\(\int \frac {(1-c^2 x^2)^{5/2}}{x (a+b \arcsin (c x))^2} \, dx\) [403]

Optimal result

Integrand size = 28, antiderivative size = 28 \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x (a+b \arcsin (c x))^2} \, dx=-\frac {\left (1-c^2 x^2\right )^3}{b c x (a+b \arcsin (c x))}-\frac {25 \cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arcsin (c x)}{b}\right )}{8 b^2}-\frac {25 \cos \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )}{16 b^2}-\frac {5 \cos \left (\frac {5 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {5 (a+b \arcsin (c x))}{b}\right )}{16 b^2}-\frac {25 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arcsin (c x)}{b}\right )}{8 b^2}-\frac {25 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )}{16 b^2}-\frac {5 \sin \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 (a+b \arcsin (c x))}{b}\right )}{16 b^2}-\frac {\text {Int}\left (\frac {\left (1-c^2 x^2\right )^2}{x^2 (a+b \arcsin (c x))},x\right )}{b c} \]

[Out]

-(-c^2*x^2+1)^3/b/c/x/(a+b*arcsin(c*x))-25/8*Ci((a+b*arcsin(c*x))/b)*cos(a/b)/b^2-25/16*Ci(3*(a+b*arcsin(c*x))
/b)*cos(3*a/b)/b^2-5/16*Ci(5*(a+b*arcsin(c*x))/b)*cos(5*a/b)/b^2-25/8*Si((a+b*arcsin(c*x))/b)*sin(a/b)/b^2-25/
16*Si(3*(a+b*arcsin(c*x))/b)*sin(3*a/b)/b^2-5/16*Si(5*(a+b*arcsin(c*x))/b)*sin(5*a/b)/b^2-Unintegrable((-c^2*x
^2+1)^2/x^2/(a+b*arcsin(c*x)),x)/b/c

Rubi [N/A]

Not integrable

Time = 0.31 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 0, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x (a+b \arcsin (c x))^2} \, dx=\int \frac {\left (1-c^2 x^2\right )^{5/2}}{x (a+b \arcsin (c x))^2} \, dx \]

[In]

Int[(1 - c^2*x^2)^(5/2)/(x*(a + b*ArcSin[c*x])^2),x]

[Out]

-((1 - c^2*x^2)^3/(b*c*x*(a + b*ArcSin[c*x]))) - (25*Cos[a/b]*CosIntegral[(a + b*ArcSin[c*x])/b])/(8*b^2) - (2
5*Cos[(3*a)/b]*CosIntegral[(3*(a + b*ArcSin[c*x]))/b])/(16*b^2) - (5*Cos[(5*a)/b]*CosIntegral[(5*(a + b*ArcSin
[c*x]))/b])/(16*b^2) - (25*Sin[a/b]*SinIntegral[(a + b*ArcSin[c*x])/b])/(8*b^2) - (25*Sin[(3*a)/b]*SinIntegral
[(3*(a + b*ArcSin[c*x]))/b])/(16*b^2) - (5*Sin[(5*a)/b]*SinIntegral[(5*(a + b*ArcSin[c*x]))/b])/(16*b^2) - Def
er[Int][(1 - c^2*x^2)^2/(x^2*(a + b*ArcSin[c*x])), x]/(b*c)

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (1-c^2 x^2\right )^3}{b c x (a+b \arcsin (c x))}-\frac {\int \frac {\left (1-c^2 x^2\right )^2}{x^2 (a+b \arcsin (c x))} \, dx}{b c}-\frac {(5 c) \int \frac {\left (1-c^2 x^2\right )^2}{a+b \arcsin (c x)} \, dx}{b} \\ & = -\frac {\left (1-c^2 x^2\right )^3}{b c x (a+b \arcsin (c x))}-\frac {5 \text {Subst}\left (\int \frac {\cos ^5\left (\frac {a}{b}-\frac {x}{b}\right )}{x} \, dx,x,a+b \arcsin (c x)\right )}{b^2}-\frac {\int \frac {\left (1-c^2 x^2\right )^2}{x^2 (a+b \arcsin (c x))} \, dx}{b c} \\ & = -\frac {\left (1-c^2 x^2\right )^3}{b c x (a+b \arcsin (c x))}-\frac {5 \text {Subst}\left (\int \left (\frac {\cos \left (\frac {5 a}{b}-\frac {5 x}{b}\right )}{16 x}+\frac {5 \cos \left (\frac {3 a}{b}-\frac {3 x}{b}\right )}{16 x}+\frac {5 \cos \left (\frac {a}{b}-\frac {x}{b}\right )}{8 x}\right ) \, dx,x,a+b \arcsin (c x)\right )}{b^2}-\frac {\int \frac {\left (1-c^2 x^2\right )^2}{x^2 (a+b \arcsin (c x))} \, dx}{b c} \\ & = -\frac {\left (1-c^2 x^2\right )^3}{b c x (a+b \arcsin (c x))}-\frac {5 \text {Subst}\left (\int \frac {\cos \left (\frac {5 a}{b}-\frac {5 x}{b}\right )}{x} \, dx,x,a+b \arcsin (c x)\right )}{16 b^2}-\frac {25 \text {Subst}\left (\int \frac {\cos \left (\frac {3 a}{b}-\frac {3 x}{b}\right )}{x} \, dx,x,a+b \arcsin (c x)\right )}{16 b^2}-\frac {25 \text {Subst}\left (\int \frac {\cos \left (\frac {a}{b}-\frac {x}{b}\right )}{x} \, dx,x,a+b \arcsin (c x)\right )}{8 b^2}-\frac {\int \frac {\left (1-c^2 x^2\right )^2}{x^2 (a+b \arcsin (c x))} \, dx}{b c} \\ & = -\frac {\left (1-c^2 x^2\right )^3}{b c x (a+b \arcsin (c x))}-\frac {\int \frac {\left (1-c^2 x^2\right )^2}{x^2 (a+b \arcsin (c x))} \, dx}{b c}-\frac {\left (25 \cos \left (\frac {a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {x}{b}\right )}{x} \, dx,x,a+b \arcsin (c x)\right )}{8 b^2}-\frac {\left (25 \cos \left (\frac {3 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {3 x}{b}\right )}{x} \, dx,x,a+b \arcsin (c x)\right )}{16 b^2}-\frac {\left (5 \cos \left (\frac {5 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {5 x}{b}\right )}{x} \, dx,x,a+b \arcsin (c x)\right )}{16 b^2}-\frac {\left (25 \sin \left (\frac {a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {x}{b}\right )}{x} \, dx,x,a+b \arcsin (c x)\right )}{8 b^2}-\frac {\left (25 \sin \left (\frac {3 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {3 x}{b}\right )}{x} \, dx,x,a+b \arcsin (c x)\right )}{16 b^2}-\frac {\left (5 \sin \left (\frac {5 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {5 x}{b}\right )}{x} \, dx,x,a+b \arcsin (c x)\right )}{16 b^2} \\ & = -\frac {\left (1-c^2 x^2\right )^3}{b c x (a+b \arcsin (c x))}-\frac {25 \cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arcsin (c x)}{b}\right )}{8 b^2}-\frac {25 \cos \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )}{16 b^2}-\frac {5 \cos \left (\frac {5 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {5 (a+b \arcsin (c x))}{b}\right )}{16 b^2}-\frac {25 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arcsin (c x)}{b}\right )}{8 b^2}-\frac {25 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )}{16 b^2}-\frac {5 \sin \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 (a+b \arcsin (c x))}{b}\right )}{16 b^2}-\frac {\int \frac {\left (1-c^2 x^2\right )^2}{x^2 (a+b \arcsin (c x))} \, dx}{b c} \\ \end{align*}

