3.271 \(\int \frac{(c e+d e x)^3}{(a+b \sin ^{-1}(c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=344 \[ \frac{4 \sqrt{\pi } e^3 \sin \left (\frac{2 a}{b}\right ) \text{FresnelC}\left (\frac{2 \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{\pi } \sqrt{b}}\right )}{3 b^{5/2} d}-\frac{4 \sqrt{2 \pi } e^3 \sin \left (\frac{4 a}{b}\right ) \text{FresnelC}\left (\frac{2 \sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{3 b^{5/2} d}+\frac{4 \sqrt{2 \pi } e^3 \cos \left (\frac{4 a}{b}\right ) S\left (\frac{2 \sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{3 b^{5/2} d}-\frac{4 \sqrt{\pi } e^3 \cos \left (\frac{2 a}{b}\right ) S\left (\frac{2 \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b} \sqrt{\pi }}\right )}{3 b^{5/2} d}+\frac{16 e^3 (c+d x)^4}{3 b^2 d \sqrt{a+b \sin ^{-1}(c+d x)}}-\frac{4 e^3 (c+d x)^2}{b^2 d \sqrt{a+b \sin ^{-1}(c+d x)}}-\frac{2 e^3 \sqrt{1-(c+d x)^2} (c+d x)^3}{3 b d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}} \]

[Out]

(-2*e^3*(c + d*x)^3*Sqrt[1 - (c + d*x)^2])/(3*b*d*(a + b*ArcSin[c + d*x])^(3/2)) - (4*e^3*(c + d*x)^2)/(b^2*d*
Sqrt[a + b*ArcSin[c + d*x]]) + (16*e^3*(c + d*x)^4)/(3*b^2*d*Sqrt[a + b*ArcSin[c + d*x]]) + (4*e^3*Sqrt[2*Pi]*
Cos[(4*a)/b]*FresnelS[(2*Sqrt[2/Pi]*Sqrt[a + b*ArcSin[c + d*x]])/Sqrt[b]])/(3*b^(5/2)*d) - (4*e^3*Sqrt[Pi]*Cos
[(2*a)/b]*FresnelS[(2*Sqrt[a + b*ArcSin[c + d*x]])/(Sqrt[b]*Sqrt[Pi])])/(3*b^(5/2)*d) + (4*e^3*Sqrt[Pi]*Fresne
lC[(2*Sqrt[a + b*ArcSin[c + d*x]])/(Sqrt[b]*Sqrt[Pi])]*Sin[(2*a)/b])/(3*b^(5/2)*d) - (4*e^3*Sqrt[2*Pi]*Fresnel
C[(2*Sqrt[2/Pi]*Sqrt[a + b*ArcSin[c + d*x]])/Sqrt[b]]*Sin[(4*a)/b])/(3*b^(5/2)*d)

________________________________________________________________________________________

Rubi [A]  time = 1.17264, antiderivative size = 344, normalized size of antiderivative = 1., number of steps used = 26, number of rules used = 11, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.44, Rules used = {4805, 12, 4633, 4719, 4635, 4406, 3306, 3305, 3351, 3304, 3352} \[ \frac{4 \sqrt{\pi } e^3 \sin \left (\frac{2 a}{b}\right ) \text{FresnelC}\left (\frac{2 \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{\pi } \sqrt{b}}\right )}{3 b^{5/2} d}-\frac{4 \sqrt{2 \pi } e^3 \sin \left (\frac{4 a}{b}\right ) \text{FresnelC}\left (\frac{2 \sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{3 b^{5/2} d}+\frac{4 \sqrt{2 \pi } e^3 \cos \left (\frac{4 a}{b}\right ) S\left (\frac{2 \sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{3 b^{5/2} d}-\frac{4 \sqrt{\pi } e^3 \cos \left (\frac{2 a}{b}\right ) S\left (\frac{2 \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b} \sqrt{\pi }}\right )}{3 b^{5/2} d}+\frac{16 e^3 (c+d x)^4}{3 b^2 d \sqrt{a+b \sin ^{-1}(c+d x)}}-\frac{4 e^3 (c+d x)^2}{b^2 d \sqrt{a+b \sin ^{-1}(c+d x)}}-\frac{2 e^3 \sqrt{1-(c+d x)^2} (c+d x)^3}{3 b d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^3/(a + b*ArcSin[c + d*x])^(5/2),x]

[Out]

(-2*e^3*(c + d*x)^3*Sqrt[1 - (c + d*x)^2])/(3*b*d*(a + b*ArcSin[c + d*x])^(3/2)) - (4*e^3*(c + d*x)^2)/(b^2*d*
Sqrt[a + b*ArcSin[c + d*x]]) + (16*e^3*(c + d*x)^4)/(3*b^2*d*Sqrt[a + b*ArcSin[c + d*x]]) + (4*e^3*Sqrt[2*Pi]*
Cos[(4*a)/b]*FresnelS[(2*Sqrt[2/Pi]*Sqrt[a + b*ArcSin[c + d*x]])/Sqrt[b]])/(3*b^(5/2)*d) - (4*e^3*Sqrt[Pi]*Cos
[(2*a)/b]*FresnelS[(2*Sqrt[a + b*ArcSin[c + d*x]])/(Sqrt[b]*Sqrt[Pi])])/(3*b^(5/2)*d) + (4*e^3*Sqrt[Pi]*Fresne
lC[(2*Sqrt[a + b*ArcSin[c + d*x]])/(Sqrt[b]*Sqrt[Pi])]*Sin[(2*a)/b])/(3*b^(5/2)*d) - (4*e^3*Sqrt[2*Pi]*Fresnel
C[(2*Sqrt[2/Pi]*Sqrt[a + b*ArcSin[c + d*x]])/Sqrt[b]]*Sin[(4*a)/b])/(3*b^(5/2)*d)

