3.488 \(\int x (d-c^2 d x^2)^{3/2} (a+b \sin ^{-1}(c x))^n \, dx\)

Optimal. Leaf size=595 \[ -\frac {d e^{-\frac {i a}{b}} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (-\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \Gamma \left (n+1,-\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 c^2 \sqrt {1-c^2 x^2}}-\frac {d 3^{-n} e^{-\frac {3 i a}{b}} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (-\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \Gamma \left (n+1,-\frac {3 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{32 c^2 \sqrt {1-c^2 x^2}}-\frac {d 5^{-n-1} e^{-\frac {5 i a}{b}} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (-\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \Gamma \left (n+1,-\frac {5 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{32 c^2 \sqrt {1-c^2 x^2}}-\frac {d e^{\frac {i a}{b}} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \Gamma \left (n+1,\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 c^2 \sqrt {1-c^2 x^2}}-\frac {d 3^{-n} e^{\frac {3 i a}{b}} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \Gamma \left (n+1,\frac {3 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{32 c^2 \sqrt {1-c^2 x^2}}-\frac {d 5^{-n-1} e^{\frac {5 i a}{b}} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \Gamma \left (n+1,\frac {5 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{32 c^2 \sqrt {1-c^2 x^2}} \]

[Out]

-1/16*d*(a+b*arcsin(c*x))^n*GAMMA(1+n,-I*(a+b*arcsin(c*x))/b)*(-c^2*d*x^2+d)^(1/2)/c^2/exp(I*a/b)/((-I*(a+b*ar
csin(c*x))/b)^n)/(-c^2*x^2+1)^(1/2)-1/16*d*exp(I*a/b)*(a+b*arcsin(c*x))^n*GAMMA(1+n,I*(a+b*arcsin(c*x))/b)*(-c
^2*d*x^2+d)^(1/2)/c^2/((I*(a+b*arcsin(c*x))/b)^n)/(-c^2*x^2+1)^(1/2)-1/32*d*(a+b*arcsin(c*x))^n*GAMMA(1+n,-3*I
*(a+b*arcsin(c*x))/b)*(-c^2*d*x^2+d)^(1/2)/(3^n)/c^2/exp(3*I*a/b)/((-I*(a+b*arcsin(c*x))/b)^n)/(-c^2*x^2+1)^(1
/2)-1/32*d*exp(3*I*a/b)*(a+b*arcsin(c*x))^n*GAMMA(1+n,3*I*(a+b*arcsin(c*x))/b)*(-c^2*d*x^2+d)^(1/2)/(3^n)/c^2/
((I*(a+b*arcsin(c*x))/b)^n)/(-c^2*x^2+1)^(1/2)-1/32*5^(-1-n)*d*(a+b*arcsin(c*x))^n*GAMMA(1+n,-5*I*(a+b*arcsin(
c*x))/b)*(-c^2*d*x^2+d)^(1/2)/c^2/exp(5*I*a/b)/((-I*(a+b*arcsin(c*x))/b)^n)/(-c^2*x^2+1)^(1/2)-1/32*5^(-1-n)*d
*exp(5*I*a/b)*(a+b*arcsin(c*x))^n*GAMMA(1+n,5*I*(a+b*arcsin(c*x))/b)*(-c^2*d*x^2+d)^(1/2)/c^2/((I*(a+b*arcsin(
c*x))/b)^n)/(-c^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.59, antiderivative size = 595, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {4725, 4723, 4406, 3308, 2181} \[ -\frac {d e^{-\frac {i a}{b}} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (-\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \text {Gamma}\left (n+1,-\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 c^2 \sqrt {1-c^2 x^2}}-\frac {d 3^{-n} e^{-\frac {3 i a}{b}} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (-\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \text {Gamma}\left (n+1,-\frac {3 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{32 c^2 \sqrt {1-c^2 x^2}}-\frac {d 5^{-n-1} e^{-\frac {5 i a}{b}} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (-\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \text {Gamma}\left (n+1,-\frac {5 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{32 c^2 \sqrt {1-c^2 x^2}}-\frac {d e^{\frac {i a}{b}} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \text {Gamma}\left (n+1,\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 c^2 \sqrt {1-c^2 x^2}}-\frac {d 3^{-n} e^{\frac {3 i a}{b}} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \text {Gamma}\left (n+1,\frac {3 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{32 c^2 \sqrt {1-c^2 x^2}}-\frac {d 5^{-n-1} e^{\frac {5 i a}{b}} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \text {Gamma}\left (n+1,\frac {5 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{32 c^2 \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x*(d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x])^n,x]

[Out]

