3.403 \(\int \frac {(1-c^2 x^2)^{5/2}}{x (a+b \sin ^{-1}(c x))^2} \, dx\)

Optimal. Leaf size=235 \[ -\frac {\text {Int}\left (\frac {\left (1-c^2 x^2\right )^2}{x^2 \left (a+b \sin ^{-1}(c x)\right )},x\right )}{b c}-\frac {25 \cos \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a+b \sin ^{-1}(c x)}{b}\right )}{8 b^2}-\frac {25 \cos \left (\frac {3 a}{b}\right ) \text {Ci}\left (\frac {3 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2}-\frac {5 \cos \left (\frac {5 a}{b}\right ) \text {Ci}\left (\frac {5 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2}-\frac {25 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \sin ^{-1}(c x)}{b}\right )}{8 b^2}-\frac {25 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2}-\frac {5 \sin \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2}-\frac {\left (1-c^2 x^2\right )^3}{b c x \left (a+b \sin ^{-1}(c x)\right )} \]

[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

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Rubi [A]  time = 0.53, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x \left (a+b \sin ^{-1}(c x)\right )^2} \, dx \]

Verification is Not applicable to the result.

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

Rubi steps

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

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Mathematica [A]  time = 14.12, size = 0, normalized size = 0.00 \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x \left (a+b \sin ^{-1}(c x)\right )^2} \, dx \]

Verification is Not applicable to the result.

[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]

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fricas [A]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c^{4} x^{4} - 2 \, c^{2} x^{2} + 1\right )} \sqrt {-c^{2} x^{2} + 1}}{b^{2} x \arcsin \left (c x\right )^{2} + 2 \, a b x \arcsin \left (c x\right ) + a^{2} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[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,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 1.02, size = 0, normalized size = 0.00 \[ \int \frac {\left (-c^{2} x^{2}+1\right )^{\frac {5}{2}}}{x \left (a +b \arcsin \left (c x \right )\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {c^{6} x^{6} - 3 \, c^{4} x^{4} + 3 \, c^{2} x^{2} - \frac {{\left (5 \, c^{6} \int \frac {x^{6}}{b x^{2} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) + a x^{2}}\,{d x} - 9 \, c^{4} \int \frac {x^{4}}{b x^{2} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) + a x^{2}}\,{d x} + 3 \, c^{2} \int \frac {x^{2}}{b x^{2} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) + a x^{2}}\,{d x} + \int \frac {1}{{\left (b \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) + a\right )} x^{2}}\,{d x}\right )} {\left (b^{2} c x \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) + a b c x\right )}}{b c} - 1}{b^{2} c x \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) + a b c x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [A]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (1-c^2\,x^2\right )}^{5/2}}{x\,{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}}}{x \left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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