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

Optimal. Leaf size=179 \[ -\frac {4 \sqrt {2 \pi } \cos \left (\frac {a}{b}\right ) C\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}-\frac {4 \sqrt {2 \pi } \sin \left (\frac {a}{b}\right ) S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}+\frac {4 (c+d x)}{3 b^2 d \sqrt {a+b \sin ^{-1}(c+d x)}}-\frac {2 \sqrt {1-(c+d x)^2}}{3 b d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}} \]

[Out]

-4/3*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-4/3*Fres
nelS(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-2/3*(1-(d*x+c)^2)
^(1/2)/b/d/(a+b*arcsin(d*x+c))^(3/2)+4/3*(d*x+c)/b^2/d/(a+b*arcsin(d*x+c))^(1/2)

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Rubi [A]  time = 0.28, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {4803, 4621, 4719, 4623, 3306, 3305, 3351, 3304, 3352} \[ -\frac {4 \sqrt {2 \pi } \cos \left (\frac {a}{b}\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}-\frac {4 \sqrt {2 \pi } \sin \left (\frac {a}{b}\right ) S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}+\frac {4 (c+d x)}{3 b^2 d \sqrt {a+b \sin ^{-1}(c+d x)}}-\frac {2 \sqrt {1-(c+d x)^2}}{3 b d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c + d*x])^(-5/2),x]

[Out]

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

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 4621

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^(n + 1))
/(b*c*(n + 1)), x] + Dist[c/(b*(n + 1)), Int[(x*(a + b*ArcSin[c*x])^(n + 1))/Sqrt[1 - c^2*x^2], x], x] /; Free
Q[{a, b, c}, x] && LtQ[n, -1]

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 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 4803

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcSin[x])^n, x],
x, c + d*x], x] /; FreeQ[{a, b, c, d, n}, x]

Rubi steps

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

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Mathematica [C]  time = 0.60, size = 238, normalized size = 1.33 \[ \frac {e^{-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \left (-2 b e^{i \sin ^{-1}(c+d x)} \left (-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )-i e^{\frac {i a}{b}} \left (-2 i b e^{\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \left (\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+2 a \left (-1+e^{2 i \sin ^{-1}(c+d x)}\right )+b \left (-2 \sin ^{-1}(c+d x)+e^{2 i \sin ^{-1}(c+d x)} \left (2 \sin ^{-1}(c+d x)-i\right )-i\right )\right )\right )}{3 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arcsin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.00, size = 355, normalized size = 1.98 \[ \frac {-\frac {4 \arcsin \left (d x +c \right ) \sqrt {a +b \arcsin \left (d x +c \right )}\, \sqrt {\frac {1}{b}}\, \cos \left (\frac {a}{b}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {2}\, \sqrt {\pi }\, b}{3}-\frac {4 \arcsin \left (d x +c \right ) \sqrt {a +b \arcsin \left (d x +c \right )}\, \sqrt {\frac {1}{b}}\, \sin \left (\frac {a}{b}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {2}\, \sqrt {\pi }\, b}{3}-\frac {4 \sqrt {a +b \arcsin \left (d x +c \right )}\, \sqrt {\frac {1}{b}}\, \cos \left (\frac {a}{b}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {2}\, \sqrt {\pi }\, a}{3}-\frac {4 \sqrt {a +b \arcsin \left (d x +c \right )}\, \sqrt {\frac {1}{b}}\, \sin \left (\frac {a}{b}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {2}\, \sqrt {\pi }\, a}{3}+\frac {4 \sin \left (\frac {a +b \arcsin \left (d x +c \right )}{b}-\frac {a}{b}\right ) \arcsin \left (d x +c \right ) b}{3}-\frac {2 \cos \left (\frac {a +b \arcsin \left (d x +c \right )}{b}-\frac {a}{b}\right ) b}{3}+\frac {4 \sin \left (\frac {a +b \arcsin \left (d x +c \right )}{b}-\frac {a}{b}\right ) a}{3}}{d \,b^{2} \left (a +b \arcsin \left (d x +c \right )\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*arcsin(d*x+c))^(5/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \arcsin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arcsin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*asin(c + d*x))^(5/2),x)

[Out]

int(1/(a + b*asin(c + d*x))^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \operatorname {asin}{\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*asin(d*x+c))**(5/2),x)

[Out]

Integral((a + b*asin(c + d*x))**(-5/2), x)

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