\(\int \frac {1}{(a+b \arcsin (1+d x^2))^{7/2}} \, dx\) [423]

Optimal result

Integrand size = 16, antiderivative size = 317 \[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^{7/2}} \, dx=-\frac {\sqrt {-2 d x^2-d^2 x^4}}{5 b d x \left (a+b \arcsin \left (1+d x^2\right )\right )^{5/2}}+\frac {x}{15 b^2 \left (a+b \arcsin \left (1+d x^2\right )\right )^{3/2}}+\frac {\sqrt {-2 d x^2-d^2 x^4}}{15 b^3 d x \sqrt {a+b \arcsin \left (1+d x^2\right )}}-\frac {\left (\frac {1}{b}\right )^{7/2} \sqrt {\pi } x \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arcsin \left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right )}{15 \left (\cos \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )\right )}+\frac {\left (\frac {1}{b}\right )^{7/2} \sqrt {\pi } x \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arcsin \left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )}{15 \left (\cos \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )\right )} \]

[Out]

1/15*x/b^2/(a+b*arcsin(d*x^2+1))^(3/2)-1/15*(1/b)^(7/2)*x*FresnelS((1/b)^(1/2)*(a+b*arcsin(d*x^2+1))^(1/2)/Pi^
(1/2))*(cos(1/2*a/b)-sin(1/2*a/b))*Pi^(1/2)/(cos(1/2*arcsin(d*x^2+1))-sin(1/2*arcsin(d*x^2+1)))+1/15*(1/b)^(7/
2)*x*FresnelC((1/b)^(1/2)*(a+b*arcsin(d*x^2+1))^(1/2)/Pi^(1/2))*(cos(1/2*a/b)+sin(1/2*a/b))*Pi^(1/2)/(cos(1/2*
arcsin(d*x^2+1))-sin(1/2*arcsin(d*x^2+1)))-1/5*(-d^2*x^4-2*d*x^2)^(1/2)/b/d/x/(a+b*arcsin(d*x^2+1))^(5/2)+1/15
*(-d^2*x^4-2*d*x^2)^(1/2)/b^3/d/x/(a+b*arcsin(d*x^2+1))^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {4912, 4906} \[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^{7/2}} \, dx=\frac {\sqrt {-d^2 x^4-2 d x^2}}{15 b^3 d x \sqrt {a+b \arcsin \left (d x^2+1\right )}}+\frac {x}{15 b^2 \left (a+b \arcsin \left (d x^2+1\right )\right )^{3/2}}-\frac {\sqrt {-d^2 x^4-2 d x^2}}{5 b d x \left (a+b \arcsin \left (d x^2+1\right )\right )^{5/2}}+\frac {\sqrt {\pi } \left (\frac {1}{b}\right )^{7/2} x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arcsin \left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{15 \left (\cos \left (\frac {1}{2} \arcsin \left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (d x^2+1\right )\right )\right )}-\frac {\sqrt {\pi } \left (\frac {1}{b}\right )^{7/2} x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arcsin \left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{15 \left (\cos \left (\frac {1}{2} \arcsin \left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (d x^2+1\right )\right )\right )} \]

[In]

Int[(a + b*ArcSin[1 + d*x^2])^(-7/2),x]

[Out]

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

Rule 4906

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(-3/2), x_Symbol] :> Simp[-Sqrt[-2*c*d*x^2 - d^2*x^4]/(b*d*x*S
qrt[a + b*ArcSin[c + d*x^2]]), x] + (-Simp[(c/b)^(3/2)*Sqrt[Pi]*x*(Cos[a/(2*b)] + c*Sin[a/(2*b)])*(FresnelC[Sq
rt[c/(Pi*b)]*Sqrt[a + b*ArcSin[c + d*x^2]]]/(Cos[(1/2)*ArcSin[c + d*x^2]] - c*Sin[ArcSin[c + d*x^2]/2])), x] +
Simp[(c/b)^(3/2)*Sqrt[Pi]*x*(Cos[a/(2*b)] - c*Sin[a/(2*b)])*(FresnelS[Sqrt[c/(Pi*b)]*Sqrt[a + b*ArcSin[c + d*
x^2]]]/(Cos[(1/2)*ArcSin[c + d*x^2]] - c*Sin[ArcSin[c + d*x^2]/2])), x]) /; FreeQ[{a, b, c, d}, x] && EqQ[c^2,
1]

Rule 4912

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(n_), x_Symbol] :> Simp[x*((a + b*ArcSin[c + d*x^2])^(n + 2)/(
4*b^2*(n + 1)*(n + 2))), x] + (-Dist[1/(4*b^2*(n + 1)*(n + 2)), Int[(a + b*ArcSin[c + d*x^2])^(n + 2), x], x]
+ Simp[Sqrt[-2*c*d*x^2 - d^2*x^4]*((a + b*ArcSin[c + d*x^2])^(n + 1)/(2*b*d*(n + 1)*x)), x]) /; FreeQ[{a, b, c
, d}, x] && EqQ[c^2, 1] && LtQ[n, -1] && NeQ[n, -2]

