Integrand size = 29, antiderivative size = 29 \[ \int \frac {\left (d-c^2 d x^2\right )^2}{x (a+b \arcsin (c x))^{3/2}} \, dx=-\frac {2 d^2 \left (1-c^2 x^2\right )^{5/2}}{b c x \sqrt {a+b \arcsin (c x)}}-\frac {d^2 \sqrt {\frac {\pi }{2}} \cos \left (\frac {4 a}{b}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \arcsin (c x)}}{\sqrt {b}}\right )}{b^{3/2}}-\frac {3 d^2 \sqrt {\pi } \cos \left (\frac {2 a}{b}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {a+b \arcsin (c x)}}{\sqrt {b} \sqrt {\pi }}\right )}{b^{3/2}}-\frac {3 d^2 \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {a+b \arcsin (c x)}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {2 a}{b}\right )}{b^{3/2}}-\frac {d^2 \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \arcsin (c x)}}{\sqrt {b}}\right ) \sin \left (\frac {4 a}{b}\right )}{b^{3/2}}-\frac {2 d^2 \text {Int}\left (\frac {1}{x^2 \sqrt {1-c^2 x^2} \sqrt {a+b \arcsin (c x)}},x\right )}{b c} \]
-1/2*d^2*cos(4*a/b)*FresnelC(2*2^(1/2)/Pi^(1/2)*(a+b*arcsin(c*x))^(1/2)/b^ (1/2))*2^(1/2)*Pi^(1/2)/b^(3/2)-1/2*d^2*FresnelS(2*2^(1/2)/Pi^(1/2)*(a+b*a rcsin(c*x))^(1/2)/b^(1/2))*sin(4*a/b)*2^(1/2)*Pi^(1/2)/b^(3/2)-3*d^2*cos(2 *a/b)*FresnelC(2*(a+b*arcsin(c*x))^(1/2)/b^(1/2)/Pi^(1/2))*Pi^(1/2)/b^(3/2 )-3*d^2*FresnelS(2*(a+b*arcsin(c*x))^(1/2)/b^(1/2)/Pi^(1/2))*sin(2*a/b)*Pi ^(1/2)/b^(3/2)-2*d^2*(-c^2*x^2+1)^(5/2)/b/c/x/(a+b*arcsin(c*x))^(1/2)-2*d^ 2*Unintegrable(1/x^2/(-c^2*x^2+1)^(1/2)/(a+b*arcsin(c*x))^(1/2),x)/b/c
Not integrable
Time = 1.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {\left (d-c^2 d x^2\right )^2}{x (a+b \arcsin (c x))^{3/2}} \, dx=\int \frac {\left (d-c^2 d x^2\right )^2}{x (a+b \arcsin (c x))^{3/2}} \, dx \]
Not integrable
Time = 2.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {5214, 5168, 3042, 3793, 2009, 5226, 2009}
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.
\(\displaystyle \int \frac {\left (d-c^2 d x^2\right )^2}{x (a+b \arcsin (c x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 5214 |
\(\displaystyle -\frac {8 c d^2 \int \frac {\left (1-c^2 x^2\right )^{3/2}}{\sqrt {a+b \arcsin (c x)}}dx}{b}-\frac {2 d^2 \int \frac {\left (1-c^2 x^2\right )^{3/2}}{x^2 \sqrt {a+b \arcsin (c x)}}dx}{b c}-\frac {2 d^2 \left (1-c^2 x^2\right )^{5/2}}{b c x \sqrt {a+b \arcsin (c x)}}\) |
\(\Big \downarrow \) 5168 |
\(\displaystyle -\frac {8 d^2 \int \frac {\cos ^4\left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right )}{\sqrt {a+b \arcsin (c x)}}d(a+b \arcsin (c x))}{b^2}-\frac {2 d^2 \int \frac {\left (1-c^2 x^2\right )^{3/2}}{x^2 \sqrt {a+b \arcsin (c x)}}dx}{b c}-\frac {2 d^2 \left (1-c^2 x^2\right )^{5/2}}{b c x \sqrt {a+b \arcsin (c x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {8 d^2 \int \frac {\sin \left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}+\frac {\pi }{2}\right )^4}{\sqrt {a+b \arcsin (c x)}}d(a+b \arcsin (c x))}{b^2}-\frac {2 d^2 \int \frac {\left (1-c^2 x^2\right )^{3/2}}{x^2 \sqrt {a+b \arcsin (c x)}}dx}{b c}-\frac {2 d^2 \left (1-c^2 x^2\right )^{5/2}}{b c x \sqrt {a+b \arcsin (c x)}}\) |
\(\Big \downarrow \) 3793 |
