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

Optimal. Leaf size=390 \[ \frac {5 \sqrt {\frac {\pi }{2}} d^2 \sin \left (\frac {a}{b}\right ) C\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {b}}\right )}{2 b^{3/2} c}+\frac {5 \sqrt {\frac {3 \pi }{2}} d^2 \sin \left (\frac {3 a}{b}\right ) C\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {b}}\right )}{4 b^{3/2} c}+\frac {\sqrt {\frac {5 \pi }{2}} d^2 \sin \left (\frac {5 a}{b}\right ) C\left (\frac {\sqrt {\frac {10}{\pi }} \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {b}}\right )}{4 b^{3/2} c}-\frac {5 \sqrt {\frac {\pi }{2}} d^2 \cos \left (\frac {a}{b}\right ) S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {b}}\right )}{2 b^{3/2} c}-\frac {5 \sqrt {\frac {3 \pi }{2}} d^2 \cos \left (\frac {3 a}{b}\right ) S\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {b}}\right )}{4 b^{3/2} c}-\frac {\sqrt {\frac {5 \pi }{2}} d^2 \cos \left (\frac {5 a}{b}\right ) S\left (\frac {\sqrt {\frac {10}{\pi }} \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {b}}\right )}{4 b^{3/2} c}-\frac {2 d^2 \left (1-c^2 x^2\right )^{5/2}}{b c \sqrt {a+b \sin ^{-1}(c x)}} \]

[Out]

-5/4*d^2*cos(a/b)*FresnelS(2^(1/2)/Pi^(1/2)*(a+b*arcsin(c*x))^(1/2)/b^(1/2))*2^(1/2)*Pi^(1/2)/b^(3/2)/c+5/4*d^
2*FresnelC(2^(1/2)/Pi^(1/2)*(a+b*arcsin(c*x))^(1/2)/b^(1/2))*sin(a/b)*2^(1/2)*Pi^(1/2)/b^(3/2)/c-5/8*d^2*cos(3
*a/b)*FresnelS(6^(1/2)/Pi^(1/2)*(a+b*arcsin(c*x))^(1/2)/b^(1/2))*6^(1/2)*Pi^(1/2)/b^(3/2)/c+5/8*d^2*FresnelC(6
^(1/2)/Pi^(1/2)*(a+b*arcsin(c*x))^(1/2)/b^(1/2))*sin(3*a/b)*6^(1/2)*Pi^(1/2)/b^(3/2)/c-1/8*d^2*cos(5*a/b)*Fres
nelS(10^(1/2)/Pi^(1/2)*(a+b*arcsin(c*x))^(1/2)/b^(1/2))*10^(1/2)*Pi^(1/2)/b^(3/2)/c+1/8*d^2*FresnelC(10^(1/2)/
Pi^(1/2)*(a+b*arcsin(c*x))^(1/2)/b^(1/2))*sin(5*a/b)*10^(1/2)*Pi^(1/2)/b^(3/2)/c-2*d^2*(-c^2*x^2+1)^(5/2)/b/c/
(a+b*arcsin(c*x))^(1/2)

