3.185 \(\int (a+b \sin ^{-1}(c x))^{5/2} \, dx\)

Optimal. Leaf size=179 \[ -\frac {15 \sqrt {\frac {\pi }{2}} b^{5/2} \sin \left (\frac {a}{b}\right ) C\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {b}}\right )}{4 c}+\frac {15 \sqrt {\frac {\pi }{2}} b^{5/2} \cos \left (\frac {a}{b}\right ) S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {b}}\right )}{4 c}-\frac {15}{4} b^2 x \sqrt {a+b \sin ^{-1}(c x)}+\frac {5 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^{3/2}}{2 c}+x \left (a+b \sin ^{-1}(c x)\right )^{5/2} \]

[Out]

x*(a+b*arcsin(c*x))^(5/2)+15/8*b^(5/2)*cos(a/b)*FresnelS(2^(1/2)/Pi^(1/2)*(a+b*arcsin(c*x))^(1/2)/b^(1/2))*2^(
1/2)*Pi^(1/2)/c-15/8*b^(5/2)*FresnelC(2^(1/2)/Pi^(1/2)*(a+b*arcsin(c*x))^(1/2)/b^(1/2))*sin(a/b)*2^(1/2)*Pi^(1
/2)/c+5/2*b*(a+b*arcsin(c*x))^(3/2)*(-c^2*x^2+1)^(1/2)/c-15/4*b^2*x*(a+b*arcsin(c*x))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.50, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {4619, 4677, 4723, 3306, 3305, 3351, 3304, 3352} \[ -\frac {15 \sqrt {\frac {\pi }{2}} b^{5/2} \sin \left (\frac {a}{b}\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {b}}\right )}{4 c}+\frac {15 \sqrt {\frac {\pi }{2}} b^{5/2} \cos \left (\frac {a}{b}\right ) S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {b}}\right )}{4 c}-\frac {15}{4} b^2 x \sqrt {a+b \sin ^{-1}(c x)}+\frac {5 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^{3/2}}{2 c}+x \left (a+b \sin ^{-1}(c x)\right )^{5/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x])^(5/2),x]

[Out]

(-15*b^2*x*Sqrt[a + b*ArcSin[c*x]])/4 + (5*b*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^(3/2))/(2*c) + x*(a + b*Arc
Sin[c*x])^(5/2) + (15*b^(5/2)*Sqrt[Pi/2]*Cos[a/b]*FresnelS[(Sqrt[2/Pi]*Sqrt[a + b*ArcSin[c*x]])/Sqrt[b]])/(4*c
) - (15*b^(5/2)*Sqrt[Pi/2]*FresnelC[(Sqrt[2/Pi]*Sqrt[a + b*ArcSin[c*x]])/Sqrt[b]]*Sin[a/b])/(4*c)

Rule 3304

Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Cos[(f*x^2)/d],
x], x, Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

Rule 3305

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Sin[(f*x^2)/d], x], x,
Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

Rule 3306

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d +
f*x]/Sqrt[c + d*x], x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/Sqrt[c + d*x], x], x] /; FreeQ[{c
, d, e, f}, x] && ComplexFreeQ[f] && NeQ[d*e - c*f, 0]

Rule 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 4619

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

Rule 4677

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

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])

