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

Optimal. Leaf size=142 \[ \frac{\sqrt{\pi } \sqrt{d} \cos \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{\pi } \sqrt{d}}\right )}{8 b^{3/2}}-\frac{\sqrt{\pi } \sqrt{d} \sin \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right )}{8 b^{3/2}}-\frac{\sqrt{c+d x} \cos (2 a+2 b x)}{4 b} \]

[Out]

-(Sqrt[c + d*x]*Cos[2*a + 2*b*x])/(4*b) + (Sqrt[d]*Sqrt[Pi]*Cos[2*a - (2*b*c)/d]*FresnelC[(2*Sqrt[b]*Sqrt[c +
d*x])/(Sqrt[d]*Sqrt[Pi])])/(8*b^(3/2)) - (Sqrt[d]*Sqrt[Pi]*FresnelS[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi
])]*Sin[2*a - (2*b*c)/d])/(8*b^(3/2))

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Rubi [A]  time = 0.22096, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {4406, 12, 3296, 3306, 3305, 3351, 3304, 3352} \[ \frac{\sqrt{\pi } \sqrt{d} \cos \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{\pi } \sqrt{d}}\right )}{8 b^{3/2}}-\frac{\sqrt{\pi } \sqrt{d} \sin \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right )}{8 b^{3/2}}-\frac{\sqrt{c+d x} \cos (2 a+2 b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[c + d*x]*Cos[2*a + 2*b*x])/(4*b) + (Sqrt[d]*Sqrt[Pi]*Cos[2*a - (2*b*c)/d]*FresnelC[(2*Sqrt[b]*Sqrt[c +
d*x])/(Sqrt[d]*Sqrt[Pi])])/(8*b^(3/2)) - (Sqrt[d]*Sqrt[Pi]*FresnelS[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi
])]*Sin[2*a - (2*b*c)/d])/(8*b^(3/2))

Rule 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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} \cos (a+b x) \sin (a+b x) \, dx &=\int \frac{1}{2} \sqrt{c+d x} \sin (2 a+2 b x) \, dx\\ &=\frac{1}{2} \int \sqrt{c+d x} \sin (2 a+2 b x) \, dx\\ &=-\frac{\sqrt{c+d x} \cos (2 a+2 b x)}{4 b}+\frac{d \int \frac{\cos (2 a+2 b x)}{\sqrt{c+d x}} \, dx}{8 b}\\ &=-\frac{\sqrt{c+d x} \cos (2 a+2 b x)}{4 b}+\frac{\left (d \cos \left (2 a-\frac{2 b c}{d}\right )\right ) \int \frac{\cos \left (\frac{2 b c}{d}+2 b x\right )}{\sqrt{c+d x}} \, dx}{8 b}-\frac{\left (d \sin \left (2 a-\frac{2 b c}{d}\right )\right ) \int \frac{\sin \left (\frac{2 b c}{d}+2 b x\right )}{\sqrt{c+d x}} \, dx}{8 b}\\ &=-\frac{\sqrt{c+d x} \cos (2 a+2 b x)}{4 b}+\frac{\cos \left (2 a-\frac{2 b c}{d}\right ) \operatorname{Subst}\left (\int \cos \left (\frac{2 b x^2}{d}\right ) \, dx,x,\sqrt{c+d x}\right )}{4 b}-\frac{\sin \left (2 a-\frac{2 b c}{d}\right ) \operatorname{Subst}\left (\int \sin \left (\frac{2 b x^2}{d}\right ) \, dx,x,\sqrt{c+d x}\right )}{4 b}\\ &=-\frac{\sqrt{c+d x} \cos (2 a+2 b x)}{4 b}+\frac{\sqrt{d} \sqrt{\pi } \cos \left (2 a-\frac{2 b c}{d}\right ) C\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right )}{8 b^{3/2}}-\frac{\sqrt{d} \sqrt{\pi } S\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right ) \sin \left (2 a-\frac{2 b c}{d}\right )}{8 b^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.026048, size = 134, normalized size = 0.94 \[ \frac{\sqrt{\pi } \cos \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{\frac{b}{d}} \sqrt{c+d x}}{\sqrt{\pi }}\right )-\sqrt{\pi } \sin \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{\frac{b}{d}} \sqrt{c+d x}}{\sqrt{\pi }}\right )-2 \sqrt{\frac{b}{d}} \sqrt{c+d x} \cos (2 (a+b x))}{8 b \sqrt{\frac{b}{d}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[b/d]*Sqrt[c + d*x]*Cos[2*(a + b*x)] + Sqrt[Pi]*Cos[2*a - (2*b*c)/d]*FresnelC[(2*Sqrt[b/d]*Sqrt[c + d*
x])/Sqrt[Pi]] - Sqrt[Pi]*FresnelS[(2*Sqrt[b/d]*Sqrt[c + d*x])/Sqrt[Pi]]*Sin[2*a - (2*b*c)/d])/(8*b*Sqrt[b/d])

