3.2.61 \(\int (a+b \text {ArcSin}(c+d x))^{5/2} \, dx\) [161]

Optimal. Leaf size=204 \[ -\frac {15 b^2 (c+d x) \sqrt {a+b \text {ArcSin}(c+d x)}}{4 d}+\frac {5 b \sqrt {1-(c+d x)^2} (a+b \text {ArcSin}(c+d x))^{3/2}}{2 d}+\frac {(c+d x) (a+b \text {ArcSin}(c+d x))^{5/2}}{d}+\frac {15 b^{5/2} \sqrt {\frac {\pi }{2}} \cos \left (\frac {a}{b}\right ) S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \text {ArcSin}(c+d x)}}{\sqrt {b}}\right )}{4 d}-\frac {15 b^{5/2} \sqrt {\frac {\pi }{2}} \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \text {ArcSin}(c+d x)}}{\sqrt {b}}\right ) \sin \left (\frac {a}{b}\right )}{4 d} \]

[Out]

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

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Rubi [A]
time = 0.27, antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {4887, 4715, 4767, 4809, 3387, 3386, 3432, 3385, 3433} \begin {gather*} -\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 \text {ArcSin}(c+d x)}}{\sqrt {b}}\right )}{4 d}+\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 \text {ArcSin}(c+d x)}}{\sqrt {b}}\right )}{4 d}-\frac {15 b^2 (c+d x) \sqrt {a+b \text {ArcSin}(c+d x)}}{4 d}+\frac {5 b \sqrt {1-(c+d x)^2} (a+b \text {ArcSin}(c+d x))^{3/2}}{2 d}+\frac {(c+d x) (a+b \text {ArcSin}(c+d x))^{5/2}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c + d*x])^(5/2),x]

[Out]

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

Rule 3385

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 3386

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 3387

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 3432

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

Rule 3433

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

Rule 4715

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 4767

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/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^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 4809

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

Rule 4887

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcSin[x])^n, x],
x, c + d*x], x] /; FreeQ[{a, b, c, d, n}, x]

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 2.10, size = 432, normalized size = 2.12 \begin {gather*} \frac {e^{-\frac {i a}{b}} \left (\frac {i \left (4 a^2+15 b^2\right ) \left (-1+e^{\frac {2 i a}{b}}\right ) \sqrt {\frac {\pi }{2}} \sqrt {a+b \text {ArcSin}(c+d x)} \text {FresnelC}\left (\sqrt {\frac {1}{b}} \sqrt {\frac {2}{\pi }} \sqrt {a+b \text {ArcSin}(c+d x)}\right )}{\sqrt {\frac {1}{b}}}+\frac {\left (4 a^2+15 b^2\right ) \left (1+e^{\frac {2 i a}{b}}\right ) \sqrt {\frac {\pi }{2}} \sqrt {a+b \text {ArcSin}(c+d x)} S\left (\sqrt {\frac {1}{b}} \sqrt {\frac {2}{\pi }} \sqrt {a+b \text {ArcSin}(c+d x)}\right )}{\sqrt {\frac {1}{b}}}+2 b \left (e^{\frac {i a}{b}} (a+b \text {ArcSin}(c+d x)) \left (-15 b (c+d x)+10 a \sqrt {1-c^2-2 c d x-d^2 x^2}+2 \left (4 a (c+d x)+5 b \sqrt {1-c^2-2 c d x-d^2 x^2}\right ) \text {ArcSin}(c+d x)+4 b (c+d x) \text {ArcSin}(c+d x)^2\right )+2 a^2 \sqrt {-\frac {i (a+b \text {ArcSin}(c+d x))}{b}} \text {Gamma}\left (\frac {3}{2},-\frac {i (a+b \text {ArcSin}(c+d x))}{b}\right )+2 a^2 e^{\frac {2 i a}{b}} \sqrt {\frac {i (a+b \text {ArcSin}(c+d x))}{b}} \text {Gamma}\left (\frac {3}{2},\frac {i (a+b \text {ArcSin}(c+d x))}{b}\right )\right )\right )}{8 d \sqrt {a+b \text {ArcSin}(c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c + d*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 + d*x]]*FresnelC[Sqrt[b^(-1)]*Sqrt[
2/Pi]*Sqrt[a + b*ArcSin[c + d*x]]])/Sqrt[b^(-1)] + ((4*a^2 + 15*b^2)*(1 + E^(((2*I)*a)/b))*Sqrt[Pi/2]*Sqrt[a +
b*ArcSin[c + d*x]]*FresnelS[Sqrt[b^(-1)]*Sqrt[2/Pi]*Sqrt[a + b*ArcSin[c + d*x]]])/Sqrt[b^(-1)] + 2*b*(E^((I*a
)/b)*(a + b*ArcSin[c + d*x])*(-15*b*(c + d*x) + 10*a*Sqrt[1 - c^2 - 2*c*d*x - d^2*x^2] + 2*(4*a*(c + d*x) + 5*
b*Sqrt[1 - c^2 - 2*c*d*x - d^2*x^2])*ArcSin[c + d*x] + 4*b*(c + d*x)*ArcSin[c + d*x]^2) + 2*a^2*Sqrt[((-I)*(a
+ b*ArcSin[c + d*x]))/b]*Gamma[3/2, ((-I)*(a + b*ArcSin[c + d*x]))/b] + 2*a^2*E^(((2*I)*a)/b)*Sqrt[(I*(a + b*A
rcSin[c + d*x]))/b]*Gamma[3/2, (I*(a + b*ArcSin[c + d*x]))/b]))/(8*d*E^((I*a)/b)*Sqrt[a + b*ArcSin[c + d*x]])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(440\) vs. \(2(164)=328\).
time = 0.30, size = 441, normalized size = 2.16

