\(\int (a \cos (c+d x)+b \sin (c+d x)) \, dx\) [35]

Optimal result

Integrand size = 17, antiderivative size = 24 \[ \int (a \cos (c+d x)+b \sin (c+d x)) \, dx=-\frac {b \cos (c+d x)}{d}+\frac {a \sin (c+d x)}{d} \]

[Out]

-b*cos(d*x+c)/d+a*sin(d*x+c)/d

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2717, 2718} \[ \int (a \cos (c+d x)+b \sin (c+d x)) \, dx=\frac {a \sin (c+d x)}{d}-\frac {b \cos (c+d x)}{d} \]

[In]

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

[Out]

-((b*Cos[c + d*x])/d) + (a*Sin[c + d*x])/d

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = a \int \cos (c+d x) \, dx+b \int \sin (c+d x) \, dx \\ & = -\frac {b \cos (c+d x)}{d}+\frac {a \sin (c+d x)}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.92 \[ \int (a \cos (c+d x)+b \sin (c+d x)) \, dx=-\frac {b \cos (c) \cos (d x)}{d}+\frac {a \cos (d x) \sin (c)}{d}+\frac {a \cos (c) \sin (d x)}{d}+\frac {b \sin (c) \sin (d x)}{d} \]

[In]

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

[Out]

-((b*Cos[c]*Cos[d*x])/d) + (a*Cos[d*x]*Sin[c])/d + (a*Cos[c]*Sin[d*x])/d + (b*Sin[c]*Sin[d*x])/d

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(\frac {-\cos \left (d x +c \right ) b +\sin \left (d x +c \right ) a}{d}\)	\(23\)
parallelrisch	\(\frac {b -\cos \left (d x +c \right ) b +\sin \left (d x +c \right ) a}{d}\)	\(24\)
default	\(-\frac {b \cos \left (d x +c \right )}{d}+\frac {a \sin \left (d x +c \right )}{d}\)	\(25\)
risch	\(-\frac {b \cos \left (d x +c \right )}{d}+\frac {a \sin \left (d x +c \right )}{d}\)	\(25\)
parts	\(-\frac {b \cos \left (d x +c \right )}{d}+\frac {a \sin \left (d x +c \right )}{d}\)	\(25\)
norman	\(\frac {\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\)	\(50\)
meijerg	\(\frac {\left (\sqrt {\pi }\, \cos \left (c \right ) a +\sqrt {\pi }\, \sin \left (c \right ) b \right ) \sin \left (d x \right )}{\sqrt {\pi }\, d}+\frac {\left (\sqrt {\pi }\, \cos \left (c \right ) b -\sqrt {\pi }\, \sin \left (c \right ) a \right ) \left (\frac {1}{\sqrt {\pi }}-\frac {\cos \left (d x \right )}{\sqrt {\pi }}\right )}{d}\)	\(61\)

[In]

int(cos(d*x+c)*a+b*sin(d*x+c),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int (a \cos (c+d x)+b \sin (c+d x)) \, dx=-\frac {b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )}{d} \]

[In]

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

[Out]

-(b*cos(d*x + c) - a*sin(d*x + c))/d

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int (a \cos (c+d x)+b \sin (c+d x)) \, dx=a \left (\begin {cases} \frac {\sin {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \cos {\left (c \right )} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} - \frac {\cos {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \sin {\left (c \right )} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

a*Piecewise((sin(c + d*x)/d, Ne(d, 0)), (x*cos(c), True)) + b*Piecewise((-cos(c + d*x)/d, Ne(d, 0)), (x*sin(c)
, True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int (a \cos (c+d x)+b \sin (c+d x)) \, dx=-\frac {b \cos \left (d x + c\right )}{d} + \frac {a \sin \left (d x + c\right )}{d} \]

[In]

integrate(a*cos(d*x+c)+b*sin(d*x+c),x, algorithm="maxima")

[Out]

-b*cos(d*x + c)/d + a*sin(d*x + c)/d

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int (a \cos (c+d x)+b \sin (c+d x)) \, dx=-\frac {b \cos \left (d x + c\right )}{d} + \frac {a \sin \left (d x + c\right )}{d} \]

[In]

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

[Out]

-b*cos(d*x + c)/d + a*sin(d*x + c)/d

Mupad [B] (verification not implemented)

Time = 20.79 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int (a \cos (c+d x)+b \sin (c+d x)) \, dx=-\frac {2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \]

[In]

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

[Out]

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