\(\int \frac {1}{\sqrt {a+b \arccos (1+d x^2)}} \, dx\) [90]

Optimal result

Integrand size = 16, antiderivative size = 145 \[ \int \frac {1}{\sqrt {a+b \arccos \left (1+d x^2\right )}} \, dx=-\frac {2 \sqrt {\frac {1}{b}} \sqrt {\pi } \cos \left (\frac {a}{2 b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \sin \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right )}{d x}-\frac {2 \sqrt {\frac {1}{b}} \sqrt {\pi } \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \sin \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right )}{d x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 145, 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 = {4904} \[ \int \frac {1}{\sqrt {a+b \arccos \left (1+d x^2\right )}} \, dx=-\frac {2 \sqrt {\pi } \sqrt {\frac {1}{b}} \cos \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \arccos \left (d x^2+1\right )\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{d x}-\frac {2 \sqrt {\pi } \sqrt {\frac {1}{b}} \sin \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \arccos \left (d x^2+1\right )\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{d x} \]

[In]

Int[1/Sqrt[a + b*ArcCos[1 + d*x^2]],x]

[Out]

(-2*Sqrt[b^(-1)]*Sqrt[Pi]*Cos[a/(2*b)]*FresnelC[(Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[1 + d*x^2]])/Sqrt[Pi]]*Sin[Arc
Cos[1 + d*x^2]/2])/(d*x) - (2*Sqrt[b^(-1)]*Sqrt[Pi]*FresnelS[(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 4904

Int[1/Sqrt[(a_.) + ArcCos[1 + (d_.)*(x_)^2]*(b_.)], x_Symbol] :> Simp[-2*Sqrt[Pi/b]*Cos[a/(2*b)]*Sin[ArcCos[1
+ d*x^2]/2]*(FresnelC[Sqrt[1/(Pi*b)]*Sqrt[a + b*ArcCos[1 + d*x^2]]]/(d*x)), x] - Simp[2*Sqrt[Pi/b]*Sin[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 {2 \sqrt {\frac {1}{b}} \sqrt {\pi } \cos \left (\frac {a}{2 b}\right ) \operatorname {FresnelC}\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 \sqrt {\frac {1}{b}} \sqrt {\pi } \operatorname {FresnelS}\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 = 0.41 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.74 \[ \int \frac {1}{\sqrt {a+b \arccos \left (1+d x^2\right )}} \, dx=-\frac {2 \sqrt {\pi } \left (\cos \left (\frac {a}{2 b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {a+b \arccos \left (1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right )+\operatorname {FresnelS}\left (\frac {\sqrt {a+b \arccos \left (1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {a}{2 b}\right )\right ) \sin \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right )}{\sqrt {b} d x} \]

[In]

Integrate[1/Sqrt[a + b*ArcCos[1 + d*x^2]],x]

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F(-2)]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

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

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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