-(d*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])^n*Gamma[1 + n, ((-I)*(a + b*ArcSin[c*x]))/b])/(16*c^2*E^((I*a)/b)*
Sqrt[1 - c^2*x^2]*(((-I)*(a + b*ArcSin[c*x]))/b)^n) - (d*E^((I*a)/b)*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])^n
*Gamma[1 + n, (I*(a + b*ArcSin[c*x]))/b])/(16*c^2*Sqrt[1 - c^2*x^2]*((I*(a + b*ArcSin[c*x]))/b)^n) - (d*Sqrt[d
- c^2*d*x^2]*(a + b*ArcSin[c*x])^n*Gamma[1 + n, ((-3*I)*(a + b*ArcSin[c*x]))/b])/(32*3^n*c^2*E^(((3*I)*a)/b)*
Sqrt[1 - c^2*x^2]*(((-I)*(a + b*ArcSin[c*x]))/b)^n) - (d*E^(((3*I)*a)/b)*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x
])^n*Gamma[1 + n, ((3*I)*(a + b*ArcSin[c*x]))/b])/(32*3^n*c^2*Sqrt[1 - c^2*x^2]*((I*(a + b*ArcSin[c*x]))/b)^n)
- (5^(-1 - n)*d*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])^n*Gamma[1 + n, ((-5*I)*(a + b*ArcSin[c*x]))/b])/(32*c
^2*E^(((5*I)*a)/b)*Sqrt[1 - c^2*x^2]*(((-I)*(a + b*ArcSin[c*x]))/b)^n) - (5^(-1 - n)*d*E^(((5*I)*a)/b)*Sqrt[d
- c^2*d*x^2]*(a + b*ArcSin[c*x])^n*Gamma[1 + n, ((5*I)*(a + b*ArcSin[c*x]))/b])/(32*c^2*Sqrt[1 - c^2*x^2]*((I*
(a + b*ArcSin[c*x]))/b)^n)

Rule 2181

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(F^(g*(e - (c*f)/d))*(c +
d*x)^FracPart[m]*Gamma[m + 1, (-((f*g*Log[F])/d))*(c + d*x)])/(d*(-((f*g*Log[F])/d))^(IntPart[m] + 1)*(-((f*g*
Log[F]*(c + d*x))/d))^FracPart[m]), x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rule 3308

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/E^(I*(e + f*x))
, x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d, e, f, m}, x]

Rule 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rule 4723

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c^(
m + 1), Subst[Int[(a + b*x)^n*Sin[x]^m*Cos[x]^(2*p + 1), x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e, n},
x] && EqQ[c^2*d + e, 0] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 0])

Rule 4725

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(d^IntPar
t[p]*(d + e*x^2)^FracPart[p])/(1 - c^2*x^2)^FracPart[p], Int[x^m*(1 - c^2*x^2)^p*(a + b*ArcSin[c*x])^n, x], x]
/; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] &&  !(Integ
erQ[p] || GtQ[d, 0])

Rubi steps

\begin {align*} \int x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^n \, dx &=\frac {\left (d \sqrt {d-c^2 d x^2}\right ) \int x \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^n \, dx}{\sqrt {1-c^2 x^2}}\\ &=\frac {\left (d \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int (a+b x)^n \cos ^4(x) \sin (x) \, dx,x,\sin ^{-1}(c x)\right )}{c^2 \sqrt {1-c^2 x^2}}\\ &=\frac {\left (d \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{8} (a+b x)^n \sin (x)+\frac {3}{16} (a+b x)^n \sin (3 x)+\frac {1}{16} (a+b x)^n \sin (5 x)\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^2 \sqrt {1-c^2 x^2}}\\ &=\frac {\left (d \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int (a+b x)^n \sin (5 x) \, dx,x,\sin ^{-1}(c x)\right )}{16 c^2 \sqrt {1-c^2 x^2}}+\frac {\left (d \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int (a+b x)^n \sin (x) \, dx,x,\sin ^{-1}(c x)\right )}{8 c^2 \sqrt {1-c^2 x^2}}+\frac {\left (3 d \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int (a+b x)^n \sin (3 x) \, dx,x,\sin ^{-1}(c x)\right )}{16 c^2 \sqrt {1-c^2 x^2}}\\ &=\frac {\left (i d \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int e^{-5 i x} (a+b x)^n \, dx,x,\sin ^{-1}(c x)\right )}{32 c^2 \sqrt {1-c^2 x^2}}-\frac {\left (i d \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int e^{5 i x} (a+b x)^n \, dx,x,\sin ^{-1}(c x)\right )}{32 c^2 \sqrt {1-c^2 x^2}}+\frac {\left (i d \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int e^{-i x} (a+b x)^n \, dx,x,\sin ^{-1}(c x)\right )}{16 c^2 \sqrt {1-c^2 x^2}}-\frac {\left (i d \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int e^{i x} (a+b x)^n \, dx,x,\sin ^{-1}(c x)\right )}{16 c^2 \sqrt {1-c^2 x^2}}+\frac {\left (3 i d \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int e^{-3 i x} (a+b x)^n \, dx,x,\sin ^{-1}(c x)\right )}{32 c^2 \sqrt {1-c^2 x^2}}-\frac {\left (3 i d \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int e^{3 i x} (a+b x)^n \, dx,x,\sin ^{-1}(c x)\right )}{32 c^2 \sqrt {1-c^2 x^2}}\\ &=-\frac {d e^{-\frac {i a}{b}} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (-\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \Gamma \left (1+n,-\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 c^2 \sqrt {1-c^2 x^2}}-\frac {d e^{\frac {i a}{b}} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \Gamma \left (1+n,\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 c^2 \sqrt {1-c^2 x^2}}-\frac {3^{-n} d e^{-\frac {3 i a}{b}} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (-\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \Gamma \left (1+n,-\frac {3 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{32 c^2 \sqrt {1-c^2 x^2}}-\frac {3^{-n} d e^{\frac {3 i a}{b}} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \Gamma \left (1+n,\frac {3 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{32 c^2 \sqrt {1-c^2 x^2}}-\frac {5^{-1-n} d e^{-\frac {5 i a}{b}} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (-\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \Gamma \left (1+n,-\frac {5 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{32 c^2 \sqrt {1-c^2 x^2}}-\frac {5^{-1-n} d e^{\frac {5 i a}{b}} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{-n} \Gamma \left (1+n,\frac {5 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{32 c^2 \sqrt {1-c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 2.41, size = 464, normalized size = 0.78 \[ -\frac {d^2 15^{-n-1} e^{-\frac {5 i a}{b}} \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^n \left (\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{b^2}\right )^{-3 n} \left (\left (-\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^n \left (2\ 15^{n+1} e^{\frac {6 i a}{b}} \left (\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{b^2}\right )^{2 n} \Gamma \left (n+1,\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )+3 \left (5^{n+1} e^{\frac {2 i a}{b}} \left (\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{2 n} \left (\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{b^2}\right )^n \Gamma \left (n+1,-\frac {3 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )+5^{n+1} e^{\frac {8 i a}{b}} \left (\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{b^2}\right )^{2 n} \Gamma \left (n+1,\frac {3 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )+3^n \left (e^{\frac {10 i a}{b}} \left (\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{b^2}\right )^{2 n} \Gamma \left (n+1,\frac {5 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )+\left (\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^{3 n} \left (-\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^n \Gamma \left (n+1,-\frac {5 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )\right )\right )\right )+2\ 15^{n+1} e^{\frac {4 i a}{b}} \left (\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )^n \left (\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{b^2}\right )^{2 n} \Gamma \left (n+1,-\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )\right )}{32 c^2 \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x])^n,x]

