3.1.43 \(\int \sqrt {c+d x} \cos (a+b x) \, dx\) [43]

3.1.43.1 Optimal result

Integrand size = 16, antiderivative size = 142 \[ \int \sqrt {c+d x} \cos (a+b x) \, dx=-\frac {\sqrt {d} \sqrt {\frac {\pi }{2}} \cos \left (a-\frac {b c}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{b^{3/2}}-\frac {\sqrt {d} \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right ) \sin \left (a-\frac {b c}{d}\right )}{b^{3/2}}+\frac {\sqrt {c+d x} \sin (a+b x)}{b} \]

output

-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
 

3.1.43.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.04 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.86 \[ \int \sqrt {c+d x} \cos (a+b x) \, dx=\frac {d e^{-\frac {i (b c+a d)}{d}} \left (e^{2 i a} \sqrt {-\frac {i b (c+d x)}{d}} \Gamma \left (\frac {3}{2},-\frac {i b (c+d x)}{d}\right )+e^{\frac {2 i b c}{d}} \sqrt {\frac {i b (c+d x)}{d}} \Gamma \left (\frac {3}{2},\frac {i b (c+d x)}{d}\right )\right )}{2 b^2 \sqrt {c+d x}} \]

input

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

output

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

3.1.43.3 Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {3042, 3777, 25, 3042, 3787, 3042, 3785, 3786, 3832, 3833}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {c+d x} \sin \left (a+b x+\frac {\pi }{2}\right )dx\)

\(\Big \downarrow \) 3777

\(\displaystyle \frac {d \int -\frac {\sin (a+b x)}{\sqrt {c+d x}}dx}{2 b}+\frac {\sqrt {c+d x} \sin (a+b x)}{b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\sqrt {c+d x} \sin (a+b x)}{b}-\frac {d \int \frac {\sin (a+b x)}{\sqrt {c+d x}}dx}{2 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {c+d x} \sin (a+b x)}{b}-\frac {d \int \frac {\sin (a+b x)}{\sqrt {c+d x}}dx}{2 b}\)

\(\Big \downarrow \) 3787

\(\displaystyle \frac {\sqrt {c+d x} \sin (a+b x)}{b}-\frac {d \left (\sin \left (a-\frac {b c}{d}\right ) \int \frac {\cos \left (\frac {b c}{d}+b x\right )}{\sqrt {c+d x}}dx+\cos \left (a-\frac {b c}{d}\right ) \int \frac {\sin \left (\frac {b c}{d}+b x\right )}{\sqrt {c+d x}}dx\right )}{2 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {c+d x} \sin (a+b x)}{b}-\frac {d \left (\sin \left (a-\frac {b c}{d}\right ) \int \frac {\sin \left (\frac {b c}{d}+b x+\frac {\pi }{2}\right )}{\sqrt {c+d x}}dx+\cos \left (a-\frac {b c}{d}\right ) \int \frac {\sin \left (\frac {b c}{d}+b x\right )}{\sqrt {c+d x}}dx\right )}{2 b}\)

\(\Big \downarrow \) 3785

\(\displaystyle \frac {\sqrt {c+d x} \sin (a+b x)}{b}-\frac {d \left (\frac {2 \sin \left (a-\frac {b c}{d}\right ) \int \cos \left (\frac {b (c+d x)}{d}\right )d\sqrt {c+d x}}{d}+\cos \left (a-\frac {b c}{d}\right ) \int \frac {\sin \left (\frac {b c}{d}+b x\right )}{\sqrt {c+d x}}dx\right )}{2 b}\)

\(\Big \downarrow \) 3786

\(\displaystyle \frac {\sqrt {c+d x} \sin (a+b x)}{b}-\frac {d \left (\frac {2 \sin \left (a-\frac {b c}{d}\right ) \int \cos \left (\frac {b (c+d x)}{d}\right )d\sqrt {c+d x}}{d}+\frac {2 \cos \left (a-\frac {b c}{d}\right ) \int \sin \left (\frac {b (c+d x)}{d}\right )d\sqrt {c+d x}}{d}\right )}{2 b}\)

\(\Big \downarrow \) 3832

\(\displaystyle \frac {\sqrt {c+d x} \sin (a+b x)}{b}-\frac {d \left (\frac {2 \sin \left (a-\frac {b c}{d}\right ) \int \cos \left (\frac {b (c+d x)}{d}\right )d\sqrt {c+d x}}{d}+\frac {\sqrt {2 \pi } \cos \left (a-\frac {b c}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{\sqrt {b} \sqrt {d}}\right )}{2 b}\)

\(\Big \downarrow \) 3833

\(\displaystyle \frac {\sqrt {c+d x} \sin (a+b x)}{b}-\frac {d \left (\frac {\sqrt {2 \pi } \sin \left (a-\frac {b c}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{\sqrt {b} \sqrt {d}}+\frac {\sqrt {2 \pi } \cos \left (a-\frac {b c}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{\sqrt {b} \sqrt {d}}\right )}{2 b}\)

input

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

output

-1/2*(d*((Sqrt[2*Pi]*Cos[a - (b*c)/d]*FresnelS[(Sqrt[b]*Sqrt[2/Pi]*Sqrt[c 
+ d*x])/Sqrt[d]])/(Sqrt[b]*Sqrt[d]) + (Sqrt[2*Pi]*FresnelC[(Sqrt[b]*Sqrt[2 
/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[a - (b*c)/d])/(Sqrt[b]*Sqrt[d])))/b + (Sq 
rt[c + d*x]*Sin[a + b*x])/b
 

3.1.43.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3777

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

rule 3785

Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> S 
imp[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 3786

