3.244 \(\int \cos ^2(a+b x) \cos ^3(c+d x) \, dx\)

Optimal. Leaf size=144 \[ \frac{\sin (2 a+x (2 b-3 d)-3 c)}{16 (2 b-3 d)}+\frac{3 \sin (2 a+x (2 b-d)-c)}{16 (2 b-d)}+\frac{3 \sin (2 a+x (2 b+d)+c)}{16 (2 b+d)}+\frac{\sin (2 a+x (2 b+3 d)+3 c)}{16 (2 b+3 d)}+\frac{3 \sin (c+d x)}{8 d}+\frac{\sin (3 c+3 d x)}{24 d} \]

[Out]

Sin[2*a - 3*c + (2*b - 3*d)*x]/(16*(2*b - 3*d)) + (3*Sin[2*a - c + (2*b - d)*x])/(16*(2*b - d)) + (3*Sin[c + d
*x])/(8*d) + Sin[3*c + 3*d*x]/(24*d) + (3*Sin[2*a + c + (2*b + d)*x])/(16*(2*b + d)) + Sin[2*a + 3*c + (2*b +
3*d)*x]/(16*(2*b + 3*d))

________________________________________________________________________________________

Rubi [A]  time = 0.0887541, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {4570, 2637} \[ \frac{\sin (2 a+x (2 b-3 d)-3 c)}{16 (2 b-3 d)}+\frac{3 \sin (2 a+x (2 b-d)-c)}{16 (2 b-d)}+\frac{3 \sin (2 a+x (2 b+d)+c)}{16 (2 b+d)}+\frac{\sin (2 a+x (2 b+3 d)+3 c)}{16 (2 b+3 d)}+\frac{3 \sin (c+d x)}{8 d}+\frac{\sin (3 c+3 d x)}{24 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^2*Cos[c + d*x]^3,x]

[Out]

Sin[2*a - 3*c + (2*b - 3*d)*x]/(16*(2*b - 3*d)) + (3*Sin[2*a - c + (2*b - d)*x])/(16*(2*b - d)) + (3*Sin[c + d
*x])/(8*d) + Sin[3*c + 3*d*x]/(24*d) + (3*Sin[2*a + c + (2*b + d)*x])/(16*(2*b + d)) + Sin[2*a + 3*c + (2*b +
3*d)*x]/(16*(2*b + 3*d))

Rule 4570

Int[Cos[v_]^(p_.)*Cos[w_]^(q_.), x_Symbol] :> Int[ExpandTrigReduce[Cos[v]^p*Cos[w]^q, x], x] /; ((PolynomialQ[
v, x] && PolynomialQ[w, x]) || (BinomialQ[{v, w}, x] && IndependentQ[Cancel[v/w], x])) && IGtQ[p, 0] && IGtQ[q
, 0]

Rule 2637

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

Rubi steps

\begin{align*} \int \cos ^2(a+b x) \cos ^3(c+d x) \, dx &=\int \left (\frac{1}{16} \cos (2 a-3 c+(2 b-3 d) x)+\frac{3}{16} \cos (2 a-c+(2 b-d) x)+\frac{3}{8} \cos (c+d x)+\frac{1}{8} \cos (3 c+3 d x)+\frac{3}{16} \cos (2 a+c+(2 b+d) x)+\frac{1}{16} \cos (2 a+3 c+(2 b+3 d) x)\right ) \, dx\\ &=\frac{1}{16} \int \cos (2 a-3 c+(2 b-3 d) x) \, dx+\frac{1}{16} \int \cos (2 a+3 c+(2 b+3 d) x) \, dx+\frac{1}{8} \int \cos (3 c+3 d x) \, dx+\frac{3}{16} \int \cos (2 a-c+(2 b-d) x) \, dx+\frac{3}{16} \int \cos (2 a+c+(2 b+d) x) \, dx+\frac{3}{8} \int \cos (c+d x) \, dx\\ &=\frac{\sin (2 a-3 c+(2 b-3 d) x)}{16 (2 b-3 d)}+\frac{3 \sin (2 a-c+(2 b-d) x)}{16 (2 b-d)}+\frac{3 \sin (c+d x)}{8 d}+\frac{\sin (3 c+3 d x)}{24 d}+\frac{3 \sin (2 a+c+(2 b+d) x)}{16 (2 b+d)}+\frac{\sin (2 a+3 c+(2 b+3 d) x)}{16 (2 b+3 d)}\\ \end{align*}

