3.276 \(\int \frac {(c e+d e x)^3}{(a+b \sin ^{-1}(c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=442 \[ \frac {32 \sqrt {2 \pi } e^3 \cos \left (\frac {4 a}{b}\right ) C\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d}-\frac {16 \sqrt {\pi } e^3 \cos \left (\frac {2 a}{b}\right ) C\left (\frac {2 \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b} \sqrt {\pi }}\right )}{15 b^{7/2} d}-\frac {16 \sqrt {\pi } e^3 \sin \left (\frac {2 a}{b}\right ) S\left (\frac {2 \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b} \sqrt {\pi }}\right )}{15 b^{7/2} d}+\frac {32 \sqrt {2 \pi } e^3 \sin \left (\frac {4 a}{b}\right ) S\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d}+\frac {128 e^3 \sqrt {1-(c+d x)^2} (c+d x)^3}{15 b^3 d \sqrt {a+b \sin ^{-1}(c+d x)}}-\frac {16 e^3 \sqrt {1-(c+d x)^2} (c+d x)}{5 b^3 d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac {4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac {2 e^3 \sqrt {1-(c+d x)^2} (c+d x)^3}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}} \]

[Out]

-4/5*e^3*(d*x+c)^2/b^2/d/(a+b*arcsin(d*x+c))^(3/2)+16/15*e^3*(d*x+c)^4/b^2/d/(a+b*arcsin(d*x+c))^(3/2)-16/15*e
^3*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-16/15*e^3*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+32/15*e^3*cos(4*a/b)*FresnelC(2*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+32/15*e^3*FresnelS(2*2^(1/2)/Pi^(1/2
)*(a+b*arcsin(d*x+c))^(1/2)/b^(1/2))*sin(4*a/b)*2^(1/2)*Pi^(1/2)/b^(7/2)/d-2/5*e^3*(d*x+c)^3*(1-(d*x+c)^2)^(1/
2)/b/d/(a+b*arcsin(d*x+c))^(5/2)-16/5*e^3*(d*x+c)*(1-(d*x+c)^2)^(1/2)/b^3/d/(a+b*arcsin(d*x+c))^(1/2)+128/15*e
^3*(d*x+c)^3*(1-(d*x+c)^2)^(1/2)/b^3/d/(a+b*arcsin(d*x+c))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 1.14, antiderivative size = 442, normalized size of antiderivative = 1.00, number of steps used = 23, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4805, 12, 4633, 4719, 4631, 3306, 3305, 3351, 3304, 3352} \[ \frac {32 \sqrt {2 \pi } e^3 \cos \left (\frac {4 a}{b}\right ) \text {FresnelC}\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d}-\frac {16 \sqrt {\pi } e^3 \cos \left (\frac {2 a}{b}\right ) \text {FresnelC}\left (\frac {2 \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {\pi } \sqrt {b}}\right )}{15 b^{7/2} d}-\frac {16 \sqrt {\pi } e^3 \sin \left (\frac {2 a}{b}\right ) S\left (\frac {2 \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b} \sqrt {\pi }}\right )}{15 b^{7/2} d}+\frac {32 \sqrt {2 \pi } e^3 \sin \left (\frac {4 a}{b}\right ) S\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d}+\frac {16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac {128 e^3 \sqrt {1-(c+d x)^2} (c+d x)^3}{15 b^3 d \sqrt {a+b \sin ^{-1}(c+d x)}}-\frac {4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac {16 e^3 \sqrt {1-(c+d x)^2} (c+d x)}{5 b^3 d \sqrt {a+b \sin ^{-1}(c+d x)}}-\frac {2 e^3 \sqrt {1-(c+d x)^2} (c+d x)^3}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^3/(a + b*ArcSin[c + d*x])^(7/2),x]

[Out]

