Integrand size = 16, antiderivative size = 221 \[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{5/2}} \, dx=\frac {\sqrt {-2 d x^2-d^2 x^4}}{3 b d x \left (a+b \arccos \left (1+d x^2\right )\right )^{3/2}}+\frac {x}{3 b^2 \sqrt {a+b \arccos \left (1+d x^2\right )}}+\frac {2 \left (\frac {1}{b}\right )^{5/2} \sqrt {\pi } \cos \left (\frac {a}{2 b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \sin \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right )}{3 d x}+\frac {2 \left (\frac {1}{b}\right )^{5/2} \sqrt {\pi } \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \sin \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right )}{3 d x} \] Output:
1/3*(-d^2*x^4-2*d*x^2)^(1/2)/b/d/x/(a+b*arccos(d*x^2+1))^(3/2)+1/3*x/b^2/( a+b*arccos(d*x^2+1))^(1/2)+2/3*(1/b)^(5/2)*Pi^(1/2)*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)) /d/x+2/3*(1/b)^(5/2)*Pi^(1/2)*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))/d/x
Time = 0.71 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{5/2}} \, dx=\frac {2 \sin \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right ) \left (\sqrt {\pi } \left (a+b \arccos \left (1+d x^2\right )\right )^{3/2} \cos \left (\frac {a}{2 b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {a+b \arccos \left (1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right )+\sqrt {\pi } \left (a+b \arccos \left (1+d x^2\right )\right )^{3/2} \operatorname {FresnelS}\left (\frac {\sqrt {a+b \arccos \left (1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {a}{2 b}\right )+\sqrt {b} \left (b \cos \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right )-\left (a+b \arccos \left (1+d x^2\right )\right ) \sin \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right )\right )\right )}{3 b^{5/2} d x \left (a+b \arccos \left (1+d x^2\right )\right )^{3/2}} \] Input:
Integrate[(a + b*ArcCos[1 + d*x^2])^(-5/2),x]
Output:
(2*Sin[ArcCos[1 + d*x^2]/2]*(Sqrt[Pi]*(a + b*ArcCos[1 + d*x^2])^(3/2)*Cos[ a/(2*b)]*FresnelC[Sqrt[a + b*ArcCos[1 + d*x^2]]/(Sqrt[b]*Sqrt[Pi])] + Sqrt [Pi]*(a + b*ArcCos[1 + d*x^2])^(3/2)*FresnelS[Sqrt[a + b*ArcCos[1 + d*x^2] ]/(Sqrt[b]*Sqrt[Pi])]*Sin[a/(2*b)] + Sqrt[b]*(b*Cos[ArcCos[1 + d*x^2]/2] - (a + b*ArcCos[1 + d*x^2])*Sin[ArcCos[1 + d*x^2]/2])))/(3*b^(5/2)*d*x*(a + b*ArcCos[1 + d*x^2])^(3/2))
Time = 0.35 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.02, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5328, 5319}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{\left (a+b \arccos \left (d x^2+1\right )\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 5328 |
\(\displaystyle -\frac {\int \frac {1}{\sqrt {a+b \arccos \left (d x^2+1\right )}}dx}{3 b^2}+\frac {x}{3 b^2 \sqrt {a+b \arccos \left (d x^2+1\right )}}+\frac {\sqrt {-d^2 x^4-2 d x^2}}{3 b d x \left (a+b \arccos \left (d x^2+1\right )\right )^{3/2}}\) |
\(\Big \downarrow \) 5319 |
\(\displaystyle -\frac {-\frac {2 \sqrt {\pi } \sqrt {\frac {1}{b}} \cos \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \arccos \left (d x^2+1\right )\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{d x}-\frac {2 \sqrt {\pi } \sqrt {\frac {1}{b}} \sin \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \arccos \left (d x^2+1\right )\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{d x}}{3 b^2}+\frac {x}{3 b^2 \sqrt {a+b \arccos \left (d x^2+1\right )}}+\frac {\sqrt {-d^2 x^4-2 d x^2}}{3 b d x \left (a+b \arccos \left (d x^2+1\right )\right )^{3/2}}\) |
Input:
Int[(a + b*ArcCos[1 + d*x^2])^(-5/2),x]
Output:
Sqrt[-2*d*x^2 - d^2*x^4]/(3*b*d*x*(a + b*ArcCos[1 + d*x^2])^(3/2)) + x/(3* b^2*Sqrt[a + b*ArcCos[1 + d*x^2]]) - ((-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))/(3*b^2)
