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

Optimal. Leaf size=618 \[ -\frac{15 b c d^2 x^{m+2} \sqrt{c^2 d x^2+d} \text{HypergeometricPFQ}\left (\left \{1,\frac{m}{2}+1,\frac{m}{2}+1\right \},\left \{\frac{m}{2}+\frac{3}{2},\frac{m}{2}+2\right \},-c^2 x^2\right )}{(m+1) (m+2)^2 (m+4) (m+6) \sqrt{c^2 x^2+1}}+\frac{15 d^2 x^{m+1} \sqrt{c^2 d x^2+d} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},-c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{(m+6) \left (m^3+7 m^2+14 m+8\right ) \sqrt{c^2 x^2+1}}+\frac{15 d^2 x^{m+1} \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{(m+6) \left (m^2+6 m+8\right )}+\frac{x^{m+1} \left (c^2 d x^2+d\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{m+6}+\frac{5 d x^{m+1} \left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{(m+4) (m+6)}-\frac{5 b c d^2 x^{m+2} \sqrt{c^2 d x^2+d}}{(m+6) \left (m^2+6 m+8\right ) \sqrt{c^2 x^2+1}}-\frac{b c d^2 x^{m+2} \sqrt{c^2 d x^2+d}}{\left (m^2+8 m+12\right ) \sqrt{c^2 x^2+1}}-\frac{15 b c d^2 x^{m+2} \sqrt{c^2 d x^2+d}}{(m+2)^2 (m+4) (m+6) \sqrt{c^2 x^2+1}}-\frac{2 b c^3 d^2 x^{m+4} \sqrt{c^2 d x^2+d}}{(m+4) (m+6) \sqrt{c^2 x^2+1}}-\frac{5 b c^3 d^2 x^{m+4} \sqrt{c^2 d x^2+d}}{(m+4)^2 (m+6) \sqrt{c^2 x^2+1}}-\frac{b c^5 d^2 x^{m+6} \sqrt{c^2 d x^2+d}}{(m+6)^2 \sqrt{c^2 x^2+1}} \]

[Out]

(-15*b*c*d^2*x^(2 + m)*Sqrt[d + c^2*d*x^2])/((2 + m)^2*(4 + m)*(6 + m)*Sqrt[1 + c^2*x^2]) - (5*b*c*d^2*x^(2 +
m)*Sqrt[d + c^2*d*x^2])/((6 + m)*(8 + 6*m + m^2)*Sqrt[1 + c^2*x^2]) - (b*c*d^2*x^(2 + m)*Sqrt[d + c^2*d*x^2])/
((12 + 8*m + m^2)*Sqrt[1 + c^2*x^2]) - (5*b*c^3*d^2*x^(4 + m)*Sqrt[d + c^2*d*x^2])/((4 + m)^2*(6 + m)*Sqrt[1 +
c^2*x^2]) - (2*b*c^3*d^2*x^(4 + m)*Sqrt[d + c^2*d*x^2])/((4 + m)*(6 + m)*Sqrt[1 + c^2*x^2]) - (b*c^5*d^2*x^(6
+ m)*Sqrt[d + c^2*d*x^2])/((6 + m)^2*Sqrt[1 + c^2*x^2]) + (15*d^2*x^(1 + m)*Sqrt[d + c^2*d*x^2]*(a + b*ArcSin
h[c*x]))/((6 + m)*(8 + 6*m + m^2)) + (5*d*x^(1 + m)*(d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/((4 + m)*(6 +
m)) + (x^(1 + m)*(d + c^2*d*x^2)^(5/2)*(a + b*ArcSinh[c*x]))/(6 + m) + (15*d^2*x^(1 + m)*Sqrt[d + c^2*d*x^2]*(
a + b*ArcSinh[c*x])*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, -(c^2*x^2)])/((6 + m)*(8 + 14*m + 7*m^2 + m^3
)*Sqrt[1 + c^2*x^2]) - (15*b*c*d^2*x^(2 + m)*Sqrt[d + c^2*d*x^2]*HypergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2
+ m/2, 2 + m/2}, -(c^2*x^2)])/((1 + m)*(2 + m)^2*(4 + m)*(6 + m)*Sqrt[1 + c^2*x^2])

