3.201 \(\int \frac{\sin (a+\frac{b}{\sqrt{c+d x}})}{(e+f x)^2} \, dx\)

Optimal. Leaf size=350 \[ -\frac{b d \cos \left (a+\frac{b \sqrt{f}}{\sqrt{c f-d e}}\right ) \text{CosIntegral}\left (\frac{b \sqrt{f}}{\sqrt{c f-d e}}-\frac{b}{\sqrt{c+d x}}\right )}{2 \sqrt{f} (c f-d e)^{3/2}}+\frac{b d \cos \left (a-\frac{b \sqrt{f}}{\sqrt{c f-d e}}\right ) \text{CosIntegral}\left (\frac{b \sqrt{f}}{\sqrt{c f-d e}}+\frac{b}{\sqrt{c+d x}}\right )}{2 \sqrt{f} (c f-d e)^{3/2}}-\frac{b d \sin \left (a+\frac{b \sqrt{f}}{\sqrt{c f-d e}}\right ) \text{Si}\left (\frac{b \sqrt{f}}{\sqrt{c f-d e}}-\frac{b}{\sqrt{c+d x}}\right )}{2 \sqrt{f} (c f-d e)^{3/2}}-\frac{b d \sin \left (a-\frac{b \sqrt{f}}{\sqrt{c f-d e}}\right ) \text{Si}\left (\frac{\sqrt{f} b}{\sqrt{c f-d e}}+\frac{b}{\sqrt{c+d x}}\right )}{2 \sqrt{f} (c f-d e)^{3/2}}+\frac{(c+d x) \sin \left (a+\frac{b}{\sqrt{c+d x}}\right )}{(e+f x) (d e-c f)} \]

[Out]

-(b*d*Cos[a + (b*Sqrt[f])/Sqrt[-(d*e) + c*f]]*CosIntegral[(b*Sqrt[f])/Sqrt[-(d*e) + c*f] - b/Sqrt[c + d*x]])/(
2*Sqrt[f]*(-(d*e) + c*f)^(3/2)) + (b*d*Cos[a - (b*Sqrt[f])/Sqrt[-(d*e) + c*f]]*CosIntegral[(b*Sqrt[f])/Sqrt[-(
d*e) + c*f] + b/Sqrt[c + d*x]])/(2*Sqrt[f]*(-(d*e) + c*f)^(3/2)) + ((c + d*x)*Sin[a + b/Sqrt[c + d*x]])/((d*e
- c*f)*(e + f*x)) - (b*d*Sin[a + (b*Sqrt[f])/Sqrt[-(d*e) + c*f]]*SinIntegral[(b*Sqrt[f])/Sqrt[-(d*e) + c*f] -
b/Sqrt[c + d*x]])/(2*Sqrt[f]*(-(d*e) + c*f)^(3/2)) - (b*d*Sin[a - (b*Sqrt[f])/Sqrt[-(d*e) + c*f]]*SinIntegral[
(b*Sqrt[f])/Sqrt[-(d*e) + c*f] + b/Sqrt[c + d*x]])/(2*Sqrt[f]*(-(d*e) + c*f)^(3/2))