Mathematica [N/A]

Not integrable

Time = 11.46 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x (a+b \arcsin (c x))^2} \, dx=\int \frac {\left (1-c^2 x^2\right )^{5/2}}{x (a+b \arcsin (c x))^2} \, dx \]

[In]

Integrate[(1 - c^2*x^2)^(5/2)/(x*(a + b*ArcSin[c*x])^2),x]

[Out]

Integrate[(1 - c^2*x^2)^(5/2)/(x*(a + b*ArcSin[c*x])^2), x]

Maple [N/A] (verified)

Not integrable

Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93

\[\int \frac {\left (-c^{2} x^{2}+1\right )^{\frac {5}{2}}}{x \left (a +b \arcsin \left (c x \right )\right )^{2}}d x\]

[In]

int((-c^2*x^2+1)^(5/2)/x/(a+b*arcsin(c*x))^2,x)

[Out]

int((-c^2*x^2+1)^(5/2)/x/(a+b*arcsin(c*x))^2,x)

Fricas [N/A]

Not integrable

Time = 0.25 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.14 \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x (a+b \arcsin (c x))^2} \, dx=\int { \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {5}{2}}}{{\left (b \arcsin \left (c x\right ) + a\right )}^{2} x} \,d x } \]

[In]

integrate((-c^2*x^2+1)^(5/2)/x/(a+b*arcsin(c*x))^2,x, algorithm="fricas")

[Out]

integral((c^4*x^4 - 2*c^2*x^2 + 1)*sqrt(-c^2*x^2 + 1)/(b^2*x*arcsin(c*x)^2 + 2*a*b*x*arcsin(c*x) + a^2*x), x)

Sympy [N/A]

Not integrable

Time = 7.62 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x (a+b \arcsin (c x))^2} \, dx=\int \frac {\left (- \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}}}{x \left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}\, dx \]

[In]

integrate((-c**2*x**2+1)**(5/2)/x/(a+b*asin(c*x))**2,x)

[Out]

Integral((-(c*x - 1)*(c*x + 1))**(5/2)/(x*(a + b*asin(c*x))**2), x)

Maxima [N/A]

Not integrable

Time = 1.04 (sec) , antiderivative size = 161, normalized size of antiderivative = 5.75 \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x (a+b \arcsin (c x))^2} \, dx=\int { \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {5}{2}}}{{\left (b \arcsin \left (c x\right ) + a\right )}^{2} x} \,d x } \]

[In]

integrate((-c^2*x^2+1)^(5/2)/x/(a+b*arcsin(c*x))^2,x, algorithm="maxima")

[Out]

(c^6*x^6 - 3*c^4*x^4 + 3*c^2*x^2 - (b^2*c*x*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + a*b*c*x)*integrate((5
*c^6*x^6 - 9*c^4*x^4 + 3*c^2*x^2 + 1)/(b^2*c*x^2*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + a*b*c*x^2), x) -
1)/(b^2*c*x*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + a*b*c*x)

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x (a+b \arcsin (c x))^2} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((-c^2*x^2+1)^(5/2)/x/(a+b*arcsin(c*x))^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [N/A]

Not integrable

Time = 0.15 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x (a+b \arcsin (c x))^2} \, dx=\int \frac {{\left (1-c^2\,x^2\right )}^{5/2}}{x\,{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2} \,d x \]

[In]

int((1 - c^2*x^2)^(5/2)/(x*(a + b*asin(c*x))^2),x)

[Out]

int((1 - c^2*x^2)^(5/2)/(x*(a + b*asin(c*x))^2), x)