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

Optimal. Leaf size=241 \[ \frac {\sqrt {\frac {\pi }{2}} d \cos \left (\frac {4 a}{b}\right ) C\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {b}}\right )}{b^{3/2} c^2}+\frac {\sqrt {\pi } d \cos \left (\frac {2 a}{b}\right ) C\left (\frac {2 \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {b} \sqrt {\pi }}\right )}{b^{3/2} c^2}+\frac {\sqrt {\pi } d \sin \left (\frac {2 a}{b}\right ) S\left (\frac {2 \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {b} \sqrt {\pi }}\right )}{b^{3/2} c^2}+\frac {\sqrt {\frac {\pi }{2}} d \sin \left (\frac {4 a}{b}\right ) S\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {b}}\right )}{b^{3/2} c^2}-\frac {2 d x \left (1-c^2 x^2\right )^{3/2}}{b c \sqrt {a+b \sin ^{-1}(c x)}} \]

[Out]

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

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Rubi [A]  time = 0.79, antiderivative size = 241, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4721, 4661, 3312, 3306, 3305, 3351, 3304, 3352, 4723, 4406} \[ \frac {\sqrt {\frac {\pi }{2}} d \cos \left (\frac {4 a}{b}\right ) \text {FresnelC}\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {b}}\right )}{b^{3/2} c^2}+\frac {\sqrt {\pi } d \cos \left (\frac {2 a}{b}\right ) \text {FresnelC}\left (\frac {2 \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {\pi } \sqrt {b}}\right )}{b^{3/2} c^2}+\frac {\sqrt {\pi } d \sin \left (\frac {2 a}{b}\right ) S\left (\frac {2 \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {b} \sqrt {\pi }}\right )}{b^{3/2} c^2}+\frac {\sqrt {\frac {\pi }{2}} d \sin \left (\frac {4 a}{b}\right ) S\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {b}}\right )}{b^{3/2} c^2}-\frac {2 d x \left (1-c^2 x^2\right )^{3/2}}{b c \sqrt {a+b \sin ^{-1}(c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d*x*(1 - c^2*x^2)^(3/2))/(b*c*Sqrt[a + b*ArcSin[c*x]]) + (d*Sqrt[Pi/2]*Cos[(4*a)/b]*FresnelC[(2*Sqrt[2/Pi]
*Sqrt[a + b*ArcSin[c*x]])/Sqrt[b]])/(b^(3/2)*c^2) + (d*Sqrt[Pi]*Cos[(2*a)/b]*FresnelC[(2*Sqrt[a + b*ArcSin[c*x
]])/(Sqrt[b]*Sqrt[Pi])])/(b^(3/2)*c^2) + (d*Sqrt[Pi]*FresnelS[(2*Sqrt[a + b*ArcSin[c*x]])/(Sqrt[b]*Sqrt[Pi])]*
Sin[(2*a)/b])/(b^(3/2)*c^2) + (d*Sqrt[Pi/2]*FresnelS[(2*Sqrt[2/Pi]*Sqrt[a + b*ArcSin[c*x]])/Sqrt[b]]*Sin[(4*a)
/b])/(b^(3/2)*c^2)

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 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

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 4661

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c, Subst[Int[(
a + b*x)^n*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] && I
GtQ[2*p, 0] && (IntegerQ[p] || GtQ[d, 0])

Rule 4721

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[
((f*x)^m*Sqrt[1 - c^2*x^2]*(d + e*x^2)^p*(a + b*ArcSin[c*x])^(n + 1))/(b*c*(n + 1)), x] + (-Dist[(f*m*d^IntPar
t[p]*(d + e*x^2)^FracPart[p])/(b*c*(n + 1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m - 1)*(1 - c^2*x^2)^(p - 1/
2)*(a + b*ArcSin[c*x])^(n + 1), x], x] + Dist[(c*(m + 2*p + 1)*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(b*f*(n +
1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 - c^2*x^2)^(p - 1/2)*(a + b*ArcSin[c*x])^(n + 1), x], x])
/; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] && LtQ[n, -1] && IGtQ[m, -3] && IGtQ[2*p, 0]

