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

Optimal. Leaf size=133 \[ \frac{\sqrt{\frac{\pi }{2}} \sqrt{b} \sin \left (\frac{a}{b}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{d}-\frac{\sqrt{\frac{\pi }{2}} \sqrt{b} \cos \left (\frac{a}{b}\right ) S\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{d}+\frac{(c+d x) \sqrt{a+b \sin ^{-1}(c+d x)}}{d} \]

[Out]

((c + d*x)*Sqrt[a + b*ArcSin[c + d*x]])/d - (Sqrt[b]*Sqrt[Pi/2]*Cos[a/b]*FresnelS[(Sqrt[2/Pi]*Sqrt[a + b*ArcSi
n[c + d*x]])/Sqrt[b]])/d + (Sqrt[b]*Sqrt[Pi/2]*FresnelC[(Sqrt[2/Pi]*Sqrt[a + b*ArcSin[c + d*x]])/Sqrt[b]]*Sin[
a/b])/d

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Rubi [A]  time = 0.272595, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.571, Rules used = {4803, 4619, 4723, 3306, 3305, 3351, 3304, 3352} \[ \frac{\sqrt{\frac{\pi }{2}} \sqrt{b} \sin \left (\frac{a}{b}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{d}-\frac{\sqrt{\frac{\pi }{2}} \sqrt{b} \cos \left (\frac{a}{b}\right ) S\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{d}+\frac{(c+d x) \sqrt{a+b \sin ^{-1}(c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*ArcSin[c + d*x]],x]

[Out]

((c + d*x)*Sqrt[a + b*ArcSin[c + d*x]])/d - (Sqrt[b]*Sqrt[Pi/2]*Cos[a/b]*FresnelS[(Sqrt[2/Pi]*Sqrt[a + b*ArcSi
n[c + d*x]])/Sqrt[b]])/d + (Sqrt[b]*Sqrt[Pi/2]*FresnelC[(Sqrt[2/Pi]*Sqrt[a + b*ArcSin[c + d*x]])/Sqrt[b]]*Sin[
a/b])/d

Rule 4803

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcSin[x])^n, x],
x, c + d*x], x] /; FreeQ[{a, b, c, d, n}, x]

Rule 4619

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcSin[c*x])^n, x] - Dist[b*c*n, Int[
(x*(a + b*ArcSin[c*x])^(n - 1))/Sqrt[1 - c^2*x^2], x], x] /; FreeQ[{a, b, c}, x] && GtQ[n, 0]

Rule 4723

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c^(
m + 1), Subst[Int[(a + b*x)^n*Sin[x]^m*Cos[x]^(2*p + 1), x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e, n},
x] && EqQ[c^2*d + e, 0] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 0])

Rule 3306

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d +
f*x]/Sqrt[c + d*x], x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/Sqrt[c + d*x], x], x] /; FreeQ[{c
, d, e, f}, x] && ComplexFreeQ[f] && NeQ[d*e - c*f, 0]

Rule 3305

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Sin[(f*x^2)/d], x], x,
Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

Rule 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3304

Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Cos[(f*x^2)/d],
x], x, Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

Rule 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rubi steps

\begin{align*} \int \sqrt{a+b \sin ^{-1}(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \sqrt{a+b \sin ^{-1}(x)} \, dx,x,c+d x\right )}{d}\\ &=\frac{(c+d x) \sqrt{a+b \sin ^{-1}(c+d x)}}{d}-\frac{b \operatorname{Subst}\left (\int \frac{x}{\sqrt{1-x^2} \sqrt{a+b \sin ^{-1}(x)}} \, dx,x,c+d x\right )}{2 d}\\ &=\frac{(c+d x) \sqrt{a+b \sin ^{-1}(c+d x)}}{d}-\frac{b \operatorname{Subst}\left (\int \frac{\sin (x)}{\sqrt{a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{2 d}\\ &=\frac{(c+d x) \sqrt{a+b \sin ^{-1}(c+d x)}}{d}-\frac{\left (b \cos \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{a}{b}+x\right )}{\sqrt{a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{2 d}+\frac{\left (b \sin \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{a}{b}+x\right )}{\sqrt{a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{2 d}\\ &=\frac{(c+d x) \sqrt{a+b \sin ^{-1}(c+d x)}}{d}-\frac{\cos \left (\frac{a}{b}\right ) \operatorname{Subst}\left (\int \sin \left (\frac{x^2}{b}\right ) \, dx,x,\sqrt{a+b \sin ^{-1}(c+d x)}\right )}{d}+\frac{\sin \left (\frac{a}{b}\right ) \operatorname{Subst}\left (\int \cos \left (\frac{x^2}{b}\right ) \, dx,x,\sqrt{a+b \sin ^{-1}(c+d x)}\right )}{d}\\ &=\frac{(c+d x) \sqrt{a+b \sin ^{-1}(c+d x)}}{d}-\frac{\sqrt{b} \sqrt{\frac{\pi }{2}} \cos \left (\frac{a}{b}\right ) S\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{d}+\frac{\sqrt{b} \sqrt{\frac{\pi }{2}} C\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right ) \sin \left (\frac{a}{b}\right )}{d}\\ \end{align*}

