\(\int \frac {x}{(a+b \arcsin (c+d x))^{7/2}} \, dx\) [170]

Optimal result

Integrand size = 16, antiderivative size = 468 \[ \int \frac {x}{(a+b \arcsin (c+d x))^{7/2}} \, dx=\frac {2 c \sqrt {1-(c+d x)^2}}{5 b d^2 (a+b \arcsin (c+d x))^{5/2}}-\frac {2 (c+d x) \sqrt {1-(c+d x)^2}}{5 b d^2 (a+b \arcsin (c+d x))^{5/2}}-\frac {4}{15 b^2 d^2 (a+b \arcsin (c+d x))^{3/2}}-\frac {4 c (c+d x)}{15 b^2 d^2 (a+b \arcsin (c+d x))^{3/2}}+\frac {8 (c+d x)^2}{15 b^2 d^2 (a+b \arcsin (c+d x))^{3/2}}-\frac {8 c \sqrt {1-(c+d x)^2}}{15 b^3 d^2 \sqrt {a+b \arcsin (c+d x)}}+\frac {32 (c+d x) \sqrt {1-(c+d x)^2}}{15 b^3 d^2 \sqrt {a+b \arcsin (c+d x)}}-\frac {32 \sqrt {\pi } \cos \left (\frac {2 a}{b}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {a+b \arcsin (c+d x)}}{\sqrt {b} \sqrt {\pi }}\right )}{15 b^{7/2} d^2}-\frac {8 c \sqrt {2 \pi } \cos \left (\frac {a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arcsin (c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d^2}+\frac {8 c \sqrt {2 \pi } \operatorname {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arcsin (c+d x)}}{\sqrt {b}}\right ) \sin \left (\frac {a}{b}\right )}{15 b^{7/2} d^2}-\frac {32 \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {a+b \arcsin (c+d x)}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {2 a}{b}\right )}{15 b^{7/2} d^2} \]

[Out]

-4/15/b^2/d^2/(a+b*arcsin(d*x+c))^(3/2)-4/15*c*(d*x+c)/b^2/d^2/(a+b*arcsin(d*x+c))^(3/2)+8/15*(d*x+c)^2/b^2/d^
2/(a+b*arcsin(d*x+c))^(3/2)-32/15*cos(2*a/b)*FresnelC(2*(a+b*arcsin(d*x+c))^(1/2)/b^(1/2)/Pi^(1/2))*Pi^(1/2)/b
^(7/2)/d^2-32/15*FresnelS(2*(a+b*arcsin(d*x+c))^(1/2)/b^(1/2)/Pi^(1/2))*sin(2*a/b)*Pi^(1/2)/b^(7/2)/d^2-8/15*c
*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)/b^(7/2)/d^2+8/15*c*Fre
snelC(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)/b^(7/2)/d^2+2/5*c*(1-(d*x+
c)^2)^(1/2)/b/d^2/(a+b*arcsin(d*x+c))^(5/2)-2/5*(d*x+c)*(1-(d*x+c)^2)^(1/2)/b/d^2/(a+b*arcsin(d*x+c))^(5/2)-8/
15*c*(1-(d*x+c)^2)^(1/2)/b^3/d^2/(a+b*arcsin(d*x+c))^(1/2)+32/15*(d*x+c)*(1-(d*x+c)^2)^(1/2)/b^3/d^2/(a+b*arcs
in(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 468, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.812, Rules used = {4889, 4829, 4717, 4807, 4809, 3387, 3386, 3432, 3385, 3433, 4729, 4727, 4737} \[ \int \frac {x}{(a+b \arcsin (c+d x))^{7/2}} \, dx=\frac {8 \sqrt {2 \pi } c \sin \left (\frac {a}{b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arcsin (c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d^2}-\frac {32 \sqrt {\pi } \cos \left (\frac {2 a}{b}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {a+b \arcsin (c+d x)}}{\sqrt {b} \sqrt {\pi }}\right )}{15 b^{7/2} d^2}-\frac {32 \sqrt {\pi } \sin \left (\frac {2 a}{b}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {a+b \arcsin (c+d x)}}{\sqrt {b} \sqrt {\pi }}\right )}{15 b^{7/2} d^2}-\frac {8 \sqrt {2 \pi } c \cos \left (\frac {a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arcsin (c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d^2}+\frac {32 \sqrt {1-(c+d x)^2} (c+d x)}{15 b^3 d^2 \sqrt {a+b \arcsin (c+d x)}}-\frac {8 c \sqrt {1-(c+d x)^2}}{15 b^3 d^2 \sqrt {a+b \arcsin (c+d x)}}+\frac {8 (c+d x)^2}{15 b^2 d^2 (a+b \arcsin (c+d x))^{3/2}}-\frac {4 c (c+d x)}{15 b^2 d^2 (a+b \arcsin (c+d x))^{3/2}}-\frac {4}{15 b^2 d^2 (a+b \arcsin (c+d x))^{3/2}}-\frac {2 \sqrt {1-(c+d x)^2} (c+d x)}{5 b d^2 (a+b \arcsin (c+d x))^{5/2}}+\frac {2 c \sqrt {1-(c+d x)^2}}{5 b d^2 (a+b \arcsin (c+d x))^{5/2}} \]

