\(\int \sqrt {a+b \arccos (1+d x^2)} \, dx\) [89]

Optimal result

Integrand size = 16, antiderivative size = 184 \[ \int \sqrt {a+b \arccos \left (1+d x^2\right )} \, dx=\frac {2 \sqrt {\pi } \cos \left (\frac {a}{2 b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \sin \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right )}{\sqrt {\frac {1}{b}} d x}-\frac {2 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \sin \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right )}{\sqrt {\frac {1}{b}} d x}-\frac {2 \sqrt {a+b \arccos \left (1+d x^2\right )} \sin ^2\left (\frac {1}{2} \arccos \left (1+d x^2\right )\right )}{d x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 184, 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 = {4896} \[ \int \sqrt {a+b \arccos \left (1+d x^2\right )} \, dx=-\frac {2 \sqrt {\pi } \sin \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \arccos \left (d x^2+1\right )\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{\sqrt {\frac {1}{b}} d x}+\frac {2 \sqrt {\pi } \cos \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \arccos \left (d x^2+1\right )\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{\sqrt {\frac {1}{b}} d x}-\frac {2 \sin ^2\left (\frac {1}{2} \arccos \left (d x^2+1\right )\right ) \sqrt {a+b \arccos \left (d x^2+1\right )}}{d x} \]

[In]

Int[Sqrt[a + b*ArcCos[1 + d*x^2]],x]

[Out]

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

Rule 4896

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

Rubi steps \begin{align*} \text {integral}& = \frac {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 )}{\sqrt {\frac {1}{b}} d x}-\frac {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 )}{\sqrt {\frac {1}{b}} d x}-\frac {2 \sqrt {a+b \arccos \left (1+d x^2\right )} \sin ^2\left (\frac {1}{2} \arccos \left (1+d x^2\right )\right )}{d x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.81 \[ \int \sqrt {a+b \arccos \left (1+d x^2\right )} \, dx=-\frac {2 \sin \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right ) \left (-\sqrt {b} \sqrt {\pi } \cos \left (\frac {a}{2 b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {a+b \arccos \left (1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right )+\sqrt {b} \sqrt {\pi } \operatorname {FresnelC}\left (\frac {\sqrt {a+b \arccos \left (1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {a}{2 b}\right )+\sqrt {a+b \arccos \left (1+d x^2\right )} \sin \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right )\right )}{d x} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F(-2)]

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

[In]

integrate((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]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate((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]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((a + b*acos(d*x^2 + 1))^(1/2),x)

[Out]

int((a + b*acos(d*x^2 + 1))^(1/2), x)