3.151 \(\int x^m \sqrt{d-c^2 d x^2} (a+b \sin ^{-1}(c x)) \, dx\)

Optimal. Leaf size=245 \[ -\frac{b c x^{m+2} \sqrt{d-c^2 d x^2} \text{HypergeometricPFQ}\left (\left \{1,\frac{m}{2}+1,\frac{m}{2}+1\right \},\left \{\frac{m}{2}+\frac{3}{2},\frac{m}{2}+2\right \},c^2 x^2\right )}{(m+1) (m+2)^2 \sqrt{1-c^2 x^2}}+\frac{x^{m+1} \sqrt{d-c^2 d x^2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{\left (m^2+3 m+2\right ) \sqrt{1-c^2 x^2}}+\frac{x^{m+1} \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{m+2}-\frac{b c x^{m+2} \sqrt{d-c^2 d x^2}}{(m+2)^2 \sqrt{1-c^2 x^2}} \]

[Out]

-((b*c*x^(2 + m)*Sqrt[d - c^2*d*x^2])/((2 + m)^2*Sqrt[1 - c^2*x^2])) + (x^(1 + m)*Sqrt[d - c^2*d*x^2]*(a + b*A
rcSin[c*x]))/(2 + m) + (x^(1 + m)*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])*Hypergeometric2F1[1/2, (1 + m)/2, (3
+ m)/2, c^2*x^2])/((2 + 3*m + m^2)*Sqrt[1 - c^2*x^2]) - (b*c*x^(2 + m)*Sqrt[d - c^2*d*x^2]*HypergeometricPFQ[
{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, c^2*x^2])/((1 + m)*(2 + m)^2*Sqrt[1 - c^2*x^2])

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Rubi [A]  time = 0.202124, antiderivative size = 245, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {4697, 4711, 30} \[ -\frac{b c x^{m+2} \sqrt{d-c^2 d x^2} \, _3F_2\left (1,\frac{m}{2}+1,\frac{m}{2}+1;\frac{m}{2}+\frac{3}{2},\frac{m}{2}+2;c^2 x^2\right )}{(m+1) (m+2)^2 \sqrt{1-c^2 x^2}}+\frac{x^{m+1} \sqrt{d-c^2 d x^2} \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{\left (m^2+3 m+2\right ) \sqrt{1-c^2 x^2}}+\frac{x^{m+1} \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{m+2}-\frac{b c x^{m+2} \sqrt{d-c^2 d x^2}}{(m+2)^2 \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*c*x^(2 + m)*Sqrt[d - c^2*d*x^2])/((2 + m)^2*Sqrt[1 - c^2*x^2])) + (x^(1 + m)*Sqrt[d - c^2*d*x^2]*(a + b*A
rcSin[c*x]))/(2 + m) + (x^(1 + m)*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])*Hypergeometric2F1[1/2, (1 + m)/2, (3
+ m)/2, c^2*x^2])/((2 + 3*m + m^2)*Sqrt[1 - c^2*x^2]) - (b*c*x^(2 + m)*Sqrt[d - c^2*d*x^2]*HypergeometricPFQ[
{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, c^2*x^2])/((1 + m)*(2 + m)^2*Sqrt[1 - c^2*x^2])

Rule 4697

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

Rule 4711

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[((f*x)
^(m + 1)*(a + b*ArcSin[c*x])*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, c^2*x^2])/(Sqrt[d]*f*(m + 1)), x] -
Simp[(b*c*(f*x)^(m + 2)*HypergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, c^2*x^2])/(Sqrt[d]*f^2*
(m + 1)*(m + 2)), x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && GtQ[d, 0] &&  !IntegerQ[m]

Rule 30

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

Rubi steps

\begin{align*} \int x^m \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac{x^{1+m} \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{2+m}+\frac{\sqrt{d-c^2 d x^2} \int \frac{x^m \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{1-c^2 x^2}} \, dx}{(2+m) \sqrt{1-c^2 x^2}}-\frac{\left (b c \sqrt{d-c^2 d x^2}\right ) \int x^{1+m} \, dx}{(2+m) \sqrt{1-c^2 x^2}}\\ &=-\frac{b c x^{2+m} \sqrt{d-c^2 d x^2}}{(2+m)^2 \sqrt{1-c^2 x^2}}+\frac{x^{1+m} \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{2+m}+\frac{x^{1+m} \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) \, _2F_1\left (\frac{1}{2},\frac{1+m}{2};\frac{3+m}{2};c^2 x^2\right )}{\left (2+3 m+m^2\right ) \sqrt{1-c^2 x^2}}-\frac{b c x^{2+m} \sqrt{d-c^2 d x^2} \, _3F_2\left (1,1+\frac{m}{2},1+\frac{m}{2};\frac{3}{2}+\frac{m}{2},2+\frac{m}{2};c^2 x^2\right )}{(1+m) (2+m)^2 \sqrt{1-c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0733063, size = 181, normalized size = 0.74 \[ \frac{x^{m+1} \sqrt{d-c^2 d x^2} \left (-b c x \text{HypergeometricPFQ}\left (\left \{1,\frac{m}{2}+1,\frac{m}{2}+1\right \},\left \{\frac{m}{2}+\frac{3}{2},\frac{m}{2}+2\right \},c^2 x^2\right )+(m+2) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )+(m+1) \left (a (m+2) \sqrt{1-c^2 x^2}+b (m+2) \sqrt{1-c^2 x^2} \sin ^{-1}(c x)-b c x\right )\right )}{(m+1) (m+2)^2 \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*Sqrt[d - c^2*d*x^2]*((1 + m)*(-(b*c*x) + a*(2 + m)*Sqrt[1 - c^2*x^2] + b*(2 + m)*Sqrt[1 - c^2*x^2]*
ArcSin[c*x]) + (2 + m)*(a + b*ArcSin[c*x])*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, c^2*x^2] - b*c*x*Hyper
geometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, c^2*x^2]))/((1 + m)*(2 + m)^2*Sqrt[1 - c^2*x^2])

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Maple [F]  time = 1.957, size = 0, normalized size = 0. \begin{align*} \int{x}^{m}\sqrt{-{c}^{2}d{x}^{2}+d} \left ( a+b\arcsin \left ( cx \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{-c^{2} d x^{2} + d}{\left (b \arcsin \left (c x\right ) + a\right )} x^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(-c^2*d*x^2 + d)*(b*arcsin(c*x) + a)*x^m, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{-c^{2} d x^{2} + d}{\left (b \arcsin \left (c x\right ) + a\right )} x^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-c^2*d*x^2 + d)*(b*arcsin(c*x) + a)*x^m, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{m} \sqrt{- d \left (c x - 1\right ) \left (c x + 1\right )} \left (a + b \operatorname{asin}{\left (c x \right )}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**m*sqrt(-d*(c*x - 1)*(c*x + 1))*(a + b*asin(c*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{-c^{2} d x^{2} + d}{\left (b \arcsin \left (c x\right ) + a\right )} x^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(-c^2*d*x^2 + d)*(b*arcsin(c*x) + a)*x^m, x)