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Rubi [A]  time = 0.82, antiderivative size = 390, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {4659, 4723, 4406, 3306, 3305, 3351, 3304, 3352} \[ \frac {5 \sqrt {\frac {\pi }{2}} d^2 \sin \left (\frac {a}{b}\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {b}}\right )}{2 b^{3/2} c}+\frac {5 \sqrt {\frac {3 \pi }{2}} d^2 \sin \left (\frac {3 a}{b}\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {b}}\right )}{4 b^{3/2} c}+\frac {\sqrt {\frac {5 \pi }{2}} d^2 \sin \left (\frac {5 a}{b}\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {10}{\pi }} \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {b}}\right )}{4 b^{3/2} c}-\frac {5 \sqrt {\frac {\pi }{2}} d^2 \cos \left (\frac {a}{b}\right ) S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {b}}\right )}{2 b^{3/2} c}-\frac {5 \sqrt {\frac {3 \pi }{2}} d^2 \cos \left (\frac {3 a}{b}\right ) S\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {b}}\right )}{4 b^{3/2} c}-\frac {\sqrt {\frac {5 \pi }{2}} d^2 \cos \left (\frac {5 a}{b}\right ) S\left (\frac {\sqrt {\frac {10}{\pi }} \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {b}}\right )}{4 b^{3/2} c}-\frac {2 d^2 \left (1-c^2 x^2\right )^{5/2}}{b c \sqrt {a+b \sin ^{-1}(c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d^2*(1 - c^2*x^2)^(5/2))/(b*c*Sqrt[a + b*ArcSin[c*x]]) - (5*d^2*Sqrt[Pi/2]*Cos[a/b]*FresnelS[(Sqrt[2/Pi]*S
qrt[a + b*ArcSin[c*x]])/Sqrt[b]])/(2*b^(3/2)*c) - (5*d^2*Sqrt[(3*Pi)/2]*Cos[(3*a)/b]*FresnelS[(Sqrt[6/Pi]*Sqrt
[a + b*ArcSin[c*x]])/Sqrt[b]])/(4*b^(3/2)*c) - (d^2*Sqrt[(5*Pi)/2]*Cos[(5*a)/b]*FresnelS[(Sqrt[10/Pi]*Sqrt[a +
b*ArcSin[c*x]])/Sqrt[b]])/(4*b^(3/2)*c) + (5*d^2*Sqrt[Pi/2]*FresnelC[(Sqrt[2/Pi]*Sqrt[a + b*ArcSin[c*x]])/Sqr
t[b]]*Sin[a/b])/(2*b^(3/2)*c) + (5*d^2*Sqrt[(3*Pi)/2]*FresnelC[(Sqrt[6/Pi]*Sqrt[a + b*ArcSin[c*x]])/Sqrt[b]]*S
in[(3*a)/b])/(4*b^(3/2)*c) + (d^2*Sqrt[(5*Pi)/2]*FresnelC[(Sqrt[10/Pi]*Sqrt[a + b*ArcSin[c*x]])/Sqrt[b]]*Sin[(
5*a)/b])/(4*b^(3/2)*c)

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 4659

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

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])