[In]

Int[x/(a + b*ArcSin[c + d*x])^(7/2),x]

[Out]

(2*c*Sqrt[1 - (c + d*x)^2])/(5*b*d^2*(a + b*ArcSin[c + d*x])^(5/2)) - (2*(c + d*x)*Sqrt[1 - (c + d*x)^2])/(5*b
*d^2*(a + b*ArcSin[c + d*x])^(5/2)) - 4/(15*b^2*d^2*(a + b*ArcSin[c + d*x])^(3/2)) - (4*c*(c + d*x))/(15*b^2*d
^2*(a + b*ArcSin[c + d*x])^(3/2)) + (8*(c + d*x)^2)/(15*b^2*d^2*(a + b*ArcSin[c + d*x])^(3/2)) - (8*c*Sqrt[1 -
(c + d*x)^2])/(15*b^3*d^2*Sqrt[a + b*ArcSin[c + d*x]]) + (32*(c + d*x)*Sqrt[1 - (c + d*x)^2])/(15*b^3*d^2*Sqr
t[a + b*ArcSin[c + d*x]]) - (32*Sqrt[Pi]*Cos[(2*a)/b]*FresnelC[(2*Sqrt[a + b*ArcSin[c + d*x]])/(Sqrt[b]*Sqrt[P
i])])/(15*b^(7/2)*d^2) - (8*c*Sqrt[2*Pi]*Cos[a/b]*FresnelS[(Sqrt[2/Pi]*Sqrt[a + b*ArcSin[c + d*x]])/Sqrt[b]])/
(15*b^(7/2)*d^2) + (8*c*Sqrt[2*Pi]*FresnelC[(Sqrt[2/Pi]*Sqrt[a + b*ArcSin[c + d*x]])/Sqrt[b]]*Sin[a/b])/(15*b^
(7/2)*d^2) - (32*Sqrt[Pi]*FresnelS[(2*Sqrt[a + b*ArcSin[c + d*x]])/(Sqrt[b]*Sqrt[Pi])]*Sin[(2*a)/b])/(15*b^(7/
2)*d^2)

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 4717

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

Rule 4727

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

Rule 4729

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

Rule 4737

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

Rule 4807

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[
((f*x)^m/(b*c*(n + 1)))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSin[c*x])^(n + 1), x] - Dist[f*(m/(b
*c*(n + 1)))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*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] && LtQ[n, -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 4829

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((d_) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(d + e
*x)^m*(a + b*ArcSin[c*x])^n, x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[m, 0] && LtQ[n, -1]

