\(\int \frac {\cos (a+b x)}{(c+d x)^{3/2}} \, dx\) [45]

Optimal result

Integrand size = 16, antiderivative size = 139 \[ \int \frac {\cos (a+b x)}{(c+d x)^{3/2}} \, dx=-\frac {2 \cos (a+b x)}{d \sqrt {c+d x}}-\frac {2 \sqrt {b} \sqrt {2 \pi } \cos \left (a-\frac {b c}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{d^{3/2}}-\frac {2 \sqrt {b} \sqrt {2 \pi } \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right ) \sin \left (a-\frac {b c}{d}\right )}{d^{3/2}} \] Output:

-2*cos(b*x+a)/d/(d*x+c)^(1/2)-2*b^(1/2)*2^(1/2)*Pi^(1/2)*cos(a-b*c/d)*Fres 
nelS(b^(1/2)*2^(1/2)/Pi^(1/2)*(d*x+c)^(1/2)/d^(1/2))/d^(3/2)-2*b^(1/2)*2^( 
1/2)*Pi^(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^(3/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.36 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.06 \[ \int \frac {\cos (a+b x)}{(c+d x)^{3/2}} \, dx=\frac {e^{-i a} \left (e^{2 i a-\frac {i b c}{d}} \sqrt {-\frac {i b (c+d x)}{d}} \Gamma \left (\frac {1}{2},-\frac {i b (c+d x)}{d}\right )+e^{-i b x} \left (-1-e^{2 i (a+b x)}+e^{\frac {i b (c+d x)}{d}} \sqrt {\frac {i b (c+d x)}{d}} \Gamma \left (\frac {1}{2},\frac {i b (c+d x)}{d}\right )\right )\right )}{d \sqrt {c+d x}} \] Input:

Integrate[Cos[a + b*x]/(c + d*x)^(3/2),x]
 

Output:

(E^((2*I)*a - (I*b*c)/d)*Sqrt[((-I)*b*(c + d*x))/d]*Gamma[1/2, ((-I)*b*(c 
+ d*x))/d] + (-1 - E^((2*I)*(a + b*x)) + E^((I*b*(c + d*x))/d)*Sqrt[(I*b*( 
c + d*x))/d]*Gamma[1/2, (I*b*(c + d*x))/d])/E^(I*b*x))/(d*E^(I*a)*Sqrt[c + 
 d*x])
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.04, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {3042, 3778, 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.

\(\displaystyle \int \frac {\cos (a+b x)}{(c+d x)^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin \left (a+b x+\frac {\pi }{2}\right )}{(c+d x)^{3/2}}dx\)

\(\Big \downarrow \) 3778

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3787

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3785

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

\(\Big \downarrow \) 3786

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

\(\Big \downarrow \) 3832

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

\(\Big \downarrow \) 3833

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

Input:

Int[Cos[a + b*x]/(c + d*x)^(3/2),x]
 

Output:

(-2*Cos[a + b*x])/(d*Sqrt[c + d*x]) - (2*b*((Sqrt[2*Pi]*Cos[a - (b*c)/d]*F 
resnelS[(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])))/d
 

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 3778

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

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]
 

Maple [A] (verified)

Time = 1.18 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.01

method	result	size
derivativedivides	\(\frac {-\frac {2 \cos \left (\frac {b \left (d x +c \right )}{d}+\frac {a d -b c}{d}\right )}{\sqrt {d x +c}}-\frac {2 b \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {a d -b c}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {a d -b c}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{d \sqrt {\frac {b}{d}}}}{d}\)	\(140\)
default	\(\frac {-\frac {2 \cos \left (\frac {b \left (d x +c \right )}{d}+\frac {a d -b c}{d}\right )}{\sqrt {d x +c}}-\frac {2 b \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {a d -b c}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {a d -b c}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{d \sqrt {\frac {b}{d}}}}{d}\)	\(140\)

Input:

int(cos(b*x+a)/(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.04 \[ \int \frac {\cos (a+b x)}{(c+d x)^{3/2}} \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (\pi d x + \pi c\right )} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {b c - a d}{d}\right ) \operatorname {S}\left (\sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) + \sqrt {2} {\left (\pi d x + \pi c\right )} \sqrt {\frac {b}{\pi d}} \operatorname {C}\left (\sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {b c - a d}{d}\right ) + \sqrt {d x + c} \cos \left (b x + a\right )\right )}}{d^{2} x + c d} \] Input:

integrate(cos(b*x+a)/(d*x+c)^(3/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {\cos (a+b x)}{(c+d x)^{3/2}} \, dx=\int \frac {\cos {\left (a + b x \right )}}{\left (c + d x\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(cos(b*x+a)/(d*x+c)**(3/2),x)
 

Output:

Integral(cos(a + b*x)/(c + d*x)**(3/2), x)
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.21 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.93 \[ \int \frac {\cos (a+b x)}{(c+d x)^{3/2}} \, dx=\frac {{\left ({\left (-\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, \frac {i \, {\left (d x + c\right )} b}{d}\right ) + \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -\frac {i \, {\left (d x + c\right )} b}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) + {\left (\left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, \frac {i \, {\left (d x + c\right )} b}{d}\right ) - \left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -\frac {i \, {\left (d x + c\right )} b}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right )\right )} \sqrt {\frac {{\left (d x + c\right )} b}{d}}}{4 \, \sqrt {d x + c} d} \] Input:

integrate(cos(b*x+a)/(d*x+c)^(3/2),x, algorithm="maxima")
 

Output:

1/4*((-(I + 1)*sqrt(2)*gamma(-1/2, I*(d*x + c)*b/d) + (I - 1)*sqrt(2)*gamm 
a(-1/2, -I*(d*x + c)*b/d))*cos(-(b*c - a*d)/d) + ((I - 1)*sqrt(2)*gamma(-1 
/2, I*(d*x + c)*b/d) - (I + 1)*sqrt(2)*gamma(-1/2, -I*(d*x + c)*b/d))*sin( 
-(b*c - a*d)/d))*sqrt((d*x + c)*b/d)/(sqrt(d*x + c)*d)
 

Giac [F]

\[ \int \frac {\cos (a+b x)}{(c+d x)^{3/2}} \, dx=\int { \frac {\cos \left (b x + a\right )}{{\left (d x + c\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(cos(b*x+a)/(d*x+c)^(3/2),x, algorithm="giac")
 

Output:

integrate(cos(b*x + a)/(d*x + c)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos (a+b x)}{(c+d x)^{3/2}} \, dx=\int \frac {\cos \left (a+b\,x\right )}{{\left (c+d\,x\right )}^{3/2}} \,d x \] Input:

int(cos(a + b*x)/(c + d*x)^(3/2),x)
 

Output:

int(cos(a + b*x)/(c + d*x)^(3/2), x)
 

Reduce [F]

\[ \int \frac {\cos (a+b x)}{(c+d x)^{3/2}} \, dx=\int \frac {\cos \left (b x +a \right )}{\sqrt {d x +c}\, c +\sqrt {d x +c}\, d x}d x \] Input:

int(cos(b*x+a)/(d*x+c)^(3/2),x)
 

Output:

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