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Maple [A]  time = 0.023, size = 142, normalized size = 1. \begin{align*} 2\,{\frac{1}{d} \left ( -1/8\,{\frac{d\sqrt{dx+c}}{b}\cos \left ( 2\,{\frac{ \left ( dx+c \right ) b}{d}}+2\,{\frac{ad-bc}{d}} \right ) }+1/16\,{\frac{d\sqrt{\pi }}{b} \left ( \cos \left ( 2\,{\frac{ad-bc}{d}} \right ){\it FresnelC} \left ( 2\,{\frac{\sqrt{dx+c}b}{d\sqrt{\pi }}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) -\sin \left ( 2\,{\frac{ad-bc}{d}} \right ){\it FresnelS} \left ( 2\,{\frac{\sqrt{dx+c}b}{d\sqrt{\pi }}{\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)*cos(b*x+a)*sin(b*x+a),x)

[Out]

2/d*(-1/8/b*d*(d*x+c)^(1/2)*cos(2/d*(d*x+c)*b+2*(a*d-b*c)/d)+1/16/b*d*Pi^(1/2)/(b/d)^(1/2)*(cos(2*(a*d-b*c)/d)
*FresnelC(2/Pi^(1/2)/(b/d)^(1/2)*(d*x+c)^(1/2)*b/d)-sin(2*(a*d-b*c)/d)*FresnelS(2/Pi^(1/2)/(b/d)^(1/2)*(d*x+c)
^(1/2)*b/d)))

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Maxima [C]  time = 1.97827, size = 788, normalized size = 5.55 \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)*cos(b*x+a)*sin(b*x+a),x, algorithm="maxima")

[Out]

-1/64*sqrt(2)*(8*sqrt(2)*sqrt(d*x + c)*d*sqrt(abs(b)/abs(d))*cos(2*((d*x + c)*b - b*c + a*d)/d) - ((sqrt(pi)*c
os(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*
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(p
i)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d*cos(-2*(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*ar
ctan2(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(-2*(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(2
*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*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(-2*(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/sq
rt(d^2))) - sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d*sin(-2*(b*c - a*d)/d))*
erf(sqrt(d*x + c)*sqrt(-2*I*b/d)))/(b*d*sqrt(abs(b)/abs(d)))

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Fricas [A]  time = 0.519941, size = 308, normalized size = 2.17 \begin{align*} \frac{\pi d \sqrt{\frac{b}{\pi d}} \cos \left (-\frac{2 \,{\left (b c - a d\right )}}{d}\right ) \operatorname{C}\left (2 \, \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) - \pi d \sqrt{\frac{b}{\pi d}} \operatorname{S}\left (2 \, \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) \sin \left (-\frac{2 \,{\left (b c - a d\right )}}{d}\right ) - 2 \,{\left (2 \, b \cos \left (b x + a\right )^{2} - b\right )} \sqrt{d x + c}}{8 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(pi*d*sqrt(b/(pi*d))*cos(-2*(b*c - a*d)/d)*fresnel_cos(2*sqrt(d*x + c)*sqrt(b/(pi*d))) - pi*d*sqrt(b/(pi*d
))*fresnel_sin(2*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-2*(b*c - a*d)/d) - 2*(2*b*cos(b*x + a)^2 - b)*sqrt(d*x + c
))/b^2

