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

Optimal. Leaf size=342 \[ -\frac {\sqrt {2 \pi } e^2 \cos \left (\frac {a}{b}\right ) C\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}+\frac {\sqrt {6 \pi } e^2 \cos \left (\frac {3 a}{b}\right ) C\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{b^{5/2} d}-\frac {\sqrt {2 \pi } e^2 \sin \left (\frac {a}{b}\right ) S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}+\frac {\sqrt {6 \pi } e^2 \sin \left (\frac {3 a}{b}\right ) S\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{b^{5/2} d}+\frac {4 e^2 (c+d x)^3}{b^2 d \sqrt {a+b \sin ^{-1}(c+d x)}}-\frac {8 e^2 (c+d x)}{3 b^2 d \sqrt {a+b \sin ^{-1}(c+d x)}}-\frac {2 e^2 \sqrt {1-(c+d x)^2} (c+d x)^2}{3 b d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}} \]

[Out]

-1/3*e^2*cos(a/b)*FresnelC(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-1/3*
e^2*FresnelS(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^(5/2)/d+e^2*cos(3
*a/b)*FresnelC(6^(1/2)/Pi^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b^(1/2))*6^(1/2)*Pi^(1/2)/b^(5/2)/d+e^2*FresnelS(6^(
1/2)/Pi^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b^(1/2))*sin(3*a/b)*6^(1/2)*Pi^(1/2)/b^(5/2)/d-2/3*e^2*(d*x+c)^2*(1-(d
*x+c)^2)^(1/2)/b/d/(a+b*arcsin(d*x+c))^(3/2)-8/3*e^2*(d*x+c)/b^2/d/(a+b*arcsin(d*x+c))^(1/2)+4*e^2*(d*x+c)^3/b
^2/d/(a+b*arcsin(d*x+c))^(1/2)

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Rubi [A]  time = 1.05, antiderivative size = 342, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 12, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {4805, 12, 4633, 4719, 4635, 4406, 3306, 3305, 3351, 3304, 3352, 4623} \[ -\frac {\sqrt {2 \pi } e^2 \cos \left (\frac {a}{b}\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}+\frac {\sqrt {6 \pi } e^2 \cos \left (\frac {3 a}{b}\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{b^{5/2} d}-\frac {\sqrt {2 \pi } e^2 \sin \left (\frac {a}{b}\right ) S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}+\frac {\sqrt {6 \pi } e^2 \sin \left (\frac {3 a}{b}\right ) S\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{b^{5/2} d}+\frac {4 e^2 (c+d x)^3}{b^2 d \sqrt {a+b \sin ^{-1}(c+d x)}}-\frac {8 e^2 (c+d x)}{3 b^2 d \sqrt {a+b \sin ^{-1}(c+d x)}}-\frac {2 e^2 \sqrt {1-(c+d x)^2} (c+d x)^2}{3 b d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*e^2*(c + d*x)^2*Sqrt[1 - (c + d*x)^2])/(3*b*d*(a + b*ArcSin[c + d*x])^(3/2)) - (8*e^2*(c + d*x))/(3*b^2*d*
Sqrt[a + b*ArcSin[c + d*x]]) + (4*e^2*(c + d*x)^3)/(b^2*d*Sqrt[a + b*ArcSin[c + d*x]]) - (e^2*Sqrt[2*Pi]*Cos[a
/b]*FresnelC[(Sqrt[2/Pi]*Sqrt[a + b*ArcSin[c + d*x]])/Sqrt[b]])/(3*b^(5/2)*d) + (e^2*Sqrt[6*Pi]*Cos[(3*a)/b]*F
resnelC[(Sqrt[6/Pi]*Sqrt[a + b*ArcSin[c + d*x]])/Sqrt[b]])/(b^(5/2)*d) - (e^2*Sqrt[2*Pi]*FresnelS[(Sqrt[2/Pi]*
Sqrt[a + b*ArcSin[c + d*x]])/Sqrt[b]]*Sin[a/b])/(3*b^(5/2)*d) + (e^2*Sqrt[6*Pi]*FresnelS[(Sqrt[6/Pi]*Sqrt[a +
b*ArcSin[c + d*x]])/Sqrt[b]]*Sin[(3*a)/b])/(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 4623