Rule 4889

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + f*(x/d))^m*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {-\frac {c}{d}+\frac {x}{d}}{(a+b \arcsin (x))^{7/2}} \, dx,x,c+d x\right )}{d} \\ & = \frac {\text {Subst}\left (\int \left (-\frac {c}{d (a+b \arcsin (x))^{7/2}}+\frac {x}{d (a+b \arcsin (x))^{7/2}}\right ) \, dx,x,c+d x\right )}{d} \\ & = \frac {\text {Subst}\left (\int \frac {x}{(a+b \arcsin (x))^{7/2}} \, dx,x,c+d x\right )}{d^2}-\frac {c \text {Subst}\left (\int \frac {1}{(a+b \arcsin (x))^{7/2}} \, dx,x,c+d x\right )}{d^2} \\ & = \frac {2 c \sqrt {1-(c+d x)^2}}{5 b d^2 (a+b \arcsin (c+d x))^{5/2}}-\frac {2 (c+d x) \sqrt {1-(c+d x)^2}}{5 b d^2 (a+b \arcsin (c+d x))^{5/2}}+\frac {2 \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} (a+b \arcsin (x))^{5/2}} \, dx,x,c+d x\right )}{5 b d^2}-\frac {4 \text {Subst}\left (\int \frac {x^2}{\sqrt {1-x^2} (a+b \arcsin (x))^{5/2}} \, dx,x,c+d x\right )}{5 b d^2}+\frac {(2 c) \text {Subst}\left (\int \frac {x}{\sqrt {1-x^2} (a+b \arcsin (x))^{5/2}} \, dx,x,c+d x\right )}{5 b d^2} \\ & = \frac {2 c \sqrt {1-(c+d x)^2}}{5 b d^2 (a+b \arcsin (c+d x))^{5/2}}-\frac {2 (c+d x) \sqrt {1-(c+d x)^2}}{5 b d^2 (a+b \arcsin (c+d x))^{5/2}}-\frac {4}{15 b^2 d^2 (a+b \arcsin (c+d x))^{3/2}}-\frac {4 c (c+d x)}{15 b^2 d^2 (a+b \arcsin (c+d x))^{3/2}}+\frac {8 (c+d x)^2}{15 b^2 d^2 (a+b \arcsin (c+d x))^{3/2}}-\frac {16 \text {Subst}\left (\int \frac {x}{(a+b \arcsin (x))^{3/2}} \, dx,x,c+d x\right )}{15 b^2 d^2}+\frac {(4 c) \text {Subst}\left (\int \frac {1}{(a+b \arcsin (x))^{3/2}} \, dx,x,c+d x\right )}{15 b^2 d^2} \\ & = \frac {2 c \sqrt {1-(c+d x)^2}}{5 b d^2 (a+b \arcsin (c+d x))^{5/2}}-\frac {2 (c+d x) \sqrt {1-(c+d x)^2}}{5 b d^2 (a+b \arcsin (c+d x))^{5/2}}-\frac {4}{15 b^2 d^2 (a+b \arcsin (c+d x))^{3/2}}-\frac {4 c (c+d x)}{15 b^2 d^2 (a+b \arcsin (c+d x))^{3/2}}+\frac {8 (c+d x)^2}{15 b^2 d^2 (a+b \arcsin (c+d x))^{3/2}}-\frac {8 c \sqrt {1-(c+d x)^2}}{15 b^3 d^2 \sqrt {a+b \arcsin (c+d x)}}+\frac {32 (c+d x) \sqrt {1-(c+d x)^2}}{15 b^3 d^2 \sqrt {a+b \arcsin (c+d x)}}-\frac {32 \text {Subst}\left (\int \frac {\cos \left (\frac {2 a}{b}-\frac {2 x}{b}\right )}{\sqrt {x}} \, dx,x,a+b \arcsin (c+d x)\right )}{15 b^4 d^2}-\frac {(8 c) \text {Subst}\left (\int \frac {x}{\sqrt {1-x^2} \sqrt {a+b \arcsin (x)}} \, dx,x,c+d x\right )}{15 b^3 d^2} \\ & = \frac {2 c \sqrt {1-(c+d x)^2}}{5 b d^2 (a+b \arcsin (c+d x))^{5/2}}-\frac {2 (c+d x) \sqrt {1-(c+d x)^2}}{5 b d^2 (a+b \arcsin (c+d x))^{5/2}}-\frac {4}{15 b^2 d^2 (a+b \arcsin (c+d x))^{3/2}}-\frac {4 c (c+d x)}{15 b^2 d^2 (a+b \arcsin (c+d x))^{3/2}}+\frac {8 (c+d x)^2}{15 b^2 d^2 (a+b \arcsin (c+d x))^{3/2}}-\frac {8 c \sqrt {1-(c+d x)^2}}{15 b^3 d^2 \sqrt {a+b \arcsin (c+d x)}}+\frac {32 (c+d x) \sqrt {1-(c+d x)^2}}{15 b^3 d^2 \sqrt {a+b \arcsin (c+d x)}}+\frac {(8 c) \text {Subst}\left (\int \frac {\sin \left (\frac {a}{b}-\frac {x}{b}\right )}{\sqrt {x}} \, dx,x,a+b \arcsin (c+d x)\right )}{15 b^4 d^2}-\frac {\left (32 \cos \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {2 x}{b}\right )}{\sqrt {x}} \, dx,x,a+b \arcsin (c+d x)\right )}{15 b^4 d^2}-\frac {\left (32 \sin \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {2 x}{b}\right )}{\sqrt {x}} \, dx,x,a+b \arcsin (c+d x)\right )}{15 b^4 d^2} \\ & = \frac {2 c \sqrt {1-(c+d x)^2}}{5 b d^2 (a+b \arcsin (c+d x))^{5/2}}-\frac {2 (c+d x) \sqrt {1-(c+d x)^2}}{5 b d^2 (a+b \arcsin (c+d x))^{5/2}}-\frac {4}{15 b^2 d^2 (a+b \arcsin (c+d x))^{3/2}}-\frac {4 c (c+d x)}{15 b^2 d^2 (a+b \arcsin (c+d x))^{3/2}}+\frac {8 (c+d x)^2}{15 b^2 d^2 (a+b \arcsin (c+d x))^{3/2}}-\frac {8 c \sqrt {1-(c+d x)^2}}{15 b^3 d^2 \sqrt {a+b \arcsin (c+d x)}}+\frac {32 (c+d x) \sqrt {1-(c+d x)^2}}{15 b^3 d^2 \sqrt {a+b \arcsin (c+d x)}}-\frac {\left (8 c \cos \left (\frac {a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {x}{b}\right )}{\sqrt {x}} \, dx,x,a+b \arcsin (c+d x)\right )}{15 b^4 d^2}-\frac {\left (64 \cos \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \cos \left (\frac {2 x^2}{b}\right ) \, dx,x,\sqrt {a+b \arcsin (c+d x)}\right )}{15 b^4 d^2}+\frac {\left (8 c \sin \left (\frac {a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {x}{b}\right )}{\sqrt {x}} \, dx,x,a+b \arcsin (c+d x)\right )}{15 b^4 d^2}-\frac {\left (64 \sin \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \sin \left (\frac {2 x^2}{b}\right ) \, dx,x,\sqrt {a+b \arcsin (c+d x)}\right )}{15 b^4 d^2} \\ & = \frac {2 c \sqrt {1-(c+d x)^2}}{5 b d^2 (a+b \arcsin (c+d x))^{5/2}}-\frac {2 (c+d x) \sqrt {1-(c+d x)^2}}{5 b d^2 (a+b \arcsin (c+d x))^{5/2}}-\frac {4}{15 b^2 d^2 (a+b \arcsin (c+d x))^{3/2}}-\frac {4 c (c+d x)}{15 b^2 d^2 (a+b \arcsin (c+d x))^{3/2}}+\frac {8 (c+d x)^2}{15 b^2 d^2 (a+b \arcsin (c+d x))^{3/2}}-\frac {8 c \sqrt {1-(c+d x)^2}}{15 b^3 d^2 \sqrt {a+b \arcsin (c+d x)}}+\frac {32 (c+d x) \sqrt {1-(c+d x)^2}}{15 b^3 d^2 \sqrt {a+b \arcsin (c+d x)}}-\frac {32 \sqrt {\pi } \cos \left (\frac {2 a}{b}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {a+b \arcsin (c+d x)}}{\sqrt {b} \sqrt {\pi }}\right )}{15 b^{7/2} d^2}-\frac {32 \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {a+b \arcsin (c+d x)}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {2 a}{b}\right )}{15 b^{7/2} d^2}-\frac {\left (16 c \cos \left (\frac {a}{b}\right )\right ) \text {Subst}\left (\int \sin \left (\frac {x^2}{b}\right ) \, dx,x,\sqrt {a+b \arcsin (c+d x)}\right )}{15 b^4 d^2}+\frac {\left (16 c \sin \left (\frac {a}{b}\right )\right ) \text {Subst}\left (\int \cos \left (\frac {x^2}{b}\right ) \, dx,x,\sqrt {a+b \arcsin (c+d x)}\right )}{15 b^4 d^2} \\ & = \frac {2 c \sqrt {1-(c+d x)^2}}{5 b d^2 (a+b \arcsin (c+d x))^{5/2}}-\frac {2 (c+d x) \sqrt {1-(c+d x)^2}}{5 b d^2 (a+b \arcsin (c+d x))^{5/2}}-\frac {4}{15 b^2 d^2 (a+b \arcsin (c+d x))^{3/2}}-\frac {4 c (c+d x)}{15 b^2 d^2 (a+b \arcsin (c+d x))^{3/2}}+\frac {8 (c+d x)^2}{15 b^2 d^2 (a+b \arcsin (c+d x))^{3/2}}-\frac {8 c \sqrt {1-(c+d x)^2}}{15 b^3 d^2 \sqrt {a+b \arcsin (c+d x)}}+\frac {32 (c+d x) \sqrt {1-(c+d x)^2}}{15 b^3 d^2 \sqrt {a+b \arcsin (c+d x)}}-\frac {32 \sqrt {\pi } \cos \left (\frac {2 a}{b}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {a+b \arcsin (c+d x)}}{\sqrt {b} \sqrt {\pi }}\right )}{15 b^{7/2} d^2}-\frac {8 c \sqrt {2 \pi } \cos \left (\frac {a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arcsin (c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d^2}+\frac {8 c \sqrt {2 \pi } \operatorname {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arcsin (c+d x)}}{\sqrt {b}}\right ) \sin \left (\frac {a}{b}\right )}{15 b^{7/2} d^2}-\frac {32 \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {a+b \arcsin (c+d x)}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {2 a}{b}\right )}{15 b^{7/2} d^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.38 (sec) , antiderivative size = 521, normalized size of antiderivative = 1.11 \[ \int \frac {x}{(a+b \arcsin (c+d x))^{7/2}} \, dx=-\frac {4 a b^{3/2} c (c+d x)+8 a^2 \sqrt {b} c \sqrt {1-(c+d x)^2}-6 b^{5/2} c \sqrt {1-(c+d x)^2}+4 b^{5/2} c (c+d x) \arcsin (c+d x)+16 a b^{3/2} c \sqrt {1-(c+d x)^2} \arcsin (c+d x)+8 b^{5/2} c \sqrt {1-(c+d x)^2} \arcsin (c+d x)^2+4 a b^{3/2} \cos (2 \arcsin (c+d x))+4 b^{5/2} \arcsin (c+d x) \cos (2 \arcsin (c+d x))+32 \sqrt {\pi } (a+b \arcsin (c+d x))^{5/2} \cos \left (\frac {2 a}{b}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {a+b \arcsin (c+d x)}}{\sqrt {b} \sqrt {\pi }}\right )+8 c \sqrt {2 \pi } (a+b \arcsin (c+d x))^{5/2} \cos \left (\frac {a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arcsin (c+d x)}}{\sqrt {b}}\right )-8 c \sqrt {2 \pi } (a+b \arcsin (c+d x))^{5/2} \operatorname {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arcsin (c+d x)}}{\sqrt {b}}\right ) \sin \left (\frac {a}{b}\right )+32 \sqrt {\pi } (a+b \arcsin (c+d x))^{5/2} \operatorname {FresnelS}\left (\frac {2 \sqrt {a+b \arcsin (c+d x)}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {2 a}{b}\right )-16 a^2 \sqrt {b} \sin (2 \arcsin (c+d x))+3 b^{5/2} \sin (2 \arcsin (c+d x))-32 a b^{3/2} \arcsin (c+d x) \sin (2 \arcsin (c+d x))-16 b^{5/2} \arcsin (c+d x)^2 \sin (2 \arcsin (c+d x))}{15 b^{7/2} d^2 (a+b \arcsin (c+d x))^{5/2}} \]

