3.279 \(\int \frac{1}{(a+b \sin ^{-1}(c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=218 \[ -\frac{8 \sqrt{2 \pi } \sin \left (\frac{a}{b}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{15 b^{7/2} d}+\frac{8 \sqrt{2 \pi } \cos \left (\frac{a}{b}\right ) S\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{15 b^{7/2} d}+\frac{4 (c+d x)}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac{8 \sqrt{1-(c+d x)^2}}{15 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}-\frac{2 \sqrt{1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}} \]

[Out]

(-2*Sqrt[1 - (c + d*x)^2])/(5*b*d*(a + b*ArcSin[c + d*x])^(5/2)) + (4*(c + d*x))/(15*b^2*d*(a + b*ArcSin[c + d
*x])^(3/2)) + (8*Sqrt[1 - (c + d*x)^2])/(15*b^3*d*Sqrt[a + b*ArcSin[c + d*x]]) + (8*Sqrt[2*Pi]*Cos[a/b]*Fresne
lS[(Sqrt[2/Pi]*Sqrt[a + b*ArcSin[c + d*x]])/Sqrt[b]])/(15*b^(7/2)*d) - (8*Sqrt[2*Pi]*FresnelC[(Sqrt[2/Pi]*Sqrt
[a + b*ArcSin[c + d*x]])/Sqrt[b]]*Sin[a/b])/(15*b^(7/2)*d)

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Rubi [A]  time = 0.451615, antiderivative size = 218, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.643, Rules used = {4803, 4621, 4719, 4723, 3306, 3305, 3351, 3304, 3352} \[ -\frac{8 \sqrt{2 \pi } \sin \left (\frac{a}{b}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{15 b^{7/2} d}+\frac{8 \sqrt{2 \pi } \cos \left (\frac{a}{b}\right ) S\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{15 b^{7/2} d}+\frac{4 (c+d x)}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac{8 \sqrt{1-(c+d x)^2}}{15 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}-\frac{2 \sqrt{1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c + d*x])^(-7/2),x]

[Out]

(-2*Sqrt[1 - (c + d*x)^2])/(5*b*d*(a + b*ArcSin[c + d*x])^(5/2)) + (4*(c + d*x))/(15*b^2*d*(a + b*ArcSin[c + d
*x])^(3/2)) + (8*Sqrt[1 - (c + d*x)^2])/(15*b^3*d*Sqrt[a + b*ArcSin[c + d*x]]) + (8*Sqrt[2*Pi]*Cos[a/b]*Fresne
lS[(Sqrt[2/Pi]*Sqrt[a + b*ArcSin[c + d*x]])/Sqrt[b]])/(15*b^(7/2)*d) - (8*Sqrt[2*Pi]*FresnelC[(Sqrt[2/Pi]*Sqrt
[a + b*ArcSin[c + d*x]])/Sqrt[b]]*Sin[a/b])/(15*b^(7/2)*d)

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]

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 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 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])

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 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 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 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 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]

Rubi steps

\begin{align*} \int \frac{1}{\left (a+b \sin ^{-1}(c+d x)\right )^{7/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (a+b \sin ^{-1}(x)\right )^{7/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac{2 \sqrt{1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}-\frac{2 \operatorname{Subst}\left (\int \frac{x}{\sqrt{1-x^2} \left (a+b \sin ^{-1}(x)\right )^{5/2}} \, dx,x,c+d x\right )}{5 b d}\\ &=-\frac{2 \sqrt{1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}+\frac{4 (c+d x)}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac{4 \operatorname{Subst}\left (\int \frac{1}{\left (a+b \sin ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{15 b^2 d}\\ &=-\frac{2 \sqrt{1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}+\frac{4 (c+d x)}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac{8 \sqrt{1-(c+d x)^2}}{15 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}+\frac{8 \operatorname{Subst}\left (\int \frac{x}{\sqrt{1-x^2} \sqrt{a+b \sin ^{-1}(x)}} \, dx,x,c+d x\right )}{15 b^3 d}\\ &=-\frac{2 \sqrt{1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}+\frac{4 (c+d x)}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac{8 \sqrt{1-(c+d x)^2}}{15 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}+\frac{8 \operatorname{Subst}\left (\int \frac{\sin (x)}{\sqrt{a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{15 b^3 d}\\ &=-\frac{2 \sqrt{1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}+\frac{4 (c+d x)}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac{8 \sqrt{1-(c+d x)^2}}{15 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}+\frac{\left (8 \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+d x)\right )}{15 b^3 d}-\frac{\left (8 \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+d x)\right )}{15 b^3 d}\\ &=-\frac{2 \sqrt{1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}+\frac{4 (c+d x)}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac{8 \sqrt{1-(c+d x)^2}}{15 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}+\frac{\left (16 \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+d x)}\right )}{15 b^4 d}-\frac{\left (16 \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+d x)}\right )}{15 b^4 d}\\ &=-\frac{2 \sqrt{1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}+\frac{4 (c+d x)}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac{8 \sqrt{1-(c+d x)^2}}{15 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}+\frac{8 \sqrt{2 \pi } \cos \left (\frac{a}{b}\right ) S\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{15 b^{7/2} d}-\frac{8 \sqrt{2 \pi } C\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right ) \sin \left (\frac{a}{b}\right )}{15 b^{7/2} d}\\ \end{align*}

