\(\int \cos ^3(c+d x) \sin (a+b x) \, dx\) [210]

Optimal result

Integrand size = 15, antiderivative size = 91 \[ \int \cos ^3(c+d x) \sin (a+b x) \, dx=-\frac {\cos (a-3 c+(b-3 d) x)}{8 (b-3 d)}-\frac {3 \cos (a-c+(b-d) x)}{8 (b-d)}-\frac {3 \cos (a+c+(b+d) x)}{8 (b+d)}-\frac {\cos (a+3 c+(b+3 d) x)}{8 (b+3 d)} \]

[Out]

-1/8*cos(a-3*c+(b-3*d)*x)/(b-3*d)-3/8*cos(a-c+(b-d)*x)/(b-d)-3/8*cos(a+c+(b+d)*x)/(b+d)-1/8*cos(a+3*c+(b+3*d)*
x)/(b+3*d)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {4670, 2718} \[ \int \cos ^3(c+d x) \sin (a+b x) \, dx=-\frac {\cos (a+x (b-3 d)-3 c)}{8 (b-3 d)}-\frac {3 \cos (a+x (b-d)-c)}{8 (b-d)}-\frac {3 \cos (a+x (b+d)+c)}{8 (b+d)}-\frac {\cos (a+x (b+3 d)+3 c)}{8 (b+3 d)} \]

[In]

Int[Cos[c + d*x]^3*Sin[a + b*x],x]

[Out]

-1/8*Cos[a - 3*c + (b - 3*d)*x]/(b - 3*d) - (3*Cos[a - c + (b - d)*x])/(8*(b - d)) - (3*Cos[a + c + (b + d)*x]
)/(8*(b + d)) - Cos[a + 3*c + (b + 3*d)*x]/(8*(b + 3*d))

Rule 2718

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

Rule 4670

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

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{8} \sin (a-3 c+(b-3 d) x)+\frac {3}{8} \sin (a-c+(b-d) x)+\frac {3}{8} \sin (a+c+(b+d) x)+\frac {1}{8} \sin (a+3 c+(b+3 d) x)\right ) \, dx \\ & = \frac {1}{8} \int \sin (a-3 c+(b-3 d) x) \, dx+\frac {1}{8} \int \sin (a+3 c+(b+3 d) x) \, dx+\frac {3}{8} \int \sin (a-c+(b-d) x) \, dx+\frac {3}{8} \int \sin (a+c+(b+d) x) \, dx \\ & = -\frac {\cos (a-3 c+(b-3 d) x)}{8 (b-3 d)}-\frac {3 \cos (a-c+(b-d) x)}{8 (b-d)}-\frac {3 \cos (a+c+(b+d) x)}{8 (b+d)}-\frac {\cos (a+3 c+(b+3 d) x)}{8 (b+3 d)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.68 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.96 \[ \int \cos ^3(c+d x) \sin (a+b x) \, dx=\frac {1}{8} \left (-\frac {\cos (a-3 c+b x-3 d x)}{b-3 d}-\frac {3 \cos (a-c+b x-d x)}{b-d}-\frac {\cos (a+3 c+b x+3 d x)}{b+3 d}-\frac {3 \cos (a+c+(b+d) x)}{b+d}\right ) \]

[In]

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

[Out]

(-(Cos[a - 3*c + b*x - 3*d*x]/(b - 3*d)) - (3*Cos[a - c + b*x - d*x])/(b - d) - Cos[a + 3*c + b*x + 3*d*x]/(b
+ 3*d) - (3*Cos[a + c + (b + d)*x])/(b + d))/8

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.92

method	result	size
default	\(-\frac {\cos \left (a -3 c +\left (b -3 d \right ) x \right )}{8 \left (b -3 d \right )}-\frac {3 \cos \left (a -c +\left (b -d \right ) x \right )}{8 \left (b -d \right )}-\frac {3 \cos \left (a +c +\left (b +d \right ) x \right )}{8 \left (b +d \right )}-\frac {\cos \left (a +3 c +\left (b +3 d \right ) x \right )}{8 \left (b +3 d \right )}\)	\(84\)
risch	\(-\frac {\cos \left (x b -3 d x +a -3 c \right )}{8 \left (b -3 d \right )}-\frac {3 \cos \left (x b -d x +a -c \right )}{8 \left (b -d \right )}-\frac {3 \cos \left (x b +d x +a +c \right )}{8 \left (b +d \right )}-\frac {\cos \left (x b +3 d x +a +3 c \right )}{8 \left (b +3 d \right )}\)	\(85\)
parallelrisch	\(\frac {-2 b \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2} \left (b^{2}-7 d^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-12 d \tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (b^{2}-3 d^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-6 b \left (-4 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2} d^{2}+b^{2}-3 d^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+24 d \tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (b^{2}+d^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-6 b \left (\left (b^{2}-3 d^{2}\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-4 d^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-12 d \tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (b^{2}-3 d^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b^{3}+14 b \,d^{2}}{\left (b -d \right ) \left (b +3 d \right ) \left (b -3 d \right ) \left (b +d \right ) \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (1+\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}\right )}\)	\(276\)