[In]

Integrate[x/(a + b*ArcSin[c + d*x])^(7/2),x]

[Out]

-1/15*(4*a*b^(3/2)*c*(c + d*x) + 8*a^2*Sqrt[b]*c*Sqrt[1 - (c + d*x)^2] - 6*b^(5/2)*c*Sqrt[1 - (c + d*x)^2] + 4
*b^(5/2)*c*(c + d*x)*ArcSin[c + d*x] + 16*a*b^(3/2)*c*Sqrt[1 - (c + d*x)^2]*ArcSin[c + d*x] + 8*b^(5/2)*c*Sqrt
[1 - (c + d*x)^2]*ArcSin[c + d*x]^2 + 4*a*b^(3/2)*Cos[2*ArcSin[c + d*x]] + 4*b^(5/2)*ArcSin[c + d*x]*Cos[2*Arc
Sin[c + d*x]] + 32*Sqrt[Pi]*(a + b*ArcSin[c + d*x])^(5/2)*Cos[(2*a)/b]*FresnelC[(2*Sqrt[a + b*ArcSin[c + d*x]]
)/(Sqrt[b]*Sqrt[Pi])] + 8*c*Sqrt[2*Pi]*(a + b*ArcSin[c + d*x])^(5/2)*Cos[a/b]*FresnelS[(Sqrt[2/Pi]*Sqrt[a + b*
ArcSin[c + d*x]])/Sqrt[b]] - 8*c*Sqrt[2*Pi]*(a + b*ArcSin[c + d*x])^(5/2)*FresnelC[(Sqrt[2/Pi]*Sqrt[a + b*ArcS
in[c + d*x]])/Sqrt[b]]*Sin[a/b] + 32*Sqrt[Pi]*(a + b*ArcSin[c + d*x])^(5/2)*FresnelS[(2*Sqrt[a + b*ArcSin[c +
d*x]])/(Sqrt[b]*Sqrt[Pi])]*Sin[(2*a)/b] - 16*a^2*Sqrt[b]*Sin[2*ArcSin[c + d*x]] + 3*b^(5/2)*Sin[2*ArcSin[c + d
*x]] - 32*a*b^(3/2)*ArcSin[c + d*x]*Sin[2*ArcSin[c + d*x]] - 16*b^(5/2)*ArcSin[c + d*x]^2*Sin[2*ArcSin[c + d*x
]])/(b^(7/2)*d^2*(a + b*ArcSin[c + d*x])^(5/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1237\) vs. \(2(384)=768\).

Time = 1.18 (sec) , antiderivative size = 1238, normalized size of antiderivative = 2.65

method	result	size
default	\(\text {Expression too large to display}\)	\(1238\)

[In]