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[1/(b*c), Subst[Int[x^n*Cos[a/b - x/b], x], x, a
+ b*ArcSin[c*x]], x] /; FreeQ[{a, b, c, n}, 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 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)^2}{\left (a+b \sin ^{-1}(c+d x)\right )^{5/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {e^2 x^2}{\left (a+b \sin ^{-1}(x)\right )^{5/2}} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^2 \operatorname {Subst}\left (\int \frac {x^2}{\left (a+b \sin ^{-1}(x)\right )^{5/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac {2 e^2 (c+d x)^2 \sqrt {1-(c+d x)^2}}{3 b d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac {\left (4 e^2\right ) \operatorname {Subst}\left (\int \frac {x}{\sqrt {1-x^2} \left (a+b \sin ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{3 b d}-\frac {\left (2 e^2\right ) \operatorname {Subst}\left (\int \frac {x^3}{\sqrt {1-x^2} \left (a+b \sin ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{b d}\\ &=-\frac {2 e^2 (c+d x)^2 \sqrt {1-(c+d x)^2}}{3 b d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac {8 e^2 (c+d x)}{3 b^2 d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {4 e^2 (c+d x)^3}{b^2 d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {\left (8 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b \sin ^{-1}(x)}} \, dx,x,c+d x\right )}{3 b^2 d}-\frac {\left (12 e^2\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+b \sin ^{-1}(x)}} \, dx,x,c+d x\right )}{b^2 d}\\ &=-\frac {2 e^2 (c+d x)^2 \sqrt {1-(c+d x)^2}}{3 b d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac {8 e^2 (c+d x)}{3 b^2 d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {4 e^2 (c+d x)^3}{b^2 d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {\left (8 e^2\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {a}{b}-\frac {x}{b}\right )}{\sqrt {x}} \, dx,x,a+b \sin ^{-1}(c+d x)\right )}{3 b^3 d}-\frac {\left (12 e^2\right ) \operatorname {Subst}\left (\int \frac {\cos (x) \sin ^2(x)}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{b^2 d}\\ &=-\frac {2 e^2 (c+d x)^2 \sqrt {1-(c+d x)^2}}{3 b d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac {8 e^2 (c+d x)}{3 b^2 d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {4 e^2 (c+d x)^3}{b^2 d \sqrt {a+b \sin ^{-1}(c+d x)}}-\frac {\left (12 e^2\right ) \operatorname {Subst}\left (\int \left (\frac {\cos (x)}{4 \sqrt {a+b x}}-\frac {\cos (3 x)}{4 \sqrt {a+b x}}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{b^2 d}+\frac {\left (8 e^2 \cos \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {x}{b}\right )}{\sqrt {x}} \, dx,x,a+b \sin ^{-1}(c+d x)\right )}{3 b^3 d}+\frac {\left (8 e^2 \sin \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {x}{b}\right )}{\sqrt {x}} \, dx,x,a+b \sin ^{-1}(c+d x)\right )}{3 b^3 d}\\ &=-\frac {2 e^2 (c+d x)^2 \sqrt {1-(c+d x)^2}}{3 b d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac {8 e^2 (c+d x)}{3 b^2 d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {4 e^2 (c+d x)^3}{b^2 d \sqrt {a+b \sin ^{-1}(c+d x)}}-\frac {\left (3 e^2\right ) \operatorname {Subst}\left (\int \frac {\cos (x)}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{b^2 d}+\frac {\left (3 e^2\right ) \operatorname {Subst}\left (\int \frac {\cos (3 x)}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{b^2 d}+\frac {\left (16 e^2 \cos \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 )}{3 b^3 d}+\frac {\left (16 e^2 \sin \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 )}{3 b^3 d}\\ &=-\frac {2 e^2 (c+d x)^2 \sqrt {1-(c+d x)^2}}{3 b d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac {8 e^2 (c+d x)}{3 b^2 d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {4 e^2 (c+d x)^3}{b^2 d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {8 e^2 \sqrt {2 \pi } \cos \left (\frac {a}{b}\right ) C\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}+\frac {8 e^2 \sqrt {2 \pi } S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right ) \sin \left (\frac {a}{b}\right )}{3 b^{5/2} d}-\frac {\left (3 e^2 \cos \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 )}{b^2 d}+\frac {\left (3 e^2 \cos \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {3 a}{b}+3 x\right )}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{b^2 d}-\frac {\left (3 e^2 \sin \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 )}{b^2 d}+\frac {\left (3 e^2 \sin \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {3 a}{b}+3 x\right )}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{b^2 d}\\ &=-\frac {2 e^2 (c+d x)^2 \sqrt {1-(c+d x)^2}}{3 b d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac {8 e^2 (c+d x)}{3 b^2 d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {4 e^2 (c+d x)^3}{b^2 d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {8 e^2 \sqrt {2 \pi } \cos \left (\frac {a}{b}\right ) C\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}+\frac {8 e^2 \sqrt {2 \pi } S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right ) \sin \left (\frac {a}{b}\right )}{3 b^{5/2} d}-\frac {\left (6 e^2 \cos \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 )}{b^3 d}+\frac {\left (6 e^2 \cos \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \cos \left (\frac {3 x^2}{b}\right ) \, dx,x,\sqrt {a+b \sin ^{-1}(c+d x)}\right )}{b^3 d}-\frac {\left (6 e^2 \sin \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 )}{b^3 d}+\frac {\left (6 e^2 \sin \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \sin \left (\frac {3 x^2}{b}\right ) \, dx,x,\sqrt {a+b \sin ^{-1}(c+d x)}\right )}{b^3 d}\\ &=-\frac {2 e^2 (c+d x)^2 \sqrt {1-(c+d x)^2}}{3 b d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac {8 e^2 (c+d x)}{3 b^2 d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {4 e^2 (c+d x)^3}{b^2 d \sqrt {a+b \sin ^{-1}(c+d x)}}-\frac {e^2 \sqrt {2 \pi } \cos \left (\frac {a}{b}\right ) C\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}+\frac {e^2 \sqrt {6 \pi } \cos \left (\frac {3 a}{b}\right ) C\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{b^{5/2} d}-\frac {e^2 \sqrt {2 \pi } S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right ) \sin \left (\frac {a}{b}\right )}{3 b^{5/2} d}+\frac {e^2 \sqrt {6 \pi } S\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right ) \sin \left (\frac {3 a}{b}\right )}{b^{5/2} d}\\ \end {align*}