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Rubi [A]  time = 0.929078, antiderivative size = 350, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {3431, 3341, 3334, 3303, 3299, 3302} \[ -\frac{b d \cos \left (a+\frac{b \sqrt{f}}{\sqrt{c f-d e}}\right ) \text{CosIntegral}\left (\frac{b \sqrt{f}}{\sqrt{c f-d e}}-\frac{b}{\sqrt{c+d x}}\right )}{2 \sqrt{f} (c f-d e)^{3/2}}+\frac{b d \cos \left (a-\frac{b \sqrt{f}}{\sqrt{c f-d e}}\right ) \text{CosIntegral}\left (\frac{b \sqrt{f}}{\sqrt{c f-d e}}+\frac{b}{\sqrt{c+d x}}\right )}{2 \sqrt{f} (c f-d e)^{3/2}}-\frac{b d \sin \left (a+\frac{b \sqrt{f}}{\sqrt{c f-d e}}\right ) \text{Si}\left (\frac{b \sqrt{f}}{\sqrt{c f-d e}}-\frac{b}{\sqrt{c+d x}}\right )}{2 \sqrt{f} (c f-d e)^{3/2}}-\frac{b d \sin \left (a-\frac{b \sqrt{f}}{\sqrt{c f-d e}}\right ) \text{Si}\left (\frac{\sqrt{f} b}{\sqrt{c f-d e}}+\frac{b}{\sqrt{c+d x}}\right )}{2 \sqrt{f} (c f-d e)^{3/2}}+\frac{(c+d x) \sin \left (a+\frac{b}{\sqrt{c+d x}}\right )}{(e+f x) (d e-c f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*d*Cos[a + (b*Sqrt[f])/Sqrt[-(d*e) + c*f]]*CosIntegral[(b*Sqrt[f])/Sqrt[-(d*e) + c*f] - b/Sqrt[c + d*x]])/(
2*Sqrt[f]*(-(d*e) + c*f)^(3/2)) + (b*d*Cos[a - (b*Sqrt[f])/Sqrt[-(d*e) + c*f]]*CosIntegral[(b*Sqrt[f])/Sqrt[-(
d*e) + c*f] + b/Sqrt[c + d*x]])/(2*Sqrt[f]*(-(d*e) + c*f)^(3/2)) + ((c + d*x)*Sin[a + b/Sqrt[c + d*x]])/((d*e
- c*f)*(e + f*x)) - (b*d*Sin[a + (b*Sqrt[f])/Sqrt[-(d*e) + c*f]]*SinIntegral[(b*Sqrt[f])/Sqrt[-(d*e) + c*f] -
b/Sqrt[c + d*x]])/(2*Sqrt[f]*(-(d*e) + c*f)^(3/2)) - (b*d*Sin[a - (b*Sqrt[f])/Sqrt[-(d*e) + c*f]]*SinIntegral[
(b*Sqrt[f])/Sqrt[-(d*e) + c*f] + b/Sqrt[c + d*x]])/(2*Sqrt[f]*(-(d*e) + c*f)^(3/2))

Rule 3431

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_Symbol] :
> Dist[1/(n*f), Subst[Int[ExpandIntegrand[(a + b*Sin[c + d*x])^p, x^(1/n - 1)*(g - (e*h)/f + (h*x^(1/n))/f)^m,
x], x], x, (e + f*x)^n], x] /; FreeQ[{a, b, c, d, e, f, g, h, m}, x] && IGtQ[p, 0] && IntegerQ[1/n]

Rule 3341

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

Rule 3334

Int[Cos[(c_.) + (d_.)*(x_)]*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[Cos[c + d*x], (a +
b*x^n)^p, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[p, 0] && IGtQ[n, 0] && (EqQ[n, 2] || EqQ[p, -1])