Mathematica [A]  time = 1.63928, size = 158, normalized size = 1.1 \[ \frac{1}{48} \left (\frac{3 \sin (2 a+2 b x-3 c-3 d x)}{2 b-3 d}+\frac{9 \sin (2 a+2 b x-c-d x)}{2 b-d}+\frac{9 \sin (2 a+2 b x+c+d x)}{2 b+d}+\frac{3 \sin (2 a+2 b x+3 c+3 d x)}{2 b+3 d}+\frac{18 \sin (c) \cos (d x)}{d}+\frac{2 \sin (3 c) \cos (3 d x)}{d}+\frac{18 \cos (c) \sin (d x)}{d}+\frac{2 \cos (3 c) \sin (3 d x)}{d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((18*Cos[d*x]*Sin[c])/d + (2*Cos[3*d*x]*Sin[3*c])/d + (18*Cos[c]*Sin[d*x])/d + (2*Cos[3*c]*Sin[3*d*x])/d + (3*
Sin[2*a - 3*c + 2*b*x - 3*d*x])/(2*b - 3*d) + (9*Sin[2*a - c + 2*b*x - d*x])/(2*b - d) + (9*Sin[2*a + c + 2*b*
x + d*x])/(2*b + d) + (3*Sin[2*a + 3*c + 2*b*x + 3*d*x])/(2*b + 3*d))/48

________________________________________________________________________________________

Maple [A]  time = 0.036, size = 133, normalized size = 0.9 \begin{align*}{\frac{\sin \left ( 2\,a-3\,c+ \left ( 2\,b-3\,d \right ) x \right ) }{32\,b-48\,d}}+{\frac{3\,\sin \left ( 2\,a-c+ \left ( 2\,b-d \right ) x \right ) }{32\,b-16\,d}}+{\frac{3\,\sin \left ( dx+c \right ) }{8\,d}}+{\frac{\sin \left ( 3\,dx+3\,c \right ) }{24\,d}}+{\frac{3\,\sin \left ( 2\,a+c+ \left ( 2\,b+d \right ) x \right ) }{32\,b+16\,d}}+{\frac{\sin \left ( 2\,a+3\,c+ \left ( 2\,b+3\,d \right ) x \right ) }{32\,b+48\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*sin(2*a-3*c+(2*b-3*d)*x)/(2*b-3*d)+3/16*sin(2*a-c+(2*b-d)*x)/(2*b-d)+3/8*sin(d*x+c)/d+1/24*sin(3*d*x+3*c)
/d+3/16*sin(2*a+c+(2*b+d)*x)/(2*b+d)+1/16*sin(2*a+3*c+(2*b+3*d)*x)/(2*b+3*d)

