3.142 \(\int \frac{(f x)^{3/2} (a+b \sin ^{-1}(c x))}{\sqrt{d-c^2 d x^2}} \, dx\)

Optimal. Leaf size=137 \[ \frac{2 \sqrt{1-c^2 x^2} (f x)^{5/2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{5}{4},\frac{9}{4},c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{5 f \sqrt{d-c^2 d x^2}}-\frac{4 b c \sqrt{1-c^2 x^2} (f x)^{7/2} \text{HypergeometricPFQ}\left (\left \{1,\frac{7}{4},\frac{7}{4}\right \},\left \{\frac{9}{4},\frac{11}{4}\right \},c^2 x^2\right )}{35 f^2 \sqrt{d-c^2 d x^2}} \]

[Out]

(2*(f*x)^(5/2)*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])*Hypergeometric2F1[1/2, 5/4, 9/4, c^2*x^2])/(5*f*Sqrt[d -
c^2*d*x^2]) - (4*b*c*(f*x)^(7/2)*Sqrt[1 - c^2*x^2]*HypergeometricPFQ[{1, 7/4, 7/4}, {9/4, 11/4}, c^2*x^2])/(35
*f^2*Sqrt[d - c^2*d*x^2])

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Rubi [A]  time = 0.216347, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {4713, 4711} \[ \frac{2 \sqrt{1-c^2 x^2} (f x)^{5/2} \, _2F_1\left (\frac{1}{2},\frac{5}{4};\frac{9}{4};c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{5 f \sqrt{d-c^2 d x^2}}-\frac{4 b c \sqrt{1-c^2 x^2} (f x)^{7/2} \, _3F_2\left (1,\frac{7}{4},\frac{7}{4};\frac{9}{4},\frac{11}{4};c^2 x^2\right )}{35 f^2 \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((f*x)^(3/2)*(a + b*ArcSin[c*x]))/Sqrt[d - c^2*d*x^2],x]

[Out]

(2*(f*x)^(5/2)*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])*Hypergeometric2F1[1/2, 5/4, 9/4, c^2*x^2])/(5*f*Sqrt[d -
c^2*d*x^2]) - (4*b*c*(f*x)^(7/2)*Sqrt[1 - c^2*x^2]*HypergeometricPFQ[{1, 7/4, 7/4}, {9/4, 11/4}, c^2*x^2])/(35
*f^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{(f x)^{3/2} \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{(f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{1-c^2 x^2}} \, dx}{\sqrt{d-c^2 d x^2}}\\ &=\frac{2 (f x)^{5/2} \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \, _2F_1\left (\frac{1}{2},\frac{5}{4};\frac{9}{4};c^2 x^2\right )}{5 f \sqrt{d-c^2 d x^2}}-\frac{4 b c (f x)^{7/2} \sqrt{1-c^2 x^2} \, _3F_2\left (1,\frac{7}{4},\frac{7}{4};\frac{9}{4},\frac{11}{4};c^2 x^2\right )}{35 f^2 \sqrt{d-c^2 d x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0419652, size = 97, normalized size = 0.71 \[ -\frac{2 x \sqrt{1-c^2 x^2} (f x)^{3/2} \left (2 b c x \text{HypergeometricPFQ}\left (\left \{1,\frac{7}{4},\frac{7}{4}\right \},\left \{\frac{9}{4},\frac{11}{4}\right \},c^2 x^2\right )-7 \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{5}{4},\frac{9}{4},c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )\right )}{35 \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((f*x)^(3/2)*(a + b*ArcSin[c*x]))/Sqrt[d - c^2*d*x^2],x]

[Out]

(-2*x*(f*x)^(3/2)*Sqrt[1 - c^2*x^2]*(-7*(a + b*ArcSin[c*x])*Hypergeometric2F1[1/2, 5/4, 9/4, c^2*x^2] + 2*b*c*
x*HypergeometricPFQ[{1, 7/4, 7/4}, {9/4, 11/4}, c^2*x^2]))/(35*Sqrt[d - c^2*d*x^2])

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Maple [F]  time = 0.529, size = 0, normalized size = 0. \begin{align*} \int{(a+b\arcsin \left ( cx \right ) ) \left ( fx \right ) ^{{\frac{3}{2}}}{\frac{1}{\sqrt{-{c}^{2}d{x}^{2}+d}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x)^(3/2)*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(1/2),x)

[Out]

int((f*x)^(3/2)*(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 (f x\right )^{\frac{3}{2}}{\left (b \arcsin \left (c x\right ) + a\right )}}{\sqrt{-c^{2} d x^{2} + d}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^(3/2)*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

integrate((f*x)^(3/2)*(b*arcsin(c*x) + a)/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 f x \arcsin \left (c x\right ) + a f x\right )} \sqrt{f x}}{c^{2} d x^{2} - d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^(3/2)*(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*f*x*arcsin(c*x) + a*f*x)*sqrt(f*x)/(c^2*d*x^2 - d), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)**(3/2)*(a+b*asin(c*x))/(-c**2*d*x**2+d)**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (f x\right )^{\frac{3}{2}}{\left (b \arcsin \left (c x\right ) + a\right )}}{\sqrt{-c^{2} d x^{2} + d}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^(3/2)*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(1/2),x, algorithm="giac")

[Out]

integrate((f*x)^(3/2)*(b*arcsin(c*x) + a)/sqrt(-c^2*d*x^2 + d), x)