[In]

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

[Out]

-1/8*cos(a-3*c+(b-3*d)*x)/(b-3*d)-3/8*cos(a-c+(b-d)*x)/(b-d)-3/8*cos(a+c+(b+d)*x)/(b+d)-1/8*cos(a+3*c+(b+3*d)*
x)/(b+3*d)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.16 \[ \int \cos ^3(c+d x) \sin (a+b x) \, dx=\frac {6 \, b d^{2} \cos \left (b x + a\right ) \cos \left (d x + c\right ) - {\left (b^{3} - b d^{2}\right )} \cos \left (b x + a\right ) \cos \left (d x + c\right )^{3} + 3 \, {\left (2 \, d^{3} - {\left (b^{2} d - d^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (b x + a\right ) \sin \left (d x + c\right )}{b^{4} - 10 \, b^{2} d^{2} + 9 \, d^{4}} \]

[In]

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

[Out]

(6*b*d^2*cos(b*x + a)*cos(d*x + c) - (b^3 - b*d^2)*cos(b*x + a)*cos(d*x + c)^3 + 3*(2*d^3 - (b^2*d - d^3)*cos(
d*x + c)^2)*sin(b*x + a)*sin(d*x + c))/(b^4 - 10*b^2*d^2 + 9*d^4)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 918 vs. \(2 (78) = 156\).

Time = 2.12 (sec) , antiderivative size = 918, normalized size of antiderivative = 10.09 \[ \int \cos ^3(c+d x) \sin (a+b x) \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise((x*sin(a)*cos(c)**3, Eq(b, 0) & Eq(d, 0)), (-3*x*sin(a - 3*d*x)*sin(c + d*x)**2*cos(c + d*x)/8 + x*s
in(a - 3*d*x)*cos(c + d*x)**3/8 - x*sin(c + d*x)**3*cos(a - 3*d*x)/8 + 3*x*sin(c + d*x)*cos(a - 3*d*x)*cos(c +
d*x)**2/8 - sin(a - 3*d*x)*sin(c + d*x)**3/(24*d) - sin(a - 3*d*x)*sin(c + d*x)*cos(c + d*x)**2/(4*d) + 3*cos
(a - 3*d*x)*cos(c + d*x)**3/(8*d), Eq(b, -3*d)), (3*x*sin(a - d*x)*sin(c + d*x)**2*cos(c + d*x)/8 + 3*x*sin(a
- d*x)*cos(c + d*x)**3/8 + 3*x*sin(c + d*x)**3*cos(a - d*x)/8 + 3*x*sin(c + d*x)*cos(a - d*x)*cos(c + d*x)**2/
8 + 3*sin(a - d*x)*sin(c + d*x)**3/(8*d) + 3*sin(a - d*x)*sin(c + d*x)*cos(c + d*x)**2/(4*d) - cos(a - d*x)*co
s(c + d*x)**3/(8*d), Eq(b, -d)), (3*x*sin(a + d*x)*sin(c + d*x)**2*cos(c + d*x)/8 + 3*x*sin(a + d*x)*cos(c + d
*x)**3/8 - 3*x*sin(c + d*x)**3*cos(a + d*x)/8 - 3*x*sin(c + d*x)*cos(a + d*x)*cos(c + d*x)**2/8 + 3*sin(a + d*
x)*sin(c + d*x)**3/(8*d) + 3*sin(a + d*x)*sin(c + d*x)*cos(c + d*x)**2/(4*d) + cos(a + d*x)*cos(c + d*x)**3/(8
*d), Eq(b, d)), (-3*x*sin(a + 3*d*x)*sin(c + d*x)**2*cos(c + d*x)/8 + x*sin(a + 3*d*x)*cos(c + d*x)**3/8 + x*s
in(c + d*x)**3*cos(a + 3*d*x)/8 - 3*x*sin(c + d*x)*cos(a + 3*d*x)*cos(c + d*x)**2/8 - sin(a + 3*d*x)*sin(c + d
*x)**3/(24*d) - sin(a + 3*d*x)*sin(c + d*x)*cos(c + d*x)**2/(4*d) - 3*cos(a + 3*d*x)*cos(c + d*x)**3/(8*d), Eq
(b, 3*d)), (-b**3*cos(a + b*x)*cos(c + d*x)**3/(b**4 - 10*b**2*d**2 + 9*d**4) - 3*b**2*d*sin(a + b*x)*sin(c +
d*x)*cos(c + d*x)**2/(b**4 - 10*b**2*d**2 + 9*d**4) + 6*b*d**2*sin(c + d*x)**2*cos(a + b*x)*cos(c + d*x)/(b**4
- 10*b**2*d**2 + 9*d**4) + 7*b*d**2*cos(a + b*x)*cos(c + d*x)**3/(b**4 - 10*b**2*d**2 + 9*d**4) + 6*d**3*sin(
a + b*x)*sin(c + d*x)**3/(b**4 - 10*b**2*d**2 + 9*d**4) + 9*d**3*sin(a + b*x)*sin(c + d*x)*cos(c + d*x)**2/(b*
*4 - 10*b**2*d**2 + 9*d**4), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 912 vs. \(2 (83) = 166\).

