3.137 \(\int x^2 (d+c^2 d x^2)^{5/2} (a+b \sinh ^{-1}(c x)) \, dx\)

Optimal. Leaf size=337 \[ \frac {5 d^2 x \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{128 c^2}+\frac {5}{64} d^2 x^3 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{8} x^3 \left (c^2 d x^2+d\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5}{48} d x^3 \left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {5 d^2 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{256 b c^3 \sqrt {c^2 x^2+1}}-\frac {5 b d^2 x^2 \sqrt {c^2 d x^2+d}}{256 c \sqrt {c^2 x^2+1}}-\frac {59 b c d^2 x^4 \sqrt {c^2 d x^2+d}}{768 \sqrt {c^2 x^2+1}}-\frac {b c^5 d^2 x^8 \sqrt {c^2 d x^2+d}}{64 \sqrt {c^2 x^2+1}}-\frac {17 b c^3 d^2 x^6 \sqrt {c^2 d x^2+d}}{288 \sqrt {c^2 x^2+1}} \]

[Out]

5/48*d*x^3*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))+1/8*x^3*(c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x))+5/128*d^2*x*(
a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)/c^2+5/64*d^2*x^3*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)-5/256*b*d^2*x^2*
(c^2*d*x^2+d)^(1/2)/c/(c^2*x^2+1)^(1/2)-59/768*b*c*d^2*x^4*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-17/288*b*c^3*
d^2*x^6*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-1/64*b*c^5*d^2*x^8*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-5/256*d
^2*(a+b*arcsinh(c*x))^2*(c^2*d*x^2+d)^(1/2)/b/c^3/(c^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.47, antiderivative size = 337, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {5744, 5742, 5758, 5675, 30, 14, 266, 43} \[ \frac {5}{64} d^2 x^3 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5 d^2 x \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{128 c^2}-\frac {5 d^2 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{256 b c^3 \sqrt {c^2 x^2+1}}+\frac {1}{8} x^3 \left (c^2 d x^2+d\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5}{48} d x^3 \left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {b c^5 d^2 x^8 \sqrt {c^2 d x^2+d}}{64 \sqrt {c^2 x^2+1}}-\frac {17 b c^3 d^2 x^6 \sqrt {c^2 d x^2+d}}{288 \sqrt {c^2 x^2+1}}-\frac {59 b c d^2 x^4 \sqrt {c^2 d x^2+d}}{768 \sqrt {c^2 x^2+1}}-\frac {5 b d^2 x^2 \sqrt {c^2 d x^2+d}}{256 c \sqrt {c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(d + c^2*d*x^2)^(5/2)*(a + b*ArcSinh[c*x]),x]

[Out]

(-5*b*d^2*x^2*Sqrt[d + c^2*d*x^2])/(256*c*Sqrt[1 + c^2*x^2]) - (59*b*c*d^2*x^4*Sqrt[d + c^2*d*x^2])/(768*Sqrt[
1 + c^2*x^2]) - (17*b*c^3*d^2*x^6*Sqrt[d + c^2*d*x^2])/(288*Sqrt[1 + c^2*x^2]) - (b*c^5*d^2*x^8*Sqrt[d + c^2*d
*x^2])/(64*Sqrt[1 + c^2*x^2]) + (5*d^2*x*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(128*c^2) + (5*d^2*x^3*Sqrt
[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/64 + (5*d*x^3*(d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/48 + (x^3*(d +
c^2*d*x^2)^(5/2)*(a + b*ArcSinh[c*x]))/8 - (5*d^2*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^2)/(256*b*c^3*Sqrt
[1 + c^2*x^2])

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 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 5675

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

Rule 5742

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(
(f*x)^(m + 1)*Sqrt[d + e*x^2]*(a + b*ArcSinh[c*x])^n)/(f*(m + 2)), x] + (Dist[Sqrt[d + e*x^2]/((m + 2)*Sqrt[1
+ c^2*x^2]), Int[((f*x)^m*(a + b*ArcSinh[c*x])^n)/Sqrt[1 + c^2*x^2], x], x] - Dist[(b*c*n*Sqrt[d + e*x^2])/(f*
(m + 2)*Sqrt[1 + c^2*x^2]), Int[(f*x)^(m + 1)*(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] &&  !LtQ[m, -1] && (RationalQ[m] || EqQ[n, 1])

