3.2.49 \(\int x^m (d-c^2 d x^2)^{5/2} (a+b \text {ArcSin}(c x)) \, dx\) [149]

Optimal. Leaf size=635 \[ -\frac {15 b c d^2 x^{2+m} \sqrt {d-c^2 d x^2}}{(2+m)^2 (4+m) (6+m) \sqrt {1-c^2 x^2}}-\frac {5 b c d^2 x^{2+m} \sqrt {d-c^2 d x^2}}{(6+m) \left (8+6 m+m^2\right ) \sqrt {1-c^2 x^2}}-\frac {b c d^2 x^{2+m} \sqrt {d-c^2 d x^2}}{\left (12+8 m+m^2\right ) \sqrt {1-c^2 x^2}}+\frac {5 b c^3 d^2 x^{4+m} \sqrt {d-c^2 d x^2}}{(4+m)^2 (6+m) \sqrt {1-c^2 x^2}}+\frac {2 b c^3 d^2 x^{4+m} \sqrt {d-c^2 d x^2}}{(4+m) (6+m) \sqrt {1-c^2 x^2}}-\frac {b c^5 d^2 x^{6+m} \sqrt {d-c^2 d x^2}}{(6+m)^2 \sqrt {1-c^2 x^2}}+\frac {15 d^2 x^{1+m} \sqrt {d-c^2 d x^2} (a+b \text {ArcSin}(c x))}{(6+m) \left (8+6 m+m^2\right )}+\frac {5 d x^{1+m} \left (d-c^2 d x^2\right )^{3/2} (a+b \text {ArcSin}(c x))}{(4+m) (6+m)}+\frac {x^{1+m} \left (d-c^2 d x^2\right )^{5/2} (a+b \text {ArcSin}(c x))}{6+m}+\frac {15 d^2 x^{1+m} \sqrt {d-c^2 d x^2} (a+b \text {ArcSin}(c x)) \text {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},c^2 x^2\right )}{(6+m) \left (8+14 m+7 m^2+m^3\right ) \sqrt {1-c^2 x^2}}-\frac {15 b c d^2 x^{2+m} \sqrt {d-c^2 d x^2} \text {HypergeometricPFQ}\left (\left \{1,1+\frac {m}{2},1+\frac {m}{2}\right \},\left \{\frac {3}{2}+\frac {m}{2},2+\frac {m}{2}\right \},c^2 x^2\right )}{(1+m) (2+m)^2 (4+m) (6+m) \sqrt {1-c^2 x^2}} \]

[Out]

5*d*x^(1+m)*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))/(4+m)/(6+m)+x^(1+m)*(-c^2*d*x^2+d)^(5/2)*(a+b*arcsin(c*x))/
(6+m)+15*d^2*x^(1+m)*(a+b*arcsin(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*arcsin(c*x))*hyp
ergeom([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)*hypergeom([1, 1+1/2*m, 1+1/2*m],[2+1/2*m, 3/2+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 = 0.38, antiderivative size = 635, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {4787, 4783, 4805, 30, 14, 276} \begin {gather*} -\frac {15 b c d^2 x^{m+2} \sqrt {d-c^2 d x^2} \, _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 {1-c^2 x^2}}+\frac {15 d^2 x^{m+1} \sqrt {d-c^2 d x^2} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};c^2 x^2\right ) (a+b \text {ArcSin}(c x))}{(m+6) \left (m^3+7 m^2+14 m+8\right ) \sqrt {1-c^2 x^2}}+\frac {15 d^2 x^{m+1} \sqrt {d-c^2 d x^2} (a+b \text {ArcSin}(c x))}{(m+6) \left (m^2+6 m+8\right )}+\frac {x^{m+1} \left (d-c^2 d x^2\right )^{5/2} (a+b \text {ArcSin}(c x))}{m+6}+\frac {5 d x^{m+1} \left (d-c^2 d x^2\right )^{3/2} (a+b \text {ArcSin}(c x))}{(m+4) (m+6)}-\frac {5 b c d^2 x^{m+2} \sqrt {d-c^2 d x^2}}{(m+6) \left (m^2+6 m+8\right ) \sqrt {1-c^2 x^2}}-\frac {b c d^2 x^{m+2} \sqrt {d-c^2 d x^2}}{\left (m^2+8 m+12\right ) \sqrt {1-c^2 x^2}}-\frac {15 b c d^2 x^{m+2} \sqrt {d-c^2 d x^2}}{(m+2)^2 (m+4) (m+6) \sqrt {1-c^2 x^2}}-\frac {b c^5 d^2 x^{m+6} \sqrt {d-c^2 d x^2}}{(m+6)^2 \sqrt {1-c^2 x^2}}+\frac {2 b c^3 d^2 x^{m+4} \sqrt {d-c^2 d x^2}}{(m+4) (m+6) \sqrt {1-c^2 x^2}}+\frac {5 b c^3 d^2 x^{m+4} \sqrt {d-c^2 d x^2}}{(m+4)^2 (m+6) \sqrt {1-c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^m*(d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[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
[c*x]))/((6 + m)*(8 + 6*m + m^2)) + (5*d*x^(1 + m)*(d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x]))/((4 + m)*(6 + m)
) + (x^(1 + m)*(d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x]))/(6 + m) + (15*d^2*x^(1 + m)*Sqrt[d - c^2*d*x^2]*(a +
b*ArcSin[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 276