Rubi steps

\begin {align*} \int \left (a+b \sin ^{-1}(c x)\right )^{5/2} \, dx &=x \left (a+b \sin ^{-1}(c x)\right )^{5/2}-\frac {1}{2} (5 b c) \int \frac {x \left (a+b \sin ^{-1}(c x)\right )^{3/2}}{\sqrt {1-c^2 x^2}} \, dx\\ &=\frac {5 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^{3/2}}{2 c}+x \left (a+b \sin ^{-1}(c x)\right )^{5/2}-\frac {1}{4} \left (15 b^2\right ) \int \sqrt {a+b \sin ^{-1}(c x)} \, dx\\ &=-\frac {15}{4} b^2 x \sqrt {a+b \sin ^{-1}(c x)}+\frac {5 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^{3/2}}{2 c}+x \left (a+b \sin ^{-1}(c x)\right )^{5/2}+\frac {1}{8} \left (15 b^3 c\right ) \int \frac {x}{\sqrt {1-c^2 x^2} \sqrt {a+b \sin ^{-1}(c x)}} \, dx\\ &=-\frac {15}{4} b^2 x \sqrt {a+b \sin ^{-1}(c x)}+\frac {5 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^{3/2}}{2 c}+x \left (a+b \sin ^{-1}(c x)\right )^{5/2}+\frac {\left (15 b^3\right ) \operatorname {Subst}\left (\int \frac {\sin (x)}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c x)\right )}{8 c}\\ &=-\frac {15}{4} b^2 x \sqrt {a+b \sin ^{-1}(c x)}+\frac {5 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^{3/2}}{2 c}+x \left (a+b \sin ^{-1}(c x)\right )^{5/2}+\frac {\left (15 b^3 \cos \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {a}{b}+x\right )}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c x)\right )}{8 c}-\frac {\left (15 b^3 \sin \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {a}{b}+x\right )}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c x)\right )}{8 c}\\ &=-\frac {15}{4} b^2 x \sqrt {a+b \sin ^{-1}(c x)}+\frac {5 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^{3/2}}{2 c}+x \left (a+b \sin ^{-1}(c x)\right )^{5/2}+\frac {\left (15 b^2 \cos \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \sin \left (\frac {x^2}{b}\right ) \, dx,x,\sqrt {a+b \sin ^{-1}(c x)}\right )}{4 c}-\frac {\left (15 b^2 \sin \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \cos \left (\frac {x^2}{b}\right ) \, dx,x,\sqrt {a+b \sin ^{-1}(c x)}\right )}{4 c}\\ &=-\frac {15}{4} b^2 x \sqrt {a+b \sin ^{-1}(c x)}+\frac {5 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^{3/2}}{2 c}+x \left (a+b \sin ^{-1}(c x)\right )^{5/2}+\frac {15 b^{5/2} \sqrt {\frac {\pi }{2}} \cos \left (\frac {a}{b}\right ) S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {b}}\right )}{4 c}-\frac {15 b^{5/2} \sqrt {\frac {\pi }{2}} C\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {b}}\right ) \sin \left (\frac {a}{b}\right )}{4 c}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 3.19, size = 379, normalized size = 2.12 \[ \frac {e^{-\frac {i a}{b}} \left (\frac {i \sqrt {\frac {\pi }{2}} \left (4 a^2+15 b^2\right ) \left (-1+e^{\frac {2 i a}{b}}\right ) \sqrt {a+b \sin ^{-1}(c x)} C\left (\sqrt {\frac {1}{b}} \sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c x)}\right )}{\sqrt {\frac {1}{b}}}+\frac {\sqrt {\frac {\pi }{2}} \left (4 a^2+15 b^2\right ) \left (1+e^{\frac {2 i a}{b}}\right ) \sqrt {a+b \sin ^{-1}(c x)} S\left (\sqrt {\frac {1}{b}} \sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c x)}\right )}{\sqrt {\frac {1}{b}}}+2 b \left (2 a^2 \sqrt {-\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}} \Gamma \left (\frac {3}{2},-\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )+2 a^2 e^{\frac {2 i a}{b}} \sqrt {\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}} \Gamma \left (\frac {3}{2},\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )+e^{\frac {i a}{b}} \left (a+b \sin ^{-1}(c x)\right ) \left (2 \sin ^{-1}(c x) \left (4 a c x+5 b \sqrt {1-c^2 x^2}\right )+10 a \sqrt {1-c^2 x^2}-15 b c x+4 b c x \sin ^{-1}(c x)^2\right )\right )\right )}{8 c \sqrt {a+b \sin ^{-1}(c x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((I*(4*a^2 + 15*b^2)*(-1 + E^(((2*I)*a)/b))*Sqrt[Pi/2]*Sqrt[a + b*ArcSin[c*x]]*FresnelC[Sqrt[b^(-1)]*Sqrt[2/Pi
]*Sqrt[a + b*ArcSin[c*x]]])/Sqrt[b^(-1)] + ((4*a^2 + 15*b^2)*(1 + E^(((2*I)*a)/b))*Sqrt[Pi/2]*Sqrt[a + b*ArcSi
n[c*x]]*FresnelS[Sqrt[b^(-1)]*Sqrt[2/Pi]*Sqrt[a + b*ArcSin[c*x]]])/Sqrt[b^(-1)] + 2*b*(E^((I*a)/b)*(a + b*ArcS
in[c*x])*(-15*b*c*x + 10*a*Sqrt[1 - c^2*x^2] + 2*(4*a*c*x + 5*b*Sqrt[1 - c^2*x^2])*ArcSin[c*x] + 4*b*c*x*ArcSi
n[c*x]^2) + 2*a^2*Sqrt[((-I)*(a + b*ArcSin[c*x]))/b]*Gamma[3/2, ((-I)*(a + b*ArcSin[c*x]))/b] + 2*a^2*E^(((2*I
)*a)/b)*Sqrt[(I*(a + b*ArcSin[c*x]))/b]*Gamma[3/2, (I*(a + b*ArcSin[c*x]))/b]))/(8*c*E^((I*a)/b)*Sqrt[a + b*Ar
cSin[c*x]])