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^2 \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac {1}{8} x^3 \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{8} (5 d) \int x^2 \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx-\frac {\left (b c d^2 \sqrt {d+c^2 d x^2}\right ) \int x^3 \left (1+c^2 x^2\right )^2 \, dx}{8 \sqrt {1+c^2 x^2}}\\ &=\frac {5}{48} d x^3 \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{8} x^3 \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{16} \left (5 d^2\right ) \int x^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx-\frac {\left (b c d^2 \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int x \left (1+c^2 x\right )^2 \, dx,x,x^2\right )}{16 \sqrt {1+c^2 x^2}}-\frac {\left (5 b c d^2 \sqrt {d+c^2 d x^2}\right ) \int x^3 \left (1+c^2 x^2\right ) \, dx}{48 \sqrt {1+c^2 x^2}}\\ &=\frac {5}{64} d^2 x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5}{48} d x^3 \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{8} x^3 \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {\left (5 d^2 \sqrt {d+c^2 d x^2}\right ) \int \frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}} \, dx}{64 \sqrt {1+c^2 x^2}}-\frac {\left (b c d^2 \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int \left (x+2 c^2 x^2+c^4 x^3\right ) \, dx,x,x^2\right )}{16 \sqrt {1+c^2 x^2}}-\frac {\left (5 b c d^2 \sqrt {d+c^2 d x^2}\right ) \int x^3 \, dx}{64 \sqrt {1+c^2 x^2}}-\frac {\left (5 b c d^2 \sqrt {d+c^2 d x^2}\right ) \int \left (x^3+c^2 x^5\right ) \, dx}{48 \sqrt {1+c^2 x^2}}\\ &=-\frac {59 b c d^2 x^4 \sqrt {d+c^2 d x^2}}{768 \sqrt {1+c^2 x^2}}-\frac {17 b c^3 d^2 x^6 \sqrt {d+c^2 d x^2}}{288 \sqrt {1+c^2 x^2}}-\frac {b c^5 d^2 x^8 \sqrt {d+c^2 d x^2}}{64 \sqrt {1+c^2 x^2}}+\frac {5 d^2 x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{128 c^2}+\frac {5}{64} d^2 x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5}{48} d x^3 \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{8} x^3 \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {\left (5 d^2 \sqrt {d+c^2 d x^2}\right ) \int \frac {a+b \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}} \, dx}{128 c^2 \sqrt {1+c^2 x^2}}-\frac {\left (5 b d^2 \sqrt {d+c^2 d x^2}\right ) \int x \, dx}{128 c \sqrt {1+c^2 x^2}}\\ &=-\frac {5 b d^2 x^2 \sqrt {d+c^2 d x^2}}{256 c \sqrt {1+c^2 x^2}}-\frac {59 b c d^2 x^4 \sqrt {d+c^2 d x^2}}{768 \sqrt {1+c^2 x^2}}-\frac {17 b c^3 d^2 x^6 \sqrt {d+c^2 d x^2}}{288 \sqrt {1+c^2 x^2}}-\frac {b c^5 d^2 x^8 \sqrt {d+c^2 d x^2}}{64 \sqrt {1+c^2 x^2}}+\frac {5 d^2 x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{128 c^2}+\frac {5}{64} d^2 x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5}{48} d x^3 \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{8} x^3 \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {5 d^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{256 b c^3 \sqrt {1+c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.94, size = 388, normalized size = 1.15 \[ \frac {d^2 \left (2880 a c x \sqrt {c^2 x^2+1} \sqrt {c^2 d x^2+d}-2880 a \sqrt {d} \sqrt {c^2 x^2+1} \log \left (\sqrt {d} \sqrt {c^2 d x^2+d}+c d x\right )+9216 a c^7 x^7 \sqrt {c^2 x^2+1} \sqrt {c^2 d x^2+d}+26112 a c^5 x^5 \sqrt {c^2 x^2+1} \sqrt {c^2 d x^2+d}+22656 a c^3 x^3 \sqrt {c^2 x^2+1} \sqrt {c^2 d x^2+d}-1440 b \sqrt {c^2 d x^2+d} \sinh ^{-1}(c x)^2+24 b \sqrt {c^2 d x^2+d} \sinh ^{-1}(c x) \left (-48 \sinh \left (2 \sinh ^{-1}(c x)\right )+24 \sinh \left (4 \sinh ^{-1}(c x)\right )+16 \sinh \left (6 \sinh ^{-1}(c x)\right )+3 \sinh \left (8 \sinh ^{-1}(c x)\right )\right )+576 b \sqrt {c^2 d x^2+d} \cosh \left (2 \sinh ^{-1}(c x)\right )-144 b \sqrt {c^2 d x^2+d} \cosh \left (4 \sinh ^{-1}(c x)\right )-64 b \sqrt {c^2 d x^2+d} \cosh \left (6 \sinh ^{-1}(c x)\right )-9 b \sqrt {c^2 d x^2+d} \cosh \left (8 \sinh ^{-1}(c x)\right )\right )}{73728 c^3 \sqrt {c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(d + c^2*d*x^2)^(5/2)*(a + b*ArcSinh[c*x]),x]

