Integrand size = 28, antiderivative size = 220 \[ \int \frac {x^2 \left (1-c^2 x^2\right )^{3/2}}{(a+b \arccos (c x))^2} \, dx=-\frac {x^2 \left (1-c^2 x^2\right )^2}{b c (a+b \arccos (c x))}+\frac {\operatorname {CosIntegral}\left (\frac {2 (a+b \arccos (c x))}{b}\right ) \sin \left (\frac {2 a}{b}\right )}{16 b^2 c^3}-\frac {\operatorname {CosIntegral}\left (\frac {4 (a+b \arccos (c x))}{b}\right ) \sin \left (\frac {4 a}{b}\right )}{4 b^2 c^3}-\frac {3 \operatorname {CosIntegral}\left (\frac {6 (a+b \arccos (c x))}{b}\right ) \sin \left (\frac {6 a}{b}\right )}{16 b^2 c^3}-\frac {\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arccos (c x))}{b}\right )}{16 b^2 c^3}+\frac {\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arccos (c x))}{b}\right )}{4 b^2 c^3}+\frac {3 \cos \left (\frac {6 a}{b}\right ) \text {Si}\left (\frac {6 (a+b \arccos (c x))}{b}\right )}{16 b^2 c^3} \] Output:
-x^2*(-c^2*x^2+1)^2/b/c/(a+b*arccos(c*x))+1/16*Ci(2*(a+b*arccos(c*x))/b)*s in(2*a/b)/b^2/c^3-1/4*Ci(4*(a+b*arccos(c*x))/b)*sin(4*a/b)/b^2/c^3-3/16*Ci (6*(a+b*arccos(c*x))/b)*sin(6*a/b)/b^2/c^3-1/16*cos(2*a/b)*Si(2*(a+b*arcco s(c*x))/b)/b^2/c^3+1/4*cos(4*a/b)*Si(4*(a+b*arccos(c*x))/b)/b^2/c^3+3/16*c os(6*a/b)*Si(6*(a+b*arccos(c*x))/b)/b^2/c^3
Time = 0.62 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.39 \[ \int \frac {x^2 \left (1-c^2 x^2\right )^{3/2}}{(a+b \arccos (c x))^2} \, dx=-\frac {-16 b c^2 x^2+32 b c^4 x^4-16 b c^6 x^6-(a+b \arccos (c x)) \operatorname {CosIntegral}\left (2 \left (\frac {a}{b}+\arccos (c x)\right )\right ) \sin \left (\frac {2 a}{b}\right )-4 (a+b \arccos (c x)) \operatorname {CosIntegral}\left (4 \left (\frac {a}{b}+\arccos (c x)\right )\right ) \sin \left (\frac {4 a}{b}\right )+3 a \operatorname {CosIntegral}\left (6 \left (\frac {a}{b}+\arccos (c x)\right )\right ) \sin \left (\frac {6 a}{b}\right )+3 b \arccos (c x) \operatorname {CosIntegral}\left (6 \left (\frac {a}{b}+\arccos (c x)\right )\right ) \sin \left (\frac {6 a}{b}\right )+a \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\arccos (c x)\right )\right )+b \arccos (c x) \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\arccos (c x)\right )\right )+4 a \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\arccos (c x)\right )\right )+4 b \arccos (c x) \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\arccos (c x)\right )\right )-3 a \cos \left (\frac {6 a}{b}\right ) \text {Si}\left (6 \left (\frac {a}{b}+\arccos (c x)\right )\right )-3 b \arccos (c x) \cos \left (\frac {6 a}{b}\right ) \text {Si}\left (6 \left (\frac {a}{b}+\arccos (c x)\right )\right )}{16 b^2 c^3 (a+b \arccos (c x))} \] Input:
Integrate[(x^2*(1 - c^2*x^2)^(3/2))/(a + b*ArcCos[c*x])^2,x]
Output:
-1/16*(-16*b*c^2*x^2 + 32*b*c^4*x^4 - 16*b*c^6*x^6 - (a + b*ArcCos[c*x])*C osIntegral[2*(a/b + ArcCos[c*x])]*Sin[(2*a)/b] - 4*(a + b*ArcCos[c*x])*Cos Integral[4*(a/b + ArcCos[c*x])]*Sin[(4*a)/b] + 3*a*CosIntegral[6*(a/b + Ar cCos[c*x])]*Sin[(6*a)/b] + 3*b*ArcCos[c*x]*CosIntegral[6*(a/b + ArcCos[c*x ])]*Sin[(6*a)/b] + a*Cos[(2*a)/b]*SinIntegral[2*(a/b + ArcCos[c*x])] + b*A rcCos[c*x]*Cos[(2*a)/b]*SinIntegral[2*(a/b + ArcCos[c*x])] + 4*a*Cos[(4*a) /b]*SinIntegral[4*(a/b + ArcCos[c*x])] + 4*b*ArcCos[c*x]*Cos[(4*a)/b]*SinI ntegral[4*(a/b + ArcCos[c*x])] - 3*a*Cos[(6*a)/b]*SinIntegral[6*(a/b + Arc Cos[c*x])] - 3*b*ArcCos[c*x]*Cos[(6*a)/b]*SinIntegral[6*(a/b + ArcCos[c*x] )])/(b^2*c^3*(a + b*ArcCos[c*x]))
Time = 0.97 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.14, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {5215, 5225, 25, 4906, 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.