Mathematica [C]  time = 0.244318, size = 287, normalized size = 1.32 \[ \frac{e^{-i \sin ^{-1}(c+d x)} \left (-8 e^{\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^2 \sqrt{\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \text{Gamma}\left (\frac{1}{2},\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+8 a^2+4 a b \left (4 \sin ^{-1}(c+d x)+i\right )+2 b^2 \left (4 \sin ^{-1}(c+d x)^2+2 i \sin ^{-1}(c+d x)-3\right )\right )+4 e^{-\frac{i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right ) \left (e^{\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \left (2 a+b \left (2 \sin ^{-1}(c+d x)-i\right )\right )-2 i b \left (-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{3/2} \text{Gamma}\left (\frac{1}{2},-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )\right )-6 b^2 e^{i \sin ^{-1}(c+d x)}}{30 b^3 d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSin[c + d*x])^(-7/2),x]

[Out]

(-6*b^2*E^(I*ArcSin[c + d*x]) + (4*(a + b*ArcSin[c + d*x])*(E^((I*(a + b*ArcSin[c + d*x]))/b)*(2*a + b*(-I + 2
*ArcSin[c + d*x])) - (2*I)*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) + (8*a^2 + 4*a*b*(I + 4*ArcSin[c + d*x]) + 2*b^2*(-3 + (2*I)*ArcSin[c + d*x] + 4*ArcSin[c +
d*x]^2) - 8*E^((I*(a + b*ArcSin[c + d*x]))/b)*(a + b*ArcSin[c + d*x])^2*Sqrt[(I*(a + b*ArcSin[c + d*x]))/b]*G
amma[1/2, (I*(a + b*ArcSin[c + d*x]))/b])/E^(I*ArcSin[c + d*x]))/(30*b^3*d*(a + b*ArcSin[c + d*x])^(5/2))

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Maple [B]  time = 0., size = 600, normalized size = 2.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*arcsin(d*x+c))^(7/2),x)

[Out]

2/15/d/b^3*(4*arcsin(d*x+c)^2*(1/b)^(1/2)*2^(1/2)*Pi^(1/2)*(a+b*arcsin(d*x+c))^(1/2)*cos(a/b)*FresnelS(2^(1/2)
/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*b^2-4*arcsin(d*x+c)^2*(1/b)^(1/2)*2^(1/2)*Pi^(1/2)*(a+b*arc
sin(d*x+c))^(1/2)*sin(a/b)*FresnelC(2^(1/2)/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*b^2+8*arcsin(d*x
+c)*(1/b)^(1/2)*2^(1/2)*Pi^(1/2)*(a+b*arcsin(d*x+c))^(1/2)*cos(a/b)*FresnelS(2^(1/2)/Pi^(1/2)/(1/b)^(1/2)*(a+b
*arcsin(d*x+c))^(1/2)/b)*a*b-8*arcsin(d*x+c)*(1/b)^(1/2)*2^(1/2)*Pi^(1/2)*(a+b*arcsin(d*x+c))^(1/2)*sin(a/b)*F
resnelC(2^(1/2)/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*a*b+4*(1/b)^(1/2)*2^(1/2)*Pi^(1/2)*(a+b*arcs
in(d*x+c))^(1/2)*cos(a/b)*FresnelS(2^(1/2)/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*a^2-4*(1/b)^(1/2)
*2^(1/2)*Pi^(1/2)*(a+b*arcsin(d*x+c))^(1/2)*sin(a/b)*FresnelC(2^(1/2)/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x+c))
^(1/2)/b)*a^2+4*arcsin(d*x+c)^2*cos((a+b*arcsin(d*x+c))/b-a/b)*b^2+8*arcsin(d*x+c)*cos((a+b*arcsin(d*x+c))/b-a
/b)*a*b+2*arcsin(d*x+c)*sin((a+b*arcsin(d*x+c))/b-a/b)*b^2+4*cos((a+b*arcsin(d*x+c))/b-a/b)*a^2-3*cos((a+b*arc
sin(d*x+c))/b-a/b)*b^2+2*sin((a+b*arcsin(d*x+c))/b-a/b)*a*b)/(a+b*arcsin(d*x+c))^(5/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \arcsin \left (d x + c\right ) + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arcsin(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

integrate((b*arcsin(d*x + c) + a)^(-7/2), x)

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arcsin(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*asin(d*x+c))**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \arcsin \left (d x + c\right ) + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arcsin(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((b*arcsin(d*x + c) + a)^(-7/2), x)