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

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Mathematica [C]  time = 2.46, size = 375, normalized size = 1.56 \[ \frac {d \left (8 \sqrt {\pi } \left (\frac {1}{b}\right )^{3/2} \cos \left (\frac {2 a}{b}\right ) C\left (\frac {2 \sqrt {\frac {1}{b}} \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {\pi }}\right )+8 \sqrt {\pi } \left (\frac {1}{b}\right )^{3/2} \sin \left (\frac {2 a}{b}\right ) S\left (\frac {2 \sqrt {\frac {1}{b}} \sqrt {a+b \sin ^{-1}(c x)}}{\sqrt {\pi }}\right )+\frac {i e^{-\frac {4 i a}{b}} \left (2 i e^{\frac {4 i a}{b}} \sin \left (2 \sin ^{-1}(c x)\right )+i e^{\frac {4 i a}{b}} \sin \left (4 \sin ^{-1}(c x)\right )+\sqrt {2} e^{\frac {2 i a}{b}} \sqrt {-\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}} \Gamma \left (\frac {1}{2},-\frac {2 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )-\sqrt {2} e^{\frac {6 i a}{b}} \sqrt {\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}} \Gamma \left (\frac {1}{2},\frac {2 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )-\sqrt {-\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}} \Gamma \left (\frac {1}{2},-\frac {4 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )+e^{\frac {8 i a}{b}} \sqrt {\frac {i \left (a+b \sin ^{-1}(c x)\right )}{b}} \Gamma \left (\frac {1}{2},\frac {4 i \left (a+b \sin ^{-1}(c x)\right )}{b}\right )\right )}{b \sqrt {a+b \sin ^{-1}(c x)}}\right )}{4 c^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(d*(8*(b^(-1))^(3/2)*Sqrt[Pi]*Cos[(2*a)/b]*FresnelC[(2*Sqrt[b^(-1)]*Sqrt[a + b*ArcSin[c*x]])/Sqrt[Pi]] + 8*(b^
(-1))^(3/2)*Sqrt[Pi]*FresnelS[(2*Sqrt[b^(-1)]*Sqrt[a + b*ArcSin[c*x]])/Sqrt[Pi]]*Sin[(2*a)/b] + (I*(Sqrt[2]*E^
(((2*I)*a)/b)*Sqrt[((-I)*(a + b*ArcSin[c*x]))/b]*Gamma[1/2, ((-2*I)*(a + b*ArcSin[c*x]))/b] - Sqrt[2]*E^(((6*I
)*a)/b)*Sqrt[(I*(a + b*ArcSin[c*x]))/b]*Gamma[1/2, ((2*I)*(a + b*ArcSin[c*x]))/b] - Sqrt[((-I)*(a + b*ArcSin[c
*x]))/b]*Gamma[1/2, ((-4*I)*(a + b*ArcSin[c*x]))/b] + E^(((8*I)*a)/b)*Sqrt[(I*(a + b*ArcSin[c*x]))/b]*Gamma[1/
2, ((4*I)*(a + b*ArcSin[c*x]))/b] + (2*I)*E^(((4*I)*a)/b)*Sin[2*ArcSin[c*x]] + I*E^(((4*I)*a)/b)*Sin[4*ArcSin[
c*x]]))/(b*E^(((4*I)*a)/b)*Sqrt[a + b*ArcSin[c*x]])))/(4*c^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(x*(-c^2*d*x^2+d)/(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 \[ \int -\frac {{\left (c^{2} d x^{2} - d\right )} x}{{\left (b \arcsin \left (c x\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.21, size = 281, normalized size = 1.17 \[ -\frac {d \left (-2 \sqrt {\pi }\, \sqrt {\frac {1}{b}}\, \sqrt {a +b \arcsin \left (c x \right )}\, \cos \left (\frac {4 a}{b}\right ) \FresnelC \left (\frac {2 \sqrt {2}\, \sqrt {a +b \arcsin \left (c x \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {2}-2 \sqrt {\pi }\, \sqrt {\frac {1}{b}}\, \sqrt {a +b \arcsin \left (c x \right )}\, \sin \left (\frac {4 a}{b}\right ) \mathrm {S}\left (\frac {2 \sqrt {2}\, \sqrt {a +b \arcsin \left (c x \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {2}-4 \sqrt {\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {a +b \arcsin \left (c x \right )}\, \cos \left (\frac {2 a}{b}\right ) \FresnelC \left (\frac {2 \sqrt {a +b \arcsin \left (c x \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right )-4 \sqrt {\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {a +b \arcsin \left (c x \right )}\, \sin \left (\frac {2 a}{b}\right ) \mathrm {S}\left (\frac {2 \sqrt {a +b \arcsin \left (c x \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right )+2 \sin \left (\frac {2 a +2 b \arcsin \left (c x \right )}{b}-\frac {2 a}{b}\right )+\sin \left (\frac {4 a +4 b \arcsin \left (c x \right )}{b}-\frac {4 a}{b}\right )\right )}{4 c^{2} b \sqrt {a +b \arcsin \left (c x \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/c^2*d/b/(a+b*arcsin(c*x))^(1/2)*(-2*Pi^(1/2)*(1/b)^(1/2)*(a+b*arcsin(c*x))^(1/2)*cos(4*a/b)*FresnelC(2*2^
(1/2)/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(c*x))^(1/2)/b)*2^(1/2)-2*Pi^(1/2)*(1/b)^(1/2)*(a+b*arcsin(c*x))^(1/2)*s
in(4*a/b)*FresnelS(2*2^(1/2)/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(c*x))^(1/2)/b)*2^(1/2)-4*(1/b)^(1/2)*Pi^(1/2)*(a
+b*arcsin(c*x))^(1/2)*cos(2*a/b)*FresnelC(2/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(c*x))^(1/2)/b)-4*(1/b)^(1/2)*Pi^(
1/2)*(a+b*arcsin(c*x))^(1/2)*sin(2*a/b)*FresnelS(2/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(c*x))^(1/2)/b)+2*sin(2*(a+
b*arcsin(c*x))/b-2*a/b)+sin(4*(a+b*arcsin(c*x))/b-4*a/b))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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