\(\int \frac {1}{(a+b \arccos (1+d x^2))^{3/2}} \, dx\) [91]

Optimal result

Integrand size = 16, antiderivative size = 190 \[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{3/2}} \, dx=\frac {\sqrt {-2 d x^2-d^2 x^4}}{b d x \sqrt {a+b \arccos \left (1+d x^2\right )}}+\frac {2 \left (\frac {1}{b}\right )^{3/2} \sqrt {\pi } \cos \left (\frac {a}{2 b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \sin \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right )}{d x}-\frac {2 \left (\frac {1}{b}\right )^{3/2} \sqrt {\pi } \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \sin \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right )}{d x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {4907} \[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{3/2}} \, dx=\frac {\sqrt {-d^2 x^4-2 d x^2}}{b d x \sqrt {a+b \arccos \left (d x^2+1\right )}}-\frac {2 \sqrt {\pi } \left (\frac {1}{b}\right )^{3/2} \sin \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \arccos \left (d x^2+1\right )\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{d x}+\frac {2 \sqrt {\pi } \left (\frac {1}{b}\right )^{3/2} \cos \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \arccos \left (d x^2+1\right )\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{d x} \]

[In]

Int[(a + b*ArcCos[1 + d*x^2])^(-3/2),x]

[Out]

Sqrt[-2*d*x^2 - d^2*x^4]/(b*d*x*Sqrt[a + b*ArcCos[1 + d*x^2]]) + (2*(b^(-1))^(3/2)*Sqrt[Pi]*Cos[a/(2*b)]*Fresn
elS[(Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[1 + d*x^2]])/Sqrt[Pi]]*Sin[ArcCos[1 + d*x^2]/2])/(d*x) - (2*(b^(-1))^(3/2)
*Sqrt[Pi]*FresnelC[(Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[1 + d*x^2]])/Sqrt[Pi]]*Sin[a/(2*b)]*Sin[ArcCos[1 + d*x^2]/2
])/(d*x)

Rule 4907

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

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

Mathematica [A] (verified)

Time = 0.99 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.87 \[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{3/2}} \, dx=\frac {\frac {\sqrt {b} \sqrt {-d x^2 \left (2+d x^2\right )}}{\sqrt {a+b \arccos \left (1+d x^2\right )}}+2 \sqrt {\pi } \cos \left (\frac {a}{2 b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {a+b \arccos \left (1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right )-2 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {\sqrt {a+b \arccos \left (1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right )}{b^{3/2} d x} \]

[In]

Integrate[(a + b*ArcCos[1 + d*x^2])^(-3/2),x]

[Out]

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

Maple [F]

\[\int \frac {1}{{\left (a +b \arccos \left (d \,x^{2}+1\right )\right )}^{\frac {3}{2}}}d x\]

[In]

int(1/(a+b*arccos(d*x^2+1))^(3/2),x)

[Out]

int(1/(a+b*arccos(d*x^2+1))^(3/2),x)

Fricas [F(-2)]

Exception generated. \[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(1/(a+b*arccos(d*x^2+1))^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{3/2}} \, dx=\int \frac {1}{\left (a + b \operatorname {acos}{\left (d x^{2} + 1 \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(1/(a+b*acos(d*x**2+1))**(3/2),x)

[Out]

Integral((a + b*acos(d*x**2 + 1))**(-3/2), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{3/2}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(1/(a+b*arccos(d*x^2+1))^(3/2),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]

\[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b \arccos \left (d x^{2} + 1\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(1/(a+b*arccos(d*x^2+1))^(3/2),x, algorithm="giac")

[Out]

integrate((b*arccos(d*x^2 + 1) + a)^(-3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{3/2}} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {acos}\left (d\,x^2+1\right )\right )}^{3/2}} \,d x \]

[In]

int(1/(a + b*acos(d*x^2 + 1))^(3/2),x)

[Out]

int(1/(a + b*acos(d*x^2 + 1))^(3/2), x)