________________________________________________________________________________________

fricas [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((a+b*arcsin(c*x))^(5/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

________________________________________________________________________________________

giac [C]  time = 4.07, size = 1179, normalized size = 6.59 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*sqrt(pi)*a^3*b^3*erf(-1/2*I*sqrt(2)*sqrt(b*arcsin(c*x) + a)/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b*arcs
in(c*x) + a)*sqrt(abs(b))/b)*e^(I*a/b)/((I*b^4/sqrt(abs(b)) + b^3*sqrt(abs(b)))*c) + 1/2*sqrt(2)*sqrt(pi)*a^3*
b^3*erf(1/2*I*sqrt(2)*sqrt(b*arcsin(c*x) + a)/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b*arcsin(c*x) + a)*sqrt(abs(b))/
b)*e^(-I*a/b)/((-I*b^4/sqrt(abs(b)) + b^3*sqrt(abs(b)))*c) + 3/2*I*sqrt(2)*sqrt(pi)*a^2*b^3*erf(-1/2*I*sqrt(2)
*sqrt(b*arcsin(c*x) + a)/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b*arcsin(c*x) + a)*sqrt(abs(b))/b)*e^(I*a/b)/((I*b^3/
sqrt(abs(b)) + b^2*sqrt(abs(b)))*c) - 3/2*I*sqrt(2)*sqrt(pi)*a^2*b^3*erf(1/2*I*sqrt(2)*sqrt(b*arcsin(c*x) + a)
/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b*arcsin(c*x) + a)*sqrt(abs(b))/b)*e^(-I*a/b)/((-I*b^3/sqrt(abs(b)) + b^2*sqr
t(abs(b)))*c) - 3/2*I*sqrt(2)*sqrt(pi)*a^2*b^2*erf(-1/2*I*sqrt(2)*sqrt(b*arcsin(c*x) + a)/sqrt(abs(b)) - 1/2*s
qrt(2)*sqrt(b*arcsin(c*x) + a)*sqrt(abs(b))/b)*e^(I*a/b)/((I*b^2/sqrt(abs(b)) + b*sqrt(abs(b)))*c) - 15/16*I*s
qrt(2)*sqrt(pi)*b^4*erf(-1/2*I*sqrt(2)*sqrt(b*arcsin(c*x) + a)/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b*arcsin(c*x) +
a)*sqrt(abs(b))/b)*e^(I*a/b)/((I*b^2/sqrt(abs(b)) + b*sqrt(abs(b)))*c) + 3/2*I*sqrt(2)*sqrt(pi)*a^2*b^2*erf(1
/2*I*sqrt(2)*sqrt(b*arcsin(c*x) + a)/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b*arcsin(c*x) + a)*sqrt(abs(b))/b)*e^(-I*
a/b)/((-I*b^2/sqrt(abs(b)) + b*sqrt(abs(b)))*c) + 15/16*I*sqrt(2)*sqrt(pi)*b^4*erf(1/2*I*sqrt(2)*sqrt(b*arcsin
(c*x) + a)/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b*arcsin(c*x) + a)*sqrt(abs(b))/b)*e^(-I*a/b)/((-I*b^2/sqrt(abs(b))
+ b*sqrt(abs(b)))*c) - 1/2*I*sqrt(b*arcsin(c*x) + a)*b^2*arcsin(c*x)^2*e^(I*arcsin(c*x))/c + 1/2*I*sqrt(b*arc
sin(c*x) + a)*b^2*arcsin(c*x)^2*e^(-I*arcsin(c*x))/c - sqrt(pi)*a^3*b*erf(-1/2*I*sqrt(2)*sqrt(b*arcsin(c*x) +
a)/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b*arcsin(c*x) + a)*sqrt(abs(b))/b)*e^(I*a/b)/((I*sqrt(2)*b^2/sqrt(abs(b)) +
sqrt(2)*b*sqrt(abs(b)))*c) - sqrt(pi)*a^3*b*erf(1/2*I*sqrt(2)*sqrt(b*arcsin(c*x) + a)/sqrt(abs(b)) - 1/2*sqrt
(2)*sqrt(b*arcsin(c*x) + a)*sqrt(abs(b))/b)*e^(-I*a/b)/((-I*sqrt(2)*b^2/sqrt(abs(b)) + sqrt(2)*b*sqrt(abs(b)))
*c) - I*sqrt(b*arcsin(c*x) + a)*a*b*arcsin(c*x)*e^(I*arcsin(c*x))/c + 5/4*sqrt(b*arcsin(c*x) + a)*b^2*arcsin(c
*x)*e^(I*arcsin(c*x))/c + I*sqrt(b*arcsin(c*x) + a)*a*b*arcsin(c*x)*e^(-I*arcsin(c*x))/c + 5/4*sqrt(b*arcsin(c
*x) + a)*b^2*arcsin(c*x)*e^(-I*arcsin(c*x))/c - 1/2*I*sqrt(b*arcsin(c*x) + a)*a^2*e^(I*arcsin(c*x))/c + 5/4*sq
rt(b*arcsin(c*x) + a)*a*b*e^(I*arcsin(c*x))/c + 15/8*I*sqrt(b*arcsin(c*x) + a)*b^2*e^(I*arcsin(c*x))/c + 1/2*I
*sqrt(b*arcsin(c*x) + a)*a^2*e^(-I*arcsin(c*x))/c + 5/4*sqrt(b*arcsin(c*x) + a)*a*b*e^(-I*arcsin(c*x))/c - 15/
8*I*sqrt(b*arcsin(c*x) + a)*b^2*e^(-I*arcsin(c*x))/c