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Rubi [A]  time = 0.556182, antiderivative size = 618, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {5744, 5742, 5762, 30, 14, 270} \[ -\frac{15 b c d^2 x^{m+2} \sqrt{c^2 d x^2+d} \, _3F_2\left (1,\frac{m}{2}+1,\frac{m}{2}+1;\frac{m}{2}+\frac{3}{2},\frac{m}{2}+2;-c^2 x^2\right )}{(m+1) (m+2)^2 (m+4) (m+6) \sqrt{c^2 x^2+1}}+\frac{15 d^2 x^{m+1} \sqrt{c^2 d x^2+d} \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};-c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{(m+6) \left (m^3+7 m^2+14 m+8\right ) \sqrt{c^2 x^2+1}}+\frac{15 d^2 x^{m+1} \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{(m+6) \left (m^2+6 m+8\right )}+\frac{x^{m+1} \left (c^2 d x^2+d\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{m+6}+\frac{5 d x^{m+1} \left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{(m+4) (m+6)}-\frac{5 b c d^2 x^{m+2} \sqrt{c^2 d x^2+d}}{(m+6) \left (m^2+6 m+8\right ) \sqrt{c^2 x^2+1}}-\frac{b c d^2 x^{m+2} \sqrt{c^2 d x^2+d}}{\left (m^2+8 m+12\right ) \sqrt{c^2 x^2+1}}-\frac{15 b c d^2 x^{m+2} \sqrt{c^2 d x^2+d}}{(m+2)^2 (m+4) (m+6) \sqrt{c^2 x^2+1}}-\frac{2 b c^3 d^2 x^{m+4} \sqrt{c^2 d x^2+d}}{(m+4) (m+6) \sqrt{c^2 x^2+1}}-\frac{5 b c^3 d^2 x^{m+4} \sqrt{c^2 d x^2+d}}{(m+4)^2 (m+6) \sqrt{c^2 x^2+1}}-\frac{b c^5 d^2 x^{m+6} \sqrt{c^2 d x^2+d}}{(m+6)^2 \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*b*c*d^2*x^(2 + m)*Sqrt[d + c^2*d*x^2])/((2 + m)^2*(4 + m)*(6 + m)*Sqrt[1 + c^2*x^2]) - (5*b*c*d^2*x^(2 +
m)*Sqrt[d + c^2*d*x^2])/((6 + m)*(8 + 6*m + m^2)*Sqrt[1 + c^2*x^2]) - (b*c*d^2*x^(2 + m)*Sqrt[d + c^2*d*x^2])/
((12 + 8*m + m^2)*Sqrt[1 + c^2*x^2]) - (5*b*c^3*d^2*x^(4 + m)*Sqrt[d + c^2*d*x^2])/((4 + m)^2*(6 + m)*Sqrt[1 +
c^2*x^2]) - (2*b*c^3*d^2*x^(4 + m)*Sqrt[d + c^2*d*x^2])/((4 + m)*(6 + m)*Sqrt[1 + c^2*x^2]) - (b*c^5*d^2*x^(6
+ m)*Sqrt[d + c^2*d*x^2])/((6 + m)^2*Sqrt[1 + c^2*x^2]) + (15*d^2*x^(1 + m)*Sqrt[d + c^2*d*x^2]*(a + b*ArcSin
h[c*x]))/((6 + m)*(8 + 6*m + m^2)) + (5*d*x^(1 + m)*(d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/((4 + m)*(6 +
m)) + (x^(1 + m)*(d + c^2*d*x^2)^(5/2)*(a + b*ArcSinh[c*x]))/(6 + m) + (15*d^2*x^(1 + m)*Sqrt[d + c^2*d*x^2]*(
a + b*ArcSinh[c*x])*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, -(c^2*x^2)])/((6 + m)*(8 + 14*m + 7*m^2 + m^3
)*Sqrt[1 + c^2*x^2]) - (15*b*c*d^2*x^(2 + m)*Sqrt[d + c^2*d*x^2]*HypergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2
+ m/2, 2 + m/2}, -(c^2*x^2)])/((1 + m)*(2 + m)^2*(4 + m)*(6 + m)*Sqrt[1 + c^2*x^2])