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Mathematica [C]  time = 1.97, size = 411, normalized size = 1.20 \[ \frac {e^2 \left (e^{3 i \sin ^{-1}(c+d x)} \left (6 i a+6 i b \sin ^{-1}(c+d x)+b\right )-i e^{i \sin ^{-1}(c+d x)} \left (2 a+2 b \sin ^{-1}(c+d x)-i b\right )-2 b e^{-\frac {i a}{b}} \left (-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+6 \sqrt {3} b e^{-\frac {3 i a}{b}} \left (-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {3 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+i e^{-i \sin ^{-1}(c+d x)} \left (2 i b e^{\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \left (\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+2 a+2 b \sin ^{-1}(c+d x)+i b\right )+6 \sqrt {3} b e^{\frac {3 i a}{b}} \left (\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},\frac {3 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )-6 i a e^{-3 i \sin ^{-1}(c+d x)}+b e^{-3 i \sin ^{-1}(c+d x)} \left (1-6 i \sin ^{-1}(c+d x)\right )\right )}{12 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(e^2*(((-6*I)*a)/E^((3*I)*ArcSin[c + d*x]) + (b*(1 - (6*I)*ArcSin[c + d*x]))/E^((3*I)*ArcSin[c + d*x]) + E^((3
*I)*ArcSin[c + d*x])*((6*I)*a + b + (6*I)*b*ArcSin[c + d*x]) - I*E^(I*ArcSin[c + d*x])*(2*a - I*b + 2*b*ArcSin
[c + d*x]) - (2*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) + (I*(2*a + I*b + 2*b*ArcSin[c + d*x] + (2*I)*b*E^((I*(a + b*ArcSin[c + d*x]))/b)*((I*(a + b*ArcSin[c +
d*x]))/b)^(3/2)*Gamma[1/2, (I*(a + b*ArcSin[c + d*x]))/b]))/E^(I*ArcSin[c + d*x]) + (6*Sqrt[3]*b*(((-I)*(a +
b*ArcSin[c + d*x]))/b)^(3/2)*Gamma[1/2, ((-3*I)*(a + b*ArcSin[c + d*x]))/b])/E^(((3*I)*a)/b) + 6*Sqrt[3]*b*E^(
((3*I)*a)/b)*((I*(a + b*ArcSin[c + d*x]))/b)^(3/2)*Gamma[1/2, ((3*I)*(a + b*ArcSin[c + d*x]))/b]))/(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)^2/(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 \[ \int \frac {{\left (d e x + c e\right )}^{2}}{{\left (b \arcsin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.34, size = 721, normalized size = 2.11 \[ \frac {e^{2} \left (6 \arcsin \left (d x +c \right ) \sqrt {2}\, \sqrt {\pi }\, \sqrt {3}\, \cos \left (\frac {3 a}{b}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {3}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {\frac {1}{b}}\, \sqrt {a +b \arcsin \left (d x +c \right )}\, b +6 \arcsin \left (d x +c \right ) \sqrt {2}\, \sqrt {\pi }\, \sqrt {3}\, \sin \left (\frac {3 a}{b}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {3}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {\frac {1}{b}}\, \sqrt {a +b \arcsin \left (d x +c \right )}\, b -2 \arcsin \left (d x +c \right ) \sqrt {a +b \arcsin \left (d x +c \right )}\, \sqrt {\frac {1}{b}}\, \cos \left (\frac {a}{b}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {2}\, \sqrt {\pi }\, b -2 \arcsin \left (d x +c \right ) \sqrt {a +b \arcsin \left (d x +c \right )}\, \sqrt {\frac {1}{b}}\, \sin \left (\frac {a}{b}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {2}\, \sqrt {\pi }\, b +6 \sqrt {2}\, \sqrt {\pi }\, \sqrt {3}\, \cos \left (\frac {3 a}{b}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {3}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {\frac {1}{b}}\, \sqrt {a +b \arcsin \left (d x +c \right )}\, a +6 \sqrt {2}\, \sqrt {\pi }\, \sqrt {3}\, \sin \left (\frac {3 a}{b}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {3}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {\frac {1}{b}}\, \sqrt {a +b \arcsin \left (d x +c \right )}\, a -2 \sqrt {a +b \arcsin \left (d x +c \right )}\, \sqrt {\frac {1}{b}}\, \cos \left (\frac {a}{b}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {2}\, \sqrt {\pi }\, a -2 \sqrt {a +b \arcsin \left (d x +c \right )}\, \sqrt {\frac {1}{b}}\, \sin \left (\frac {a}{b}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {2}\, \sqrt {\pi }\, a +2 \sin \left (\frac {a +b \arcsin \left (d x +c \right )}{b}-\frac {a}{b}\right ) \arcsin \left (d x +c \right ) b -6 \sin \left (\frac {3 a +3 b \arcsin \left (d x +c \right )}{b}-\frac {3 a}{b}\right ) \arcsin \left (d x +c \right ) b -\cos \left (\frac {a +b \arcsin \left (d x +c \right )}{b}-\frac {a}{b}\right ) b +2 \sin \left (\frac {a +b \arcsin \left (d x +c \right )}{b}-\frac {a}{b}\right ) a +\cos \left (\frac {3 a +3 b \arcsin \left (d x +c \right )}{b}-\frac {3 a}{b}\right ) b -6 \sin \left (\frac {3 a +3 b \arcsin \left (d x +c \right )}{b}-\frac {3 a}{b}\right ) a \right )}{6 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)^2/(a+b*arcsin(d*x+c))^(5/2),x)