Rule 3303

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

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

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

Rubi steps

\begin{align*} \int \frac{\sin \left (a+\frac{b}{\sqrt{c+d x}}\right )}{(e+f x)^2} \, dx &=-\frac{2 \operatorname{Subst}\left (\int \frac{x \sin (a+b x)}{\left (\frac{f}{d}+\left (e-\frac{c f}{d}\right ) x^2\right )^2} \, dx,x,\frac{1}{\sqrt{c+d x}}\right )}{d}\\ &=\frac{(c+d x) \sin \left (a+\frac{b}{\sqrt{c+d x}}\right )}{(d e-c f) (e+f x)}-\frac{b \operatorname{Subst}\left (\int \frac{\cos (a+b x)}{\frac{f}{d}+\left (e-\frac{c f}{d}\right ) x^2} \, dx,x,\frac{1}{\sqrt{c+d x}}\right )}{d e-c f}\\ &=\frac{(c+d x) \sin \left (a+\frac{b}{\sqrt{c+d x}}\right )}{(d e-c f) (e+f x)}-\frac{b \operatorname{Subst}\left (\int \left (\frac{d \cos (a+b x)}{2 \sqrt{f} \left (\sqrt{f}-\sqrt{-d e+c f} x\right )}+\frac{d \cos (a+b x)}{2 \sqrt{f} \left (\sqrt{f}+\sqrt{-d e+c f} x\right )}\right ) \, dx,x,\frac{1}{\sqrt{c+d x}}\right )}{d e-c f}\\ &=\frac{(c+d x) \sin \left (a+\frac{b}{\sqrt{c+d x}}\right )}{(d e-c f) (e+f x)}-\frac{(b d) \operatorname{Subst}\left (\int \frac{\cos (a+b x)}{\sqrt{f}-\sqrt{-d e+c f} x} \, dx,x,\frac{1}{\sqrt{c+d x}}\right )}{2 \sqrt{f} (d e-c f)}-\frac{(b d) \operatorname{Subst}\left (\int \frac{\cos (a+b x)}{\sqrt{f}+\sqrt{-d e+c f} x} \, dx,x,\frac{1}{\sqrt{c+d x}}\right )}{2 \sqrt{f} (d e-c f)}\\ &=\frac{(c+d x) \sin \left (a+\frac{b}{\sqrt{c+d x}}\right )}{(d e-c f) (e+f x)}-\frac{\left (b d \cos \left (a-\frac{b \sqrt{f}}{\sqrt{-d e+c f}}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{b \sqrt{f}}{\sqrt{-d e+c f}}+b x\right )}{\sqrt{f}+\sqrt{-d e+c f} x} \, dx,x,\frac{1}{\sqrt{c+d x}}\right )}{2 \sqrt{f} (d e-c f)}-\frac{\left (b d \cos \left (a+\frac{b \sqrt{f}}{\sqrt{-d e+c f}}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{b \sqrt{f}}{\sqrt{-d e+c f}}-b x\right )}{\sqrt{f}-\sqrt{-d e+c f} x} \, dx,x,\frac{1}{\sqrt{c+d x}}\right )}{2 \sqrt{f} (d e-c f)}+\frac{\left (b d \sin \left (a-\frac{b \sqrt{f}}{\sqrt{-d e+c f}}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{b \sqrt{f}}{\sqrt{-d e+c f}}+b x\right )}{\sqrt{f}+\sqrt{-d e+c f} x} \, dx,x,\frac{1}{\sqrt{c+d x}}\right )}{2 \sqrt{f} (d e-c f)}-\frac{\left (b d \sin \left (a+\frac{b \sqrt{f}}{\sqrt{-d e+c f}}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{b \sqrt{f}}{\sqrt{-d e+c f}}-b x\right )}{\sqrt{f}-\sqrt{-d e+c f} x} \, dx,x,\frac{1}{\sqrt{c+d x}}\right )}{2 \sqrt{f} (d e-c f)}\\ &=-\frac{b d \cos \left (a+\frac{b \sqrt{f}}{\sqrt{-d e+c f}}\right ) \text{Ci}\left (\frac{b \sqrt{f}}{\sqrt{-d e+c f}}-\frac{b}{\sqrt{c+d x}}\right )}{2 \sqrt{f} (-d e+c f)^{3/2}}+\frac{b d \cos \left (a-\frac{b \sqrt{f}}{\sqrt{-d e+c f}}\right ) \text{Ci}\left (\frac{b \sqrt{f}}{\sqrt{-d e+c f}}+\frac{b}{\sqrt{c+d x}}\right )}{2 \sqrt{f} (-d e+c f)^{3/2}}+\frac{(c+d x) \sin \left (a+\frac{b}{\sqrt{c+d x}}\right )}{(d e-c f) (e+f x)}-\frac{b d \sin \left (a+\frac{b \sqrt{f}}{\sqrt{-d e+c f}}\right ) \text{Si}\left (\frac{b \sqrt{f}}{\sqrt{-d e+c f}}-\frac{b}{\sqrt{c+d x}}\right )}{2 \sqrt{f} (-d e+c f)^{3/2}}-\frac{b d \sin \left (a-\frac{b \sqrt{f}}{\sqrt{-d e+c f}}\right ) \text{Si}\left (\frac{b \sqrt{f}}{\sqrt{-d e+c f}}+\frac{b}{\sqrt{c+d x}}\right )}{2 \sqrt{f} (-d e+c f)^{3/2}}\\ \end{align*}

Mathematica [F]  time = 180.033, size = 0, normalized size = 0. \[ \text{\$Aborted} \]

Verification is Not applicable to the result.

[In]

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

[Out]

$Aborted

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Maple [B]  time = 0.052, size = 2724, normalized size = 7.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a+b/(d*x+c)^(1/2))/(f*x+e)^2,x)

[Out]

