3.170 \(\int \frac {x}{(a+b \sin ^{-1}(c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=468 \[ \frac {8 \sqrt {2 \pi } c \sin \left (\frac {a}{b}\right ) C\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d^2}-\frac {32 \sqrt {\pi } \cos \left (\frac {2 a}{b}\right ) C\left (\frac {2 \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b} \sqrt {\pi }}\right )}{15 b^{7/2} d^2}-\frac {32 \sqrt {\pi } \sin \left (\frac {2 a}{b}\right ) S\left (\frac {2 \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b} \sqrt {\pi }}\right )}{15 b^{7/2} d^2}-\frac {8 \sqrt {2 \pi } c \cos \left (\frac {a}{b}\right ) S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d^2}+\frac {32 \sqrt {1-(c+d x)^2} (c+d x)}{15 b^3 d^2 \sqrt {a+b \sin ^{-1}(c+d x)}}-\frac {8 c \sqrt {1-(c+d x)^2}}{15 b^3 d^2 \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {8 (c+d x)^2}{15 b^2 d^2 \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac {4 c (c+d x)}{15 b^2 d^2 \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac {4}{15 b^2 d^2 \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac {2 \sqrt {1-(c+d x)^2} (c+d x)}{5 b d^2 \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}+\frac {2 c \sqrt {1-(c+d x)^2}}{5 b d^2 \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}} \]

[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]  time = 1.06, antiderivative size = 468, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 13, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.812, Rules used = {4805, 4745, 4621, 4719, 4723, 3306, 3305, 3351, 3304, 3352, 4633, 4631, 4641} \[ \frac {8 \sqrt {2 \pi } c \sin \left (\frac {a}{b}\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d^2}-\frac {32 \sqrt {\pi } \cos \left (\frac {2 a}{b}\right ) \text {FresnelC}\left (\frac {2 \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {\pi } \sqrt {b}}\right )}{15 b^{7/2} d^2}-\frac {32 \sqrt {\pi } \sin \left (\frac {2 a}{b}\right ) S\left (\frac {2 \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b} \sqrt {\pi }}\right )}{15 b^{7/2} d^2}-\frac {8 \sqrt {2 \pi } c \cos \left (\frac {a}{b}\right ) S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d^2}+\frac {8 (c+d x)^2}{15 b^2 d^2 \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac {32 \sqrt {1-(c+d x)^2} (c+d x)}{15 b^3 d^2 \sqrt {a+b \sin ^{-1}(c+d x)}}-\frac {4 c (c+d x)}{15 b^2 d^2 \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac {8 c \sqrt {1-(c+d x)^2}}{15 b^3 d^2 \sqrt {a+b \sin ^{-1}(c+d x)}}-\frac {4}{15 b^2 d^2 \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac {2 \sqrt {1-(c+d x)^2} (c+d x)}{5 b d^2 \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}+\frac {2 c \sqrt {1-(c+d x)^2}}{5 b d^2 \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}} \]

Antiderivative was successfully verified.

[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 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 4621

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 4631

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*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[(a + b*x)^(n + 1)
, Sin[x]^(m - 1)*(m - (m + 1)*Sin[x]^2), x], x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && G
eQ[n, -2] && LtQ[n, -1]

Rule 4633

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 4641

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

Rule 4719

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[
((f*x)^m*(a + b*ArcSin[c*x])^(n + 1))/(b*c*Sqrt[d]*(n + 1)), x] - Dist[(f*m)/(b*c*Sqrt[d]*(n + 1)), 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] && GtQ[d, 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 4745