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[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 3787

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Cos 
[(d*e - c*f)/d]   Int[Sin[c*(f/d) + f*x]/Sqrt[c + d*x], x], x] + Simp[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 3832

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ 
d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
 

rule 3833

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ 
d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
 

3.1.43.4 Maple [A] (verified)

Time = 0.70 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.01

method	result	size
derivativedivides	\(\frac {\frac {d \sqrt {d x +c}\, \sin \left (\frac {b \left (d x +c \right )}{d}+\frac {a d -b c}{d}\right )}{b}-\frac {d \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {a d -b c}{d}\right ) \operatorname {S}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {a d -b c}{d}\right ) \operatorname {C}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{2 b \sqrt {\frac {b}{d}}}}{d}\)	\(144\)
default	\(\frac {\frac {d \sqrt {d x +c}\, \sin \left (\frac {b \left (d x +c \right )}{d}+\frac {a d -b c}{d}\right )}{b}-\frac {d \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {a d -b c}{d}\right ) \operatorname {S}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {a d -b c}{d}\right ) \operatorname {C}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{2 b \sqrt {\frac {b}{d}}}}{d}\)	\(144\)

input

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

output

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

3.1.43.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.89 \[ \int \sqrt {c+d x} \cos (a+b x) \, dx=-\frac {\sqrt {2} \pi d \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {b c - a d}{d}\right ) \operatorname {S}\left (\sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) + \sqrt {2} \pi d \sqrt {\frac {b}{\pi d}} \operatorname {C}\left (\sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {b c - a d}{d}\right ) - 2 \, \sqrt {d x + c} b \sin \left (b x + a\right )}{2 \, b^{2}} \]

input

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

output

-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(sq 
rt(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
 

3.1.43.6 Sympy [F]

\[ \int \sqrt {c+d x} \cos (a+b x) \, dx=\int \sqrt {c + d x} \cos {\left (a + b x \right )}\, dx \]

input

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

output

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

3.1.43.7 Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.24 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.38 \[ \int \sqrt {c+d x} \cos (a+b x) \, dx=\frac {\sqrt {2} {\left (4 \, \sqrt {2} \sqrt {d x + c} b \sin \left (\frac {{\left (d x + c\right )} b - b c + a d}{d}\right ) + {\left (-\left (i + 1\right ) \, \sqrt {\pi } d \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {b c - a d}{d}\right ) + \left (i - 1\right ) \, \sqrt {\pi } d \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {b c - a d}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {i \, b}{d}}\right ) + {\left (\left (i - 1\right ) \, \sqrt {\pi } d \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {b c - a d}{d}\right ) - \left (i + 1\right ) \, \sqrt {\pi } d \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {b c - a d}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {i \, b}{d}}\right )\right )}}{8 \, b^{2}} \]

input

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

output

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
 

3.1.43.8 Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.34 (sec) , antiderivative size = 426, normalized size of antiderivative = 3.00 \[ \int \sqrt {c+d x} \cos (a+b x) \, dx=-\frac {-\frac {i \, \sqrt {2} \sqrt {\pi } {\left (2 \, b c + i \, d\right )} d \operatorname {erf}\left (\frac {i \, \sqrt {2} \sqrt {b d} \sqrt {d x + c} {\left (-\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}{2 \, d}\right ) e^{\left (\frac {i \, b c - i \, a d}{d}\right )}}{\sqrt {b d} {\left (-\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )} b} + \frac {i \, \sqrt {2} \sqrt {\pi } {\left (2 \, b c - i \, d\right )} d \operatorname {erf}\left (-\frac {i \, \sqrt {2} \sqrt {b d} \sqrt {d x + c} {\left (\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}{2 \, d}\right ) e^{\left (\frac {-i \, b c + i \, a d}{d}\right )}}{\sqrt {b d} {\left (\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )} b} + 2 \, {\left (\frac {i \, \sqrt {2} \sqrt {\pi } d \operatorname {erf}\left (\frac {i \, \sqrt {2} \sqrt {b d} \sqrt {d x + c} {\left (-\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}{2 \, d}\right ) e^{\left (\frac {i \, b c - i \, a d}{d}\right )}}{\sqrt {b d} {\left (-\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}} - \frac {i \, \sqrt {2} \sqrt {\pi } d \operatorname {erf}\left (-\frac {i \, \sqrt {2} \sqrt {b d} \sqrt {d x + c} {\left (\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}{2 \, d}\right ) e^{\left (\frac {-i \, b c + i \, a d}{d}\right )}}{\sqrt {b d} {\left (\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}\right )} c + \frac {2 i \, \sqrt {d x + c} d e^{\left (\frac {i \, {\left (d x + c\right )} b - i \, b c + i \, a d}{d}\right )}}{b} - \frac {2 i \, \sqrt {d x + c} d e^{\left (\frac {-i \, {\left (d x + c\right )} b + i \, b c - i \, a d}{d}\right )}}{b}}{4 \, d} \]

input

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

output

-1/4*(-I*sqrt(2)*sqrt(pi)*(2*b*c + I*d)*d*erf(1/2*I*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) + I*sqrt(2)*sqrt(pi)*(2*b*c - I*d)*d*erf(-1/2 
*I*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*(I*sqrt(2)*sqrt(p 
i)*d*erf(1/2*I*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)) - I*sqrt(2 
)*sqrt(pi)*d*erf(-1/2*I*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*sqrt(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
 

3.1.43.9 Mupad [F(-1)]

Timed out. \[ \int \sqrt {c+d x} \cos (a+b x) \, dx=\int \cos \left (a+b\,x\right )\,\sqrt {c+d\,x} \,d x \]

input

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

output

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