-2*d*b^2*(sin(a+b/(d*x+c)^(1/2))*(-1/2*a/b^2/f*(a+b/(d*x+c)^(1/2))+1/2*(a^2*c*f-a^2*d*e-b^2*f)/b^2/f/(c*f-d*e)
)/(c*f*(a+b/(d*x+c)^(1/2))^2-d*e*(a+b/(d*x+c)^(1/2))^2-2*(a+b/(d*x+c)^(1/2))*a*c*f+2*(a+b/(d*x+c)^(1/2))*a*d*e
+a^2*c*f-a^2*d*e-b^2*f)-1/4*a/b^2/f/((a*c*f-a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e)*c*f-(a*c*f-a*d*e+(b^2
*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e)*d*e-a*c*f+a*d*e)*(Si(b/(d*x+c)^(1/2)+a-(a*c*f-a*d*e+(b^2*c*f^2-b^2*d*e*f)^(
1/2))/(c*f-d*e))*cos((a*c*f-a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e))+Ci(b/(d*x+c)^(1/2)+a-(a*c*f-a*d*e+(b
^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e))*sin((a*c*f-a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e)))-1/4*a/b^2/f/(-
(-a*c*f+a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e)*c*f+(-a*c*f+a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e)*
d*e-a*c*f+a*d*e)*(Si(b/(d*x+c)^(1/2)+a+(-a*c*f+a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e))*cos((-a*c*f+a*d*e
+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e))-Ci(b/(d*x+c)^(1/2)+a+(-a*c*f+a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-
d*e))*sin((-a*c*f+a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e)))+1/4*((a*c*f-a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2)
)/(c*f-d*e)*a*c*f-(a*c*f-a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e)*a*d*e-a^2*c*f+a^2*d*e+b^2*f)/b^2/f/(c*f-
d*e)/((a*c*f-a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e)*c*f-(a*c*f-a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d
*e)*d*e-a*c*f+a*d*e)*(-Si(b/(d*x+c)^(1/2)+a-(a*c*f-a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e))*sin((a*c*f-a*
d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e))+Ci(b/(d*x+c)^(1/2)+a-(a*c*f-a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*
f-d*e))*cos((a*c*f-a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e)))+1/4*(-(-a*c*f+a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1
/2))/(c*f-d*e)*a*c*f+(-a*c*f+a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e)*a*d*e-a^2*c*f+a^2*d*e+b^2*f)/b^2/f/(
c*f-d*e)/(-(-a*c*f+a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e)*c*f+(-a*c*f+a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))
/(c*f-d*e)*d*e-a*c*f+a*d*e)*(Si(b/(d*x+c)^(1/2)+a+(-a*c*f+a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e))*sin((-
a*c*f+a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e))+Ci(b/(d*x+c)^(1/2)+a+(-a*c*f+a*d*e+(b^2*c*f^2-b^2*d*e*f)^(
1/2))/(c*f-d*e))*cos((-a*c*f+a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e)))-a*(sin(a+b/(d*x+c)^(1/2))*(-1/2/b^
2/f*(a+b/(d*x+c)^(1/2))+1/2*a/b^2/f)/(c*f*(a+b/(d*x+c)^(1/2))^2-d*e*(a+b/(d*x+c)^(1/2))^2-2*(a+b/(d*x+c)^(1/2)
)*a*c*f+2*(a+b/(d*x+c)^(1/2))*a*d*e+a^2*c*f-a^2*d*e-b^2*f)-1/4/b^2/f/((a*c*f-a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2)
)/(c*f-d*e)*c*f-(a*c*f-a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e)*d*e-a*c*f+a*d*e)*(Si(b/(d*x+c)^(1/2)+a-(a*
c*f-a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e))*cos((a*c*f-a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e))+Ci(
b/(d*x+c)^(1/2)+a-(a*c*f-a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e))*sin((a*c*f-a*d*e+(b^2*c*f^2-b^2*d*e*f)^
(1/2))/(c*f-d*e)))-1/4/b^2/f/(-(-a*c*f+a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e)*c*f+(-a*c*f+a*d*e+(b^2*c*f
^2-b^2*d*e*f)^(1/2))/(c*f-d*e)*d*e-a*c*f+a*d*e)*(Si(b/(d*x+c)^(1/2)+a+(-a*c*f+a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2
))/(c*f-d*e))*cos((-a*c*f+a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e))-Ci(b/(d*x+c)^(1/2)+a+(-a*c*f+a*d*e+(b^
2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e))*sin((-a*c*f+a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e)))+1/4/b^2/f/(c*f
-d*e)*(-Si(b/(d*x+c)^(1/2)+a-(a*c*f-a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e))*sin((a*c*f-a*d*e+(b^2*c*f^2-
b^2*d*e*f)^(1/2))/(c*f-d*e))+Ci(b/(d*x+c)^(1/2)+a-(a*c*f-a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e))*cos((a*
c*f-a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e)))+1/4/b^2/f/(c*f-d*e)*(Si(b/(d*x+c)^(1/2)+a+(-a*c*f+a*d*e+(b^
2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e))*sin((-a*c*f+a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e))+Ci(b/(d*x+c)^(1
/2)+a+(-a*c*f+a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e))*cos((-a*c*f+a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*
f-d*e)))))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(a + b/sqrt(d*x + c))/(f*x + e)^2, x)

