3.268 \(\int \frac{c e+d e x}{(a+b \sin ^{-1}(c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=144 \[ \frac{2 \sqrt{\pi } e \cos \left (\frac{2 a}{b}\right ) \text{FresnelC}\left (\frac{2 \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{\pi } \sqrt{b}}\right )}{b^{3/2} d}+\frac{2 \sqrt{\pi } e \sin \left (\frac{2 a}{b}\right ) S\left (\frac{2 \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b} \sqrt{\pi }}\right )}{b^{3/2} d}-\frac{2 e \sqrt{1-(c+d x)^2} (c+d x)}{b d \sqrt{a+b \sin ^{-1}(c+d x)}} \]

[Out]

(-2*e*(c + d*x)*Sqrt[1 - (c + d*x)^2])/(b*d*Sqrt[a + b*ArcSin[c + d*x]]) + (2*e*Sqrt[Pi]*Cos[(2*a)/b]*FresnelC
[(2*Sqrt[a + b*ArcSin[c + d*x]])/(Sqrt[b]*Sqrt[Pi])])/(b^(3/2)*d) + (2*e*Sqrt[Pi]*FresnelS[(2*Sqrt[a + b*ArcSi
n[c + d*x]])/(Sqrt[b]*Sqrt[Pi])]*Sin[(2*a)/b])/(b^(3/2)*d)

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Rubi [A]  time = 0.230992, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {4805, 12, 4631, 3306, 3305, 3351, 3304, 3352} \[ \frac{2 \sqrt{\pi } e \cos \left (\frac{2 a}{b}\right ) \text{FresnelC}\left (\frac{2 \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{\pi } \sqrt{b}}\right )}{b^{3/2} d}+\frac{2 \sqrt{\pi } e \sin \left (\frac{2 a}{b}\right ) S\left (\frac{2 \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b} \sqrt{\pi }}\right )}{b^{3/2} d}-\frac{2 e \sqrt{1-(c+d x)^2} (c+d x)}{b d \sqrt{a+b \sin ^{-1}(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)/(a + b*ArcSin[c + d*x])^(3/2),x]

[Out]

(-2*e*(c + d*x)*Sqrt[1 - (c + d*x)^2])/(b*d*Sqrt[a + b*ArcSin[c + d*x]]) + (2*e*Sqrt[Pi]*Cos[(2*a)/b]*FresnelC
[(2*Sqrt[a + b*ArcSin[c + d*x]])/(Sqrt[b]*Sqrt[Pi])])/(b^(3/2)*d) + (2*e*Sqrt[Pi]*FresnelS[(2*Sqrt[a + b*ArcSi
n[c + d*x]])/(Sqrt[b]*Sqrt[Pi])]*Sin[(2*a)/b])/(b^(3/2)*d)

Rule 4805

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 4631

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[(x^m*Sqrt[1 - c^2*x^2]*(a + b*ArcSin
[c*x])^(n + 1))/(b*c*(n + 1)), x] - Dist[1/(b*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[(a + b*x)^(n + 1)
, Sin[x]^(m - 1)*(m - (m + 1)*Sin[x]^2), x], x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && G
eQ[n, -2] && LtQ[n, -1]