________________________________________________________________________________________

Maxima [B]  time = 1.57257, size = 1839, normalized size = 12.77 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(3*(8*b^3*d*sin(3*c) - 12*b^2*d^2*sin(3*c) - 2*b*d^3*sin(3*c) + 3*d^4*sin(3*c))*cos((2*b + 3*d)*x + 2*a
+ 6*c) - 3*(8*b^3*d*sin(3*c) - 12*b^2*d^2*sin(3*c) - 2*b*d^3*sin(3*c) + 3*d^4*sin(3*c))*cos((2*b + 3*d)*x + 2*
a) + 9*(8*b^3*d*sin(3*c) - 4*b^2*d^2*sin(3*c) - 18*b*d^3*sin(3*c) + 9*d^4*sin(3*c))*cos((2*b + d)*x + 2*a + 4*
c) - 9*(8*b^3*d*sin(3*c) - 4*b^2*d^2*sin(3*c) - 18*b*d^3*sin(3*c) + 9*d^4*sin(3*c))*cos((2*b + d)*x + 2*a - 2*
c) - 9*(8*b^3*d*sin(3*c) + 4*b^2*d^2*sin(3*c) - 18*b*d^3*sin(3*c) - 9*d^4*sin(3*c))*cos(-(2*b - d)*x - 2*a + 4
*c) + 9*(8*b^3*d*sin(3*c) + 4*b^2*d^2*sin(3*c) - 18*b*d^3*sin(3*c) - 9*d^4*sin(3*c))*cos(-(2*b - d)*x - 2*a -
2*c) - 3*(8*b^3*d*sin(3*c) + 12*b^2*d^2*sin(3*c) - 2*b*d^3*sin(3*c) - 3*d^4*sin(3*c))*cos(-(2*b - 3*d)*x - 2*a
+ 6*c) + 3*(8*b^3*d*sin(3*c) + 12*b^2*d^2*sin(3*c) - 2*b*d^3*sin(3*c) - 3*d^4*sin(3*c))*cos(-(2*b - 3*d)*x -
2*a) - 2*(16*b^4*sin(3*c) - 40*b^2*d^2*sin(3*c) + 9*d^4*sin(3*c))*cos(3*d*x) + 2*(16*b^4*sin(3*c) - 40*b^2*d^2
*sin(3*c) + 9*d^4*sin(3*c))*cos(3*d*x + 6*c) + 18*(16*b^4*sin(3*c) - 40*b^2*d^2*sin(3*c) + 9*d^4*sin(3*c))*cos
(d*x + 4*c) - 18*(16*b^4*sin(3*c) - 40*b^2*d^2*sin(3*c) + 9*d^4*sin(3*c))*cos(d*x - 2*c) - 3*(8*b^3*d*cos(3*c)
- 12*b^2*d^2*cos(3*c) - 2*b*d^3*cos(3*c) + 3*d^4*cos(3*c))*sin((2*b + 3*d)*x + 2*a + 6*c) - 3*(8*b^3*d*cos(3*
c) - 12*b^2*d^2*cos(3*c) - 2*b*d^3*cos(3*c) + 3*d^4*cos(3*c))*sin((2*b + 3*d)*x + 2*a) - 9*(8*b^3*d*cos(3*c) -
4*b^2*d^2*cos(3*c) - 18*b*d^3*cos(3*c) + 9*d^4*cos(3*c))*sin((2*b + d)*x + 2*a + 4*c) - 9*(8*b^3*d*cos(3*c) -
4*b^2*d^2*cos(3*c) - 18*b*d^3*cos(3*c) + 9*d^4*cos(3*c))*sin((2*b + d)*x + 2*a - 2*c) + 9*(8*b^3*d*cos(3*c) +
4*b^2*d^2*cos(3*c) - 18*b*d^3*cos(3*c) - 9*d^4*cos(3*c))*sin(-(2*b - d)*x - 2*a + 4*c) + 9*(8*b^3*d*cos(3*c)
+ 4*b^2*d^2*cos(3*c) - 18*b*d^3*cos(3*c) - 9*d^4*cos(3*c))*sin(-(2*b - d)*x - 2*a - 2*c) + 3*(8*b^3*d*cos(3*c)
+ 12*b^2*d^2*cos(3*c) - 2*b*d^3*cos(3*c) - 3*d^4*cos(3*c))*sin(-(2*b - 3*d)*x - 2*a + 6*c) + 3*(8*b^3*d*cos(3
*c) + 12*b^2*d^2*cos(3*c) - 2*b*d^3*cos(3*c) - 3*d^4*cos(3*c))*sin(-(2*b - 3*d)*x - 2*a) - 2*(16*b^4*cos(3*c)
- 40*b^2*d^2*cos(3*c) + 9*d^4*cos(3*c))*sin(3*d*x) - 2*(16*b^4*cos(3*c) - 40*b^2*d^2*cos(3*c) + 9*d^4*cos(3*c)
)*sin(3*d*x + 6*c) - 18*(16*b^4*cos(3*c) - 40*b^2*d^2*cos(3*c) + 9*d^4*cos(3*c))*sin(d*x + 4*c) - 18*(16*b^4*c
os(3*c) - 40*b^2*d^2*cos(3*c) + 9*d^4*cos(3*c))*sin(d*x - 2*c))/(9*(cos(3*c)^2 + sin(3*c)^2)*d^5 - 40*(b^2*cos
(3*c)^2 + b^2*sin(3*c)^2)*d^3 + 16*(b^4*cos(3*c)^2 + b^4*sin(3*c)^2)*d)

