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

Optimal. Leaf size=109 \[ \frac{16 b^2 c^2 (d x)^{7/2} \text{HypergeometricPFQ}\left (\left \{1,\frac{7}{4},\frac{7}{4}\right \},\left \{\frac{9}{4},\frac{11}{4}\right \},c^2 x^2\right )}{105 d^3}-\frac{8 b c (d x)^{5/2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{5}{4},\frac{9}{4},c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{15 d^2}+\frac{2 (d x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2}{3 d} \]

[Out]

(2*(d*x)^(3/2)*(a + b*ArcSin[c*x])^2)/(3*d) - (8*b*c*(d*x)^(5/2)*(a + b*ArcSin[c*x])*Hypergeometric2F1[1/2, 5/
4, 9/4, c^2*x^2])/(15*d^2) + (16*b^2*c^2*(d*x)^(7/2)*HypergeometricPFQ[{1, 7/4, 7/4}, {9/4, 11/4}, c^2*x^2])/(
105*d^3)

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Rubi [A]  time = 0.142631, 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} \[ \frac{16 b^2 c^2 (d x)^{7/2} \, _3F_2\left (1,\frac{7}{4},\frac{7}{4};\frac{9}{4},\frac{11}{4};c^2 x^2\right )}{105 d^3}-\frac{8 b c (d x)^{5/2} \, _2F_1\left (\frac{1}{2},\frac{5}{4};\frac{9}{4};c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{15 d^2}+\frac{2 (d x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(d*x)^(3/2)*(a + b*ArcSin[c*x])^2)/(3*d) - (8*b*c*(d*x)^(5/2)*(a + b*ArcSin[c*x])*Hypergeometric2F1[1/2, 5/
4, 9/4, c^2*x^2])/(15*d^2) + (16*b^2*c^2*(d*x)^(7/2)*HypergeometricPFQ[{1, 7/4, 7/4}, {9/4, 11/4}, c^2*x^2])/(
105*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 \sqrt{d x} \left (a+b \sin ^{-1}(c x)\right )^2 \, dx &=\frac{2 (d x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2}{3 d}-\frac{(4 b c) \int \frac{(d x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{1-c^2 x^2}} \, dx}{3 d}\\ &=\frac{2 (d x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2}{3 d}-\frac{8 b c (d x)^{5/2} \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 )}{15 d^2}+\frac{16 b^2 c^2 (d x)^{7/2} \, _3F_2\left (1,\frac{7}{4},\frac{7}{4};\frac{9}{4},\frac{11}{4};c^2 x^2\right )}{105 d^3}\\ \end{align*}

Mathematica [A]  time = 0.0519041, size = 90, normalized size = 0.83 \[ \frac{2}{105} x \sqrt{d x} \left (8 b^2 c^2 x^2 \text{HypergeometricPFQ}\left (\left \{1,\frac{7}{4},\frac{7}{4}\right \},\left \{\frac{9}{4},\frac{11}{4}\right \},c^2 x^2\right )+7 \left (a+b \sin ^{-1}(c x)\right ) \left (5 \left (a+b \sin ^{-1}(c x)\right )-4 b c x \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{5}{4},\frac{9}{4},c^2 x^2\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*Sqrt[d*x]*(7*(a + b*ArcSin[c*x])*(5*(a + b*ArcSin[c*x]) - 4*b*c*x*Hypergeometric2F1[1/2, 5/4, 9/4, c^2*x^
2]) + 8*b^2*c^2*x^2*HypergeometricPFQ[{1, 7/4, 7/4}, {9/4, 11/4}, c^2*x^2]))/105

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Maple [F]  time = 0.184, size = 0, normalized size = 0. \begin{align*} \int \sqrt{dx} \left ( a+b\arcsin \left ( cx \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((d*x)^(1/2)*(a+b*arcsin(c*x))^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((d*x)^(1/2)*(a+b*arcsin(c*x))^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 ({\left (b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}\right )} \sqrt{d x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{d x} \left (a + b \operatorname{asin}{\left (c x \right )}\right )^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(d*x)*(a + b*asin(c*x))**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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