[Out]

(d^2*(2880*a*c*x*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x^2] + 22656*a*c^3*x^3*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x^2]
+ 26112*a*c^5*x^5*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x^2] + 9216*a*c^7*x^7*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x^2
] - 1440*b*Sqrt[d + c^2*d*x^2]*ArcSinh[c*x]^2 + 576*b*Sqrt[d + c^2*d*x^2]*Cosh[2*ArcSinh[c*x]] - 144*b*Sqrt[d
+ c^2*d*x^2]*Cosh[4*ArcSinh[c*x]] - 64*b*Sqrt[d + c^2*d*x^2]*Cosh[6*ArcSinh[c*x]] - 9*b*Sqrt[d + c^2*d*x^2]*Co
sh[8*ArcSinh[c*x]] - 2880*a*Sqrt[d]*Sqrt[1 + c^2*x^2]*Log[c*d*x + Sqrt[d]*Sqrt[d + c^2*d*x^2]] + 24*b*Sqrt[d +
c^2*d*x^2]*ArcSinh[c*x]*(-48*Sinh[2*ArcSinh[c*x]] + 24*Sinh[4*ArcSinh[c*x]] + 16*Sinh[6*ArcSinh[c*x]] + 3*Sin
h[8*ArcSinh[c*x]])))/(73728*c^3*Sqrt[1 + c^2*x^2])

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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a c^{4} d^{2} x^{6} + 2 \, a c^{2} d^{2} x^{4} + a d^{2} x^{2} + {\left (b c^{4} d^{2} x^{6} + 2 \, b c^{2} d^{2} x^{4} + b d^{2} x^{2}\right )} \operatorname {arsinh}\left (c x\right )\right )} \sqrt {c^{2} d x^{2} + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x)),x, algorithm="fricas")

[Out]

integral((a*c^4*d^2*x^6 + 2*a*c^2*d^2*x^4 + a*d^2*x^2 + (b*c^4*d^2*x^6 + 2*b*c^2*d^2*x^4 + b*d^2*x^2)*arcsinh(
c*x))*sqrt(c^2*d*x^2 + d), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x)),x, algorithm="giac")

[Out]

integrate((c^2*d*x^2 + d)^(5/2)*(b*arcsinh(c*x) + a)*x^2, x)

