3.214 \(\int \frac{(a+b \sin ^{-1}(c x))^2}{(d x)^{5/2}} \, dx\)

Optimal. Leaf size=109 \[ \frac{16 b^2 c^2 \sqrt{d x} \text{HypergeometricPFQ}\left (\left \{\frac{1}{4},\frac{1}{4},1\right \},\left \{\frac{3}{4},\frac{5}{4}\right \},c^2 x^2\right )}{3 d^3}-\frac{8 b c \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 \sqrt{d x}}-\frac{2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 d (d x)^{3/2}} \]

[Out]

(-2*(a + b*ArcSin[c*x])^2)/(3*d*(d*x)^(3/2)) - (8*b*c*(a + b*ArcSin[c*x])*Hypergeometric2F1[-1/4, 1/2, 3/4, c^
2*x^2])/(3*d^2*Sqrt[d*x]) + (16*b^2*c^2*Sqrt[d*x]*HypergeometricPFQ[{1/4, 1/4, 1}, {3/4, 5/4}, c^2*x^2])/(3*d^
3)

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Rubi [A]  time = 0.14545, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {4627, 4711} \[ \frac{16 b^2 c^2 \sqrt{d x} \, _3F_2\left (\frac{1}{4},\frac{1}{4},1;\frac{3}{4},\frac{5}{4};c^2 x^2\right )}{3 d^3}-\frac{8 b c \, _2F_1\left (-\frac{1}{4},\frac{1}{2};\frac{3}{4};c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 \sqrt{d x}}-\frac{2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 d (d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x])^2/(d*x)^(5/2),x]

[Out]

(-2*(a + b*ArcSin[c*x])^2)/(3*d*(d*x)^(3/2)) - (8*b*c*(a + b*ArcSin[c*x])*Hypergeometric2F1[-1/4, 1/2, 3/4, c^
2*x^2])/(3*d^2*Sqrt[d*x]) + (16*b^2*c^2*Sqrt[d*x]*HypergeometricPFQ[{1/4, 1/4, 1}, {3/4, 5/4}, c^2*x^2])/(3*d^
3)

Rule 4627

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcSi
n[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1))/Sqrt[1
- c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -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{\left (a+b \sin ^{-1}(c x)\right )^2}{(d x)^{5/2}} \, dx &=-\frac{2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 d (d x)^{3/2}}+\frac{(4 b c) \int \frac{a+b \sin ^{-1}(c x)}{(d x)^{3/2} \sqrt{1-c^2 x^2}} \, dx}{3 d}\\ &=-\frac{2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 d (d x)^{3/2}}-\frac{8 b c \left (a+b \sin ^{-1}(c x)\right ) \, _2F_1\left (-\frac{1}{4},\frac{1}{2};\frac{3}{4};c^2 x^2\right )}{3 d^2 \sqrt{d x}}+\frac{16 b^2 c^2 \sqrt{d x} \, _3F_2\left (\frac{1}{4},\frac{1}{4},1;\frac{3}{4},\frac{5}{4};c^2 x^2\right )}{3 d^3}\\ \end{align*}

Mathematica [A]  time = 0.0541036, size = 87, normalized size = 0.8 \[ \frac{x \left (16 b^2 c^2 x^2 \text{HypergeometricPFQ}\left (\left \{\frac{1}{4},\frac{1}{4},1\right \},\left \{\frac{3}{4},\frac{5}{4}\right \},c^2 x^2\right )-2 \left (a+b \sin ^{-1}(c x)\right ) \left (4 b c x \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},c^2 x^2\right )+a+b \sin ^{-1}(c x)\right )\right )}{3 (d x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c*x])^2/(d*x)^(5/2),x]

[Out]

(x*(-2*(a + b*ArcSin[c*x])*(a + b*ArcSin[c*x] + 4*b*c*x*Hypergeometric2F1[-1/4, 1/2, 3/4, c^2*x^2]) + 16*b^2*c
^2*x^2*HypergeometricPFQ[{1/4, 1/4, 1}, {3/4, 5/4}, c^2*x^2]))/(3*(d*x)^(5/2))

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Maple [F]  time = 0.144, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+b\arcsin \left ( cx \right ) \right ) ^{2} \left ( dx \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(c*x))^2/(d*x)^(5/2),x)

[Out]

int((a+b*arcsin(c*x))^2/(d*x)^(5/2),x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))^2/(d*x)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))^2/(d*x)^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(c*x))**2/(d*x)**(5/2),x)

[Out]

Exception raised: TypeError

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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 )}^{2}}{\left (d x\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))^2/(d*x)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsin(c*x) + a)^2/(d*x)^(5/2), x)