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

Optimal. Leaf size=107 \[ \frac {16 b^2 c^2 (d x)^{5/2} \, _3F_2\left (1,\frac {5}{4},\frac {5}{4};\frac {7}{4},\frac {9}{4};c^2 x^2\right )}{15 d^3}-\frac {8 b c (d x)^{3/2} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 d^2}+\frac {2 \sqrt {d x} \left (a+b \sin ^{-1}(c x)\right )^2}{d} \]

[Out]

-8/3*b*c*(d*x)^(3/2)*(a+b*arcsin(c*x))*hypergeom([1/2, 3/4],[7/4],c^2*x^2)/d^2+16/15*b^2*c^2*(d*x)^(5/2)*Hyper
geometricPFQ([1, 5/4, 5/4],[7/4, 9/4],c^2*x^2)/d^3+2*(a+b*arcsin(c*x))^2*(d*x)^(1/2)/d

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Rubi [A]  time = 0.13, antiderivative size = 107, normalized size of antiderivative = 1.00, 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} \[ \frac {16 b^2 c^2 (d x)^{5/2} \, _3F_2\left (1,\frac {5}{4},\frac {5}{4};\frac {7}{4},\frac {9}{4};c^2 x^2\right )}{15 d^3}-\frac {8 b c (d x)^{3/2} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 d^2}+\frac {2 \sqrt {d x} \left (a+b \sin ^{-1}(c x)\right )^2}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[d*x]*(a + b*ArcSin[c*x])^2)/d - (8*b*c*(d*x)^(3/2)*(a + b*ArcSin[c*x])*Hypergeometric2F1[1/2, 3/4, 7/4
, c^2*x^2])/(3*d^2) + (16*b^2*c^2*(d*x)^(5/2)*HypergeometricPFQ[{1, 5/4, 5/4}, {7/4, 9/4}, c^2*x^2])/(15*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}{\sqrt {d x}} \, dx &=\frac {2 \sqrt {d x} \left (a+b \sin ^{-1}(c x)\right )^2}{d}-\frac {(4 b c) \int \frac {\sqrt {d x} \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx}{d}\\ &=\frac {2 \sqrt {d x} \left (a+b \sin ^{-1}(c x)\right )^2}{d}-\frac {8 b c (d x)^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};c^2 x^2\right )}{3 d^2}+\frac {16 b^2 c^2 (d x)^{5/2} \, _3F_2\left (1,\frac {5}{4},\frac {5}{4};\frac {7}{4},\frac {9}{4};c^2 x^2\right )}{15 d^3}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 90, normalized size = 0.84 \[ \frac {2 x \left (8 b^2 c^2 x^2 \, _3F_2\left (1,\frac {5}{4},\frac {5}{4};\frac {7}{4},\frac {9}{4};c^2 x^2\right )+5 \left (a+b \sin ^{-1}(c x)\right ) \left (3 \left (a+b \sin ^{-1}(c x)\right )-4 b c x \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};c^2 x^2\right )\right )\right )}{15 \sqrt {d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}\right )} \sqrt {d x}}{d x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arcsin \left (c x \right )\right )^{2}}{\sqrt {d x}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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