(-2*e^3*(c + d*x)^3*Sqrt[1 - (c + d*x)^2])/(5*b*d*(a + b*ArcSin[c + d*x])^(5/2)) - (4*e^3*(c + d*x)^2)/(5*b^2*
d*(a + b*ArcSin[c + d*x])^(3/2)) + (16*e^3*(c + d*x)^4)/(15*b^2*d*(a + b*ArcSin[c + d*x])^(3/2)) - (16*e^3*(c
+ d*x)*Sqrt[1 - (c + d*x)^2])/(5*b^3*d*Sqrt[a + b*ArcSin[c + d*x]]) + (128*e^3*(c + d*x)^3*Sqrt[1 - (c + d*x)^
2])/(15*b^3*d*Sqrt[a + b*ArcSin[c + d*x]]) + (32*e^3*Sqrt[2*Pi]*Cos[(4*a)/b]*FresnelC[(2*Sqrt[2/Pi]*Sqrt[a + b
*ArcSin[c + d*x]])/Sqrt[b]])/(15*b^(7/2)*d) - (16*e^3*Sqrt[Pi]*Cos[(2*a)/b]*FresnelC[(2*Sqrt[a + b*ArcSin[c +
d*x]])/(Sqrt[b]*Sqrt[Pi])])/(15*b^(7/2)*d) - (16*e^3*Sqrt[Pi]*FresnelS[(2*Sqrt[a + b*ArcSin[c + d*x]])/(Sqrt[b
]*Sqrt[Pi])]*Sin[(2*a)/b])/(15*b^(7/2)*d) + (32*e^3*Sqrt[2*Pi]*FresnelS[(2*Sqrt[2/Pi]*Sqrt[a + b*ArcSin[c + d*
x]])/Sqrt[b]]*Sin[(4*a)/b])/(15*b^(7/2)*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 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 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 {(c e+d e x)^3}{\left (a+b \sin ^{-1}(c+d x)\right )^{7/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {e^3 x^3}{\left (a+b \sin ^{-1}(x)\right )^{7/2}} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^3 \operatorname {Subst}\left (\int \frac {x^3}{\left (a+b \sin ^{-1}(x)\right )^{7/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac {2 e^3 (c+d x)^3 \sqrt {1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}+\frac {\left (6 e^3\right ) \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}-\frac {\left (8 e^3\right ) \operatorname {Subst}\left (\int \frac {x^4}{\sqrt {1-x^2} \left (a+b \sin ^{-1}(x)\right )^{5/2}} \, dx,x,c+d x\right )}{5 b d}\\ &=-\frac {2 e^3 (c+d x)^3 \sqrt {1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}-\frac {4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac {16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac {\left (8 e^3\right ) \operatorname {Subst}\left (\int \frac {x}{\left (a+b \sin ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{5 b^2 d}-\frac {\left (64 e^3\right ) \operatorname {Subst}\left (\int \frac {x^3}{\left (a+b \sin ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{15 b^2 d}\\ &=-\frac {2 e^3 (c+d x)^3 \sqrt {1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}-\frac {4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac {16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac {16 e^3 (c+d x) \sqrt {1-(c+d x)^2}}{5 b^3 d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {128 e^3 (c+d x)^3 \sqrt {1-(c+d x)^2}}{15 b^3 d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {\left (16 e^3\right ) \operatorname {Subst}\left (\int \frac {\cos (2 x)}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{5 b^3 d}-\frac {\left (128 e^3\right ) \operatorname {Subst}\left (\int \left (\frac {\cos (2 x)}{2 \sqrt {a+b x}}-\frac {\cos (4 x)}{2 \sqrt {a+b x}}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{15 b^3 d}\\ &=-\frac {2 e^3 (c+d x)^3 \sqrt {1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}-\frac {4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac {16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac {16 e^3 (c+d x) \sqrt {1-(c+d x)^2}}{5 b^3 d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {128 e^3 (c+d x)^3 \sqrt {1-(c+d x)^2}}{15 b^3 d \sqrt {a+b \sin ^{-1}(c+d x)}}-\frac {\left (64 e^3\right ) \operatorname {Subst}\left (\int \frac {\cos (2 x)}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{15 b^3 d}+\frac {\left (64 e^3\right ) \operatorname {Subst}\left (\int \frac {\cos (4 x)}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{15 b^3 d}+\frac {\left (16 e^3 \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 )}{5 b^3 d}+\frac {\left (16 e^3 \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 )}{5 b^3 d}\\ &=-\frac {2 e^3 (c+d x)^3 \sqrt {1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}-\frac {4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac {16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac {16 e^3 (c+d x) \sqrt {1-(c+d x)^2}}{5 b^3 d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {128 e^3 (c+d x)^3 \sqrt {1-(c+d x)^2}}{15 b^3 d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {\left (32 e^3 \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 )}{5 b^4 d}-\frac {\left (64 e^3 \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}+\frac {\left (64 e^3 \cos \left (\frac {4 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {4 a}{b}+4 x\right )}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{15 b^3 d}+\frac {\left (32 e^3 \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 )}{5 b^4 d}-\frac {\left (64 e^3 \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}+\frac {\left (64 e^3 \sin \left (\frac {4 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {4 a}{b}+4 x\right )}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{15 b^3 d}\\ &=-\frac {2 e^3 (c+d x)^3 \sqrt {1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}-\frac {4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac {16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac {16 e^3 (c+d x) \sqrt {1-(c+d x)^2}}{5 b^3 d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {128 e^3 (c+d x)^3 \sqrt {1-(c+d x)^2}}{15 b^3 d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {16 e^3 \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 )}{5 b^{7/2} d}+\frac {16 e^3 \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 )}{5 b^{7/2} d}-\frac {\left (128 e^3 \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}+\frac {\left (128 e^3 \cos \left (\frac {4 a}{b}\right )\right ) \operatorname {Subst}\left (\int \cos \left (\frac {4 x^2}{b}\right ) \, dx,x,\sqrt {a+b \sin ^{-1}(c+d x)}\right )}{15 b^4 d}-\frac {\left (128 e^3 \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}+\frac {\left (128 e^3 \sin \left (\frac {4 a}{b}\right )\right ) \operatorname {Subst}\left (\int \sin \left (\frac {4 x^2}{b}\right ) \, dx,x,\sqrt {a+b \sin ^{-1}(c+d x)}\right )}{15 b^4 d}\\ &=-\frac {2 e^3 (c+d x)^3 \sqrt {1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}-\frac {4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac {16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac {16 e^3 (c+d x) \sqrt {1-(c+d x)^2}}{5 b^3 d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {128 e^3 (c+d x)^3 \sqrt {1-(c+d x)^2}}{15 b^3 d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {32 e^3 \sqrt {2 \pi } \cos \left (\frac {4 a}{b}\right ) C\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d}-\frac {16 e^3 \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}-\frac {16 e^3 \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}+\frac {32 e^3 \sqrt {2 \pi } S\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right ) \sin \left (\frac {4 a}{b}\right )}{15 b^{7/2} d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 2.69, size = 445, normalized size = 1.01 \[ \frac {e^3 \left (-4 \left (a+b \sin ^{-1}(c+d x)\right ) \left (e^{2 i \sin ^{-1}(c+d x)} \left (4 i a+4 i b \sin ^{-1}(c+d x)+b\right )+4 \sqrt {2} b e^{-\frac {2 i a}{b}} \left (-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {2 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+e^{-2 i \sin ^{-1}(c+d x)} \left (4 \sqrt {2} b e^{\frac {2 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \left (\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},\frac {2 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )-4 i a-4 i b \sin ^{-1}(c+d x)+b\right )\right )+4 \left (a+b \sin ^{-1}(c+d x)\right ) \left (e^{-4 i \sin ^{-1}(c+d x)} \left (-8 i a-8 i b \sin ^{-1}(c+d x)+b\right )+e^{4 i \sin ^{-1}(c+d x)} \left (8 i a+8 i b \sin ^{-1}(c+d x)+b\right )+16 b e^{-\frac {4 i a}{b}} \left (-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {4 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+16 b e^{\frac {4 i a}{b}} \left (\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},\frac {4 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )\right )-6 b^2 \sin \left (2 \sin ^{-1}(c+d x)\right )+3 b^2 \sin \left (4 \sin ^{-1}(c+d x)\right )\right )}{60 b^3 d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c*e + d*e*x)^3/(a + b*ArcSin[c + d*x])^(7/2),x]

