\(\int \frac {(1-c^2 x^2)^{5/2}}{x (a+b \arcsin (c x))} \, dx\) [337]

Optimal result

Integrand size = 28, antiderivative size = 28 \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x (a+b \arcsin (c x))} \, dx=\frac {11 \operatorname {CosIntegral}\left (\frac {a+b \arcsin (c x)}{b}\right ) \sin \left (\frac {a}{b}\right )}{8 b}+\frac {7 \operatorname {CosIntegral}\left (\frac {3 (a+b \arcsin (c x))}{b}\right ) \sin \left (\frac {3 a}{b}\right )}{16 b}+\frac {\operatorname {CosIntegral}\left (\frac {5 (a+b \arcsin (c x))}{b}\right ) \sin \left (\frac {5 a}{b}\right )}{16 b}-\frac {11 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arcsin (c x)}{b}\right )}{8 b}-\frac {7 \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )}{16 b}-\frac {\cos \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 (a+b \arcsin (c x))}{b}\right )}{16 b}+\text {Int}\left (\frac {1}{x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))},x\right ) \]

[Out]

-11/8*cos(a/b)*Si((a+b*arcsin(c*x))/b)/b-7/16*cos(3*a/b)*Si(3*(a+b*arcsin(c*x))/b)/b-1/16*cos(5*a/b)*Si(5*(a+b
*arcsin(c*x))/b)/b+11/8*Ci((a+b*arcsin(c*x))/b)*sin(a/b)/b+7/16*Ci(3*(a+b*arcsin(c*x))/b)*sin(3*a/b)/b+1/16*Ci
(5*(a+b*arcsin(c*x))/b)*sin(5*a/b)/b+Unintegrable(1/x/(a+b*arcsin(c*x))/(-c^2*x^2+1)^(1/2),x)

Rubi [N/A]

Not integrable

Time = 0.69 (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 = {} \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x (a+b \arcsin (c x))} \, dx=\int \frac {\left (1-c^2 x^2\right )^{5/2}}{x (a+b \arcsin (c x))} \, dx \]

[In]

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

[Out]