Mathematica [C]  time = 0.0567237, size = 129, normalized size = 0.97 \[ \frac{b e^{-\frac{i a}{b}} \left (\sqrt{-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \text{Gamma}\left (\frac{3}{2},-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+e^{\frac{2 i a}{b}} \sqrt{\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \text{Gamma}\left (\frac{3}{2},\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )\right )}{2 d \sqrt{a+b \sin ^{-1}(c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(b*(Sqrt[((-I)*(a + b*ArcSin[c + d*x]))/b]*Gamma[3/2, ((-I)*(a + b*ArcSin[c + d*x]))/b] + E^(((2*I)*a)/b)*Sqrt
[(I*(a + b*ArcSin[c + d*x]))/b]*Gamma[3/2, (I*(a + b*ArcSin[c + d*x]))/b]))/(2*d*E^((I*a)/b)*Sqrt[a + b*ArcSin
[c + d*x]])

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Maple [A]  time = 0., size = 194, normalized size = 1.5 \begin{align*}{\frac{1}{2\,d} \left ( -\sqrt{2}\sqrt{\pi }\sqrt{{b}^{-1}}\sqrt{a+b\arcsin \left ( dx+c \right ) }\cos \left ({\frac{a}{b}} \right ){\it FresnelS} \left ({\frac{\sqrt{2}}{\sqrt{\pi }b}\sqrt{a+b\arcsin \left ( dx+c \right ) }{\frac{1}{\sqrt{{b}^{-1}}}}} \right ) b+\sqrt{2}\sqrt{\pi }\sqrt{{b}^{-1}}\sqrt{a+b\arcsin \left ( dx+c \right ) }\sin \left ({\frac{a}{b}} \right ){\it FresnelC} \left ({\frac{\sqrt{2}}{\sqrt{\pi }b}\sqrt{a+b\arcsin \left ( dx+c \right ) }{\frac{1}{\sqrt{{b}^{-1}}}}} \right ) b+2\,\arcsin \left ( dx+c \right ) \sin \left ({\frac{a+b\arcsin \left ( dx+c \right ) }{b}}-{\frac{a}{b}} \right ) b+2\,\sin \left ({\frac{a+b\arcsin \left ( dx+c \right ) }{b}}-{\frac{a}{b}} \right ) a \right ){\frac{1}{\sqrt{a+b\arcsin \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d/(a+b*arcsin(d*x+c))^(1/2)*(-2^(1/2)*Pi^(1/2)*(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)*cos(a/b)*FresnelS(2^(
1/2)/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*b+2^(1/2)*Pi^(1/2)*(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2
)*sin(a/b)*FresnelC(2^(1/2)/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*b+2*arcsin(d*x+c)*sin((a+b*arcsi
n(d*x+c))/b-a/b)*b+2*sin((a+b*arcsin(d*x+c))/b-a/b)*a)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: UnboundLocalError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.46163, size = 301, normalized size = 2.26 \begin{align*} \frac{\sqrt{2} \sqrt{\pi } b i \operatorname{erf}\left (-\frac{\sqrt{2} \sqrt{b \arcsin \left (d x + c\right ) + a} i}{2 \, \sqrt{{\left | b \right |}}} - \frac{\sqrt{2} \sqrt{b \arcsin \left (d x + c\right ) + a} \sqrt{{\left | b \right |}}}{2 \, b}\right ) e^{\left (\frac{a i}{b}\right )}}{4 \,{\left (\frac{b i}{\sqrt{{\left | b \right |}}} + \sqrt{{\left | b \right |}}\right )} d} + \frac{\sqrt{2} \sqrt{\pi } b i \operatorname{erf}\left (\frac{\sqrt{2} \sqrt{b \arcsin \left (d x + c\right ) + a} i}{2 \, \sqrt{{\left | b \right |}}} - \frac{\sqrt{2} \sqrt{b \arcsin \left (d x + c\right ) + a} \sqrt{{\left | b \right |}}}{2 \, b}\right ) e^{\left (-\frac{a i}{b}\right )}}{4 \,{\left (\frac{b i}{\sqrt{{\left | b \right |}}} - \sqrt{{\left | b \right |}}\right )} d} - \frac{\sqrt{b \arcsin \left (d x + c\right ) + a} i e^{\left (i \arcsin \left (d x + c\right )\right )}}{2 \, d} + \frac{\sqrt{b \arcsin \left (d x + c\right ) + a} i e^{\left (-i \arcsin \left (d x + c\right )\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(2)*sqrt(pi)*b*i*erf(-1/2*sqrt(2)*sqrt(b*arcsin(d*x + c) + a)*i/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b*arcs
in(d*x + c) + a)*sqrt(abs(b))/b)*e^(a*i/b)/((b*i/sqrt(abs(b)) + sqrt(abs(b)))*d) + 1/4*sqrt(2)*sqrt(pi)*b*i*er
f(1/2*sqrt(2)*sqrt(b*arcsin(d*x + c) + a)*i/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b*arcsin(d*x + c) + a)*sqrt(abs(b)
)/b)*e^(-a*i/b)/((b*i/sqrt(abs(b)) - sqrt(abs(b)))*d) - 1/2*sqrt(b*arcsin(d*x + c) + a)*i*e^(i*arcsin(d*x + c)
)/d + 1/2*sqrt(b*arcsin(d*x + c) + a)*i*e^(-i*arcsin(d*x + c))/d