[Out]

(e^3*(-4*(a + b*ArcSin[c + d*x])*(E^((2*I)*ArcSin[c + d*x])*((4*I)*a + b + (4*I)*b*ArcSin[c + d*x]) + (4*Sqrt[
2]*b*(((-I)*(a + b*ArcSin[c + d*x]))/b)^(3/2)*Gamma[1/2, ((-2*I)*(a + b*ArcSin[c + d*x]))/b])/E^(((2*I)*a)/b)
+ ((-4*I)*a + b - (4*I)*b*ArcSin[c + d*x] + 4*Sqrt[2]*b*E^(((2*I)*(a + b*ArcSin[c + d*x]))/b)*((I*(a + b*ArcSi
n[c + d*x]))/b)^(3/2)*Gamma[1/2, ((2*I)*(a + b*ArcSin[c + d*x]))/b])/E^((2*I)*ArcSin[c + d*x])) + 4*(a + b*Arc
Sin[c + d*x])*(((-8*I)*a + b - (8*I)*b*ArcSin[c + d*x])/E^((4*I)*ArcSin[c + d*x]) + E^((4*I)*ArcSin[c + d*x])*
((8*I)*a + b + (8*I)*b*ArcSin[c + d*x]) + (16*b*(((-I)*(a + b*ArcSin[c + d*x]))/b)^(3/2)*Gamma[1/2, ((-4*I)*(a
+ b*ArcSin[c + d*x]))/b])/E^(((4*I)*a)/b) + 16*b*E^(((4*I)*a)/b)*((I*(a + b*ArcSin[c + d*x]))/b)^(3/2)*Gamma[
1/2, ((4*I)*(a + b*ArcSin[c + d*x]))/b]) - 6*b^2*Sin[2*ArcSin[c + d*x]] + 3*b^2*Sin[4*ArcSin[c + d*x]]))/(60*b
^3*d*(a + b*ArcSin[c + d*x])^(5/2))

________________________________________________________________________________________

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((d*e*x+c*e)^3/(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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d e x + c e\right )}^{3}}{{\left (b \arcsin \left (d x + c\right ) + a\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^3/(a+b*arcsin(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((d*e*x + c*e)^3/(b*arcsin(d*x + c) + a)^(7/2), x)

________________________________________________________________________________________

maple [B]  time = 0.31, size = 1181, normalized size = 2.67 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^3/(a+b*arcsin(d*x+c))^(7/2),x)

[Out]

1/60/d*e^3/b^3*(128*arcsin(d*x+c)^2*Pi^(1/2)*cos(4*a/b)*FresnelC(2*2^(1/2)/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*
x+c))^(1/2)/b)*(a+b*arcsin(d*x+c))^(1/2)*(1/b)^(1/2)*2^(1/2)*b^2+128*arcsin(d*x+c)^2*Pi^(1/2)*sin(4*a/b)*Fresn
elS(2*2^(1/2)/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*(a+b*arcsin(d*x+c))^(1/2)*(1/b)^(1/2)*2^(1/2)*
b^2+256*arcsin(d*x+c)*Pi^(1/2)*cos(4*a/b)*FresnelC(2*2^(1/2)/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)
*(a+b*arcsin(d*x+c))^(1/2)*(1/b)^(1/2)*2^(1/2)*a*b+256*arcsin(d*x+c)*Pi^(1/2)*sin(4*a/b)*FresnelS(2*2^(1/2)/Pi
^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*(a+b*arcsin(d*x+c))^(1/2)*(1/b)^(1/2)*2^(1/2)*a*b-64*(a+b*arcs
in(d*x+c))^(1/2)*arcsin(d*x+c)^2*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)*b^2-64*(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+128*Pi^(1/2)*cos(4*a/b)*FresnelC(2*2^(1/2)/Pi^(1/2)/(1/b)^
(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*(a+b*arcsin(d*x+c))^(1/2)*(1/b)^(1/2)*2^(1/2)*a^2+128*Pi^(1/2)*sin(4*a/b)*F
resnelS(2*2^(1/2)/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*(a+b*arcsin(d*x+c))^(1/2)*(1/b)^(1/2)*2^(1
/2)*a^2-128*(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-128*(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-64*(a+b*arcsin(d*x+c))^(1/2)*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^2-64*(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+32*arcsi
n(d*x+c)^2*sin(2*(a+b*arcsin(d*x+c))/b-2*a/b)*b^2-64*arcsin(d*x+c)^2*sin(4*(a+b*arcsin(d*x+c))/b-4*a/b)*b^2+64
*arcsin(d*x+c)*sin(2*(a+b*arcsin(d*x+c))/b-2*a/b)*a*b-8*arcsin(d*x+c)*cos(2*(a+b*arcsin(d*x+c))/b-2*a/b)*b^2-1
28*arcsin(d*x+c)*sin(4*(a+b*arcsin(d*x+c))/b-4*a/b)*a*b+8*arcsin(d*x+c)*cos(4*(a+b*arcsin(d*x+c))/b-4*a/b)*b^2
+32*sin(2*(a+b*arcsin(d*x+c))/b-2*a/b)*a^2-6*sin(2*(a+b*arcsin(d*x+c))/b-2*a/b)*b^2-8*cos(2*(a+b*arcsin(d*x+c)
)/b-2*a/b)*a*b-64*sin(4*(a+b*arcsin(d*x+c))/b-4*a/b)*a^2+3*sin(4*(a+b*arcsin(d*x+c))/b-4*a/b)*b^2+8*cos(4*(a+b
*arcsin(d*x+c))/b-4*a/b)*a*b)/(a+b*arcsin(d*x+c))^(5/2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d e x + c e\right )}^{3}}{{\left (b \arcsin \left (d x + c\right ) + a\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^3/(a+b*arcsin(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