Rubi steps

\begin {align*} \int \frac {\left (d-c^2 d x^2\right )^2}{\left (a+b \sin ^{-1}(c x)\right )^{3/2}} \, dx &=-\frac {2 d^2 \left (1-c^2 x^2\right )^{5/2}}{b c \sqrt {a+b \sin ^{-1}(c x)}}-\frac {\left (10 c d^2\right ) \int \frac {x \left (1-c^2 x^2\right )^{3/2}}{\sqrt {a+b \sin ^{-1}(c x)}} \, dx}{b}\\ &=-\frac {2 d^2 \left (1-c^2 x^2\right )^{5/2}}{b c \sqrt {a+b \sin ^{-1}(c x)}}-\frac {\left (10 d^2\right ) \operatorname {Subst}\left (\int \frac {\cos ^4(x) \sin (x)}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c x)\right )}{b c}\\ &=-\frac {2 d^2 \left (1-c^2 x^2\right )^{5/2}}{b c \sqrt {a+b \sin ^{-1}(c x)}}-\frac {\left (10 d^2\right ) \operatorname {Subst}\left (\int \left (\frac {\sin (x)}{8 \sqrt {a+b x}}+\frac {3 \sin (3 x)}{16 \sqrt {a+b x}}+\frac {\sin (5 x)}{16 \sqrt {a+b x}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{b c}\\ &=-\frac {2 d^2 \left (1-c^2 x^2\right )^{5/2}}{b c \sqrt {a+b \sin ^{-1}(c x)}}-\frac {\left (5 d^2\right ) \operatorname {Subst}\left (\int \frac {\sin (5 x)}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c x)\right )}{8 b c}-\frac {\left (5 d^2\right ) \operatorname {Subst}\left (\int \frac {\sin (x)}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c}-\frac {\left (15 d^2\right ) \operatorname {Subst}\left (\int \frac {\sin (3 x)}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c x)\right )}{8 b c}\\ &=-\frac {2 d^2 \left (1-c^2 x^2\right )^{5/2}}{b c \sqrt {a+b \sin ^{-1}(c x)}}-\frac {\left (5 d^2 \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 x)\right )}{4 b c}-\frac {\left (15 d^2 \cos \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 x)\right )}{8 b c}-\frac {\left (5 d^2 \cos \left (\frac {5 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {5 a}{b}+5 x\right )}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c x)\right )}{8 b c}+\frac {\left (5 d^2 \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 x)\right )}{4 b c}+\frac {\left (15 d^2 \sin \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 x)\right )}{8 b c}+\frac {\left (5 d^2 \sin \left (\frac {5 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {5 a}{b}+5 x\right )}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c x)\right )}{8 b c}\\ &=-\frac {2 d^2 \left (1-c^2 x^2\right )^{5/2}}{b c \sqrt {a+b \sin ^{-1}(c x)}}-\frac {\left (5 d^2 \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 x)}\right )}{2 b^2 c}-\frac {\left (15 d^2 \cos \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 x)}\right )}{4 b^2 c}-\frac {\left (5 d^2 \cos \left (\frac {5 a}{b}\right )\right ) \operatorname {Subst}\left (\int \sin \left (\frac {5 x^2}{b}\right ) \, dx,x,\sqrt {a+b \sin ^{-1}(c x)}\right )}{4 b^2 c}+\frac {\left (5 d^2 \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 x)}\right )}{2 b^2 c}+\frac {\left (15 d^2 \sin \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 x)}\right )}{4 b^2 c}+\frac {\left (5 d^2 \sin \left (\frac {5 a}{b}\right )\right ) \operatorname {Subst}\left (\int \cos \left (\frac {5 x^2}{b}\right ) \, dx,x,\sqrt {a+b \sin ^{-1}(c x)}\right )}{4 b^2 c}\\ &=-\frac {2 d^2 \left (1-c^2 x^2\right )^{5/2}}{b c \sqrt {a+b \sin ^{-1}(c x)}}-\frac {5 d^2 \sqrt {\frac {\pi }{2}} \cos \left (\frac {a}{b}\right ) S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {b}}\right )}{2 b^{3/2} c}-\frac {5 d^2 \sqrt {\frac {3 \pi }{2}} \cos \left (\frac {3 a}{b}\right ) S\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {b}}\right )}{4 b^{3/2} c}-\frac {d^2 \sqrt {\frac {5 \pi }{2}} \cos \left (\frac {5 a}{b}\right ) S\left (\frac {\sqrt {\frac {10}{\pi }} \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {b}}\right )}{4 b^{3/2} c}+\frac {5 d^2 \sqrt {\frac {\pi }{2}} C\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {b}}\right ) \sin \left (\frac {a}{b}\right )}{2 b^{3/2} c}+\frac {5 d^2 \sqrt {\frac {3 \pi }{2}} C\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {b}}\right ) \sin \left (\frac {3 a}{b}\right )}{4 b^{3/2} c}+\frac {d^2 \sqrt {\frac {5 \pi }{2}} C\left (\frac {\sqrt {\frac {10}{\pi }} \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {b}}\right ) \sin \left (\frac {5 a}{b}\right )}{4 b^{3/2} c}\\ \end {align*}