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]

Rule 12

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

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 4635

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

Rule 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 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 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 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 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 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]

Rubi steps

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

Mathematica [C]  time = 2.02095, size = 351, normalized size = 1.02 \[ \frac{e^3 \left (-4 \left (a+b \sin ^{-1}(c+d x)\right ) \left (-\sqrt{2} e^{-\frac{2 i a}{b}} \sqrt{-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \text{Gamma}\left (\frac{1}{2},-\frac{2 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )-\sqrt{2} e^{\frac{2 i a}{b}} \sqrt{\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \text{Gamma}\left (\frac{1}{2},\frac{2 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+e^{-2 i \sin ^{-1}(c+d x)}+e^{2 i \sin ^{-1}(c+d x)}\right )+4 \left (a+b \sin ^{-1}(c+d x)\right ) \left (-2 e^{-\frac{4 i a}{b}} \sqrt{-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \text{Gamma}\left (\frac{1}{2},-\frac{4 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )-2 e^{\frac{4 i a}{b}} \sqrt{\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \text{Gamma}\left (\frac{1}{2},\frac{4 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+e^{-4 i \sin ^{-1}(c+d x)}+e^{4 i \sin ^{-1}(c+d x)}\right )-2 b \sin \left (2 \sin ^{-1}(c+d x)\right )+b \sin \left (4 \sin ^{-1}(c+d x)\right )\right )}{12 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c*e + d*e*x)^3/(a + b*ArcSin[c + d*x])^(5/2),x]

[Out]

(e^3*(-4*(a + b*ArcSin[c + d*x])*(E^((-2*I)*ArcSin[c + d*x]) + E^((2*I)*ArcSin[c + d*x]) - (Sqrt[2]*Sqrt[((-I)
*(a + b*ArcSin[c + d*x]))/b]*Gamma[1/2, ((-2*I)*(a + b*ArcSin[c + d*x]))/b])/E^(((2*I)*a)/b) - Sqrt[2]*E^(((2*
I)*a)/b)*Sqrt[(I*(a + b*ArcSin[c + d*x]))/b]*Gamma[1/2, ((2*I)*(a + b*ArcSin[c + d*x]))/b]) + 4*(a + b*ArcSin[
c + d*x])*(E^((-4*I)*ArcSin[c + d*x]) + E^((4*I)*ArcSin[c + d*x]) - (2*Sqrt[((-I)*(a + b*ArcSin[c + d*x]))/b]*
Gamma[1/2, ((-4*I)*(a + b*ArcSin[c + d*x]))/b])/E^(((4*I)*a)/b) - 2*E^(((4*I)*a)/b)*Sqrt[(I*(a + b*ArcSin[c +
d*x]))/b]*Gamma[1/2, ((4*I)*(a + b*ArcSin[c + d*x]))/b]) - 2*b*Sin[2*ArcSin[c + d*x]] + b*Sin[4*ArcSin[c + d*x
]]))/(12*b^2*d*(a + b*ArcSin[c + d*x])^(3/2))

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Maple [B]  time = 0.117, size = 689, normalized size = 2. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^3/(a+b*arcsin(d*x+c))^(5/2),x)

[Out]

-1/12/d*e^3/b^2*(-16*arcsin(d*x+c)*(a+b*arcsin(d*x+c))^(1/2)*2^(1/2)*(1/b)^(1/2)*Pi^(1/2)*cos(4*a/b)*FresnelS(
2*2^(1/2)/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*b+16*arcsin(d*x+c)*(a+b*arcsin(d*x+c))^(1/2)*2^(1/
2)*(1/b)^(1/2)*Pi^(1/2)*sin(4*a/b)*FresnelC(2*2^(1/2)/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*b+16*a
rcsin(d*x+c)*Pi^(1/2)*(1/b)^(1/2)*cos(2*a/b)*FresnelS(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)*b-16*arcsin(d*x+c)*Pi^(1/2)*(1/b)^(1/2)*sin(2*a/b)*FresnelC(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)*b-16*(a+b*arcsin(d*x+c))^(1/2)*2^(1/2)*(1/b)^(1/2)*Pi^(1/2)*
cos(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+16*(a+b*arcsin(d*x+c))^(1/2)
*2^(1/2)*(1/b)^(1/2)*Pi^(1/2)*sin(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+16*Pi^(1/2)*(1/b)^(1/2)*cos(2*a/b)*FresnelS(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)*a-16*Pi^(1/2)*(1/b)^(1/2)*sin(2*a/b)*FresnelC(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)*a+8*arcsin(d*x+c)*cos(2*(a+b*arcsin(d*x+c))/b-2*a/b)*b-8*arcsin(d*x+c)*cos(4*(a+b*
arcsin(d*x+c))/b-4*a/b)*b+2*sin(2*(a+b*arcsin(d*x+c))/b-2*a/b)*b+8*cos(2*(a+b*arcsin(d*x+c))/b-2*a/b)*a-sin(4*
(a+b*arcsin(d*x+c))/b-4*a/b)*b-8*cos(4*(a+b*arcsin(d*x+c))/b-4*a/b)*a)/(a+b*arcsin(d*x+c))^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d e x + c e\right )}^{3}}{{\left (b \arcsin \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^3/(a+b*arcsin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((d*e*x + c*e)^3/(b*arcsin(d*x + c) + a)^(5/2), x)

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^3/(a+b*arcsin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**3/(a+b*asin(d*x+c))**(5/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d e x + c e\right )}^{3}}{{\left (b \arcsin \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^3/(a+b*arcsin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((d*e*x + c*e)^3/(b*arcsin(d*x + c) + a)^(5/2), x)