\(\displaystyle -\frac {8 d^2 \int \left (\frac {\cos \left (\frac {4 a}{b}-\frac {4 (a+b \arcsin (c x))}{b}\right )}{8 \sqrt {a+b \arcsin (c x)}}+\frac {\cos \left (\frac {2 a}{b}-\frac {2 (a+b \arcsin (c x))}{b}\right )}{2 \sqrt {a+b \arcsin (c x)}}+\frac {3}{8 \sqrt {a+b \arcsin (c x)}}\right )d(a+b \arcsin (c x))}{b^2}-\frac {2 d^2 \int \frac {\left (1-c^2 x^2\right )^{3/2}}{x^2 \sqrt {a+b \arcsin (c x)}}dx}{b c}-\frac {2 d^2 \left (1-c^2 x^2\right )^{5/2}}{b c x \sqrt {a+b \arcsin (c x)}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 d^2 \int \frac {\left (1-c^2 x^2\right )^{3/2}}{x^2 \sqrt {a+b \arcsin (c x)}}dx}{b c}-\frac {8 d^2 \left (\frac {1}{8} \sqrt {\frac {\pi }{2}} \sqrt {b} \cos \left (\frac {4 a}{b}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \arcsin (c x)}}{\sqrt {b}}\right )+\frac {1}{2} \sqrt {\pi } \sqrt {b} \cos \left (\frac {2 a}{b}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {a+b \arcsin (c x)}}{\sqrt {b} \sqrt {\pi }}\right )+\frac {1}{2} \sqrt {\pi } \sqrt {b} \sin \left (\frac {2 a}{b}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {a+b \arcsin (c x)}}{\sqrt {b} \sqrt {\pi }}\right )+\frac {1}{8} \sqrt {\frac {\pi }{2}} \sqrt {b} \sin \left (\frac {4 a}{b}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \arcsin (c x)}}{\sqrt {b}}\right )+\frac {3}{4} \sqrt {a+b \arcsin (c x)}\right )}{b^2}-\frac {2 d^2 \left (1-c^2 x^2\right )^{5/2}}{b c x \sqrt {a+b \arcsin (c x)}}\) |
\(\Big \downarrow \) 5226 |
\(\displaystyle -\frac {2 d^2 \int \left (\frac {x^2 c^4}{\sqrt {1-c^2 x^2} \sqrt {a+b \arcsin (c x)}}-\frac {2 c^2}{\sqrt {1-c^2 x^2} \sqrt {a+b \arcsin (c x)}}+\frac {1}{x^2 \sqrt {1-c^2 x^2} \sqrt {a+b \arcsin (c x)}}\right )dx}{b c}-\frac {8 d^2 \left (\frac {1}{8} \sqrt {\frac {\pi }{2}} \sqrt {b} \cos \left (\frac {4 a}{b}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \arcsin (c x)}}{\sqrt {b}}\right )+\frac {1}{2} \sqrt {\pi } \sqrt {b} \cos \left (\frac {2 a}{b}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {a+b \arcsin (c x)}}{\sqrt {b} \sqrt {\pi }}\right )+\frac {1}{2} \sqrt {\pi } \sqrt {b} \sin \left (\frac {2 a}{b}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {a+b \arcsin (c x)}}{\sqrt {b} \sqrt {\pi }}\right )+\frac {1}{8} \sqrt {\frac {\pi }{2}} \sqrt {b} \sin \left (\frac {4 a}{b}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \arcsin (c x)}}{\sqrt {b}}\right )+\frac {3}{4} \sqrt {a+b \arcsin (c x)}\right )}{b^2}-\frac {2 d^2 \left (1-c^2 x^2\right )^{5/2}}{b c x \sqrt {a+b \arcsin (c x)}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 d^2 \left (\int \frac {1}{x^2 \sqrt {1-c^2 x^2} \sqrt {a+b \arcsin (c x)}}dx-\frac {\sqrt {\pi } c \cos \left (\frac {2 a}{b}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {a+b \arcsin (c x)}}{\sqrt {b} \sqrt {\pi }}\right )}{2 \sqrt {b}}-\frac {\sqrt {\pi } c \sin \left (\frac {2 a}{b}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {a+b \arcsin (c x)}}{\sqrt {b} \sqrt {\pi }}\right )}{2 \sqrt {b}}-\frac {3 c \sqrt {a+b \arcsin (c x)}}{b}\right )}{b c}-\frac {8 d^2 \left (\frac {1}{8} \sqrt {\frac {\pi }{2}} \sqrt {b} \cos \left (\frac {4 a}{b}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \arcsin (c x)}}{\sqrt {b}}\right )+\frac {1}{2} \sqrt {\pi } \sqrt {b} \cos \left (\frac {2 a}{b}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {a+b \arcsin (c x)}}{\sqrt {b} \sqrt {\pi }}\right )+\frac {1}{2} \sqrt {\pi } \sqrt {b} \sin \left (\frac {2 a}{b}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {a+b \arcsin (c x)}}{\sqrt {b} \sqrt {\pi }}\right )+\frac {1}{8} \sqrt {\frac {\pi }{2}} \sqrt {b} \sin \left (\frac {4 a}{b}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \arcsin (c x)}}{\sqrt {b}}\right )+\frac {3}{4} \sqrt {a+b \arcsin (c x)}\right )}{b^2}-\frac {2 d^2 \left (1-c^2 x^2\right )^{5/2}}{b c x \sqrt {a+b \arcsin (c x)}}\) |
3.5.40.3.1 Defintions of rubi rules used
Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f , m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1]))