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Mathematica [C]  time = 2.74, size = 522, normalized size = 1.34 \[ \frac {d^2 e^{-\frac {5 i \left (a+b \sin ^{-1}(c x)\right )}{b}} \left (-5 e^{\frac {5 i a}{b}+2 i \sin ^{-1}(c x)}-10 e^{\frac {5 i a}{b}+4 i \sin ^{-1}(c x)}-10 e^{\frac {5 i a}{b}+6 i \sin ^{-1}(c x)}-5 e^{\frac {5 i a}{b}+8 i \sin ^{-1}(c x)}-e^{\frac {5 i \left (a+2 b \sin ^{-1}(c x)\right )}{b}}+10 e^{\frac {4 i a}{b}+5 i \sin ^{-1}(c x)} \sqrt {-\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}} \Gamma \left (\frac {1}{2},-\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )+10 e^{\frac {6 i a}{b}+5 i \sin ^{-1}(c x)} \sqrt {\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}} \Gamma \left (\frac {1}{2},\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )+5 \sqrt {3} e^{\frac {2 i a}{b}+5 i \sin ^{-1}(c x)} \sqrt {-\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}} \Gamma \left (\frac {1}{2},-\frac {3 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )+5 \sqrt {3} e^{\frac {8 i a}{b}+5 i \sin ^{-1}(c x)} \sqrt {\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}} \Gamma \left (\frac {1}{2},\frac {3 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )+\sqrt {5} e^{5 i \sin ^{-1}(c x)} \sqrt {-\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}} \Gamma \left (\frac {1}{2},-\frac {5 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )+\sqrt {5} e^{\frac {5 i \left (2 a+b \sin ^{-1}(c x)\right )}{b}} \sqrt {\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}} \Gamma \left (\frac {1}{2},\frac {5 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )-e^{\frac {5 i a}{b}}\right )}{16 b c \sqrt {a+b \sin ^{-1}(c x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(d^2*(-E^(((5*I)*a)/b) - 5*E^(((5*I)*a)/b + (2*I)*ArcSin[c*x]) - 10*E^(((5*I)*a)/b + (4*I)*ArcSin[c*x]) - 10*E
^(((5*I)*a)/b + (6*I)*ArcSin[c*x]) - 5*E^(((5*I)*a)/b + (8*I)*ArcSin[c*x]) - E^(((5*I)*(a + 2*b*ArcSin[c*x]))/
b) + 10*E^(((4*I)*a)/b + (5*I)*ArcSin[c*x])*Sqrt[((-I)*(a + b*ArcSin[c*x]))/b]*Gamma[1/2, ((-I)*(a + b*ArcSin[
c*x]))/b] + 10*E^(((6*I)*a)/b + (5*I)*ArcSin[c*x])*Sqrt[(I*(a + b*ArcSin[c*x]))/b]*Gamma[1/2, (I*(a + b*ArcSin
[c*x]))/b] + 5*Sqrt[3]*E^(((2*I)*a)/b + (5*I)*ArcSin[c*x])*Sqrt[((-I)*(a + b*ArcSin[c*x]))/b]*Gamma[1/2, ((-3*
I)*(a + b*ArcSin[c*x]))/b] + 5*Sqrt[3]*E^(((8*I)*a)/b + (5*I)*ArcSin[c*x])*Sqrt[(I*(a + b*ArcSin[c*x]))/b]*Gam
ma[1/2, ((3*I)*(a + b*ArcSin[c*x]))/b] + Sqrt[5]*E^((5*I)*ArcSin[c*x])*Sqrt[((-I)*(a + b*ArcSin[c*x]))/b]*Gamm
a[1/2, ((-5*I)*(a + b*ArcSin[c*x]))/b] + Sqrt[5]*E^(((5*I)*(2*a + b*ArcSin[c*x]))/b)*Sqrt[(I*(a + b*ArcSin[c*x
]))/b]*Gamma[1/2, ((5*I)*(a + b*ArcSin[c*x]))/b]))/(16*b*c*E^(((5*I)*(a + b*ArcSin[c*x]))/b)*Sqrt[a + b*ArcSin
[c*x]])

