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 \[ \frac {2 \sqrt {1-c^2 x^2} (f x)^{5/2} \, _2F_1\left (\frac {1}{2},\frac {5}{4};\frac {9}{4};c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{5 f \sqrt {d-c^2 d x^2}}-\frac {4 b c \sqrt {1-c^2 x^2} (f x)^{7/2} \, _3F_2\left (1,\frac {7}{4},\frac {7}{4};\frac {9}{4},\frac {11}{4};c^2 x^2\right )}{35 f^2 \sqrt {d-c^2 d x^2}} \]

[Out]

2/5*(f*x)^(5/2)*(a+b*arcsin(c*x))*hypergeom([1/2, 5/4],[9/4],c^2*x^2)*(-c^2*x^2+1)^(1/2)/f/(-c^2*d*x^2+d)^(1/2
)-4/35*b*c*(f*x)^(7/2)*HypergeometricPFQ([1, 7/4, 7/4],[9/4, 11/4],c^2*x^2)*(-c^2*x^2+1)^(1/2)/f^2/(-c^2*d*x^2
+d)^(1/2)

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Rubi [A]  time = 0.22, antiderivative size = 137, normalized size of antiderivative = 1.00, 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 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 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])

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*}

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Mathematica [A]  time = 0.04, size = 97, normalized size = 0.71 \[ -\frac {2 x \sqrt {1-c^2 x^2} (f x)^{3/2} \left (2 b c x \, _3F_2\left (1,\frac {7}{4},\frac {7}{4};\frac {9}{4},\frac {11}{4};c^2 x^2\right )-7 \, _2F_1\left (\frac {1}{2},\frac {5}{4};\frac {9}{4};c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )\right )}{35 \sqrt {d-c^2 d x^2}} \]

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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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-c^{2} d x^{2} + d} {\left (b f x \arcsin \left (c x\right ) + a f x\right )} \sqrt {f x}}{c^{2} d x^{2} - d}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x\right )^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}}{\sqrt {-c^{2} d x^{2} + d}}\,{d x} \]

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)

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maple [F]  time = 0.83, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x \right )^{\frac {3}{2}} \left (a +b \arcsin \left (c x \right )\right )}{\sqrt {-c^{2} d \,x^{2}+d}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x\right )^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}}{\sqrt {-c^{2} d x^{2} + d}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (f\,x\right )}^{3/2}}{\sqrt {d-c^2\,d\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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