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 4783

Int[((a_.) + ArcSin[(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*ArcSin[c*x])^n/(f*(m + 2))), x] + (Dist[(1/(m + 2))*Simp[Sqrt[d + e*x^2]/S
qrt[1 - c^2*x^2]], Int[(f*x)^m*((a + b*ArcSin[c*x])^n/Sqrt[1 - c^2*x^2]), x], x] - Dist[b*c*(n/(f*(m + 2)))*Si
mp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2]], Int[(f*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b,
c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && (IGtQ[m, -2] || EqQ[n, 1])

Rule 4787

Int[((a_.) + ArcSin[(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*ArcSin[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*ArcSin[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*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b
, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[p, 0] &&  !LtQ[m, -1]

Rule 4805

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

Rubi steps

\begin {align*} \int x^m \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac {x^{1+m} \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-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 \sin ^{-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 \sin ^{-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 \sin ^{-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 \sin ^{-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 \sin ^{-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 \sin ^{-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 \sin ^{-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 \sin ^{-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 \sin ^{-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 \sin ^{-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 \sin ^{-1}(c x)\right )}{6+m}+\frac {15 d^2 x^{1+m} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-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 = 0.83, size = 338, normalized size = 0.53 \begin {gather*} \frac {d^2 x^{1+m} \sqrt {d-c^2 d x^2} \left (-b c (1+m) (2+m) (4+m) x \left ((4+m) (6+m)-2 c^2 (2+m) (6+m) x^2+c^4 (2+m) (4+m) x^4\right )+(1+m) (2+m)^2 (4+m)^2 (6+m) \left (1-c^2 x^2\right )^{5/2} (a+b \text {ArcSin}(c x))-5 (6+m) \left (b c (1+m) (2+m) x \left (4+m-c^2 (2+m) x^2\right )-(1+m) (2+m)^2 (4+m) \left (1-c^2 x^2\right )^{3/2} (a+b \text {ArcSin}(c x))+3 (4+m) \left (b c (1+m) x-(1+m) (2+m) \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))-(2+m) (a+b \text {ArcSin}(c x)) \text {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},c^2 x^2\right )+b c x \text {HypergeometricPFQ}\left (\left \{1,1+\frac {m}{2},1+\frac {m}{2}\right \},\left \{\frac {3}{2}+\frac {m}{2},2+\frac {m}{2}\right \},c^2 x^2\right )\right )\right )\right )}{(1+m) (2+m)^2 (4+m)^2 (6+m)^2 \sqrt {1-c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*x^(1 + m)*Sqrt[d - c^2*d*x^2]*(-(b*c*(1 + m)*(2 + m)*(4 + m)*x*((4 + m)*(6 + m) - 2*c^2*(2 + m)*(6 + m)*x
^2 + c^4*(2 + m)*(4 + m)*x^4)) + (1 + m)*(2 + m)^2*(4 + m)^2*(6 + m)*(1 - c^2*x^2)^(5/2)*(a + b*ArcSin[c*x]) -
5*(6 + m)*(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*ArcSin[c*x]) + 3*(4 + m)*(b*c*(1 + m)*x - (1 + m)*(2 + m)*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]) - (2 + m)
*(a + b*ArcSin[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*(6 + m)^2*Sqrt[1 - c^2*x^2])

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Maple [F]
time = 3.98, size = 0, normalized size = 0.00 \[\int x^{m} \left (-c^{2} d \,x^{2}+d \right )^{\frac {5}{2}} \left (a +b \arcsin \left (c x \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*(-c^2*d*x^2+d)^(5/2)*(a+b*arcsin(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)*arcsin(c*x))*sqr
t(-c^2*d*x^2 + d)*x^m, x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const 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^m\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d-c^2\,d\,x^2\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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