[Out]

1/6/d*e^2/b^2*(6*arcsin(d*x+c)*2^(1/2)*Pi^(1/2)*3^(1/2)*cos(3*a/b)*FresnelC(2^(1/2)/Pi^(1/2)*3^(1/2)/(1/b)^(1/
2)*(a+b*arcsin(d*x+c))^(1/2)/b)*(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)*b+6*arcsin(d*x+c)*2^(1/2)*Pi^(1/2)*3^(1/
2)*sin(3*a/b)*FresnelS(2^(1/2)/Pi^(1/2)*3^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*(1/b)^(1/2)*(a+b*arcs
in(d*x+c))^(1/2)*b-2*arcsin(d*x+c)*(a+b*arcsin(d*x+c))^(1/2)*(1/b)^(1/2)*cos(a/b)*FresnelC(2^(1/2)/Pi^(1/2)/(1
/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*2^(1/2)*Pi^(1/2)*b-2*arcsin(d*x+c)*(a+b*arcsin(d*x+c))^(1/2)*(1/b)^(1/2
)*sin(a/b)*FresnelS(2^(1/2)/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*2^(1/2)*Pi^(1/2)*b+6*2^(1/2)*Pi^
(1/2)*3^(1/2)*cos(3*a/b)*FresnelC(2^(1/2)/Pi^(1/2)*3^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*(1/b)^(1/2
)*(a+b*arcsin(d*x+c))^(1/2)*a+6*2^(1/2)*Pi^(1/2)*3^(1/2)*sin(3*a/b)*FresnelS(2^(1/2)/Pi^(1/2)*3^(1/2)/(1/b)^(1
/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)*a-2*(a+b*arcsin(d*x+c))^(1/2)*(1/b)^(1/
2)*cos(a/b)*FresnelC(2^(1/2)/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*2^(1/2)*Pi^(1/2)*a-2*(a+b*arcsi
n(d*x+c))^(1/2)*(1/b)^(1/2)*sin(a/b)*FresnelS(2^(1/2)/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*2^(1/2
)*Pi^(1/2)*a+2*sin((a+b*arcsin(d*x+c))/b-a/b)*arcsin(d*x+c)*b-6*sin(3*(a+b*arcsin(d*x+c))/b-3*a/b)*arcsin(d*x+
c)*b-cos((a+b*arcsin(d*x+c))/b-a/b)*b+2*sin((a+b*arcsin(d*x+c))/b-a/b)*a+cos(3*(a+b*arcsin(d*x+c))/b-3*a/b)*b-
6*sin(3*(a+b*arcsin(d*x+c))/b-3*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 \[ \int \frac {{\left (d e x + c e\right )}^{2}}{{\left (b \arcsin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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