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

Optimal. Leaf size=238 \[ \frac {\sin \left (a-\frac {b \sqrt {c f-d e}}{\sqrt {f}}\right ) \text {Ci}\left (\frac {\sqrt {c f-d e} b}{\sqrt {f}}+\sqrt {c+d x} b\right )}{f}+\frac {\sin \left (a+\frac {b \sqrt {c f-d e}}{\sqrt {f}}\right ) \text {Ci}\left (\frac {b \sqrt {c f-d e}}{\sqrt {f}}-b \sqrt {c+d x}\right )}{f}-\frac {\cos \left (a+\frac {b \sqrt {c f-d e}}{\sqrt {f}}\right ) \text {Si}\left (\frac {b \sqrt {c f-d e}}{\sqrt {f}}-b \sqrt {c+d x}\right )}{f}+\frac {\cos \left (a-\frac {b \sqrt {c f-d e}}{\sqrt {f}}\right ) \text {Si}\left (\frac {\sqrt {c f-d e} b}{\sqrt {f}}+\sqrt {c+d x} b\right )}{f} \]

[Out]

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

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Rubi [A]  time = 0.75, antiderivative size = 238, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3431, 3303, 3299, 3302} \[ \frac {\sin \left (a-\frac {b \sqrt {c f-d e}}{\sqrt {f}}\right ) \text {CosIntegral}\left (\frac {b \sqrt {c f-d e}}{\sqrt {f}}+b \sqrt {c+d x}\right )}{f}+\frac {\sin \left (a+\frac {b \sqrt {c f-d e}}{\sqrt {f}}\right ) \text {CosIntegral}\left (\frac {b \sqrt {c f-d e}}{\sqrt {f}}-b \sqrt {c+d x}\right )}{f}-\frac {\cos \left (a+\frac {b \sqrt {c f-d e}}{\sqrt {f}}\right ) \text {Si}\left (\frac {b \sqrt {c f-d e}}{\sqrt {f}}-b \sqrt {c+d x}\right )}{f}+\frac {\cos \left (a-\frac {b \sqrt {c f-d e}}{\sqrt {f}}\right ) \text {Si}\left (\frac {\sqrt {c f-d e} b}{\sqrt {f}}+\sqrt {c+d x} b\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [C]  time = 1.55, size = 238, normalized size = 1.00 \[ \frac {i e^{-i \left (a+\frac {b \sqrt {c f-d e}}{\sqrt {f}}\right )} \left (-e^{2 i \left (a+\frac {b \sqrt {c f-d e}}{\sqrt {f}}\right )} \text {Ei}\left (i b \left (\sqrt {c+d x}-\frac {\sqrt {c f-d e}}{\sqrt {f}}\right )\right )-e^{2 i a} \text {Ei}\left (i b \left (\frac {\sqrt {c f-d e}}{\sqrt {f}}+\sqrt {c+d x}\right )\right )+\text {Ei}\left (-i b \left (\sqrt {c+d x}-\frac {\sqrt {c f-d e}}{\sqrt {f}}\right )\right )+e^{\frac {2 i b \sqrt {c f-d e}}{\sqrt {f}}} \text {Ei}\left (-i b \left (\frac {\sqrt {c f-d e}}{\sqrt {f}}+\sqrt {c+d x}\right )\right )\right )}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*(ExpIntegralEi[(-I)*b*(-(Sqrt[-(d*e) + c*f]/Sqrt[f]) + Sqrt[c + d*x])] - E^((2*I)*(a + (b*Sqrt[-(d*e) +
c*f])/Sqrt[f]))*ExpIntegralEi[I*b*(-(Sqrt[-(d*e) + c*f]/Sqrt[f]) + Sqrt[c + d*x])] + E^(((2*I)*b*Sqrt[-(d*e)
+ c*f])/Sqrt[f])*ExpIntegralEi[(-I)*b*(Sqrt[-(d*e) + c*f]/Sqrt[f] + Sqrt[c + d*x])] - E^((2*I)*a)*ExpIntegralE
i[I*b*(Sqrt[-(d*e) + c*f]/Sqrt[f] + Sqrt[c + d*x])]))/(E^(I*(a + (b*Sqrt[-(d*e) + c*f])/Sqrt[f]))*f)