[Out]

-1/32*(15^(-1 - n)*d^2*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^n*(2*15^(1 + n)*E^(((4*I)*a)/b)*((I*(a + b*ArcSin
[c*x]))/b)^n*((a + b*ArcSin[c*x])^2/b^2)^(2*n)*Gamma[1 + n, ((-I)*(a + b*ArcSin[c*x]))/b] + (((-I)*(a + b*ArcS
in[c*x]))/b)^n*(2*15^(1 + n)*E^(((6*I)*a)/b)*((a + b*ArcSin[c*x])^2/b^2)^(2*n)*Gamma[1 + n, (I*(a + b*ArcSin[c
*x]))/b] + 3*(5^(1 + n)*E^(((2*I)*a)/b)*((I*(a + b*ArcSin[c*x]))/b)^(2*n)*((a + b*ArcSin[c*x])^2/b^2)^n*Gamma[
1 + n, ((-3*I)*(a + b*ArcSin[c*x]))/b] + 5^(1 + n)*E^(((8*I)*a)/b)*((a + b*ArcSin[c*x])^2/b^2)^(2*n)*Gamma[1 +
n, ((3*I)*(a + b*ArcSin[c*x]))/b] + 3^n*((((-I)*(a + b*ArcSin[c*x]))/b)^n*((I*(a + b*ArcSin[c*x]))/b)^(3*n)*G
amma[1 + n, ((-5*I)*(a + b*ArcSin[c*x]))/b] + E^(((10*I)*a)/b)*((a + b*ArcSin[c*x])^2/b^2)^(2*n)*Gamma[1 + n,
((5*I)*(a + b*ArcSin[c*x]))/b])))))/(c^2*E^(((5*I)*a)/b)*Sqrt[d - c^2*d*x^2]*((a + b*ArcSin[c*x])^2/b^2)^(3*n)
)

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (c^{2} d x^{3} - d x\right )} \sqrt {-c^{2} d x^{2} + d} {\left (b \arcsin \left (c x\right ) + a\right )}^{n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))^n,x, algorithm="fricas")

[Out]

integral(-(c^2*d*x^3 - d*x)*sqrt(-c^2*d*x^2 + d)*(b*arcsin(c*x) + a)^n, 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*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))^n,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 = 0.19, size = 0, normalized size = 0.00 \[ \int x \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \arcsin \left (c x \right )\right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))^n,x)

[Out]

int(x*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))^n,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{n} x\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))^n,x, algorithm="maxima")

[Out]

integrate((-c^2*d*x^2 + d)^(3/2)*(b*arcsin(c*x) + a)^n*x, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x\,{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^n\,{\left (d-c^2\,d\,x^2\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*asin(c*x))^n*(d - c^2*d*x^2)^(3/2),x)

[Out]

int(x*(a + b*asin(c*x))^n*(d - c^2*d*x^2)^(3/2), 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*(-c**2*d*x**2+d)**(3/2)*(a+b*asin(c*x))**n,x)

[Out]

Timed out

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