method	result	size
default	\(-\frac {15 \sqrt {a +b \arcsin \left (d x +c \right )}\, \cos \left (\frac {a}{b}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right ) \sqrt {-\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {2}\, b^{3}+15 \sqrt {a +b \arcsin \left (d x +c \right )}\, \sin \left (\frac {a}{b}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right ) \sqrt {-\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {2}\, b^{3}+8 \arcsin \left (d x +c \right )^{3} \sin \left (-\frac {a +b \arcsin \left (d x +c \right )}{b}+\frac {a}{b}\right ) b^{3}+24 \arcsin \left (d x +c \right )^{2} \sin \left (-\frac {a +b \arcsin \left (d x +c \right )}{b}+\frac {a}{b}\right ) a \,b^{2}-20 \arcsin \left (d x +c \right )^{2} \cos \left (-\frac {a +b \arcsin \left (d x +c \right )}{b}+\frac {a}{b}\right ) b^{3}+24 \arcsin \left (d x +c \right ) \sin \left (-\frac {a +b \arcsin \left (d x +c \right )}{b}+\frac {a}{b}\right ) a^{2} b -30 \arcsin \left (d x +c \right ) \sin \left (-\frac {a +b \arcsin \left (d x +c \right )}{b}+\frac {a}{b}\right ) b^{3}-40 \arcsin \left (d x +c \right ) \cos \left (-\frac {a +b \arcsin \left (d x +c \right )}{b}+\frac {a}{b}\right ) a \,b^{2}+8 \sin \left (-\frac {a +b \arcsin \left (d x +c \right )}{b}+\frac {a}{b}\right ) a^{3}-30 \sin \left (-\frac {a +b \arcsin \left (d x +c \right )}{b}+\frac {a}{b}\right ) a \,b^{2}-20 \cos \left (-\frac {a +b \arcsin \left (d x +c \right )}{b}+\frac {a}{b}\right ) a^{2} b}{8 d \sqrt {a +b \arcsin \left (d x +c \right )}}\)	\(441\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