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.23, size = 446, normalized size = 1.14 \[ -\frac {d^{2} \left (5 \sqrt {3}\, \sqrt {\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arcsin \left (c x \right )}\, \cos \left (\frac {3 a}{b}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {3}\, \sqrt {a +b \arcsin \left (c x \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right )-5 \sqrt {3}\, \sqrt {\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arcsin \left (c x \right )}\, \sin \left (\frac {3 a}{b}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {3}\, \sqrt {a +b \arcsin \left (c x \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right )+\sqrt {5}\, \sqrt {\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arcsin \left (c x \right )}\, \cos \left (\frac {5 a}{b}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {5}\, \sqrt {a +b \arcsin \left (c x \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right )-\sqrt {5}\, \sqrt {\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arcsin \left (c x \right )}\, \sin \left (\frac {5 a}{b}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {5}\, \sqrt {a +b \arcsin \left (c x \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right )+10 \sqrt {\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arcsin \left (c x \right )}\, \cos \left (\frac {a}{b}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (c x \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right )-10 \sqrt {\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arcsin \left (c x \right )}\, \sin \left (\frac {a}{b}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (c x \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right )+10 \cos \left (\frac {a +b \arcsin \left (c x \right )}{b}-\frac {a}{b}\right )+5 \cos \left (\frac {3 a +3 b \arcsin \left (c x \right )}{b}-\frac {3 a}{b}\right )+\cos \left (\frac {5 a +5 b \arcsin \left (c x \right )}{b}-\frac {5 a}{b}\right )\right )}{8 c b \sqrt {a +b \arcsin \left (c x \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/c*d^2/b*(5*3^(1/2)*(1/b)^(1/2)*Pi^(1/2)*2^(1/2)*(a+b*arcsin(c*x))^(1/2)*cos(3*a/b)*FresnelS(2^(1/2)/Pi^(1
/2)*3^(1/2)/(1/b)^(1/2)*(a+b*arcsin(c*x))^(1/2)/b)-5*3^(1/2)*(1/b)^(1/2)*Pi^(1/2)*2^(1/2)*(a+b*arcsin(c*x))^(1
/2)*sin(3*a/b)*FresnelC(2^(1/2)/Pi^(1/2)*3^(1/2)/(1/b)^(1/2)*(a+b*arcsin(c*x))^(1/2)/b)+5^(1/2)*(1/b)^(1/2)*Pi
^(1/2)*2^(1/2)*(a+b*arcsin(c*x))^(1/2)*cos(5*a/b)*FresnelS(2^(1/2)/Pi^(1/2)*5^(1/2)/(1/b)^(1/2)*(a+b*arcsin(c*
x))^(1/2)/b)-5^(1/2)*(1/b)^(1/2)*Pi^(1/2)*2^(1/2)*(a+b*arcsin(c*x))^(1/2)*sin(5*a/b)*FresnelC(2^(1/2)/Pi^(1/2)
*5^(1/2)/(1/b)^(1/2)*(a+b*arcsin(c*x))^(1/2)/b)+10*(1/b)^(1/2)*Pi^(1/2)*2^(1/2)*(a+b*arcsin(c*x))^(1/2)*cos(a/
b)*FresnelS(2^(1/2)/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(c*x))^(1/2)/b)-10*(1/b)^(1/2)*Pi^(1/2)*2^(1/2)*(a+b*arcsi
n(c*x))^(1/2)*sin(a/b)*FresnelC(2^(1/2)/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(c*x))^(1/2)/b)+10*cos((a+b*arcsin(c*x
))/b-a/b)+5*cos(3*(a+b*arcsin(c*x))/b-3*a/b)+cos(5*(a+b*arcsin(c*x))/b-5*a/b))/(a+b*arcsin(c*x))^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ d^{2} \left (\int \left (- \frac {2 c^{2} x^{2}}{a \sqrt {a + b \operatorname {asin}{\left (c x \right )}} + b \sqrt {a + b \operatorname {asin}{\left (c x \right )}} \operatorname {asin}{\left (c x \right )}}\right )\, dx + \int \frac {c^{4} x^{4}}{a \sqrt {a + b \operatorname {asin}{\left (c x \right )}} + b \sqrt {a + b \operatorname {asin}{\left (c x \right )}} \operatorname {asin}{\left (c x \right )}}\, dx + \int \frac {1}{a \sqrt {a + b \operatorname {asin}{\left (c x \right )}} + b \sqrt {a + b \operatorname {asin}{\left (c x \right )}} \operatorname {asin}{\left (c x \right )}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**2*(Integral(-2*c**2*x**2/(a*sqrt(a + b*asin(c*x)) + b*sqrt(a + b*asin(c*x))*asin(c*x)), x) + Integral(c**4*
x**4/(a*sqrt(a + b*asin(c*x)) + b*sqrt(a + b*asin(c*x))*asin(c*x)), x) + Integral(1/(a*sqrt(a + b*asin(c*x)) +
b*sqrt(a + b*asin(c*x))*asin(c*x)), x))

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