3.40 \(\int \sqrt{c+d x} \sin (a+b x) \, dx\)

Optimal. Leaf size=142 \[ \frac{\sqrt{\frac{\pi }{2}} \sqrt{d} \cos \left (a-\frac{b c}{d}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{b^{3/2}}-\frac{\sqrt{\frac{\pi }{2}} \sqrt{d} \sin \left (a-\frac{b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{b^{3/2}}-\frac{\sqrt{c+d x} \cos (a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.175986, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {3296, 3306, 3305, 3351, 3304, 3352} \[ \frac{\sqrt{\frac{\pi }{2}} \sqrt{d} \cos \left (a-\frac{b c}{d}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{b^{3/2}}-\frac{\sqrt{\frac{\pi }{2}} \sqrt{d} \sin \left (a-\frac{b c}{d}\right ) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{b^{3/2}}-\frac{\sqrt{c+d x} \cos (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x]*Sin[a + b*x],x]

[Out]

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

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

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

Mathematica [C]  time = 0.0938248, size = 123, normalized size = 0.87 \[ \frac{\sqrt{c+d x} e^{-\frac{i (a d+b c)}{d}} \left (-\frac{e^{2 i a} \text{Gamma}\left (\frac{3}{2},-\frac{i b (c+d x)}{d}\right )}{\sqrt{-\frac{i b (c+d x)}{d}}}-\frac{e^{\frac{2 i b c}{d}} \text{Gamma}\left (\frac{3}{2},\frac{i b (c+d x)}{d}\right )}{\sqrt{\frac{i b (c+d x)}{d}}}\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c + d*x]*Sin[a + b*x],x]

[Out]

(Sqrt[c + d*x]*(-((E^((2*I)*a)*Gamma[3/2, ((-I)*b*(c + d*x))/d])/Sqrt[((-I)*b*(c + d*x))/d]) - (E^(((2*I)*b*c)
/d)*Gamma[3/2, (I*b*(c + d*x))/d])/Sqrt[(I*b*(c + d*x))/d]))/(2*b*E^((I*(b*c + a*d))/d))

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Maple [A]  time = 0.007, size = 145, normalized size = 1. \begin{align*} 2\,{\frac{1}{d} \left ( -1/2\,{\frac{d\sqrt{dx+c}}{b}\cos \left ({\frac{ \left ( dx+c \right ) b}{d}}+{\frac{da-cb}{d}} \right ) }+1/4\,{\frac{d\sqrt{2}\sqrt{\pi }}{b} \left ( \cos \left ({\frac{da-cb}{d}} \right ){\it FresnelC} \left ({\frac{\sqrt{2}\sqrt{dx+c}b}{\sqrt{\pi }d}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) -\sin \left ({\frac{da-cb}{d}} \right ){\it FresnelS} \left ({\frac{\sqrt{2}\sqrt{dx+c}b}{\sqrt{\pi }d}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) \right ){\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*(-1/2/b*d*(d*x+c)^(1/2)*cos(1/d*(d*x+c)*b+(a*d-b*c)/d)+1/4/b*d*2^(1/2)*Pi^(1/2)/(b/d)^(1/2)*(cos((a*d-b*c)
/d)*FresnelC(2^(1/2)/Pi^(1/2)/(b/d)^(1/2)*(d*x+c)^(1/2)*b/d)-sin((a*d-b*c)/d)*FresnelS(2^(1/2)/Pi^(1/2)/(b/d)^
(1/2)*(d*x+c)^(1/2)*b/d)))

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Maxima [C]  time = 1.83438, size = 779, normalized size = 5.49 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(8*sqrt(d*x + c)*d*sqrt(abs(b)/abs(d))*cos(((d*x + c)*b - b*c + a*d)/d) - ((sqrt(pi)*cos(1/4*pi + 1/2*arc
tan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d
^2))) - I*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + I*sqrt(pi)*sin(-1/4*pi + 1/
2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d*cos(-(b*c - a*d)/d) - (I*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0
, b) + 1/2*arctan2(0, d/sqrt(d^2))) + I*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))
) + sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - sqrt(pi)*sin(-1/4*pi + 1/2*arctan
2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d*sin(-(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(I*b/d)) - ((sqrt(pi)*cos
(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*ar
ctan2(0, d/sqrt(d^2))) + I*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - I*sqrt(pi)
*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d*cos(-(b*c - a*d)/d) - (-I*sqrt(pi)*cos(1/4*
pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - I*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arcta
n2(0, d/sqrt(d^2))) + sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - sqrt(pi)*sin(-1
/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d*sin(-(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(-I*b/d
)))/(b*d*sqrt(abs(b)/abs(d)))

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Fricas [A]  time = 1.76481, size = 327, normalized size = 2.3 \begin{align*} \frac{\sqrt{2} \pi d \sqrt{\frac{b}{\pi d}} \cos \left (-\frac{b c - a d}{d}\right ) \operatorname{C}\left (\sqrt{2} \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) - \sqrt{2} \pi d \sqrt{\frac{b}{\pi d}} \operatorname{S}\left (\sqrt{2} \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) \sin \left (-\frac{b c - a d}{d}\right ) - 2 \, \sqrt{d x + c} b \cos \left (b x + a\right )}{2 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(sqrt(2)*pi*d*sqrt(b/(pi*d))*cos(-(b*c - a*d)/d)*fresnel_cos(sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d))) - sqrt(
2)*pi*d*sqrt(b/(pi*d))*fresnel_sin(sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-(b*c - a*d)/d) - 2*sqrt(d*x + c)
*b*cos(b*x + a))/b^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C]  time = 1.18217, size = 332, normalized size = 2.34 \begin{align*} -\frac{\frac{\sqrt{2} \sqrt{\pi } d^{2} \operatorname{erf}\left (-\frac{\sqrt{2} \sqrt{b d} \sqrt{d x + c}{\left (\frac{i \, b d}{\sqrt{b^{2} d^{2}}} + 1\right )}}{2 \, d}\right ) e^{\left (\frac{i \, b c - i \, a d}{d}\right )}}{\sqrt{b d}{\left (\frac{i \, b d}{\sqrt{b^{2} d^{2}}} + 1\right )} b} + \frac{\sqrt{2} \sqrt{\pi } d^{2} \operatorname{erf}\left (-\frac{\sqrt{2} \sqrt{b d} \sqrt{d x + c}{\left (-\frac{i \, b d}{\sqrt{b^{2} d^{2}}} + 1\right )}}{2 \, d}\right ) e^{\left (\frac{-i \, b c + i \, a d}{d}\right )}}{\sqrt{b d}{\left (-\frac{i \, b d}{\sqrt{b^{2} d^{2}}} + 1\right )} b} + \frac{2 \, \sqrt{d x + c} d e^{\left (\frac{i \,{\left (d x + c\right )} b - i \, b c + i \, a d}{d}\right )}}{b} + \frac{2 \, \sqrt{d x + c} d e^{\left (\frac{-i \,{\left (d x + c\right )} b + i \, b c - i \, a d}{d}\right )}}{b}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(sqrt(2)*sqrt(pi)*d^2*erf(-1/2*sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((I*b*c - I
*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)*b) + sqrt(2)*sqrt(pi)*d^2*erf(-1/2*sqrt(2)*sqrt(b*d)*sqrt(d*x +
c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((-I*b*c + I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)*b) + 2*sqrt(d*x
+ c)*d*e^((I*(d*x + c)*b - I*b*c + I*a*d)/d)/b + 2*sqrt(d*x + c)*d*e^((-I*(d*x + c)*b + I*b*c - I*a*d)/d)/b)/d