\(\displaystyle \int \frac {x^2 \left (1-c^2 x^2\right )^{3/2}}{(a+b \arccos (c x))^2} \, dx\) |
\(\Big \downarrow \) 5215 |
\(\displaystyle -\frac {2 \int \frac {x \left (1-c^2 x^2\right )}{a+b \arccos (c x)}dx}{b c}+\frac {6 c \int \frac {x^3 \left (1-c^2 x^2\right )}{a+b \arccos (c x)}dx}{b}+\frac {x^2 \left (1-c^2 x^2\right )^2}{b c (a+b \arccos (c x))}\) |
\(\Big \downarrow \) 5225 |
\(\displaystyle -\frac {6 \int -\frac {\cos ^3\left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right ) \sin ^3\left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b^2 c^3}+\frac {2 \int -\frac {\cos \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right ) \sin ^3\left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b^2 c^3}+\frac {x^2 \left (1-c^2 x^2\right )^2}{b c (a+b \arccos (c x))}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {6 \int \frac {\cos ^3\left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right ) \sin ^3\left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b^2 c^3}-\frac {2 \int \frac {\cos \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right ) \sin ^3\left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b^2 c^3}+\frac {x^2 \left (1-c^2 x^2\right )^2}{b c (a+b \arccos (c x))}\) |
\(\Big \downarrow \) 4906 |
\(\displaystyle \frac {6 \int \left (\frac {3 \sin \left (\frac {2 a}{b}-\frac {2 (a+b \arccos (c x))}{b}\right )}{32 (a+b \arccos (c x))}-\frac {\sin \left (\frac {6 a}{b}-\frac {6 (a+b \arccos (c x))}{b}\right )}{32 (a+b \arccos (c x))}\right )d(a+b \arccos (c x))}{b^2 c^3}-\frac {2 \int \left (\frac {\sin \left (\frac {2 a}{b}-\frac {2 (a+b \arccos (c x))}{b}\right )}{4 (a+b \arccos (c x))}-\frac {\sin \left (\frac {4 a}{b}-\frac {4 (a+b \arccos (c x))}{b}\right )}{8 (a+b \arccos (c x))}\right )d(a+b \arccos (c x))}{b^2 c^3}+\frac {x^2 \left (1-c^2 x^2\right )^2}{b c (a+b \arccos (c x))}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \left (-\frac {1}{4} \sin \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {2 (a+b \arccos (c x))}{b}\right )+\frac {1}{8} \sin \left (\frac {4 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {4 (a+b \arccos (c x))}{b}\right )+\frac {1}{4} \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arccos (c x))}{b}\right )-\frac {1}{8} \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arccos (c x))}{b}\right )\right )}{b^2 c^3}-\frac {6 \left (-\frac {3}{32} \sin \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {2 (a+b \arccos (c x))}{b}\right )+\frac {1}{32} \sin \left (\frac {6 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {6 (a+b \arccos (c x))}{b}\right )+\frac {3}{32} \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arccos (c x))}{b}\right )-\frac {1}{32} \cos \left (\frac {6 a}{b}\right ) \text {Si}\left (\frac {6 (a+b \arccos (c x))}{b}\right )\right )}{b^2 c^3}+\frac {x^2 \left (1-c^2 x^2\right )^2}{b c (a+b \arccos (c x))}\) |