Rubi steps \begin{align*} \text {integral}& = -\frac {\sqrt {-2 d x^2-d^2 x^4}}{5 b d x \left (a+b \arcsin \left (1+d x^2\right )\right )^{5/2}}+\frac {x}{15 b^2 \left (a+b \arcsin \left (1+d x^2\right )\right )^{3/2}}-\frac {\int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^{3/2}} \, dx}{15 b^2} \\ & = -\frac {\sqrt {-2 d x^2-d^2 x^4}}{5 b d x \left (a+b \arcsin \left (1+d x^2\right )\right )^{5/2}}+\frac {x}{15 b^2 \left (a+b \arcsin \left (1+d x^2\right )\right )^{3/2}}+\frac {\sqrt {-2 d x^2-d^2 x^4}}{15 b^3 d x \sqrt {a+b \arcsin \left (1+d x^2\right )}}-\frac {\left (\frac {1}{b}\right )^{7/2} \sqrt {\pi } x \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arcsin \left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right )}{15 \left (\cos \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )\right )}+\frac {\left (\frac {1}{b}\right )^{7/2} \sqrt {\pi } x \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arcsin \left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )}{15 \left (\cos \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 297, normalized size of antiderivative = 0.94 \[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^{7/2}} \, dx=\frac {\frac {-\frac {3 b \sqrt {-d x^2 \left (2+d x^2\right )}}{d}+x^2 \left (a+b \arcsin \left (1+d x^2\right )\right )+\frac {\sqrt {-d x^2 \left (2+d x^2\right )} \left (a+b \arcsin \left (1+d x^2\right )\right )^2}{b d}}{x \left (a+b \arcsin \left (1+d x^2\right )\right )^{5/2}}-\frac {\left (\frac {1}{b}\right )^{3/2} \sqrt {\pi } x \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arcsin \left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right )}{\cos \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )}+\frac {\left (\frac {1}{b}\right )^{3/2} \sqrt {\pi } x \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arcsin \left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )}{\cos \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )}}{15 b^2} \]

[In]

Integrate[(a + b*ArcSin[1 + d*x^2])^(-7/2),x]

[Out]

(((-3*b*Sqrt[-(d*x^2*(2 + d*x^2))])/d + x^2*(a + b*ArcSin[1 + d*x^2]) + (Sqrt[-(d*x^2*(2 + d*x^2))]*(a + b*Arc
Sin[1 + d*x^2])^2)/(b*d))/(x*(a + b*ArcSin[1 + d*x^2])^(5/2)) - ((b^(-1))^(3/2)*Sqrt[Pi]*x*FresnelS[(Sqrt[b^(-
1)]*Sqrt[a + b*ArcSin[1 + d*x^2]])/Sqrt[Pi]]*(Cos[a/(2*b)] - Sin[a/(2*b)]))/(Cos[ArcSin[1 + d*x^2]/2] - Sin[Ar
cSin[1 + d*x^2]/2]) + ((b^(-1))^(3/2)*Sqrt[Pi]*x*FresnelC[(Sqrt[b^(-1)]*Sqrt[a + b*ArcSin[1 + d*x^2]])/Sqrt[Pi
]]*(Cos[a/(2*b)] + Sin[a/(2*b)]))/(Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSin[1 + d*x^2]/2]))/(15*b^2)

Maple [F]

\[\int \frac {1}{{\left (a +b \arcsin \left (d \,x^{2}+1\right )\right )}^{\frac {7}{2}}}d x\]

[In]

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

[Out]

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

Fricas [F(-2)]

Exception generated. \[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^{7/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

Sympy [F]

\[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^{7/2}} \, dx=\int \frac {1}{\left (a + b \operatorname {asin}{\left (d x^{2} + 1 \right )}\right )^{\frac {7}{2}}}\, dx \]

[In]

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

[Out]

Integral((a + b*asin(d*x**2 + 1))**(-7/2), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^{7/2}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: sign: argument cannot be imaginary; found sqrt((-_SAGE_VAR_d*_SAGE
_VAR_x^2)-2)

Giac [F]

\[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^{7/2}} \, dx=\int { \frac {1}{{\left (b \arcsin \left (d x^{2} + 1\right ) + a\right )}^{\frac {7}{2}}} \,d x } \]

[In]

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

[Out]

integrate((b*arcsin(d*x^2 + 1) + a)^(-7/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^{7/2}} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {asin}\left (d\,x^2+1\right )\right )}^{7/2}} \,d x \]

[In]

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

[Out]

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