________________________________________________________________________________________

maple [B]  time = 0.10, size = 393, normalized size = 2.20 \[ \frac {15 \sqrt {2}\, \sqrt {\pi }\, \sqrt {a +b \arcsin \left (c x \right )}\, \cos \left (\frac {a}{b}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (c x \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {\frac {1}{b}}\, b^{3}-15 \sqrt {2}\, \sqrt {\pi }\, \sqrt {a +b \arcsin \left (c x \right )}\, \sin \left (\frac {a}{b}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (c x \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {\frac {1}{b}}\, b^{3}+8 \arcsin \left (c x \right )^{3} \sin \left (\frac {a +b \arcsin \left (c x \right )}{b}-\frac {a}{b}\right ) b^{3}+24 \arcsin \left (c x \right )^{2} \sin \left (\frac {a +b \arcsin \left (c x \right )}{b}-\frac {a}{b}\right ) a \,b^{2}+20 \arcsin \left (c x \right )^{2} \cos \left (\frac {a +b \arcsin \left (c x \right )}{b}-\frac {a}{b}\right ) b^{3}+24 \arcsin \left (c x \right ) \sin \left (\frac {a +b \arcsin \left (c x \right )}{b}-\frac {a}{b}\right ) a^{2} b -30 \arcsin \left (c x \right ) \sin \left (\frac {a +b \arcsin \left (c x \right )}{b}-\frac {a}{b}\right ) b^{3}+40 \arcsin \left (c x \right ) \cos \left (\frac {a +b \arcsin \left (c x \right )}{b}-\frac {a}{b}\right ) a \,b^{2}+8 \sin \left (\frac {a +b \arcsin \left (c x \right )}{b}-\frac {a}{b}\right ) a^{3}-30 \sin \left (\frac {a +b \arcsin \left (c x \right )}{b}-\frac {a}{b}\right ) a \,b^{2}+20 \cos \left (\frac {a +b \arcsin \left (c x \right )}{b}-\frac {a}{b}\right ) a^{2} b}{8 c \sqrt {a +b \arcsin \left (c x \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/c/(a+b*arcsin(c*x))^(1/2)*(15*2^(1/2)*Pi^(1/2)*(a+b*arcsin(c*x))^(1/2)*cos(a/b)*FresnelS(2^(1/2)/Pi^(1/2)/
(1/b)^(1/2)*(a+b*arcsin(c*x))^(1/2)/b)*(1/b)^(1/2)*b^3-15*2^(1/2)*Pi^(1/2)*(a+b*arcsin(c*x))^(1/2)*sin(a/b)*Fr
esnelC(2^(1/2)/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(c*x))^(1/2)/b)*(1/b)^(1/2)*b^3+8*arcsin(c*x)^3*sin((a+b*arcsin
(c*x))/b-a/b)*b^3+24*arcsin(c*x)^2*sin((a+b*arcsin(c*x))/b-a/b)*a*b^2+20*arcsin(c*x)^2*cos((a+b*arcsin(c*x))/b
-a/b)*b^3+24*arcsin(c*x)*sin((a+b*arcsin(c*x))/b-a/b)*a^2*b-30*arcsin(c*x)*sin((a+b*arcsin(c*x))/b-a/b)*b^3+40
*arcsin(c*x)*cos((a+b*arcsin(c*x))/b-a/b)*a*b^2+8*sin((a+b*arcsin(c*x))/b-a/b)*a^3-30*sin((a+b*arcsin(c*x))/b-
a/b)*a*b^2+20*cos((a+b*arcsin(c*x))/b-a/b)*a^2*b)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c*x))^(5/2),x)

[Out]

int((a + b*asin(c*x))^(5/2), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asin(c*x))**(5/2), x)

________________________________________________________________________________________