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 4805

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

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Mathematica [C]  time = 2.01, size = 524, normalized size = 1.12 \[ \frac {-2 \left (-16 a^2 \sin \left (2 \sin ^{-1}(c+d x)\right )+32 \sqrt {\pi } \sqrt {\frac {1}{b}} \cos \left (\frac {2 a}{b}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^{5/2} C\left (\frac {2 \sqrt {\frac {1}{b}} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {\pi }}\right )+32 \sqrt {\pi } \sqrt {\frac {1}{b}} \sin \left (\frac {2 a}{b}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^{5/2} S\left (\frac {2 \sqrt {\frac {1}{b}} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {\pi }}\right )-32 a b \sin ^{-1}(c+d x) \sin \left (2 \sin ^{-1}(c+d x)\right )+4 a b \cos \left (2 \sin ^{-1}(c+d x)\right )+3 b^2 \sin \left (2 \sin ^{-1}(c+d x)\right )-16 b^2 \sin ^{-1}(c+d x)^2 \sin \left (2 \sin ^{-1}(c+d x)\right )+4 b^2 \sin ^{-1}(c+d x) \cos \left (2 \sin ^{-1}(c+d x)\right )\right )-c \left (e^{-i \sin ^{-1}(c+d x)} \left (8 a^2+4 a b \left (4 \sin ^{-1}(c+d x)+i\right )-8 e^{\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^2 \sqrt {\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \Gamma \left (\frac {1}{2},\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+2 b^2 \left (4 \sin ^{-1}(c+d x)^2+2 i \sin ^{-1}(c+d x)-3\right )\right )+4 e^{-\frac {i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right ) \left (e^{\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \left (2 a+2 b \sin ^{-1}(c+d x)-i b\right )-2 i b \left (-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )\right )-6 b^2 e^{i \sin ^{-1}(c+d x)}\right )}{30 b^3 d^2 \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-(c*(-6*b^2*E^(I*ArcSin[c + d*x]) + (4*(a + b*ArcSin[c + d*x])*(E^((I*(a + b*ArcSin[c + d*x]))/b)*(2*a - I*b
+ 2*b*ArcSin[c + d*x]) - (2*I)*b*(((-I)*(a + b*ArcSin[c + d*x]))/b)^(3/2)*Gamma[1/2, ((-I)*(a + b*ArcSin[c + d
*x]))/b]))/E^((I*a)/b) + (8*a^2 + 4*a*b*(I + 4*ArcSin[c + d*x]) + 2*b^2*(-3 + (2*I)*ArcSin[c + d*x] + 4*ArcSin
[c + d*x]^2) - 8*E^((I*(a + b*ArcSin[c + d*x]))/b)*(a + b*ArcSin[c + d*x])^2*Sqrt[(I*(a + b*ArcSin[c + d*x]))/
b]*Gamma[1/2, (I*(a + b*ArcSin[c + d*x]))/b])/E^(I*ArcSin[c + d*x]))) - 2*(4*a*b*Cos[2*ArcSin[c + d*x]] + 4*b^
2*ArcSin[c + d*x]*Cos[2*ArcSin[c + d*x]] + 32*Sqrt[b^(-1)]*Sqrt[Pi]*(a + b*ArcSin[c + d*x])^(5/2)*Cos[(2*a)/b]
*FresnelC[(2*Sqrt[b^(-1)]*Sqrt[a + b*ArcSin[c + d*x]])/Sqrt[Pi]] + 32*Sqrt[b^(-1)]*Sqrt[Pi]*(a + b*ArcSin[c +
d*x])^(5/2)*FresnelS[(2*Sqrt[b^(-1)]*Sqrt[a + b*ArcSin[c + d*x]])/Sqrt[Pi]]*Sin[(2*a)/b] - 16*a^2*Sin[2*ArcSin
[c + d*x]] + 3*b^2*Sin[2*ArcSin[c + d*x]] - 32*a*b*ArcSin[c + d*x]*Sin[2*ArcSin[c + d*x]] - 16*b^2*ArcSin[c +
d*x]^2*Sin[2*ArcSin[c + d*x]]))/(30*b^3*d^2*(a + b*ArcSin[c + d*x])^(5/2))

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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(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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{{\left (b \arcsin \left (d x + c\right ) + a\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [B]  time = 0.46, size = 1172, normalized size = 2.50 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15/d^2/b^3*(-8*2^(1/2)*arcsin(d*x+c)^2*(1/b)^(1/2)*Pi^(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*2^(1/2)*arcsin(d*x+c)^2*(1/b)^(1/2)*Pi^(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)*b^2*c-16*2^
(1/2)*arcsin(d*x+c)*(1/b)^(1/2)*Pi^(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*b*c+16*2^(1/2)*arcsin(d*x+c)*(1/b)^(1/2)*Pi^(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-8*2^(1/2)*(1/b)^(1/2)*Pi^(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*2^(1/2)*(1/b)^(1/2)*Pi^(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-32*(a+b*arcsin(d*x+c))^(1/2)*arcsin(d*x+c)^2*Pi^(1/2)*(1/b)^(1/2)*cos(2*a/b)*Fr
esnelC(2/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*b^2-32*(a+b*arcsin(d*x+c))^(1/2)*arcsin(d*x+c)^2*Pi
^(1/2)*(1/b)^(1/2)*sin(2*a/b)*FresnelS(2/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*b^2-64*(a+b*arcsin(
d*x+c))^(1/2)*arcsin(d*x+c)*Pi^(1/2)*(1/b)^(1/2)*cos(2*a/b)*FresnelC(2/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x+c)
)^(1/2)/b)*a*b-64*(a+b*arcsin(d*x+c))^(1/2)*arcsin(d*x+c)*Pi^(1/2)*(1/b)^(1/2)*sin(2*a/b)*FresnelS(2/Pi^(1/2)/
(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*a*b-32*(a+b*arcsin(d*x+c))^(1/2)*Pi^(1/2)*(1/b)^(1/2)*cos(2*a/b)*Fres
nelC(2/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*a^2-32*(a+b*arcsin(d*x+c))^(1/2)*Pi^(1/2)*(1/b)^(1/2)
*sin(2*a/b)*FresnelS(2/Pi^(1/2)/(1/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*arcs
in(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*arcsi
n(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)/(a+b*arcsi
n(d*x+c))^(5/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{{\left (b \arcsin \left (d x + c\right ) + a\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x}{{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*asin(c + d*x))^(7/2),x)

[Out]

int(x/(a + b*asin(c + d*x))^(7/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\left (a + b \operatorname {asin}{\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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