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Sympy [B]  time = 5.84613, size = 389, normalized size = 2.74 \begin{align*} - \frac{b^{\frac{3}{2}} \sqrt{\frac{d}{b}} \left (c + d x\right )^{\frac{5}{2}} \cos{\left (2 a - \frac{2 b c}{d} \right )} \Gamma \left (\frac{3}{4}\right ) \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{3}\left (\begin{matrix} \frac{3}{4}, \frac{5}{4} \\ \frac{3}{2}, \frac{7}{4}, \frac{9}{4} \end{matrix}\middle |{- \frac{b^{2} \left (c + d x\right )^{2}}{d^{2}}} \right )}}{4 d^{\frac{5}{2}} \Gamma \left (\frac{7}{4}\right ) \Gamma \left (\frac{9}{4}\right )} - \frac{\sqrt{b} \sqrt{\frac{d}{b}} \left (c + d x\right )^{\frac{3}{2}} \sin{\left (2 a - \frac{2 b c}{d} \right )} \Gamma \left (\frac{1}{4}\right ) \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{3}\left (\begin{matrix} \frac{1}{4}, \frac{3}{4} \\ \frac{1}{2}, \frac{5}{4}, \frac{7}{4} \end{matrix}\middle |{- \frac{b^{2} \left (c + d x\right )^{2}}{d^{2}}} \right )}}{8 d^{\frac{3}{2}} \Gamma \left (\frac{5}{4}\right ) \Gamma \left (\frac{7}{4}\right )} + \frac{\sqrt{\pi } c \sqrt{\frac{d}{b}} \sin{\left (2 a - \frac{2 b c}{d} \right )} C\left (\frac{2 b \sqrt{c + d x}}{\sqrt{\pi } d \sqrt{\frac{b}{d}}}\right )}{2 d} + \frac{\sqrt{\pi } c \sqrt{\frac{d}{b}} \cos{\left (2 a - \frac{2 b c}{d} \right )} S\left (\frac{2 b \sqrt{c + d x}}{\sqrt{\pi } d \sqrt{\frac{b}{d}}}\right )}{2 d} + \frac{\sqrt{\pi } x \sqrt{\frac{d}{b}} \sin{\left (2 a - \frac{2 b c}{d} \right )} C\left (\frac{2 b \sqrt{c + d x}}{\sqrt{\pi } d \sqrt{\frac{b}{d}}}\right )}{2} + \frac{\sqrt{\pi } x \sqrt{\frac{d}{b}} \cos{\left (2 a - \frac{2 b c}{d} \right )} S\left (\frac{2 b \sqrt{c + d x}}{\sqrt{\pi } d \sqrt{\frac{b}{d}}}\right )}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b**(3/2)*sqrt(d/b)*(c + d*x)**(5/2)*cos(2*a - 2*b*c/d)*gamma(3/4)*gamma(5/4)*hyper((3/4, 5/4), (3/2, 7/4, 9/4
), -b**2*(c + d*x)**2/d**2)/(4*d**(5/2)*gamma(7/4)*gamma(9/4)) - sqrt(b)*sqrt(d/b)*(c + d*x)**(3/2)*sin(2*a -
2*b*c/d)*gamma(1/4)*gamma(3/4)*hyper((1/4, 3/4), (1/2, 5/4, 7/4), -b**2*(c + d*x)**2/d**2)/(8*d**(3/2)*gamma(5
/4)*gamma(7/4)) + sqrt(pi)*c*sqrt(d/b)*sin(2*a - 2*b*c/d)*fresnelc(2*b*sqrt(c + d*x)/(sqrt(pi)*d*sqrt(b/d)))/(
2*d) + sqrt(pi)*c*sqrt(d/b)*cos(2*a - 2*b*c/d)*fresnels(2*b*sqrt(c + d*x)/(sqrt(pi)*d*sqrt(b/d)))/(2*d) + sqrt
(pi)*x*sqrt(d/b)*sin(2*a - 2*b*c/d)*fresnelc(2*b*sqrt(c + d*x)/(sqrt(pi)*d*sqrt(b/d)))/2 + sqrt(pi)*x*sqrt(d/b
)*cos(2*a - 2*b*c/d)*fresnels(2*b*sqrt(c + d*x)/(sqrt(pi)*d*sqrt(b/d)))/2

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Giac [C]  time = 1.15843, size = 316, normalized size = 2.23 \begin{align*} -\frac{\frac{\sqrt{\pi } d^{2} \operatorname{erf}\left (-\frac{\sqrt{b d} \sqrt{d x + c}{\left (\frac{i \, b d}{\sqrt{b^{2} d^{2}}} + 1\right )}}{d}\right ) e^{\left (\frac{2 i \, b c - 2 i \, a d}{d}\right )}}{\sqrt{b d}{\left (\frac{i \, b d}{\sqrt{b^{2} d^{2}}} + 1\right )} b} + \frac{\sqrt{\pi } d^{2} \operatorname{erf}\left (-\frac{\sqrt{b d} \sqrt{d x + c}{\left (-\frac{i \, b d}{\sqrt{b^{2} d^{2}}} + 1\right )}}{d}\right ) e^{\left (\frac{-2 i \, b c + 2 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{2 i \,{\left (d x + c\right )} b - 2 i \, b c + 2 i \, a d}{d}\right )}}{b} + \frac{2 \, \sqrt{d x + c} d e^{\left (\frac{-2 i \,{\left (d x + c\right )} b + 2 i \, b c - 2 i \, a d}{d}\right )}}{b}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(sqrt(pi)*d^2*erf(-sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((2*I*b*c - 2*I*a*d)/d)/(sqrt(
b*d)*(I*b*d/sqrt(b^2*d^2) + 1)*b) + sqrt(pi)*d^2*erf(-sqrt(b*d)*sqrt(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^
((-2*I*b*c + 2*I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)*b) + 2*sqrt(d*x + c)*d*e^((2*I*(d*x + c)*b - 2*
I*b*c + 2*I*a*d)/d)/b + 2*sqrt(d*x + c)*d*e^((-2*I*(d*x + c)*b + 2*I*b*c - 2*I*a*d)/d)/b)/d