Int[1/Sqrt[(a_.) + ArcCos[1 + (d_.)*(x_)^2]*(b_.)], x_Symbol] :> Simp[-2*Sq rt[Pi/b]*Cos[a/(2*b)]*Sin[ArcCos[1 + d*x^2]/2]*(FresnelC[Sqrt[1/(Pi*b)]*Sqr t[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]
Int[((a_.) + ArcCos[(c_) + (d_.)*(x_)^2]*(b_.))^(n_), x_Symbol] :> Simp[x*( (a + b*ArcCos[c + d*x^2])^(n + 2)/(4*b^2*(n + 1)*(n + 2))), x] + (-Simp[Sqr t[-2*c*d*x^2 - d^2*x^4]*((a + b*ArcCos[c + d*x^2])^(n + 1)/(2*b*d*(n + 1)*x )), x] - Simp[1/(4*b^2*(n + 1)*(n + 2)) Int[(a + b*ArcCos[c + d*x^2])^(n + 2), x], x]) /; FreeQ[{a, b, c, d}, x] && EqQ[c^2, 1] && LtQ[n, -1] && NeQ [n, -2]
\[\int \frac {1}{{\left (a +b \arccos \left (d \,x^{2}+1\right )\right )}^{\frac {5}{2}}}d x\]
Input:
int(1/(a+b*arccos(d*x^2+1))^(5/2),x)
Output:
int(1/(a+b*arccos(d*x^2+1))^(5/2),x)
Exception generated. \[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/(a+b*arccos(d*x^2+1))^(5/2),x, algorithm="fricas")
Output:
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
\[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{5/2}} \, dx=\int \frac {1}{\left (a + b \operatorname {acos}{\left (d x^{2} + 1 \right )}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(1/(a+b*acos(d*x**2+1))**(5/2),x)
Output:
Integral((a + b*acos(d*x**2 + 1))**(-5/2), x)
Exception generated. \[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{5/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(1/(a+b*arccos(d*x^2+1))^(5/2),x, algorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: sign: argument cannot be imagi nary; found sqrt((-_SAGE_VAR_d*_SAGE_VAR_x^2)-2)
\[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{5/2}} \, dx=\int { \frac {1}{{\left (b \arccos \left (d x^{2} + 1\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(1/(a+b*arccos(d*x^2+1))^(5/2),x, algorithm="giac")
Output:
integrate((b*arccos(d*x^2 + 1) + a)^(-5/2), x)
Timed out. \[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{5/2}} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {acos}\left (d\,x^2+1\right )\right )}^{5/2}} \,d x \] Input:
int(1/(a + b*acos(d*x^2 + 1))^(5/2),x)
Output:
int(1/(a + b*acos(d*x^2 + 1))^(5/2), x)
\[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{5/2}} \, dx=\text {too large to display} \] Input:
int(1/(a+b*acos(d*x^2+1))^(5/2),x)
Output:
(sqrt(d)*acos(d*x**2 + 1)**2*int((sqrt(acos(d*x**2 + 1)*b + a)*sqrt( - d*x **2 - 2)*acos(d*x**2 + 1)*x)/(acos(d*x**2 + 1)**3*b**3*d*x**2 + 2*acos(d*x **2 + 1)**3*b**3 + 3*acos(d*x**2 + 1)**2*a*b**2*d*x**2 + 6*acos(d*x**2 + 1 )**2*a*b**2 + 3*acos(d*x**2 + 1)*a**2*b*d*x**2 + 6*acos(d*x**2 + 1)*a**2*b + a**3*d*x**2 + 2*a**3),x)*a*b**2*d + sqrt(d)*acos(d*x**2 + 1)**2*int((sq rt(acos(d*x**2 + 1)*b + a)*sqrt( - d*x**2 - 2)*acos(d*x**2 + 1)**2*x)/(aco s(d*x**2 + 1)**3*b**3*d*x**2 + 2*acos(d*x**2 + 1)**3*b**3 + 3*acos(d*x**2 + 1)**2*a*b**2*d*x**2 + 6*acos(d*x**2 + 1)**2*a*b**2 + 3*acos(d*x**2 + 1)* a**2*b*d*x**2 + 6*acos(d*x**2 + 1)*a**2*b + a**3*d*x**2 + 2*a**3),x)*b**3* d + acos(d*x**2 + 1)**2*int((sqrt(acos(d*x**2 + 1)*b + a)*acos(d*x**2 + 1) *x**2)/(acos(d*x**2 + 1)**3*b**3*d*x**2 + 2*acos(d*x**2 + 1)**3*b**3 + 3*a cos(d*x**2 + 1)**2*a*b**2*d*x**2 + 6*acos(d*x**2 + 1)**2*a*b**2 + 3*acos(d *x**2 + 1)*a**2*b*d*x**2 + 6*acos(d*x**2 + 1)*a**2*b + a**3*d*x**2 + 2*a** 3),x)*b**3*d**2 + 2*acos(d*x**2 + 1)**2*int((sqrt(acos(d*x**2 + 1)*b + a)* acos(d*x**2 + 1))/(acos(d*x**2 + 1)**3*b**3*d*x**2 + 2*acos(d*x**2 + 1)**3 *b**3 + 3*acos(d*x**2 + 1)**2*a*b**2*d*x**2 + 6*acos(d*x**2 + 1)**2*a*b**2 + 3*acos(d*x**2 + 1)*a**2*b*d*x**2 + 6*acos(d*x**2 + 1)*a**2*b + a**3*d*x **2 + 2*a**3),x)*b**3*d - sqrt(d)*sqrt(acos(d*x**2 + 1)*b + a)*sqrt( - d*x **2 - 2)*acos(d*x**2 + 1) + 2*sqrt(d)*acos(d*x**2 + 1)*int((sqrt(acos(d*x* *2 + 1)*b + a)*sqrt( - d*x**2 - 2)*acos(d*x**2 + 1)*x)/(acos(d*x**2 + 1...