Input:
Int[(x^2*(1 - c^2*x^2)^(3/2))/(a + b*ArcCos[c*x])^2,x]
Output:
(x^2*(1 - c^2*x^2)^2)/(b*c*(a + b*ArcCos[c*x])) + (2*(-1/4*(CosIntegral[(2 *(a + b*ArcCos[c*x]))/b]*Sin[(2*a)/b]) + (CosIntegral[(4*(a + b*ArcCos[c*x ]))/b]*Sin[(4*a)/b])/8 + (Cos[(2*a)/b]*SinIntegral[(2*(a + b*ArcCos[c*x])) /b])/4 - (Cos[(4*a)/b]*SinIntegral[(4*(a + b*ArcCos[c*x]))/b])/8))/(b^2*c^ 3) - (6*((-3*CosIntegral[(2*(a + b*ArcCos[c*x]))/b]*Sin[(2*a)/b])/32 + (Co sIntegral[(6*(a + b*ArcCos[c*x]))/b]*Sin[(6*a)/b])/32 + (3*Cos[(2*a)/b]*Si nIntegral[(2*(a + b*ArcCos[c*x]))/b])/32 - (Cos[(6*a)/b]*SinIntegral[(6*(a + b*ArcCos[c*x]))/b])/32))/(b^2*c^3)
Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b _.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin[a + b*x ]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IG tQ[p, 0]
Int[((a_.) + ArcCos[(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*ArcCos[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*ArcCos[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*ArcCos[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_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^ 2)^(p_.), x_Symbol] :> Simp[(-(b*c^(m + 1))^(-1))*Simp[(d + e*x^2)^p/(1 - c ^2*x^2)^p] Subst[Int[x^n*Cos[-a/b + x/b]^m*Sin[-a/b + x/b]^(2*p + 1), x], x, a + b*ArcCos[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e , 0] && IGtQ[2*p + 2, 0] && IGtQ[m, 0]
Time = 0.16 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.65
method | result | size |
default | \(-\frac {8 \arccos \left (c x \right ) \operatorname {Si}\left (4 \arccos \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) b -8 \arccos \left (c x \right ) \operatorname {Ci}\left (4 \arccos \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right ) b +2 \arccos \left (c x \right ) \operatorname {Si}\left (2 \arccos \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) b -2 \arccos \left (c x \right ) \operatorname {Ci}\left (2 \arccos \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) b -6 \arccos \left (c x \right ) \operatorname {Si}\left (6 \arccos \left (c x \right )+\frac {6 a}{b}\right ) \cos \left (\frac {6 a}{b}\right ) b +6 \arccos \left (c x \right ) \operatorname {Ci}\left (6 \arccos \left (c x \right )+\frac {6 a}{b}\right ) \sin \left (\frac {6 a}{b}\right ) b +8 \,\operatorname {Si}\left (4 \arccos \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) a -8 \,\operatorname {Ci}\left (4 \arccos \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right ) a +2 \,\operatorname {Si}\left (2 \arccos \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) a -2 \,\operatorname {Ci}\left (2 \arccos \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) a -6 \,\operatorname {Si}\left (6 \arccos \left (c x \right )+\frac {6 a}{b}\right ) \cos \left (\frac {6 a}{b}\right ) a +6 \,\operatorname {Ci}\left (6 \arccos \left (c x \right )+\frac {6 a}{b}\right ) \sin \left (\frac {6 a}{b}\right ) a +2 \cos \left (4 \arccos \left (c x \right )\right ) b +\cos \left (2 \arccos \left (c x \right )\right ) b -\cos \left (6 \arccos \left (c x \right )\right ) b -2 b}{32 c^{3} \left (a +b \arccos \left (c x \right )\right ) b^{2}}\) | \(364\) |
Input:
int(x^2*(-c^2*x^2+1)^(3/2)/(a+b*arccos(c*x))^2,x,method=_RETURNVERBOSE)
Output:
-1/32/c^3*(8*arccos(c*x)*Si(4*arccos(c*x)+4*a/b)*cos(4*a/b)*b-8*arccos(c*x )*Ci(4*arccos(c*x)+4*a/b)*sin(4*a/b)*b+2*arccos(c*x)*Si(2*arccos(c*x)+2*a/ b)*cos(2*a/b)*b-2*arccos(c*x)*Ci(2*arccos(c*x)+2*a/b)*sin(2*a/b)*b-6*arcco s(c*x)*Si(6*arccos(c*x)+6*a/b)*cos(6*a/b)*b+6*arccos(c*x)*Ci(6*arccos(c*x) +6*a/b)*sin(6*a/b)*b+8*Si(4*arccos(c*x)+4*a/b)*cos(4*a/b)*a-8*Ci(4*arccos( c*x)+4*a/b)*sin(4*a/b)*a+2*Si(2*arccos(c*x)+2*a/b)*cos(2*a/b)*a-2*Ci(2*arc cos(c*x)+2*a/b)*sin(2*a/b)*a-6*Si(6*arccos(c*x)+6*a/b)*cos(6*a/b)*a+6*Ci(6 *arccos(c*x)+6*a/b)*sin(6*a/b)*a+2*cos(4*arccos(c*x))*b+cos(2*arccos(c*x)) *b-cos(6*arccos(c*x))*b-2*b)/(a+b*arccos(c*x))/b^2
\[ \int \frac {x^2 \left (1-c^2 x^2\right )^{3/2}}{(a+b \arccos (c x))^2} \, dx=\int { \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} x^{2}}{{\left (b \arccos \left (c x\right ) + a\right )}^{2}} \,d x } \] Input:
integrate(x^2*(-c^2*x^2+1)^(3/2)/(a+b*arccos(c*x))^2,x, algorithm="fricas" )
Output:
integral(-(c^2*x^4 - x^2)*sqrt(-c^2*x^2 + 1)/(b^2*arccos(c*x)^2 + 2*a*b*ar ccos(c*x) + a^2), x)
\[ \int \frac {x^2 \left (1-c^2 x^2\right )^{3/2}}{(a+b \arccos (c x))^2} \, dx=\int \frac {x^{2} \left (- \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {3}{2}}}{\left (a + b \operatorname {acos}{\left (c x \right )}\right )^{2}}\, dx \] Input:
integrate(x**2*(-c**2*x**2+1)**(3/2)/(a+b*acos(c*x))**2,x)
Output:
Integral(x**2*(-(c*x - 1)*(c*x + 1))**(3/2)/(a + b*acos(c*x))**2, x)
\[ \int \frac {x^2 \left (1-c^2 x^2\right )^{3/2}}{(a+b \arccos (c x))^2} \, dx=\int { \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} x^{2}}{{\left (b \arccos \left (c x\right ) + a\right )}^{2}} \,d x } \] Input:
integrate(x^2*(-c^2*x^2+1)^(3/2)/(a+b*arccos(c*x))^2,x, algorithm="maxima" )
Output:
(c^4*x^6 - 2*c^2*x^4 + x^2 - (b^2*c*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x) + a*b*c)*integrate(2*(3*c^4*x^5 - 4*c^2*x^3 + x)/(b^2*c*arctan2(sqrt( c*x + 1)*sqrt(-c*x + 1), c*x) + a*b*c), x))/(b^2*c*arctan2(sqrt(c*x + 1)*s qrt(-c*x + 1), c*x) + a*b*c)
Leaf count of result is larger than twice the leaf count of optimal. 1574 vs. \(2 (207) = 414\).