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 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 5762

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[((f*x
)^(m + 1)*(a + b*ArcSinh[c*x])*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, -(c^2*x^2)])/(Sqrt[d]*f*(m + 1)),
x] - Simp[(b*c*(f*x)^(m + 2)*HypergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, -(c^2*x^2)])/(Sqrt
[d]*f^2*(m + 1)*(m + 2)), x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && GtQ[d, 0] &&  !IntegerQ[m]

Rule 30

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

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 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]

Rubi steps

\begin{align*} \int x^m \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac{x^{1+m} \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{6+m}+\frac{(5 d) \int x^m \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{6+m}-\frac{\left (b c d^2 \sqrt{d+c^2 d x^2}\right ) \int x^{1+m} \left (1+c^2 x^2\right )^2 \, dx}{(6+m) \sqrt{1+c^2 x^2}}\\ &=\frac{5 d x^{1+m} \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{(4+m) (6+m)}+\frac{x^{1+m} \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{6+m}+\frac{\left (15 d^2\right ) \int x^m \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{(4+m) (6+m)}-\frac{\left (b c d^2 \sqrt{d+c^2 d x^2}\right ) \int \left (x^{1+m}+2 c^2 x^{3+m}+c^4 x^{5+m}\right ) \, dx}{(6+m) \sqrt{1+c^2 x^2}}-\frac{\left (5 b c d^2 \sqrt{d+c^2 d x^2}\right ) \int x^{1+m} \left (1+c^2 x^2\right ) \, dx}{(4+m) (6+m) \sqrt{1+c^2 x^2}}\\ &=-\frac{b c d^2 x^{2+m} \sqrt{d+c^2 d x^2}}{\left (12+8 m+m^2\right ) \sqrt{1+c^2 x^2}}-\frac{2 b c^3 d^2 x^{4+m} \sqrt{d+c^2 d x^2}}{(4+m) (6+m) \sqrt{1+c^2 x^2}}-\frac{b c^5 d^2 x^{6+m} \sqrt{d+c^2 d x^2}}{(6+m)^2 \sqrt{1+c^2 x^2}}+\frac{15 d^2 x^{1+m} \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{(2+m) (4+m) (6+m)}+\frac{5 d x^{1+m} \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{(4+m) (6+m)}+\frac{x^{1+m} \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{6+m}-\frac{\left (5 b c d^2 \sqrt{d+c^2 d x^2}\right ) \int \left (x^{1+m}+c^2 x^{3+m}\right ) \, dx}{(4+m) (6+m) \sqrt{1+c^2 x^2}}+\frac{\left (15 d^2 \sqrt{d+c^2 d x^2}\right ) \int \frac{x^m \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}} \, dx}{(2+m) (4+m) (6+m) \sqrt{1+c^2 x^2}}-\frac{\left (15 b c d^2 \sqrt{d+c^2 d x^2}\right ) \int x^{1+m} \, dx}{(2+m) (4+m) (6+m) \sqrt{1+c^2 x^2}}\\ &=-\frac{15 b c d^2 x^{2+m} \sqrt{d+c^2 d x^2}}{(2+m)^2 (4+m) (6+m) \sqrt{1+c^2 x^2}}-\frac{5 b c d^2 x^{2+m} \sqrt{d+c^2 d x^2}}{(2+m) (4+m) (6+m) \sqrt{1+c^2 x^2}}-\frac{b c d^2 x^{2+m} \sqrt{d+c^2 d x^2}}{\left (12+8 m+m^2\right ) \sqrt{1+c^2 x^2}}-\frac{5 b c^3 d^2 x^{4+m} \sqrt{d+c^2 d x^2}}{(4+m)^2 (6+m) \sqrt{1+c^2 x^2}}-\frac{2 b c^3 d^2 x^{4+m} \sqrt{d+c^2 d x^2}}{(4+m) (6+m) \sqrt{1+c^2 x^2}}-\frac{b c^5 d^2 x^{6+m} \sqrt{d+c^2 d x^2}}{(6+m)^2 \sqrt{1+c^2 x^2}}+\frac{15 d^2 x^{1+m} \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{(2+m) (4+m) (6+m)}+\frac{5 