integrate((d*e*x + c*e)^3/(b*arcsin(d*x + c) + a)^(7/2), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c\,e+d\,e\,x\right )}^3}{{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*e + d*e*x)^3/(a + b*asin(c + d*x))^(7/2),x)

[Out]

int((c*e + d*e*x)^3/(a + b*asin(c + d*x))^(7/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ e^{3} \left (\int \frac {c^{3}}{a^{3} \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} + 3 a^{2} b \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} \operatorname {asin}{\left (c + d x \right )} + 3 a b^{2} \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} \operatorname {asin}^{2}{\left (c + d x \right )} + b^{3} \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} \operatorname {asin}^{3}{\left (c + d x \right )}}\, dx + \int \frac {d^{3} x^{3}}{a^{3} \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} + 3 a^{2} b \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} \operatorname {asin}{\left (c + d x \right )} + 3 a b^{2} \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} \operatorname {asin}^{2}{\left (c + d x \right )} + b^{3} \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} \operatorname {asin}^{3}{\left (c + d x \right )}}\, dx + \int \frac {3 c d^{2} x^{2}}{a^{3} \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} + 3 a^{2} b \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} \operatorname {asin}{\left (c + d x \right )} + 3 a b^{2} \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} \operatorname {asin}^{2}{\left (c + d x \right )} + b^{3} \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} \operatorname {asin}^{3}{\left (c + d x \right )}}\, dx + \int \frac {3 c^{2} d x}{a^{3} \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} + 3 a^{2} b \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} \operatorname {asin}{\left (c + d x \right )} + 3 a b^{2} \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} \operatorname {asin}^{2}{\left (c + d x \right )} + b^{3} \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} \operatorname {asin}^{3}{\left (c + d x \right )}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**3/(a+b*asin(d*x+c))**(7/2),x)

[Out]

e**3*(Integral(c**3/(a**3*sqrt(a + b*asin(c + d*x)) + 3*a**2*b*sqrt(a + b*asin(c + d*x))*asin(c + d*x) + 3*a*b
**2*sqrt(a + b*asin(c + d*x))*asin(c + d*x)**2 + b**3*sqrt(a + b*asin(c + d*x))*asin(c + d*x)**3), x) + Integr
al(d**3*x**3/(a**3*sqrt(a + b*asin(c + d*x)) + 3*a**2*b*sqrt(a + b*asin(c + d*x))*asin(c + d*x) + 3*a*b**2*sqr
t(a + b*asin(c + d*x))*asin(c + d*x)**2 + b**3*sqrt(a + b*asin(c + d*x))*asin(c + d*x)**3), x) + Integral(3*c*
d**2*x**2/(a**3*sqrt(a + b*asin(c + d*x)) + 3*a**2*b*sqrt(a + b*asin(c + d*x))*asin(c + d*x) + 3*a*b**2*sqrt(a
+ b*asin(c + d*x))*asin(c + d*x)**2 + b**3*sqrt(a + b*asin(c + d*x))*asin(c + d*x)**3), x) + Integral(3*c**2*
d*x/(a**3*sqrt(a + b*asin(c + d*x)) + 3*a**2*b*sqrt(a + b*asin(c + d*x))*asin(c + d*x) + 3*a*b**2*sqrt(a + b*a
sin(c + d*x))*asin(c + d*x)**2 + b**3*sqrt(a + b*asin(c + d*x))*asin(c + d*x)**3), x))

________________________________________________________________________________________