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

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

[Out]

-1/2*cos(a-b*c/d)*FresnelS(b^(1/2)*2^(1/2)/Pi^(1/2)*(d*x+c)^(1/2)/d^(1/2))*d^(1/2)*2^(1/2)*Pi^(1/2)/b^(3/2)-1/
2*FresnelC(b^(1/2)*2^(1/2)/Pi^(1/2)*(d*x+c)^(1/2)/d^(1/2))*sin(a-b*c/d)*d^(1/2)*2^(1/2)*Pi^(1/2)/b^(3/2)+sin(b
*x+a)*(d*x+c)^(1/2)/b

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Rubi [A]  time = 0.17, antiderivative size = 142, normalized size of antiderivative = 1.00, 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} \[ -\frac {\sqrt {\frac {\pi }{2}} \sqrt {d} \sin \left (a-\frac {b c}{d}\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{b^{3/2}}-\frac {\sqrt {\frac {\pi }{2}} \sqrt {d} \cos \left (a-\frac {b c}{d}\right ) S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{b^{3/2}}+\frac {\sqrt {c+d x} \sin (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x]*Cos[a + b*x],x]

[Out]

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

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

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Mathematica [C]  time = 0.09, size = 124, normalized size = 0.87 \[ -\frac {i \sqrt {c+d x} e^{-\frac {i (a d+b c)}{d}} \left (\frac {e^{2 i a} \Gamma \left (\frac {3}{2},-\frac {i b (c+d x)}{d}\right )}{\sqrt {-\frac {i b (c+d x)}{d}}}-\frac {e^{\frac {2 i b c}{d}} \Gamma \left (\frac {3}{2},\frac {i b (c+d x)}{d}\right )}{\sqrt {\frac {i b (c+d x)}{d}}}\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/2*I)*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]))/(b*E^((I*(b*c + a*d))/d))

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fricas [A]  time = 0.66, size = 126, normalized size = 0.89 \[ -\frac {\sqrt {2} \pi d \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {b c - a d}{d}\right ) \operatorname {S}\left (\sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) + \sqrt {2} \pi d \sqrt {\frac {b}{\pi d}} \operatorname {C}\left (\sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {b c - a d}{d}\right ) - 2 \, \sqrt {d x + c} b \sin \left (b x + a\right )}{2 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [C]  time = 0.51, size = 422, normalized size = 2.97 \[ \frac {\frac {\sqrt {2} \sqrt {\pi } {\left (2 \, b c + i \, d\right )} d \operatorname {erf}\left (-\frac {\sqrt {2} \sqrt {b d} \sqrt {d x + c} {\left (\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}{2 \, d}\right ) e^{\left (\frac {i \, b c - i \, a d}{d}\right )}}{\sqrt {b d} {\left (\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )} b} + \frac {\sqrt {2} \sqrt {\pi } {\left (2 \, b c - i \, d\right )} d \operatorname {erf}\left (-\frac {\sqrt {2} \sqrt {b d} \sqrt {d x + c} {\left (-\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}{2 \, d}\right ) e^{\left (\frac {-i \, b c + i \, a d}{d}\right )}}{\sqrt {b d} {\left (-\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )} b} - 2 \, {\left (\frac {\sqrt {2} \sqrt {\pi } d \operatorname {erf}\left (-\frac {\sqrt {2} \sqrt {b d} \sqrt {d x + c} {\left (\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}{2 \, d}\right ) e^{\left (\frac {i \, b c - i \, a d}{d}\right )}}{\sqrt {b d} {\left (\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}} + \frac {\sqrt {2} \sqrt {\pi } d \operatorname {erf}\left (-\frac {\sqrt {2} \sqrt {b d} \sqrt {d x + c} {\left (-\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}{2 \, d}\right ) e^{\left (\frac {-i \, b c + i \, a d}{d}\right )}}{\sqrt {b d} {\left (-\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}\right )} c - \frac {2 i \, \sqrt {d x + c} d e^{\left (\frac {i \, {\left (d x + c\right )} b - i \, b c + i \, a d}{d}\right )}}{b} + \frac {2 i \, \sqrt {d x + c} d e^{\left (\frac {-i \, {\left (d x + c\right )} b + i \, b c - i \, a d}{d}\right )}}{b}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 144, normalized size = 1.01 \[ \frac {\frac {d \sqrt {d x +c}\, \sin \left (\frac {\left (d x +c \right ) b}{d}+\frac {d a -c b}{d}\right )}{b}-\frac {d \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {d a -c b}{d}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {d a -c b}{d}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{2 b \sqrt {\frac {b}{d}}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 1.07, size = 196, normalized size = 1.38 \[ \frac {\sqrt {2} {\left (4 \, \sqrt {2} \sqrt {d x + c} b \sin \left (\frac {{\left (d x + c\right )} b - b c + a d}{d}\right ) + {\left (-\left (i + 1\right ) \, \sqrt {\pi } d \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {b c - a d}{d}\right ) + \left (i - 1\right ) \, \sqrt {\pi } d \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {b c - a d}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {i \, b}{d}}\right ) + {\left (\left (i - 1\right ) \, \sqrt {\pi } d \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {b c - a d}{d}\right ) - \left (i + 1\right ) \, \sqrt {\pi } d \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {b c - a d}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {i \, b}{d}}\right )\right )}}{8 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(2)*(4*sqrt(2)*sqrt(d*x + c)*b*sin(((d*x + c)*b - b*c + a*d)/d) + (-(I + 1)*sqrt(pi)*d*(b^2/d^2)^(1/4)
*cos(-(b*c - a*d)/d) + (I - 1)*sqrt(pi)*d*(b^2/d^2)^(1/4)*sin(-(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(I*b/d))
+ ((I - 1)*sqrt(pi)*d*(b^2/d^2)^(1/4)*cos(-(b*c - a*d)/d) - (I + 1)*sqrt(pi)*d*(b^2/d^2)^(1/4)*sin(-(b*c - a*d
)/d))*erf(sqrt(d*x + c)*sqrt(-I*b/d)))/b^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \cos \left (a+b\,x\right )\,\sqrt {c+d\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)*(c + d*x)^(1/2),x)

[Out]

int(cos(a + b*x)*(c + d*x)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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