int(x/(a+b*arcsin(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

-1/15/d^2/b^3/(a+b*arcsin(d*x+c))^(5/2)*(-8*arcsin(d*x+c)^2*(-1/b)^(1/2)*Pi^(1/2)*2^(1/2)*(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)*b^2*c-8*arcsin(d*x+c)^2*(-1
/b)^(1/2)*Pi^(1/2)*2^(1/2)*(a+b*arcsin(d*x+c))^(1/2)*sin(a/b)*FresnelC(2^(1/2)/Pi^(1/2)/(-1/b)^(1/2)*(a+b*arcs
in(d*x+c))^(1/2)/b)*b^2*c-16*arcsin(d*x+c)*(-1/b)^(1/2)*Pi^(1/2)*2^(1/2)*(a+b*arcsin(d*x+c))^(1/2)*cos(a/b)*Fr
esnelS(2^(1/2)/Pi^(1/2)/(-1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*a*b*c-16*arcsin(d*x+c)*(-1/b)^(1/2)*Pi^(1/2)
*2^(1/2)*(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
)*a*b*c+32*arcsin(d*x+c)^2*(-1/b)^(1/2)*Pi^(1/2)*(a+b*arcsin(d*x+c))^(1/2)*cos(2*a/b)*FresnelC(2*2^(1/2)/Pi^(1
/2)/(-2/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*b^2-32*arcsin(d*x+c)^2*(-1/b)^(1/2)*Pi^(1/2)*(a+b*arcsin(d*x+c))
^(1/2)*sin(2*a/b)*FresnelS(2*2^(1/2)/Pi^(1/2)/(-2/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*b^2-8*(-1/b)^(1/2)*Pi^
(1/2)*2^(1/2)*(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)*a^2*c-8*(-1/b)^(1/2)*Pi^(1/2)*2^(1/2)*(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)*a^2*c+64*arcsin(d*x+c)*(-1/b)^(1/2)*Pi^(1/2)*(a+b*arcsin(d*x+c))^(1/2)*c
os(2*a/b)*FresnelC(2*2^(1/2)/Pi^(1/2)/(-2/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*a*b-64*arcsin(d*x+c)*(-1/b)^(1
/2)*Pi^(1/2)*(a+b*arcsin(d*x+c))^(1/2)*sin(2*a/b)*FresnelS(2*2^(1/2)/Pi^(1/2)/(-2/b)^(1/2)*(a+b*arcsin(d*x+c))
^(1/2)/b)*a*b+32*(-1/b)^(1/2)*Pi^(1/2)*(a+b*arcsin(d*x+c))^(1/2)*cos(2*a/b)*FresnelC(2*2^(1/2)/Pi^(1/2)/(-2/b)
^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*a^2-32*(-1/b)^(1/2)*Pi^(1/2)*(a+b*arcsin(d*x+c))^(1/2)*sin(2*a/b)*FresnelS
(2*2^(1/2)/Pi^(1/2)/(-2/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*a^2+8*arcsin(d*x+c)^2*cos(-(a+b*arcsin(d*x+c))/b
+a/b)*b^2*c+16*arcsin(d*x+c)^2*sin(-2*(a+b*arcsin(d*x+c))/b+2*a/b)*b^2+16*arcsin(d*x+c)*cos(-(a+b*arcsin(d*x+c
))/b+a/b)*a*b*c-4*arcsin(d*x+c)*sin(-(a+b*arcsin(d*x+c))/b+a/b)*b^2*c+32*arcsin(d*x+c)*sin(-2*(a+b*arcsin(d*x+
c))/b+2*a/b)*a*b+4*arcsin(d*x+c)*cos(-2*(a+b*arcsin(d*x+c))/b+2*a/b)*b^2+8*cos(-(a+b*arcsin(d*x+c))/b+a/b)*a^2
*c-6*cos(-(a+b*arcsin(d*x+c))/b+a/b)*b^2*c-4*sin(-(a+b*arcsin(d*x+c))/b+a/b)*a*b*c+16*sin(-2*(a+b*arcsin(d*x+c
))/b+2*a/b)*a^2-3*sin(-2*(a+b*arcsin(d*x+c))/b+2*a/b)*b^2+4*cos(-2*(a+b*arcsin(d*x+c))/b+2*a/b)*a*b)

Fricas [F(-2)]

Exception generated. \[ \int \frac {x}{(a+b \arcsin (c+d x))^{7/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x/(a+b*arcsin(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {x}{(a+b \arcsin (c+d x))^{7/2}} \, dx=\int \frac {x}{\left (a + b \operatorname {asin}{\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx \]

[In]

integrate(x/(a+b*asin(d*x+c))**(7/2),x)

[Out]

Integral(x/(a + b*asin(c + d*x))**(7/2), x)

Maxima [F]

\[ \int \frac {x}{(a+b \arcsin (c+d x))^{7/2}} \, dx=\int { \frac {x}{{\left (b \arcsin \left (d x + c\right ) + a\right )}^{\frac {7}{2}}} \,d x } \]

[In]

integrate(x/(a+b*arcsin(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

integrate(x/(b*arcsin(d*x + c) + a)^(7/2), x)

Giac [F]

\[ \int \frac {x}{(a+b \arcsin (c+d x))^{7/2}} \, dx=\int { \frac {x}{{\left (b \arcsin \left (d x + c\right ) + a\right )}^{\frac {7}{2}}} \,d x } \]

[In]

integrate(x/(a+b*arcsin(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate(x/(b*arcsin(d*x + c) + a)^(7/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x}{(a+b \arcsin (c+d x))^{7/2}} \, dx=\int \frac {x}{{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^{7/2}} \,d x \]

[In]

int(x/(a + b*asin(c + d*x))^(7/2),x)

[Out]

int(x/(a + b*asin(c + d*x))^(7/2), x)