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

Optimal. Leaf size=89 \[ \frac {2 \sqrt {d x} \left (a+b \sin ^{-1}(c x)\right )}{d}+\frac {4 b F\left (\left .\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{\sqrt {c} \sqrt {d}}-\frac {4 b E\left (\left .\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{\sqrt {c} \sqrt {d}} \]

[Out]

-4*b*EllipticE(c^(1/2)*(d*x)^(1/2)/d^(1/2),I)/c^(1/2)/d^(1/2)+4*b*EllipticF(c^(1/2)*(d*x)^(1/2)/d^(1/2),I)/c^(
1/2)/d^(1/2)+2*(a+b*arcsin(c*x))*(d*x)^(1/2)/d

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Rubi [A]  time = 0.08, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {4627, 329, 307, 221, 1199, 424} \[ \frac {2 \sqrt {d x} \left (a+b \sin ^{-1}(c x)\right )}{d}+\frac {4 b F\left (\left .\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{\sqrt {c} \sqrt {d}}-\frac {4 b E\left (\left .\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{\sqrt {c} \sqrt {d}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x])/Sqrt[d*x],x]

[Out]

(2*Sqrt[d*x]*(a + b*ArcSin[c*x]))/d - (4*b*EllipticE[ArcSin[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]], -1])/(Sqrt[c]*Sqrt[d
]) + (4*b*EllipticF[ArcSin[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]], -1])/(Sqrt[c]*Sqrt[d])

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 307

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-(b/a), 2]}, -Dist[q^(-1), Int[1/Sqrt[a + b*x^
4], x], x] + Dist[1/q, Int[(1 + q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && NegQ[b/a]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 1199

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Dist[d/Sqrt[a], Int[Sqrt[1 + (e*x^2)/d]/Sqrt
[1 - (e*x^2)/d], x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && EqQ[c*d^2 + a*e^2, 0] && GtQ[a, 0]

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]

Rubi steps

\begin {align*} \int \frac {a+b \sin ^{-1}(c x)}{\sqrt {d x}} \, dx &=\frac {2 \sqrt {d x} \left (a+b \sin ^{-1}(c x)\right )}{d}-\frac {(2 b c) \int \frac {\sqrt {d x}}{\sqrt {1-c^2 x^2}} \, dx}{d}\\ &=\frac {2 \sqrt {d x} \left (a+b \sin ^{-1}(c x)\right )}{d}-\frac {(4 b c) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1-\frac {c^2 x^4}{d^2}}} \, dx,x,\sqrt {d x}\right )}{d^2}\\ &=\frac {2 \sqrt {d x} \left (a+b \sin ^{-1}(c x)\right )}{d}+\frac {(4 b) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-\frac {c^2 x^4}{d^2}}} \, dx,x,\sqrt {d x}\right )}{d}-\frac {(4 b) \operatorname {Subst}\left (\int \frac {1+\frac {c x^2}{d}}{\sqrt {1-\frac {c^2 x^4}{d^2}}} \, dx,x,\sqrt {d x}\right )}{d}\\ &=\frac {2 \sqrt {d x} \left (a+b \sin ^{-1}(c x)\right )}{d}+\frac {4 b F\left (\left .\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{\sqrt {c} \sqrt {d}}-\frac {(4 b) \operatorname {Subst}\left (\int \frac {\sqrt {1+\frac {c x^2}{d}}}{\sqrt {1-\frac {c x^2}{d}}} \, dx,x,\sqrt {d x}\right )}{d}\\ &=\frac {2 \sqrt {d x} \left (a+b \sin ^{-1}(c x)\right )}{d}-\frac {4 b E\left (\left .\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{\sqrt {c} \sqrt {d}}+\frac {4 b F\left (\left .\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{\sqrt {c} \sqrt {d}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 45, normalized size = 0.51 \[ \frac {2 x \left (3 \left (a+b \sin ^{-1}(c x)\right )-2 b c x \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};c^2 x^2\right )\right )}{3 \sqrt {d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(3*(a + b*ArcSin[c*x]) - 2*b*c*x*Hypergeometric2F1[1/2, 3/4, 7/4, c^2*x^2]))/(3*Sqrt[d*x])

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d x} {\left (b \arcsin \left (c x\right ) + a\right )}}{d x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 98, normalized size = 1.10 \[ \frac {2 a \sqrt {d x}+2 b \left (\sqrt {d x}\, \arcsin \left (c x \right )+\frac {2 \sqrt {-c x +1}\, \sqrt {c x +1}\, \left (\EllipticF \left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )-\EllipticE \left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )\right )}{\sqrt {\frac {c}{d}}\, \sqrt {-c^{2} x^{2}+1}}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*(a*(d*x)^(1/2)+b*((d*x)^(1/2)*arcsin(c*x)+2/(c/d)^(1/2)*(-c*x+1)^(1/2)*(c*x+1)^(1/2)/(-c^2*x^2+1)^(1/2)*(E
llipticF((d*x)^(1/2)*(c/d)^(1/2),I)-EllipticE((d*x)^(1/2)*(c/d)^(1/2),I))))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {2 \, {\left (b \sqrt {d} \sqrt {x} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) + {\left (b c \int \frac {\sqrt {-c x + 1} \sqrt {x}}{\sqrt {c x + 1} c x - \sqrt {c x + 1}}\,{d x} + a \sqrt {x}\right )} \sqrt {d}\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(b*sqrt(d)*sqrt(x)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + (b*c*d*integrate(sqrt(c*x + 1)*sqrt(-c*x + 1
)*sqrt(x)/(c^2*d*x^2 - d), x) + a*sqrt(x))*sqrt(d))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c*x))/(d*x)^(1/2),x)

[Out]

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

[Out]

Exception raised: TypeError

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