d x^{1+m} \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{(4+m) (6+m)}+\frac{x^{1+m} \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{6+m}+\frac{15 d^2 x^{1+m} \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, _2F_1\left (\frac{1}{2},\frac{1+m}{2};\frac{3+m}{2};-c^2 x^2\right )}{(1+m) (2+m) (4+m) (6+m) \sqrt{1+c^2 x^2}}-\frac{15 b c d^2 x^{2+m} \sqrt{d+c^2 d x^2} \, _3F_2\left (1,1+\frac{m}{2},1+\frac{m}{2};\frac{3}{2}+\frac{m}{2},2+\frac{m}{2};-c^2 x^2\right )}{(1+m) (2+m)^2 (4+m) (6+m) \sqrt{1+c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 1.31886, size = 332, normalized size = 0.54 \[ \frac{d^2 x^{m+1} \sqrt{c^2 d x^2+d} \left (-\frac{5 \left (3 (m+4) \left (b c x \text{HypergeometricPFQ}\left (\left \{1,\frac{m}{2}+1,\frac{m}{2}+1\right \},\left \{\frac{m}{2}+\frac{3}{2},\frac{m}{2}+2\right \},-c^2 x^2\right )-(m+2) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},-c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )-(m+1) (m+2) \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )+b c (m+1) x\right )-(m+1) (m+4) (m+2)^2 \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+b c (m+1) (m+2) x \left (c^2 (m+2) x^2+m+4\right )\right )}{(m+1) (m+2)^2 (m+4)^2 \sqrt{c^2 x^2+1}}+\left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )-\frac{b c x \left (c^4 (m+2) (m+4) x^4+2 c^2 (m+2) (m+6) x^2+(m+4) (m+6)\right )}{(m+2) (m+4) (m+6) \sqrt{c^2 x^2+1}}\right )}{m+6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*x^(1 + m)*Sqrt[d + c^2*d*x^2]*(-((b*c*x*((4 + m)*(6 + m) + 2*c^2*(2 + m)*(6 + m)*x^2 + c^4*(2 + m)*(4 + m
)*x^4))/((2 + m)*(4 + m)*(6 + m)*Sqrt[1 + c^2*x^2])) + (1 + c^2*x^2)^2*(a + b*ArcSinh[c*x]) - (5*(b*c*(1 + m)*
(2 + m)*x*(4 + m + c^2*(2 + m)*x^2) - (1 + m)*(2 + m)^2*(4 + m)*(1 + c^2*x^2)^(3/2)*(a + b*ArcSinh[c*x]) + 3*(
4 + m)*(b*c*(1 + m)*x - (1 + m)*(2 + m)*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]) - (2 + m)*(a + b*ArcSinh[c*x])*
Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, -(c^2*x^2)] + b*c*x*HypergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2
+ m/2, 2 + m/2}, -(c^2*x^2)])))/((1 + m)*(2 + m)^2*(4 + m)^2*Sqrt[1 + c^2*x^2])))/(6 + m)

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Maple [F]  time = 1.312, size = 0, normalized size = 0. \begin{align*} \int{x}^{m} \left ({c}^{2}d{x}^{2}+d \right ) ^{{\frac{5}{2}}} \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c^{2} d x^{2} + d\right )}^{\frac{5}{2}}{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )} x^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a c^{4} d^{2} x^{4} + 2 \, a c^{2} d^{2} x^{2} + a d^{2} +{\left (b c^{4} d^{2} x^{4} + 2 \, b c^{2} d^{2} x^{2} + b d^{2}\right )} \operatorname{arsinh}\left (c x\right )\right )} \sqrt{c^{2} d x^{2} + d} x^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*c^4*d^2*x^4 + 2*a*c^2*d^2*x^2 + a*d^2 + (b*c^4*d^2*x^4 + 2*b*c^2*d^2*x^2 + b*d^2)*arcsinh(c*x))*sq
rt(c^2*d*x^2 + d)*x^m, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out