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x _Symbol] :> Simp[(1/(b*c))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p] Subst[Int[ x^n*Cos[-a/b + x/b]^(2*p + 1), x], x, a + b*ArcSin[c*x]], x] /; FreeQ[{a, b , c, d, e, n}, x] && EqQ[c^2*d + e, 0] && IGtQ[2*p, 0]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.)*((d_) + (e_. )*(x_)^2)^(p_.), x_Symbol] :> Simp[(f*x)^m*Sqrt[1 - c^2*x^2]*(d + e*x^2)^p* ((a + b*ArcSin[c*x])^(n + 1)/(b*c*(n + 1))), x] + (-Simp[f*(m/(b*c*(n + 1)) )*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p] Int[(f*x)^(m - 1)*(1 - c^2*x^2)^(p - 1/2)*(a + b*ArcSin[c*x])^(n + 1), x], x] + Simp[c*((m + 2*p + 1)/(b*f*(n + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p] Int[(f*x)^(m + 1)*(1 - c^2*x^2 )^(p - 1/2)*(a + b*ArcSin[c*x])^(n + 1), x], x]) /; FreeQ[{a, b, c, d, e, f }, x] && EqQ[c^2*d + e, 0] && LtQ[n, -1] && IGtQ[2*p, 0] && NeQ[m + 2*p + 1 , 0] && IGtQ[m, -3]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_. )*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*ArcSin[c*x])^n/Sqrt [d + e*x^2], (f*x)^m*(d + e*x^2)^(p + 1/2), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[c^2*d + e, 0] && IGtQ[p + 1/2, 0] && !IGtQ[(m + 1)/2, 0] && (EqQ[m, -1] || EqQ[m, -2])
Not integrable
Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93
\[\int \frac {\left (-c^{2} d \,x^{2}+d \right )^{2}}{x \left (a +b \arcsin \left (c x \right )\right )^{\frac {3}{2}}}d x\]
Exception generated. \[ \int \frac {\left (d-c^2 d x^2\right )^2}{x (a+b \arcsin (c x))^{3/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
Not integrable
Time = 5.70 (sec) , antiderivative size = 133, normalized size of antiderivative = 4.59 \[ \int \frac {\left (d-c^2 d x^2\right )^2}{x (a+b \arcsin (c x))^{3/2}} \, dx=d^{2} \left (\int \left (- \frac {2 c^{2} x^{2}}{a x \sqrt {a + b \operatorname {asin}{\left (c x \right )}} + b x \sqrt {a + b \operatorname {asin}{\left (c x \right )}} \operatorname {asin}{\left (c x \right )}}\right )\, dx + \int \frac {c^{4} x^{4}}{a x \sqrt {a + b \operatorname {asin}{\left (c x \right )}} + b x \sqrt {a + b \operatorname {asin}{\left (c x \right )}} \operatorname {asin}{\left (c x \right )}}\, dx + \int \frac {1}{a x \sqrt {a + b \operatorname {asin}{\left (c x \right )}} + b x \sqrt {a + b \operatorname {asin}{\left (c x \right )}} \operatorname {asin}{\left (c x \right )}}\, dx\right ) \]
d**2*(Integral(-2*c**2*x**2/(a*x*sqrt(a + b*asin(c*x)) + b*x*sqrt(a + b*as in(c*x))*asin(c*x)), x) + Integral(c**4*x**4/(a*x*sqrt(a + b*asin(c*x)) + b*x*sqrt(a + b*asin(c*x))*asin(c*x)), x) + Integral(1/(a*x*sqrt(a + b*asin (c*x)) + b*x*sqrt(a + b*asin(c*x))*asin(c*x)), x))
Not integrable
Time = 1.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {\left (d-c^2 d x^2\right )^2}{x (a+b \arcsin (c x))^{3/2}} \, dx=\int { \frac {{\left (c^{2} d x^{2} - d\right )}^{2}}{{\left (b \arcsin \left (c x\right ) + a\right )}^{\frac {3}{2}} x} \,d x } \]
Exception generated. \[ \int \frac {\left (d-c^2 d x^2\right )^2}{x (a+b \arcsin (c x))^{3/2}} \, dx=\text {Exception raised: RuntimeError} \]
Exception raised: RuntimeError >> an error occurred running a Giac command :INPUT:sage2OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const ve cteur & l) Error: Bad Argument Value
Not integrable
Time = 0.16 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {\left (d-c^2 d x^2\right )^2}{x (a+b \arcsin (c x))^{3/2}} \, dx=\int \frac {{\left (d-c^2\,d\,x^2\right )}^2}{x\,{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^{3/2}} \,d x \]