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 \frac{c e+d e x}{\left (a+b \sin ^{-1}(c+d x)\right )^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{e x}{\left (a+b \sin ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{d}\\ &=\frac{e \operatorname{Subst}\left (\int \frac{x}{\left (a+b \sin ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac{2 e (c+d x) \sqrt{1-(c+d x)^2}}{b d \sqrt{a+b \sin ^{-1}(c+d x)}}+\frac{(2 e) \operatorname{Subst}\left (\int \frac{\cos (2 x)}{\sqrt{a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{b d}\\ &=-\frac{2 e (c+d x) \sqrt{1-(c+d x)^2}}{b d \sqrt{a+b \sin ^{-1}(c+d x)}}+\frac{\left (2 e \cos \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{2 a}{b}+2 x\right )}{\sqrt{a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{b d}+\frac{\left (2 e \sin \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{2 a}{b}+2 x\right )}{\sqrt{a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{b d}\\ &=-\frac{2 e (c+d x) \sqrt{1-(c+d x)^2}}{b d \sqrt{a+b \sin ^{-1}(c+d x)}}+\frac{\left (4 e \cos \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \cos \left (\frac{2 x^2}{b}\right ) \, dx,x,\sqrt{a+b \sin ^{-1}(c+d x)}\right )}{b^2 d}+\frac{\left (4 e \sin \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \sin \left (\frac{2 x^2}{b}\right ) \, dx,x,\sqrt{a+b \sin ^{-1}(c+d x)}\right )}{b^2 d}\\ &=-\frac{2 e (c+d x) \sqrt{1-(c+d x)^2}}{b d \sqrt{a+b \sin ^{-1}(c+d x)}}+\frac{2 e \sqrt{\pi } \cos \left (\frac{2 a}{b}\right ) C\left (\frac{2 \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b} \sqrt{\pi }}\right )}{b^{3/2} d}+\frac{2 e \sqrt{\pi } S\left (\frac{2 \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b} \sqrt{\pi }}\right ) \sin \left (\frac{2 a}{b}\right )}{b^{3/2} d}\\ \end{align*}

Mathematica [C]  time = 0.147742, size = 168, normalized size = 1.17 \[ \frac{i e e^{-\frac{2 i a}{b}} \left (-\sqrt{2} \sqrt{-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \text{Gamma}\left (\frac{1}{2},-\frac{2 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+\sqrt{2} e^{\frac{4 i a}{b}} \sqrt{\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \text{Gamma}\left (\frac{1}{2},\frac{2 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+2 i e^{\frac{2 i a}{b}} \sin \left (2 \sin ^{-1}(c+d x)\right )\right )}{2 b d \sqrt{a+b \sin ^{-1}(c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c*e + d*e*x)/(a + b*ArcSin[c + d*x])^(3/2),x]

[Out]

((I/2)*e*(-(Sqrt[2]*Sqrt[((-I)*(a + b*ArcSin[c + d*x]))/b]*Gamma[1/2, ((-2*I)*(a + b*ArcSin[c + d*x]))/b]) + S
qrt[2]*E^(((4*I)*a)/b)*Sqrt[(I*(a + b*ArcSin[c + d*x]))/b]*Gamma[1/2, ((2*I)*(a + b*ArcSin[c + d*x]))/b] + (2*
I)*E^(((2*I)*a)/b)*Sin[2*ArcSin[c + d*x]]))/(b*d*E^(((2*I)*a)/b)*Sqrt[a + b*ArcSin[c + d*x]])

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Maple [A]  time = 0.073, size = 156, normalized size = 1.1 \begin{align*} -{\frac{e}{bd} \left ( -2\,\sqrt{{b}^{-1}}\sqrt{a+b\arcsin \left ( dx+c \right ) }\cos \left ( 2\,{\frac{a}{b}} \right ){\it FresnelC} \left ( 2\,{\frac{\sqrt{a+b\arcsin \left ( dx+c \right ) }}{\sqrt{\pi }\sqrt{{b}^{-1}}b}} \right ) \sqrt{\pi }-2\,\sqrt{{b}^{-1}}\sqrt{a+b\arcsin \left ( dx+c \right ) }\sin \left ( 2\,{\frac{a}{b}} \right ){\it FresnelS} \left ( 2\,{\frac{\sqrt{a+b\arcsin \left ( dx+c \right ) }}{\sqrt{\pi }\sqrt{{b}^{-1}}b}} \right ) \sqrt{\pi }+\sin \left ( 2\,{\frac{a+b\arcsin \left ( dx+c \right ) }{b}}-2\,{\frac{a}{b}} \right ) \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((d*e*x+c*e)/(a+b*arcsin(d*x+c))^(3/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)/(a+b*arcsin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((d*e*x + c*e)/(b*arcsin(d*x + c) + a)^(3/2), 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((d*e*x+c*e)/(a+b*arcsin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)/(a+b*asin(d*x+c))**(3/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)/(a+b*arcsin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((d*e*x + c*e)/(b*arcsin(d*x + c) + a)^(3/2), x)