Time = 0.26 (sec) , antiderivative size = 1574, normalized size of antiderivative = 7.15 \[ \int \frac {x^2 \left (1-c^2 x^2\right )^{3/2}}{(a+b \arccos (c x))^2} \, dx=\text {Too large to display} \] Input:
integrate(x^2*(-c^2*x^2+1)^(3/2)/(a+b*arccos(c*x))^2,x, algorithm="giac")
Output:
b*c^6*x^6/(b^3*c^3*arccos(c*x) + a*b^2*c^3) - 2*b*c^4*x^4/(b^3*c^3*arccos( c*x) + a*b^2*c^3) - 6*b*arccos(c*x)*cos(a/b)^5*cos_integral(6*a/b + 6*arcc os(c*x))*sin(a/b)/(b^3*c^3*arccos(c*x) + a*b^2*c^3) + 6*b*arccos(c*x)*cos( a/b)^6*sin_integral(6*a/b + 6*arccos(c*x))/(b^3*c^3*arccos(c*x) + a*b^2*c^ 3) - 6*a*cos(a/b)^5*cos_integral(6*a/b + 6*arccos(c*x))*sin(a/b)/(b^3*c^3* arccos(c*x) + a*b^2*c^3) + 6*a*cos(a/b)^6*sin_integral(6*a/b + 6*arccos(c* x))/(b^3*c^3*arccos(c*x) + a*b^2*c^3) + 6*b*arccos(c*x)*cos(a/b)^3*cos_int egral(6*a/b + 6*arccos(c*x))*sin(a/b)/(b^3*c^3*arccos(c*x) + a*b^2*c^3) + 2*b*arccos(c*x)*cos(a/b)^3*cos_integral(4*a/b + 4*arccos(c*x))*sin(a/b)/(b ^3*c^3*arccos(c*x) + a*b^2*c^3) - 9*b*arccos(c*x)*cos(a/b)^4*sin_integral( 6*a/b + 6*arccos(c*x))/(b^3*c^3*arccos(c*x) + a*b^2*c^3) - 2*b*arccos(c*x) *cos(a/b)^4*sin_integral(4*a/b + 4*arccos(c*x))/(b^3*c^3*arccos(c*x) + a*b ^2*c^3) + 6*a*cos(a/b)^3*cos_integral(6*a/b + 6*arccos(c*x))*sin(a/b)/(b^3 *c^3*arccos(c*x) + a*b^2*c^3) + 2*a*cos(a/b)^3*cos_integral(4*a/b + 4*arcc os(c*x))*sin(a/b)/(b^3*c^3*arccos(c*x) + a*b^2*c^3) - 9*a*cos(a/b)^4*sin_i ntegral(6*a/b + 6*arccos(c*x))/(b^3*c^3*arccos(c*x) + a*b^2*c^3) - 2*a*cos (a/b)^4*sin_integral(4*a/b + 4*arccos(c*x))/(b^3*c^3*arccos(c*x) + a*b^2*c ^3) + b*c^2*x^2/(b^3*c^3*arccos(c*x) + a*b^2*c^3) - 9/8*b*arccos(c*x)*cos( a/b)*cos_integral(6*a/b + 6*arccos(c*x))*sin(a/b)/(b^3*c^3*arccos(c*x) + a *b^2*c^3) - b*arccos(c*x)*cos(a/b)*cos_integral(4*a/b + 4*arccos(c*x))*...
Timed out. \[ \int \frac {x^2 \left (1-c^2 x^2\right )^{3/2}}{(a+b \arccos (c x))^2} \, dx=\int \frac {x^2\,{\left (1-c^2\,x^2\right )}^{3/2}}{{\left (a+b\,\mathrm {acos}\left (c\,x\right )\right )}^2} \,d x \] Input:
int((x^2*(1 - c^2*x^2)^(3/2))/(a + b*acos(c*x))^2,x)
Output:
int((x^2*(1 - c^2*x^2)^(3/2))/(a + b*acos(c*x))^2, x)
\[ \int \frac {x^2 \left (1-c^2 x^2\right )^{3/2}}{(a+b \arccos (c x))^2} \, dx=-\left (\int \frac {\sqrt {-c^{2} x^{2}+1}\, x^{4}}{\mathit {acos} \left (c x \right )^{2} b^{2}+2 \mathit {acos} \left (c x \right ) a b +a^{2}}d x \right ) c^{2}+\int \frac {\sqrt {-c^{2} x^{2}+1}\, x^{2}}{\mathit {acos} \left (c x \right )^{2} b^{2}+2 \mathit {acos} \left (c x \right ) a b +a^{2}}d x \] Input:
int(x^2*(-c^2*x^2+1)^(3/2)/(a+b*acos(c*x))^2,x)
Output:
- int((sqrt( - c**2*x**2 + 1)*x**4)/(acos(c*x)**2*b**2 + 2*acos(c*x)*a*b + a**2),x)*c**2 + int((sqrt( - c**2*x**2 + 1)*x**2)/(acos(c*x)**2*b**2 + 2 *acos(c*x)*a*b + a**2),x)