(11*CosIntegral[(a + b*ArcSin[c*x])/b]*Sin[a/b])/(8*b) + (7*CosIntegral[(3*(a + b*ArcSin[c*x]))/b]*Sin[(3*a)/b
])/(16*b) + (CosIntegral[(5*(a + b*ArcSin[c*x]))/b]*Sin[(5*a)/b])/(16*b) - (11*Cos[a/b]*SinIntegral[(a + b*Arc
Sin[c*x])/b])/(8*b) - (7*Cos[(3*a)/b]*SinIntegral[(3*(a + b*ArcSin[c*x]))/b])/(16*b) - (Cos[(5*a)/b]*SinIntegr
al[(5*(a + b*ArcSin[c*x]))/b])/(16*b) + Defer[Int][1/(x*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}-\frac {3 c^2 x}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))}+\frac {3 c^4 x^3}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))}-\frac {c^6 x^5}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))}\right ) \, dx \\ & = -\left (\left (3 c^2\right ) \int \frac {x}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx\right )+\left (3 c^4\right ) \int \frac {x^3}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx-c^6 \int \frac {x^5}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx+\int \frac {1}{x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx \\ & = \frac {\text {Subst}\left (\int \frac {\sin ^5\left (\frac {a}{b}-\frac {x}{b}\right )}{x} \, dx,x,a+b \arcsin (c x)\right )}{b}+\frac {3 \text {Subst}\left (\int \frac {\sin \left (\frac {a}{b}-\frac {x}{b}\right )}{x} \, dx,x,a+b \arcsin (c x)\right )}{b}-\frac {3 \text {Subst}\left (\int \frac {\sin ^3\left (\frac {a}{b}-\frac {x}{b}\right )}{x} \, dx,x,a+b \arcsin (c x)\right )}{b}+\int \frac {1}{x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx \\ & = \frac {\text {Subst}\left (\int \left (\frac {\sin \left (\frac {5 a}{b}-\frac {5 x}{b}\right )}{16 x}-\frac {5 \sin \left (\frac {3 a}{b}-\frac {3 x}{b}\right )}{16 x}+\frac {5 \sin \left (\frac {a}{b}-\frac {x}{b}\right )}{8 x}\right ) \, dx,x,a+b \arcsin (c x)\right )}{b}-\frac {3 \text {Subst}\left (\int \left (-\frac {\sin \left (\frac {3 a}{b}-\frac {3 x}{b}\right )}{4 x}+\frac {3 \sin \left (\frac {a}{b}-\frac {x}{b}\right )}{4 x}\right ) \, dx,x,a+b \arcsin (c x)\right )}{b}-\frac {\left (3 \cos \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 )}{b}+\frac {\left (3 \sin \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 )}{b}+\int \frac {1}{x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx \\ & = \frac {3 \operatorname {CosIntegral}\left (\frac {a+b \arcsin (c x)}{b}\right ) \sin \left (\frac {a}{b}\right )}{b}-\frac {3 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arcsin (c x)}{b}\right )}{b}+\frac {\text {Subst}\left (\int \frac {\sin \left (\frac {5 a}{b}-\frac {5 x}{b}\right )}{x} \, dx,x,a+b \arcsin (c x)\right )}{16 b}-\frac {5 \text {Subst}\left (\int \frac {\sin \left (\frac {3 a}{b}-\frac {3 x}{b}\right )}{x} \, dx,x,a+b \arcsin (c x)\right )}{16 b}+\frac {5 \text {Subst}\left (\int \frac {\sin \left (\frac {a}{b}-\frac {x}{b}\right )}{x} \, dx,x,a+b \arcsin (c x)\right )}{8 b}+\frac {3 \text {Subst}\left (\int \frac {\sin \left (\frac {3 a}{b}-\frac {3 x}{b}\right )}{x} \, dx,x,a+b \arcsin (c x)\right )}{4 b}-\frac {9 \text {Subst}\left (\int \frac {\sin \left (\frac {a}{b}-\frac {x}{b}\right )}{x} \, dx,x,a+b \arcsin (c x)\right )}{4 b}+\int \frac {1}{x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx \\ & = \frac {3 \operatorname {CosIntegral}\left (\frac {a+b \arcsin (c x)}{b}\right ) \sin \left (\frac {a}{b}\right )}{b}-\frac {3 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arcsin (c x)}{b}\right )}{b}-\frac {\left (5 \cos \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}+\frac {\left (9 \cos \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 )}{4 b}+\frac {\left (5 \cos \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}-\frac {\left (3 \cos \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 )}{4 b}-\frac {\cos \left (\frac {5 a}{b}\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {5 x}{b}\right )}{x} \, dx,x,a+b \arcsin (c x)\right )}{16 b}+\frac {\left (5 \sin \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}-\frac {\left (9 \sin \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 )}{4 b}-\frac {\left (5 \sin \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}+\frac {\left (3 \sin \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 )}{4 b}+\frac {\sin \left (\frac {5 a}{b}\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {5 x}{b}\right )}{x} \, dx,x,a+b \arcsin (c x)\right )}{16 b}+\int \frac {1}{x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx \\ & = \frac {11 \operatorname {CosIntegral}\left (\frac {a+b \arcsin (c x)}{b}\right ) \sin \left (\frac {a}{b}\right )}{8 b}+\frac {7 \operatorname {CosIntegral}\left (\frac {3 (a+b \arcsin (c x))}{b}\right ) \sin \left (\frac {3 a}{b}\right )}{16 b}+\frac {\operatorname {CosIntegral}\left (\frac {5 (a+b \arcsin (c x))}{b}\right ) \sin \left (\frac {5 a}{b}\right )}{16 b}-\frac {11 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arcsin (c x)}{b}\right )}{8 b}-\frac {7 \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )}{16 b}-\frac {\cos \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 (a+b \arcsin (c x))}{b}\right )}{16 b}+\int \frac {1}{x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx \\ \end{align*}

Mathematica [N/A]

Not integrable

Time = 3.22 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x (a+b \arcsin (c x))} \, dx=\int \frac {\left (1-c^2 x^2\right )^{5/2}}{x (a+b \arcsin (c x))} \, dx \]

[In]

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

[Out]

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

Maple [N/A] (verified)

Not integrable

Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93

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

[In]

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

[Out]

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

Fricas [N/A]

Not integrable

Time = 0.25 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.61 \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x (a+b \arcsin (c x))} \, dx=\int { \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {5}{2}}}{{\left (b \arcsin \left (c x\right ) + a\right )} x} \,d x } \]

[In]

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

[Out]

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

Sympy [N/A]

Not integrable

Time = 5.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x (a+b \arcsin (c x))} \, dx=\int \frac {\left (- \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}}}{x \left (a + b \operatorname {asin}{\left (c x \right )}\right )}\, dx \]

[In]

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

[Out]

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

Maxima [N/A]

Not integrable

Time = 0.50 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x (a+b \arcsin (c x))} \, dx=\int { \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {5}{2}}}{{\left (b \arcsin \left (c x\right ) + a\right )} x} \,d x } \]

[In]

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

[Out]

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

Giac [F(-2)]

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

[In]

integrate((-c^2*x^2+1)^(5/2)/x/(a+b*arcsin(c*x)),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 \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x (a+b \arcsin (c x))} \, dx=\int \frac {{\left (1-c^2\,x^2\right )}^{5/2}}{x\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )} \,d x \]

[In]

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

[Out]

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