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

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

[Out]

(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)

________________________________________________________________________________________

Rubi [A]  time = 0.93, 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 = {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 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]

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 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 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 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]

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 = 180.03, size = 0, normalized size = 0.00 \[ \text {\$Aborted} \]

Verification is Not applicable to the result.

[In]

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

[Out]

$Aborted

________________________________________________________________________________________

fricas [C]  time = 0.75, size = 454, normalized size = 1.30 \[ -\frac {{\left (i \, d f x + i \, d e\right )} \sqrt {\frac {b^{2} f}{d e - c f}} {\rm Ei}\left (-\frac {2 \, \sqrt {\frac {b^{2} f}{d e - c f}} {\left (d x + c\right )} - 2 i \, \sqrt {d x + c} b}{2 \, {\left (d x + c\right )}}\right ) e^{\left (i \, a + \sqrt {\frac {b^{2} f}{d e - c f}}\right )} + {\left (-i \, d f x - i \, d e\right )} \sqrt {\frac {b^{2} f}{d e - c f}} {\rm Ei}\left (\frac {2 \, \sqrt {\frac {b^{2} f}{d e - c f}} {\left (d x + c\right )} + 2 i \, \sqrt {d x + c} b}{2 \, {\left (d x + c\right )}}\right ) e^{\left (i \, a - \sqrt {\frac {b^{2} f}{d e - c f}}\right )} + {\left (-i \, d f x - i \, d e\right )} \sqrt {\frac {b^{2} f}{d e - c f}} {\rm Ei}\left (-\frac {2 \, \sqrt {\frac {b^{2} f}{d e - c f}} {\left (d x + c\right )} + 2 i \, \sqrt {d x + c} b}{2 \, {\left (d x + c\right )}}\right ) e^{\left (-i \, a + \sqrt {\frac {b^{2} f}{d e - c f}}\right )} + {\left (i \, d f x + i \, d e\right )} \sqrt {\frac {b^{2} f}{d e - c f}} {\rm Ei}\left (\frac {2 \, \sqrt {\frac {b^{2} f}{d e - c f}} {\left (d x + c\right )} - 2 i \, \sqrt {d x + c} b}{2 \, {\left (d x + c\right )}}\right ) e^{\left (-i \, a - \sqrt {\frac {b^{2} f}{d e - c f}}\right )} - 4 \, {\left (d f x + c f\right )} \sin \left (\frac {a d x + a c + \sqrt {d x + c} b}{d x + c}\right )}{4 \, {\left (d e^{2} f - c e f^{2} + {\left (d e f^{2} - c f^{3}\right )} x\right )}} \]

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

maple [B]  time = 0.09, size = 2724, normalized size = 7.78 \[ \text {result too large to display} \]

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-e*(a*c*f-a*d*e+(b
^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e)*d-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+e*(-a*c*f+a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e
)*d-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-e*(a*c*f-a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f
-d*e)*d-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+e*(-a*c*f+a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2
))/(c*f-d*e)*d-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-e*(a*c*f-a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e)*d-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+e*(-a*c*f+a*d*e+(b^2*c
*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e)*d-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)))))

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________