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maple [A]  time = 0.34, size = 537, normalized size = 1.59 \[ \frac {a x \left (c^{2} d \,x^{2}+d \right )^{\frac {7}{2}}}{8 c^{2} d}-\frac {a x \left (c^{2} d \,x^{2}+d \right )^{\frac {5}{2}}}{48 c^{2}}-\frac {5 a d x \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{192 c^{2}}-\frac {5 a \,d^{2} x \sqrt {c^{2} d \,x^{2}+d}}{128 c^{2}}-\frac {5 a \,d^{3} \ln \left (\frac {x \,c^{2} d}{\sqrt {c^{2} d}}+\sqrt {c^{2} d \,x^{2}+d}\right )}{128 c^{2} \sqrt {c^{2} d}}-\frac {5 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right )^{2} d^{2}}{256 \sqrt {c^{2} x^{2}+1}\, c^{3}}+\frac {359 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, d^{2}}{73728 c^{3} \sqrt {c^{2} x^{2}+1}}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, d^{2} c^{6} \arcsinh \left (c x \right ) x^{9}}{8 c^{2} x^{2}+8}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, d^{2} c^{5} x^{8}}{64 \sqrt {c^{2} x^{2}+1}}+\frac {23 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, d^{2} c^{4} \arcsinh \left (c x \right ) x^{7}}{48 \left (c^{2} x^{2}+1\right )}-\frac {17 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, d^{2} c^{3} x^{6}}{288 \sqrt {c^{2} x^{2}+1}}+\frac {127 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, d^{2} c^{2} \arcsinh \left (c x \right ) x^{5}}{192 \left (c^{2} x^{2}+1\right )}-\frac {59 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, d^{2} c \,x^{4}}{768 \sqrt {c^{2} x^{2}+1}}+\frac {133 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, d^{2} \arcsinh \left (c x \right ) x^{3}}{384 \left (c^{2} x^{2}+1\right )}-\frac {5 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, d^{2} x^{2}}{256 c \sqrt {c^{2} x^{2}+1}}+\frac {5 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, d^{2} \arcsinh \left (c x \right ) x}{128 c^{2} \left (c^{2} x^{2}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x)),x)

[Out]

1/8*a*x*(c^2*d*x^2+d)^(7/2)/c^2/d-1/48*a/c^2*x*(c^2*d*x^2+d)^(5/2)-5/192*a/c^2*d*x*(c^2*d*x^2+d)^(3/2)-5/128*a
/c^2*d^2*x*(c^2*d*x^2+d)^(1/2)-5/128*a/c^2*d^3*ln(x*c^2*d/(c^2*d)^(1/2)+(c^2*d*x^2+d)^(1/2))/(c^2*d)^(1/2)-5/2
56*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^3*arcsinh(c*x)^2*d^2+359/73728*b*(d*(c^2*x^2+1))^(1/2)*d^2/c^3/
(c^2*x^2+1)^(1/2)+1/8*b*(d*(c^2*x^2+1))^(1/2)*d^2*c^6/(c^2*x^2+1)*arcsinh(c*x)*x^9-1/64*b*(d*(c^2*x^2+1))^(1/2
)*d^2*c^5/(c^2*x^2+1)^(1/2)*x^8+23/48*b*(d*(c^2*x^2+1))^(1/2)*d^2*c^4/(c^2*x^2+1)*arcsinh(c*x)*x^7-17/288*b*(d
*(c^2*x^2+1))^(1/2)*d^2*c^3/(c^2*x^2+1)^(1/2)*x^6+127/192*b*(d*(c^2*x^2+1))^(1/2)*d^2*c^2/(c^2*x^2+1)*arcsinh(
c*x)*x^5-59/768*b*(d*(c^2*x^2+1))^(1/2)*d^2*c/(c^2*x^2+1)^(1/2)*x^4+133/384*b*(d*(c^2*x^2+1))^(1/2)*d^2/(c^2*x
^2+1)*arcsinh(c*x)*x^3-5/256*b*(d*(c^2*x^2+1))^(1/2)*d^2/c/(c^2*x^2+1)^(1/2)*x^2+5/128*b*(d*(c^2*x^2+1))^(1/2)
*d^2/c^2/(c^2*x^2+1)*arcsinh(c*x)*x

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maxima [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^2*(c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^2\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,{\left (d\,c^2\,x^2+d\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*asinh(c*x))*(d + c^2*d*x^2)^(5/2),x)

[Out]

int(x^2*(a + b*asinh(c*x))*(d + c^2*d*x^2)^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(c**2*d*x**2+d)**(5/2)*(a+b*asinh(c*x)),x)

[Out]

Timed out

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