3.3.1 \(\int \frac {\sin (a+\frac {b}{\sqrt {c+d x}})}{(e+f x)^2} \, dx\) [201]

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

[Out]

(d*x+c)*sin(a+b/(d*x+c)^(1/2))/(-c*f+d*e)/(f*x+e)+1/2*b*d*Ci(b*f^(1/2)/(c*f-d*e)^(1/2)+b/(d*x+c)^(1/2))*cos(a-
b*f^(1/2)/(c*f-d*e)^(1/2))/(c*f-d*e)^(3/2)/f^(1/2)-1/2*b*d*Ci(b*f^(1/2)/(c*f-d*e)^(1/2)-b/(d*x+c)^(1/2))*cos(a
+b*f^(1/2)/(c*f-d*e)^(1/2))/(c*f-d*e)^(3/2)/f^(1/2)-1/2*b*d*Si(b*f^(1/2)/(c*f-d*e)^(1/2)+b/(d*x+c)^(1/2))*sin(
a-b*f^(1/2)/(c*f-d*e)^(1/2))/(c*f-d*e)^(3/2)/f^(1/2)-1/2*b*d*Si(b*f^(1/2)/(c*f-d*e)^(1/2)-b/(d*x+c)^(1/2))*sin
(a+b*f^(1/2)/(c*f-d*e)^(1/2))/(c*f-d*e)^(3/2)/f^(1/2)

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Rubi [A]
time = 0.74, antiderivative size = 350, normalized size of antiderivative = 1.00, 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 = {3512, 3422, 3415, 3384, 3380, 3383} \begin {gather*} -\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)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(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]
])/(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]]*SinIntegra
l[(b*Sqrt[f])/Sqrt[-(d*e) + c*f] + b/Sqrt[c + d*x]])/(2*Sqrt[f]*(-(d*e) + c*f)^(3/2))

Rule 3380

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 3383

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]

Rule 3384

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 3415

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 3422

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 3512

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]

Rubi steps

\begin {align*} \int \frac {\sin \left (a+\frac {b}{\sqrt {c+d x}}\right )}{(e+f x)^2} \, dx &=-\frac {2 \text {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 \text {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 \text {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) \text {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) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [F]
time = 180.04, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(2733\) vs. \(2(284)=568\).
time = 0.06, size = 2734, normalized size = 7.81

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(2734\)
default	\(\text {Expression too large to display}\)	\(2734\)

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,method=_RETURNVERBOSE)

[Out]

-2*d*b^2*(sin(a+b/(d*x+c)^(1/2))*(-1/2*a/f/b^2*(a+b/(d*x+c)^(1/2))+1/2*(a^2*c*f-a^2*d*e-b^2*f)/f/b^2/(c*f-d*e)
)/(a^2*c*f-a^2*d*e-2*a*c*f*(a+b/(d*x+c)^(1/2))+2*a*d*e*(a+b/(d*x+c)^(1/2))+c*f*(a+b/(d*x+c)^(1/2))^2-d*e*(a+b/
(d*x+c)^(1/2))^2-f*b^2)+1/4*a/f/b^2/(a*c*f-a*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+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(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))*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/f/b^2/
(a*c*f-a*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+(b^2*c*f^2-b^2*d*e*f)^
(1/2))/(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))*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^2*c*f-a^2*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*c*f-a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e)-f*b^2)/f/b^2/(c*f
-d*e)/(a*c*f-a*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+(b^2*c*f^2-b^2*d*e
*f)^(1/2))/(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*(a^2*c*f-a^2*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*c*f+a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e)-f*b^2)/f/b^2/(
c*f-d*e)/(a*c*f-a*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+(b^2*c*f^2-b^
2*d*e*f)^(1/2))/(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)))-a*(sin(a+b/(d*x+c)^(1/2))*(-1/2/b^2
*(a+b/(d*x+c)^(1/2))/f+1/2*a/f/b^2)/(a^2*c*f-a^2*d*e-2*a*c*f*(a+b/(d*x+c)^(1/2))+2*a*d*e*(a+b/(d*x+c)^(1/2))+c
*f*(a+b/(d*x+c)^(1/2))^2-d*e*(a+b/(d*x+c)^(1/2))^2-f*b^2)+1/4/f/b^2/(a*c*f-a*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+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(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))*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/f/b^2/(a*c*f-a*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+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(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))*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/f/b^2/(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/f/b^2/(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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] Result contains complex when optimal does not.
time = 0.48, size = 477, normalized size = 1.36 \begin {gather*} -\frac {{\left (-i \, d f x - i \, d e\right )} \sqrt {-\frac {b^{2} f}{c f - d e}} {\rm Ei}\left (-\frac {\sqrt {-\frac {b^{2} f}{c f - d e}} {\left (d x + c\right )} - i \, \sqrt {d x + c} b}{d x + c}\right ) e^{\left (i \, a + \sqrt {-\frac {b^{2} f}{c f - d e}}\right )} + {\left (i \, d f x + i \, d e\right )} \sqrt {-\frac {b^{2} f}{c f - d e}} {\rm Ei}\left (\frac {\sqrt {-\frac {b^{2} f}{c f - d e}} {\left (d x + c\right )} + i \, \sqrt {d x + c} b}{d x + c}\right ) e^{\left (i \, a - \sqrt {-\frac {b^{2} f}{c f - d e}}\right )} + {\left (i \, d f x + i \, d e\right )} \sqrt {-\frac {b^{2} f}{c f - d e}} {\rm Ei}\left (-\frac {\sqrt {-\frac {b^{2} f}{c f - d e}} {\left (d x + c\right )} + i \, \sqrt {d x + c} b}{d x + c}\right ) e^{\left (-i \, a + \sqrt {-\frac {b^{2} f}{c f - d e}}\right )} + {\left (-i \, d f x - i \, d e\right )} \sqrt {-\frac {b^{2} f}{c f - d e}} {\rm Ei}\left (\frac {\sqrt {-\frac {b^{2} f}{c f - d e}} {\left (d x + c\right )} - i \, \sqrt {d x + c} b}{d x + c}\right ) e^{\left (-i \, a - \sqrt {-\frac {b^{2} f}{c f - d e}}\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 (c f^{3} x - d f e^{2} - {\left (d f^{2} x - c f^{2}\right )} e\right )}} \end {gather*}

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sin {\left (a + \frac {b}{\sqrt {c + d x}} \right )}}{\left (e + f x\right )^{2}}\, dx \end {gather*}

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]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sin \left (a+\frac {b}{\sqrt {c+d\,x}}\right )}{{\left (e+f\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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