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

Optimal. Leaf size=163 \[ \frac{\sqrt{1-c^2 x^2} x^{m+1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{(m+1) \sqrt{d-c^2 d x^2}}-\frac{b c \sqrt{1-c^2 x^2} x^{m+2} \text{HypergeometricPFQ}\left (\left \{1,\frac{m}{2}+1,\frac{m}{2}+1\right \},\left \{\frac{m}{2}+\frac{3}{2},\frac{m}{2}+2\right \},c^2 x^2\right )}{\left (m^2+3 m+2\right ) \sqrt{d-c^2 d x^2}} \]

[Out]

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

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Rubi [A]  time = 0.196726, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.074, Rules used = {4713, 4711} \[ \frac{\sqrt{1-c^2 x^2} x^{m+1} \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{(m+1) \sqrt{d-c^2 d x^2}}-\frac{b c \sqrt{1-c^2 x^2} x^{m+2} \, _3F_2\left (1,\frac{m}{2}+1,\frac{m}{2}+1;\frac{m}{2}+\frac{3}{2},\frac{m}{2}+2;c^2 x^2\right )}{\left (m^2+3 m+2\right ) \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4713

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

Rubi steps

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

Mathematica [A]  time = 0.0630509, size = 129, normalized size = 0.79 \[ \frac{\sqrt{1-c^2 x^2} x^{m+1} \left ((m+2) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )-b c x \text{HypergeometricPFQ}\left (\left \{1,\frac{m}{2}+1,\frac{m}{2}+1\right \},\left \{\frac{m}{2}+\frac{3}{2},\frac{m}{2}+2\right \},c^2 x^2\right )\right )}{(m+1) (m+2) \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.905, size = 0, normalized size = 0. \begin{align*} \int{{x}^{m} \left ( a+b\arcsin \left ( cx \right ) \right ){\frac{1}{\sqrt{-{c}^{2}d{x}^{2}+d}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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