Integrand size = 16, antiderivative size = 269 \[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{7/2}} \, dx=\frac {\sqrt {-2 d x^2-d^2 x^4}}{5 b d x \left (a+b \arccos \left (1+d x^2\right )\right )^{5/2}}+\frac {x}{15 b^2 \left (a+b \arccos \left (1+d x^2\right )\right )^{3/2}}-\frac {\sqrt {-2 d x^2-d^2 x^4}}{15 b^3 d x \sqrt {a+b \arccos \left (1+d x^2\right )}}-\frac {2 \left (\frac {1}{b}\right )^{7/2} \sqrt {\pi } \cos \left (\frac {a}{2 b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \sin \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right )}{15 d x}+\frac {2 \left (\frac {1}{b}\right )^{7/2} \sqrt {\pi } \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \sin \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right )}{15 d x} \]
1/15*x/b^2/(a+b*arccos(d*x^2+1))^(3/2)-2/15*(1/b)^(7/2)*cos(1/2*a/b)*Fresn elS((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+2/15*(1/b)^(7/2)*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+1/5 *(-d^2*x^4-2*d*x^2)^(1/2)/b/d/x/(a+b*arccos(d*x^2+1))^(5/2)-1/15*(-d^2*x^4 -2*d*x^2)^(1/2)/b^3/d/x/(a+b*arccos(d*x^2+1))^(1/2)
Time = 1.05 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.87 \[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{7/2}} \, dx=\frac {2 \sin \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right ) \left (-\sqrt {\pi } \left (a+b \arccos \left (1+d x^2\right )\right )^{5/2} \cos \left (\frac {a}{2 b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {a+b \arccos \left (1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right )+\sqrt {\pi } \left (a+b \arccos \left (1+d x^2\right )\right )^{5/2} \operatorname {FresnelC}\left (\frac {\sqrt {a+b \arccos \left (1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {a}{2 b}\right )+\sqrt {b} \left (-\left (\left (-3 b^2+\left (a+b \arccos \left (1+d x^2\right )\right )^2\right ) \cos \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right )\right )-b \left (a+b \arccos \left (1+d x^2\right )\right ) \sin \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right )\right )\right )}{15 b^{7/2} d x \left (a+b \arccos \left (1+d x^2\right )\right )^{5/2}} \]
(2*Sin[ArcCos[1 + d*x^2]/2]*(-(Sqrt[Pi]*(a + b*ArcCos[1 + d*x^2])^(5/2)*Co s[a/(2*b)]*FresnelS[Sqrt[a + b*ArcCos[1 + d*x^2]]/(Sqrt[b]*Sqrt[Pi])]) + S qrt[Pi]*(a + b*ArcCos[1 + d*x^2])^(5/2)*FresnelC[Sqrt[a + b*ArcCos[1 + d*x ^2]]/(Sqrt[b]*Sqrt[Pi])]*Sin[a/(2*b)] + Sqrt[b]*(-((-3*b^2 + (a + b*ArcCos [1 + d*x^2])^2)*Cos[ArcCos[1 + d*x^2]/2]) - b*(a + b*ArcCos[1 + d*x^2])*Si n[ArcCos[1 + d*x^2]/2])))/(15*b^(7/2)*d*x*(a + b*ArcCos[1 + d*x^2])^(5/2))
Time = 0.34 (sec) , antiderivative size = 270, 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 = {5328, 5322}
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.
\(\displaystyle \int \frac {1}{\left (a+b \arccos \left (d x^2+1\right )\right )^{7/2}} \, dx\) |
\(\Big \downarrow \) 5328 |
\(\displaystyle -\frac {\int \frac {1}{\left (a+b \arccos \left (d x^2+1\right )\right )^{3/2}}dx}{15 b^2}+\frac {x}{15 b^2 \left (a+b \arccos \left (d x^2+1\right )\right )^{3/2}}+\frac {\sqrt {-d^2 x^4-2 d x^2}}{5 b d x \left (a+b \arccos \left (d x^2+1\right )\right )^{5/2}}\) |
\(\Big \downarrow \) 5322 |
\(\displaystyle -\frac {\frac {\sqrt {-d^2 x^4-2 d x^2}}{b d x \sqrt {a+b \arccos \left (d x^2+1\right )}}-\frac {2 \sqrt {\pi } \left (\frac {1}{b}\right )^{3/2} \sin \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \arccos \left (d x^2+1\right )\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{d x}+\frac {2 \sqrt {\pi } \left (\frac {1}{b}\right )^{3/2} \cos \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \arccos \left (d x^2+1\right )\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{d x}}{15 b^2}+\frac {x}{15 b^2 \left (a+b \arccos \left (d x^2+1\right )\right )^{3/2}}+\frac {\sqrt {-d^2 x^4-2 d x^2}}{5 b d x \left (a+b \arccos \left (d x^2+1\right )\right )^{5/2}}\) |
Sqrt[-2*d*x^2 - d^2*x^4]/(5*b*d*x*(a + b*ArcCos[1 + d*x^2])^(5/2)) + x/(15 *b^2*(a + b*ArcCos[1 + d*x^2])^(3/2)) - (Sqrt[-2*d*x^2 - d^2*x^4]/(b*d*x*S qrt[a + b*ArcCos[1 + d*x^2]]) + (2*(b^(-1))^(3/2)*Sqrt[Pi]*Cos[a/(2*b)]*Fr esnelS[(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)]*Sq rt[a + b*ArcCos[1 + d*x^2]])/Sqrt[Pi]]*Sin[a/(2*b)]*Sin[ArcCos[1 + d*x^2]/ 2])/(d*x))/(15*b^2)
3.1.93.3.1 Defintions of rubi rules used
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[Sqrt[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]
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]
\[\int \frac {1}{{\left (a +b \arccos \left (d \,x^{2}+1\right )\right )}^{\frac {7}{2}}}d x\]
Exception generated. \[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{7/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
\[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{7/2}} \, dx=\int \frac {1}{\left (a + b \operatorname {acos}{\left (d x^{2} + 1 \right )}\right )^{\frac {7}{2}}}\, dx \]
Exception generated. \[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{7/2}} \, dx=\text {Exception raised: RuntimeError} \]
Exception raised: RuntimeError >> ECL says: sign: argument cannot be imagi nary; found sqrt((-_SAGE_VAR_d*_SAGE_VAR_x^2)-2)
\[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{7/2}} \, dx=\int { \frac {1}{{\left (b \arccos \left (d x^{2} + 1\right ) + a\right )}^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{7/2}} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {acos}\left (d\,x^2+1\right )\right )}^{7/2}} \,d x \]