________________________________________________________________________________________

Fricas [A]  time = 0.533443, size = 370, normalized size = 2.57 \begin{align*} -\frac{6 \,{\left (6 \, b d^{3} \cos \left (b x + a\right ) \cos \left (d x + c\right ) -{\left (4 \, b^{3} d - b d^{3}\right )} \cos \left (b x + a\right ) \cos \left (d x + c\right )^{3}\right )} \sin \left (b x + a\right ) -{\left (18 \, d^{4} \cos \left (b x + a\right )^{2} + 16 \, b^{4} - 40 \, b^{2} d^{2} +{\left (8 \, b^{4} - 2 \, b^{2} d^{2} - 9 \,{\left (4 \, b^{2} d^{2} - d^{4}\right )} \cos \left (b x + a\right )^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \,{\left (16 \, b^{4} d - 40 \, b^{2} d^{3} + 9 \, d^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(6*(6*b*d^3*cos(b*x + a)*cos(d*x + c) - (4*b^3*d - b*d^3)*cos(b*x + a)*cos(d*x + c)^3)*sin(b*x + a) - (18
*d^4*cos(b*x + a)^2 + 16*b^4 - 40*b^2*d^2 + (8*b^4 - 2*b^2*d^2 - 9*(4*b^2*d^2 - d^4)*cos(b*x + a)^2)*cos(d*x +
c)^2)*sin(d*x + c))/(16*b^4*d - 40*b^2*d^3 + 9*d^5)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.12546, size = 174, normalized size = 1.21 \begin{align*} \frac{\sin \left (2 \, b x + 3 \, d x + 2 \, a + 3 \, c\right )}{16 \,{\left (2 \, b + 3 \, d\right )}} + \frac{3 \, \sin \left (2 \, b x + d x + 2 \, a + c\right )}{16 \,{\left (2 \, b + d\right )}} + \frac{3 \, \sin \left (2 \, b x - d x + 2 \, a - c\right )}{16 \,{\left (2 \, b - d\right )}} + \frac{\sin \left (2 \, b x - 3 \, d x + 2 \, a - 3 \, c\right )}{16 \,{\left (2 \, b - 3 \, d\right )}} + \frac{\sin \left (3 \, d x + 3 \, c\right )}{24 \, d} + \frac{3 \, \sin \left (d x + c\right )}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*sin(2*b*x + 3*d*x + 2*a + 3*c)/(2*b + 3*d) + 3/16*sin(2*b*x + d*x + 2*a + c)/(2*b + d) + 3/16*sin(2*b*x -
d*x + 2*a - c)/(2*b - d) + 1/16*sin(2*b*x - 3*d*x + 2*a - 3*c)/(2*b - 3*d) + 1/24*sin(3*d*x + 3*c)/d + 3/8*si
n(d*x + c)/d