3.100 \(\int \sqrt {\pi -c^2 \pi x^2} (a+b \sin ^{-1}(c x)) \, dx\)

Optimal. Leaf size=68 \[ \frac {1}{2} x \sqrt {\pi -\pi c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac {\sqrt {\pi } \left (a+b \sin ^{-1}(c x)\right )^2}{4 b c}-\frac {1}{4} \sqrt {\pi } b c x^2 \]

[Out]

-1/4*b*c*x^2*Pi^(1/2)+1/4*(a+b*arcsin(c*x))^2*Pi^(1/2)/b/c+1/2*x*(a+b*arcsin(c*x))*(-Pi*c^2*x^2+Pi)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 116, normalized size of antiderivative = 1.71, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {4647, 4641, 30} \[ \frac {1}{2} x \sqrt {\pi -\pi c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac {\sqrt {\pi -\pi c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b c \sqrt {1-c^2 x^2}}-\frac {b c x^2 \sqrt {\pi -\pi c^2 x^2}}{4 \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Pi - c^2*Pi*x^2]*(a + b*ArcSin[c*x]),x]

[Out]

-(b*c*x^2*Sqrt[Pi - c^2*Pi*x^2])/(4*Sqrt[1 - c^2*x^2]) + (x*Sqrt[Pi - c^2*Pi*x^2]*(a + b*ArcSin[c*x]))/2 + (Sq
rt[Pi - c^2*Pi*x^2]*(a + b*ArcSin[c*x])^2)/(4*b*c*Sqrt[1 - c^2*x^2])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 4641

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(a + b*ArcSin[c*x])^
(n + 1)/(b*c*Sqrt[d]*(n + 1)), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] && GtQ[d, 0] && NeQ[n,
-1]

Rule 4647

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(x*Sqrt[d + e*x^2]*(
a + b*ArcSin[c*x])^n)/2, x] + (Dist[Sqrt[d + e*x^2]/(2*Sqrt[1 - c^2*x^2]), Int[(a + b*ArcSin[c*x])^n/Sqrt[1 -
c^2*x^2], x], x] - Dist[(b*c*n*Sqrt[d + e*x^2])/(2*Sqrt[1 - c^2*x^2]), Int[x*(a + b*ArcSin[c*x])^(n - 1), x],
x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0]

Rubi steps

\begin {align*} \int \sqrt {\pi -c^2 \pi x^2} \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac {1}{2} x \sqrt {\pi -c^2 \pi x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac {\sqrt {\pi -c^2 \pi x^2} \int \frac {a+b \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}} \, dx}{2 \sqrt {1-c^2 x^2}}-\frac {\left (b c \sqrt {\pi -c^2 \pi x^2}\right ) \int x \, dx}{2 \sqrt {1-c^2 x^2}}\\ &=-\frac {b c x^2 \sqrt {\pi -c^2 \pi x^2}}{4 \sqrt {1-c^2 x^2}}+\frac {1}{2} x \sqrt {\pi -c^2 \pi x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac {\sqrt {\pi -c^2 \pi x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b c \sqrt {1-c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 87, normalized size = 1.28 \[ \frac {\sqrt {\pi } \left (a^2+2 a b c x \sqrt {1-c^2 x^2}+2 b \sin ^{-1}(c x) \left (a+b c x \sqrt {1-c^2 x^2}\right )-b^2 c^2 x^2+b^2 \sin ^{-1}(c x)^2\right )}{4 b c} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Pi - c^2*Pi*x^2]*(a + b*ArcSin[c*x]),x]

[Out]

(Sqrt[Pi]*(a^2 - b^2*c^2*x^2 + 2*a*b*c*x*Sqrt[1 - c^2*x^2] + 2*b*(a + b*c*x*Sqrt[1 - c^2*x^2])*ArcSin[c*x] + b
^2*ArcSin[c*x]^2))/(4*b*c)

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fricas [F]  time = 0.83, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {\pi - \pi c^{2} x^{2}} {\left (b \arcsin \left (c x\right ) + a\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(pi - pi*c^2*x^2)*(b*arcsin(c*x) + a), 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((a+b*arcsin(c*x))*(-pi*c^2*x^2+pi)^(1/2),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 = 0.06, size = 101, normalized size = 1.49 \[ \frac {a x \sqrt {-\pi \,c^{2} x^{2}+\pi }}{2}+\frac {a \pi \arctan \left (\frac {\sqrt {\pi \,c^{2}}\, x}{\sqrt {-\pi \,c^{2} x^{2}+\pi }}\right )}{2 \sqrt {\pi \,c^{2}}}+\frac {b \sqrt {\pi }\, \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, x}{2}-\frac {b c \,x^{2} \sqrt {\pi }}{4}+\frac {b \sqrt {\pi }\, \arcsin \left (c x \right )^{2}}{4 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*x*(-Pi*c^2*x^2+Pi)^(1/2)+1/2*a*Pi/(Pi*c^2)^(1/2)*arctan((Pi*c^2)^(1/2)*x/(-Pi*c^2*x^2+Pi)^(1/2))+1/2*b*P
i^(1/2)*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*x-1/4*b*c*x^2*Pi^(1/2)+1/4*b*Pi^(1/2)/c*arcsin(c*x)^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \sqrt {\pi } b \int \sqrt {c x + 1} \sqrt {-c x + 1} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )\,{d x} + \frac {1}{2} \, {\left (\sqrt {\pi - \pi c^{2} x^{2}} x + \frac {\sqrt {\pi } \arcsin \left (c x\right )}{c}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(pi)*b*integrate(sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)), x) + 1/2*(sqrt(p
i - pi*c^2*x^2)*x + sqrt(pi)*arcsin(c*x)/c)*a

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\sqrt {\Pi -\Pi \,c^2\,x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(pi)*(Integral(a*sqrt(-c**2*x**2 + 1), x) + Integral(b*sqrt(-c**2*x**2 + 1)*asin(c*x), x))

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