2*d*sqrt(c**2*x**2 + 1)*as
inh(c*x)/(3675*c**5), Ne(c, 0)), (a**2*d*x**5/5, True))

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