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

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C] Result contains complex when optimal does not.
time = 1.19, size = 1279, normalized size = 6.27 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x+c))^(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(d*x + c) + a)/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b*
arcsin(d*x + c) + a)*sqrt(abs(b))/b)*e^(I*a/b)/((I*b^4/sqrt(abs(b)) + b^3*sqrt(abs(b)))*d) + 1/2*sqrt(2)*sqrt(
pi)*a^3*b^3*erf(1/2*I*sqrt(2)*sqrt(b*arcsin(d*x + c) + a)/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b*arcsin(d*x + c) +
a)*sqrt(abs(b))/b)*e^(-I*a/b)/((-I*b^4/sqrt(abs(b)) + b^3*sqrt(abs(b)))*d) + 3/2*I*sqrt(2)*sqrt(pi)*a^2*b^3*er
f(-1/2*I*sqrt(2)*sqrt(b*arcsin(d*x + c) + a)/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b*arcsin(d*x + c) + a)*sqrt(abs(b
))/b)*e^(I*a/b)/((I*b^3/sqrt(abs(b)) + b^2*sqrt(abs(b)))*d) - 3/2*I*sqrt(2)*sqrt(pi)*a^2*b^3*erf(1/2*I*sqrt(2)
*sqrt(b*arcsin(d*x + c) + a)/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b*arcsin(d*x + c) + a)*sqrt(abs(b))/b)*e^(-I*a/b)
/((-I*b^3/sqrt(abs(b)) + b^2*sqrt(abs(b)))*d) - 3/2*I*sqrt(2)*sqrt(pi)*a^2*b^2*erf(-1/2*I*sqrt(2)*sqrt(b*arcsi
n(d*x + c) + a)/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b*arcsin(d*x + c) + a)*sqrt(abs(b))/b)*e^(I*a/b)/((I*b^2/sqrt(
abs(b)) + b*sqrt(abs(b)))*d) - 15/16*I*sqrt(2)*sqrt(pi)*b^4*erf(-1/2*I*sqrt(2)*sqrt(b*arcsin(d*x + c) + a)/sqr
t(abs(b)) - 1/2*sqrt(2)*sqrt(b*arcsin(d*x + c) + a)*sqrt(abs(b))/b)*e^(I*a/b)/((I*b^2/sqrt(abs(b)) + b*sqrt(ab
s(b)))*d) + 3/2*I*sqrt(2)*sqrt(pi)*a^2*b^2*erf(1/2*I*sqrt(2)*sqrt(b*arcsin(d*x + c) + a)/sqrt(abs(b)) - 1/2*sq
rt(2)*sqrt(b*arcsin(d*x + c) + a)*sqrt(abs(b))/b)*e^(-I*a/b)/((-I*b^2/sqrt(abs(b)) + b*sqrt(abs(b)))*d) + 15/1
6*I*sqrt(2)*sqrt(pi)*b^4*erf(1/2*I*sqrt(2)*sqrt(b*arcsin(d*x + c) + a)/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b*arcsi
n(d*x + c) + a)*sqrt(abs(b))/b)*e^(-I*a/b)/((-I*b^2/sqrt(abs(b)) + b*sqrt(abs(b)))*d) - 1/2*I*sqrt(b*arcsin(d*
x + c) + a)*b^2*arcsin(d*x + c)^2*e^(I*arcsin(d*x + c))/d + 1/2*I*sqrt(b*arcsin(d*x + c) + a)*b^2*arcsin(d*x +
c)^2*e^(-I*arcsin(d*x + c))/d - sqrt(pi)*a^3*b*erf(-1/2*I*sqrt(2)*sqrt(b*arcsin(d*x + c) + a)/sqrt(abs(b)) -
1/2*sqrt(2)*sqrt(b*arcsin(d*x + c) + a)*sqrt(abs(b))/b)*e^(I*a/b)/((I*sqrt(2)*b^2/sqrt(abs(b)) + sqrt(2)*b*sqr
t(abs(b)))*d) - sqrt(pi)*a^3*b*erf(1/2*I*sqrt(2)*sqrt(b*arcsin(d*x + c) + a)/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b
*arcsin(d*x + c) + a)*sqrt(abs(b))/b)*e^(-I*a/b)/((-I*sqrt(2)*b^2/sqrt(abs(b)) + sqrt(2)*b*sqrt(abs(b)))*d) -
I*sqrt(b*arcsin(d*x + c) + a)*a*b*arcsin(d*x + c)*e^(I*arcsin(d*x + c))/d + 5/4*sqrt(b*arcsin(d*x + c) + a)*b^
2*arcsin(d*x + c)*e^(I*arcsin(d*x + c))/d + I*sqrt(b*arcsin(d*x + c) + a)*a*b*arcsin(d*x + c)*e^(-I*arcsin(d*x
+ c))/d + 5/4*sqrt(b*arcsin(d*x + c) + a)*b^2*arcsin(d*x + c)*e^(-I*arcsin(d*x + c))/d - 1/2*I*sqrt(b*arcsin(
d*x + c) + a)*a^2*e^(I*arcsin(d*x + c))/d + 5/4*sqrt(b*arcsin(d*x + c) + a)*a*b*e^(I*arcsin(d*x + c))/d + 15/8
*I*sqrt(b*arcsin(d*x + c) + a)*b^2*e^(I*arcsin(d*x + c))/d + 1/2*I*sqrt(b*arcsin(d*x + c) + a)*a^2*e^(-I*arcsi
n(d*x + c))/d + 5/4*sqrt(b*arcsin(d*x + c) + a)*a*b*e^(-I*arcsin(d*x + c))/d - 15/8*I*sqrt(b*arcsin(d*x + c) +
a)*b^2*e^(-I*arcsin(d*x + c))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c + d*x))^(5/2),x)

[Out]

int((a + b*asin(c + d*x))^(5/2), x)

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