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Fricas [C]  time = 2.43122, size = 1003, normalized size = 2.87 \begin{align*} -\frac{{\left (i \, d f x + i \, d e\right )} \sqrt{\frac{b^{2} f}{d e - c f}}{\rm Ei}\left (-\frac{2 \, \sqrt{\frac{b^{2} f}{d e - c f}}{\left (d x + c\right )} - 2 i \, \sqrt{d x + c} b}{2 \,{\left (d x + c\right )}}\right ) e^{\left (i \, a + \sqrt{\frac{b^{2} f}{d e - c f}}\right )} +{\left (-i \, d f x - i \, d e\right )} \sqrt{\frac{b^{2} f}{d e - c f}}{\rm Ei}\left (\frac{2 \, \sqrt{\frac{b^{2} f}{d e - c f}}{\left (d x + c\right )} + 2 i \, \sqrt{d x + c} b}{2 \,{\left (d x + c\right )}}\right ) e^{\left (i \, a - \sqrt{\frac{b^{2} f}{d e - c f}}\right )} +{\left (-i \, d f x - i \, d e\right )} \sqrt{\frac{b^{2} f}{d e - c f}}{\rm Ei}\left (-\frac{2 \, \sqrt{\frac{b^{2} f}{d e - c f}}{\left (d x + c\right )} + 2 i \, \sqrt{d x + c} b}{2 \,{\left (d x + c\right )}}\right ) e^{\left (-i \, a + \sqrt{\frac{b^{2} f}{d e - c f}}\right )} +{\left (i \, d f x + i \, d e\right )} \sqrt{\frac{b^{2} f}{d e - c f}}{\rm Ei}\left (\frac{2 \, \sqrt{\frac{b^{2} f}{d e - c f}}{\left (d x + c\right )} - 2 i \, \sqrt{d x + c} b}{2 \,{\left (d x + c\right )}}\right ) e^{\left (-i \, a - \sqrt{\frac{b^{2} f}{d e - c f}}\right )} - 4 \,{\left (d f x + c f\right )} \sin \left (\frac{a d x + a c + \sqrt{d x + c} b}{d x + c}\right )}{4 \,{\left (d e^{2} f - c e f^{2} +{\left (d e f^{2} - c f^{3}\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*((I*d*f*x + I*d*e)*sqrt(b^2*f/(d*e - c*f))*Ei(-1/2*(2*sqrt(b^2*f/(d*e - c*f))*(d*x + c) - 2*I*sqrt(d*x +
c)*b)/(d*x + c))*e^(I*a + sqrt(b^2*f/(d*e - c*f))) + (-I*d*f*x - I*d*e)*sqrt(b^2*f/(d*e - c*f))*Ei(1/2*(2*sqrt
(b^2*f/(d*e - c*f))*(d*x + c) + 2*I*sqrt(d*x + c)*b)/(d*x + c))*e^(I*a - sqrt(b^2*f/(d*e - c*f))) + (-I*d*f*x
- I*d*e)*sqrt(b^2*f/(d*e - c*f))*Ei(-1/2*(2*sqrt(b^2*f/(d*e - c*f))*(d*x + c) + 2*I*sqrt(d*x + c)*b)/(d*x + c)
)*e^(-I*a + sqrt(b^2*f/(d*e - c*f))) + (I*d*f*x + I*d*e)*sqrt(b^2*f/(d*e - c*f))*Ei(1/2*(2*sqrt(b^2*f/(d*e - c
*f))*(d*x + c) - 2*I*sqrt(d*x + c)*b)/(d*x + c))*e^(-I*a - sqrt(b^2*f/(d*e - c*f))) - 4*(d*f*x + c*f)*sin((a*d
*x + a*c + sqrt(d*x + c)*b)/(d*x + c)))/(d*e^2*f - c*e*f^2 + (d*e*f^2 - c*f^3)*x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/(d*x+c)**(1/2))/(f*x+e)**2,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(a + b/sqrt(d*x + c))/(f*x + e)^2, x)