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fricas [C]  time = 0.87, size = 250, normalized size = 1.05 \[ \frac {-i \, {\rm Ei}\left (i \, \sqrt {d x + c} b - \sqrt {\frac {b^{2} d e - b^{2} c f}{f}}\right ) e^{\left (i \, a + \sqrt {\frac {b^{2} d e - b^{2} c f}{f}}\right )} - i \, {\rm Ei}\left (i \, \sqrt {d x + c} b + \sqrt {\frac {b^{2} d e - b^{2} c f}{f}}\right ) e^{\left (i \, a - \sqrt {\frac {b^{2} d e - b^{2} c f}{f}}\right )} + i \, {\rm Ei}\left (-i \, \sqrt {d x + c} b - \sqrt {\frac {b^{2} d e - b^{2} c f}{f}}\right ) e^{\left (-i \, a + \sqrt {\frac {b^{2} d e - b^{2} c f}{f}}\right )} + i \, {\rm Ei}\left (-i \, \sqrt {d x + c} b + \sqrt {\frac {b^{2} d e - b^{2} c f}{f}}\right ) e^{\left (-i \, a - \sqrt {\frac {b^{2} d e - b^{2} c f}{f}}\right )}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-I*Ei(I*sqrt(d*x + c)*b - sqrt((b^2*d*e - b^2*c*f)/f))*e^(I*a + sqrt((b^2*d*e - b^2*c*f)/f)) - I*Ei(I*sqr
t(d*x + c)*b + sqrt((b^2*d*e - b^2*c*f)/f))*e^(I*a - sqrt((b^2*d*e - b^2*c*f)/f)) + I*Ei(-I*sqrt(d*x + c)*b -
sqrt((b^2*d*e - b^2*c*f)/f))*e^(-I*a + sqrt((b^2*d*e - b^2*c*f)/f)) + I*Ei(-I*sqrt(d*x + c)*b + sqrt((b^2*d*e
- b^2*c*f)/f))*e^(-I*a - sqrt((b^2*d*e - b^2*c*f)/f)))/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.04, size = 785, normalized size = 3.30 \[ \frac {\frac {\left (a f +\sqrt {b^{2} c \,f^{2}-b^{2} d e f}\right ) b^{2} \left (\Si \left (b \sqrt {d x +c}+a -\frac {a f +\sqrt {b^{2} c \,f^{2}-b^{2} d e f}}{f}\right ) \cos \left (\frac {a f +\sqrt {b^{2} c \,f^{2}-b^{2} d e f}}{f}\right )+\Ci \left (b \sqrt {d x +c}+a -\frac {a f +\sqrt {b^{2} c \,f^{2}-b^{2} d e f}}{f}\right ) \sin \left (\frac {a f +\sqrt {b^{2} c \,f^{2}-b^{2} d e f}}{f}\right )\right )}{f^{2} \left (\frac {a f +\sqrt {b^{2} c \,f^{2}-b^{2} d e f}}{f}-a \right )}-\frac {\left (-a f +\sqrt {b^{2} c \,f^{2}-b^{2} d e f}\right ) b^{2} \left (\Si \left (b \sqrt {d x +c}+a +\frac {-a f +\sqrt {b^{2} c \,f^{2}-b^{2} d e f}}{f}\right ) \cos \left (\frac {-a f +\sqrt {b^{2} c \,f^{2}-b^{2} d e f}}{f}\right )-\Ci \left (b \sqrt {d x +c}+a +\frac {-a f +\sqrt {b^{2} c \,f^{2}-b^{2} d e f}}{f}\right ) \sin \left (\frac {-a f +\sqrt {b^{2} c \,f^{2}-b^{2} d e f}}{f}\right )\right )}{f^{2} \left (-\frac {-a f +\sqrt {b^{2} c \,f^{2}-b^{2} d e f}}{f}-a \right )}-2 a \,b^{2} \left (\frac {\Si \left (b \sqrt {d x +c}+a -\frac {a f +\sqrt {b^{2} c \,f^{2}-b^{2} d e f}}{f}\right ) \cos \left (\frac {a f +\sqrt {b^{2} c \,f^{2}-b^{2} d e f}}{f}\right )+\Ci \left (b \sqrt {d x +c}+a -\frac {a f +\sqrt {b^{2} c \,f^{2}-b^{2} d e f}}{f}\right ) \sin \left (\frac {a f +\sqrt {b^{2} c \,f^{2}-b^{2} d e f}}{f}\right )}{2 \left (\frac {a f +\sqrt {b^{2} c \,f^{2}-b^{2} d e f}}{f}-a \right ) f}+\frac {\Si \left (b \sqrt {d x +c}+a +\frac {-a f +\sqrt {b^{2} c \,f^{2}-b^{2} d e f}}{f}\right ) \cos \left (\frac {-a f +\sqrt {b^{2} c \,f^{2}-b^{2} d e f}}{f}\right )-\Ci \left (b \sqrt {d x +c}+a +\frac {-a f +\sqrt {b^{2} c \,f^{2}-b^{2} d e f}}{f}\right ) \sin \left (\frac {-a f +\sqrt {b^{2} c \,f^{2}-b^{2} d e f}}{f}\right )}{2 \left (-\frac {-a f +\sqrt {b^{2} c \,f^{2}-b^{2} d e f}}{f}-a \right ) f}\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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