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

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

[Out]

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

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Rubi [A]  time = 1.17, antiderivative size = 344, normalized size of antiderivative = 1.00, number of steps used = 26, number of rules used = 11, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, 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 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 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 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 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 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 )^{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*}

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Mathematica [C]  time = 2.34, size = 351, normalized size = 1.02 \[ \frac {e^3 \left (-4 \left (a+b \sin ^{-1}(c+d x)\right ) \left (-\sqrt {2} e^{-\frac {2 i a}{b}} \sqrt {-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \Gamma \left (\frac {1}{2},-\frac {2 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )-\sqrt {2} e^{\frac {2 i a}{b}} \sqrt {\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \Gamma \left (\frac {1}{2},\frac {2 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+e^{-2 i \sin ^{-1}(c+d x)}+e^{2 i \sin ^{-1}(c+d x)}\right )+4 \left (a+b \sin ^{-1}(c+d x)\right ) \left (-2 e^{-\frac {4 i a}{b}} \sqrt {-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \Gamma \left (\frac {1}{2},-\frac {4 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )-2 e^{\frac {4 i a}{b}} \sqrt {\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \Gamma \left (\frac {1}{2},\frac {4 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+e^{-4 i \sin ^{-1}(c+d x)}+e^{4 i \sin ^{-1}(c+d x)}\right )-2 b \sin \left (2 \sin ^{-1}(c+d x)\right )+b \sin \left (4 \sin ^{-1}(c+d x)\right )\right )}{12 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}} \]

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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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))^(5/2),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 \[ \int \frac {{\left (d e x + c e\right )}^{3}}{{\left (b \arcsin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.28, size = 689, normalized size = 2.00 \[ -\frac {e^{3} \left (-16 \arcsin \left (d x +c \right ) \sqrt {2}\, \sqrt {\pi }\, \sqrt {\frac {1}{b}}\, \cos \left (\frac {4 a}{b}\right ) \mathrm {S}\left (\frac {2 \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {a +b \arcsin \left (d x +c \right )}\, b +16 \arcsin \left (d x +c \right ) \sqrt {2}\, \sqrt {\pi }\, \sqrt {\frac {1}{b}}\, \sin \left (\frac {4 a}{b}\right ) \FresnelC \left (\frac {2 \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {a +b \arcsin \left (d x +c \right )}\, b +16 \arcsin \left (d x +c \right ) \sqrt {\pi }\, \sqrt {\frac {1}{b}}\, \cos \left (\frac {2 a}{b}\right ) \mathrm {S}\left (\frac {2 \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {a +b \arcsin \left (d x +c \right )}\, b -16 \arcsin \left (d x +c \right ) \sqrt {\pi }\, \sqrt {\frac {1}{b}}\, \sin \left (\frac {2 a}{b}\right ) \FresnelC \left (\frac {2 \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {a +b \arcsin \left (d x +c \right )}\, b -16 \sqrt {2}\, \sqrt {\pi }\, \sqrt {\frac {1}{b}}\, \cos \left (\frac {4 a}{b}\right ) \mathrm {S}\left (\frac {2 \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {a +b \arcsin \left (d x +c \right )}\, a +16 \sqrt {2}\, \sqrt {\pi }\, \sqrt {\frac {1}{b}}\, \sin \left (\frac {4 a}{b}\right ) \FresnelC \left (\frac {2 \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {a +b \arcsin \left (d x +c \right )}\, a +16 \sqrt {\pi }\, \sqrt {\frac {1}{b}}\, \cos \left (\frac {2 a}{b}\right ) \mathrm {S}\left (\frac {2 \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {a +b \arcsin \left (d x +c \right )}\, a -16 \sqrt {\pi }\, \sqrt {\frac {1}{b}}\, \sin \left (\frac {2 a}{b}\right ) \FresnelC \left (\frac {2 \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {a +b \arcsin \left (d x +c \right )}\, a +8 \arcsin \left (d x +c \right ) \cos \left (\frac {2 a +2 b \arcsin \left (d x +c \right )}{b}-\frac {2 a}{b}\right ) b -8 \arcsin \left (d x +c \right ) \cos \left (\frac {4 a +4 b \arcsin \left (d x +c \right )}{b}-\frac {4 a}{b}\right ) b +2 \sin \left (\frac {2 a +2 b \arcsin \left (d x +c \right )}{b}-\frac {2 a}{b}\right ) b +8 \cos \left (\frac {2 a +2 b \arcsin \left (d x +c \right )}{b}-\frac {2 a}{b}\right ) a -\sin \left (\frac {4 a +4 b \arcsin \left (d x +c \right )}{b}-\frac {4 a}{b}\right ) b -8 \cos \left (\frac {4 a +4 b \arcsin \left (d x +c \right )}{b}-\frac {4 a}{b}\right ) a \right )}{12 d \,b^{2} \left (a +b \arcsin \left (d x +c \right )\right )^{\frac {3}{2}}} \]

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)*2^(1/2)*Pi^(1/2)*(1/b)^(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+b*arcsin(d*x+c))^(1/2)*b+16*arcsin(d*x+c)*2^(1/2)*Pi^(1/2)*(1/b)^(1/2)*si
n(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)*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*2^(1/2)*Pi^(1/2)*(1/b)^(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+b*arcsin(d*x+c))^(1/2)*a+16*2^(1/2)*Pi^(1/2)*(1/b)^(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+b*arcsin(d*x+c))^(1/2)*
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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d e x + c e\right )}^{3}}{{\left (b \arcsin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^3/(a+b*arcsin(d*x+c))^(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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c\,e+d\,e\,x\right )}^3}{{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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