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

Optimal. Leaf size=618 \[ -\frac {15 b c d^2 x^{m+2} \sqrt {c^2 d x^2+d} \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;-c^2 x^2\right )}{(m+1) (m+2)^2 (m+4) (m+6) \sqrt {c^2 x^2+1}}+\frac {15 d^2 x^{m+1} \sqrt {c^2 d x^2+d} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};-c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{(m+6) \left (m^3+7 m^2+14 m+8\right ) \sqrt {c^2 x^2+1}}+\frac {15 d^2 x^{m+1} \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{(m+6) \left (m^2+6 m+8\right )}+\frac {x^{m+1} \left (c^2 d x^2+d\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{m+6}+\frac {5 d x^{m+1} \left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{(m+4) (m+6)}-\frac {5 b c d^2 x^{m+2} \sqrt {c^2 d x^2+d}}{(m+6) \left (m^2+6 m+8\right ) \sqrt {c^2 x^2+1}}-\frac {b c d^2 x^{m+2} \sqrt {c^2 d x^2+d}}{\left (m^2+8 m+12\right ) \sqrt {c^2 x^2+1}}-\frac {15 b c d^2 x^{m+2} \sqrt {c^2 d x^2+d}}{(m+2)^2 (m+4) (m+6) \sqrt {c^2 x^2+1}}-\frac {b c^5 d^2 x^{m+6} \sqrt {c^2 d x^2+d}}{(m+6)^2 \sqrt {c^2 x^2+1}}-\frac {2 b c^3 d^2 x^{m+4} \sqrt {c^2 d x^2+d}}{(m+4) (m+6) \sqrt {c^2 x^2+1}}-\frac {5 b c^3 d^2 x^{m+4} \sqrt {c^2 d x^2+d}}{(m+4)^2 (m+6) \sqrt {c^2 x^2+1}} \]

[Out]

5*d*x^(1+m)*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))/(4+m)/(6+m)+x^(1+m)*(c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x))/
(6+m)+15*d^2*x^(1+m)*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)/(6+m)/(m^2+6*m+8)-15*b*c*d^2*x^(2+m)*(c^2*d*x^2+d)
^(1/2)/(2+m)^2/(4+m)/(6+m)/(c^2*x^2+1)^(1/2)-5*b*c*d^2*x^(2+m)*(c^2*d*x^2+d)^(1/2)/(6+m)/(m^2+6*m+8)/(c^2*x^2+
1)^(1/2)-b*c*d^2*x^(2+m)*(c^2*d*x^2+d)^(1/2)/(m^2+8*m+12)/(c^2*x^2+1)^(1/2)-5*b*c^3*d^2*x^(4+m)*(c^2*d*x^2+d)^
(1/2)/(4+m)^2/(6+m)/(c^2*x^2+1)^(1/2)-2*b*c^3*d^2*x^(4+m)*(c^2*d*x^2+d)^(1/2)/(4+m)/(6+m)/(c^2*x^2+1)^(1/2)-b*
c^5*d^2*x^(6+m)*(c^2*d*x^2+d)^(1/2)/(6+m)^2/(c^2*x^2+1)^(1/2)+15*d^2*x^(1+m)*(a+b*arcsinh(c*x))*hypergeom([1/2
, 1/2+1/2*m],[3/2+1/2*m],-c^2*x^2)*(c^2*d*x^2+d)^(1/2)/(6+m)/(m^3+7*m^2+14*m+8)/(c^2*x^2+1)^(1/2)-15*b*c*d^2*x
^(2+m)*HypergeometricPFQ([1, 1+1/2*m, 1+1/2*m],[3/2+1/2*m, 2+1/2*m],-c^2*x^2)*(c^2*d*x^2+d)^(1/2)/(2+m)^2/(6+m
)/(m^2+5*m+4)/(c^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.56, antiderivative size = 618, normalized size of antiderivative = 1.00, 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 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 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]

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 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]

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*}

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Mathematica [A]  time = 1.28, size = 332, normalized size = 0.54 \[ \frac {d^2 x^{m+1} \sqrt {c^2 d x^2+d} \left (-\frac {5 \left (3 (m+4) \left (b c x \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;-c^2 x^2\right )-(m+2) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};-c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )-(m+1) (m+2) \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )+b c (m+1) x\right )-\left ((m+1) (m+4) (m+2)^2 \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )\right )+b c (m+1) (m+2) x \left (c^2 (m+2) x^2+m+4\right )\right )}{(m+1) (m+2)^2 (m+4)^2 \sqrt {c^2 x^2+1}}+\left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )-\frac {b c x \left (c^4 (m+2) (m+4) x^4+2 c^2 (m+2) (m+6) x^2+(m+4) (m+6)\right )}{(m+2) (m+4) (m+6) \sqrt {c^2 x^2+1}}\right )}{m+6} \]

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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fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a c^{4} d^{2} x^{4} + 2 \, a c^{2} d^{2} x^{2} + a d^{2} + {\left (b c^{4} d^{2} x^{4} + 2 \, b c^{2} d^{2} x^{2} + b d^{2}\right )} \operatorname {arsinh}\left (c x\right )\right )} \sqrt {c^{2} d x^{2} + d} x^{m}, x\right ) \]

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

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [F]  time = 1.64, size = 0, normalized size = 0.00 \[ \int x^{m} \left (c^{2} d \,x^{2}+d \right )^{\frac {5}{2}} \left (a +b \arcsinh \left (c x \right )\right )\, dx \]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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