Time = 0.25 (sec) , antiderivative size = 912, normalized size of antiderivative = 10.02 \[ \int \cos ^3(c+d x) \sin (a+b x) \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/16*((b^3*cos(3*c) - 3*b^2*d*cos(3*c) - b*d^2*cos(3*c) + 3*d^3*cos(3*c))*cos((b + 3*d)*x + a + 6*c) + (b^3*c
os(3*c) - 3*b^2*d*cos(3*c) - b*d^2*cos(3*c) + 3*d^3*cos(3*c))*cos((b + 3*d)*x + a) + 3*(b^3*cos(3*c) - b^2*d*c
os(3*c) - 9*b*d^2*cos(3*c) + 9*d^3*cos(3*c))*cos((b + d)*x + a + 4*c) + 3*(b^3*cos(3*c) - b^2*d*cos(3*c) - 9*b
*d^2*cos(3*c) + 9*d^3*cos(3*c))*cos((b + d)*x + a - 2*c) + 3*(b^3*cos(3*c) + b^2*d*cos(3*c) - 9*b*d^2*cos(3*c)
- 9*d^3*cos(3*c))*cos(-(b - d)*x - a + 4*c) + 3*(b^3*cos(3*c) + b^2*d*cos(3*c) - 9*b*d^2*cos(3*c) - 9*d^3*cos
(3*c))*cos(-(b - d)*x - a - 2*c) + (b^3*cos(3*c) + 3*b^2*d*cos(3*c) - b*d^2*cos(3*c) - 3*d^3*cos(3*c))*cos(-(b
- 3*d)*x - a + 6*c) + (b^3*cos(3*c) + 3*b^2*d*cos(3*c) - b*d^2*cos(3*c) - 3*d^3*cos(3*c))*cos(-(b - 3*d)*x -
a) + (b^3*sin(3*c) - 3*b^2*d*sin(3*c) - b*d^2*sin(3*c) + 3*d^3*sin(3*c))*sin((b + 3*d)*x + a + 6*c) - (b^3*sin
(3*c) - 3*b^2*d*sin(3*c) - b*d^2*sin(3*c) + 3*d^3*sin(3*c))*sin((b + 3*d)*x + a) + 3*(b^3*sin(3*c) - b^2*d*sin
(3*c) - 9*b*d^2*sin(3*c) + 9*d^3*sin(3*c))*sin((b + d)*x + a + 4*c) - 3*(b^3*sin(3*c) - b^2*d*sin(3*c) - 9*b*d
^2*sin(3*c) + 9*d^3*sin(3*c))*sin((b + d)*x + a - 2*c) + 3*(b^3*sin(3*c) + b^2*d*sin(3*c) - 9*b*d^2*sin(3*c) -
9*d^3*sin(3*c))*sin(-(b - d)*x - a + 4*c) - 3*(b^3*sin(3*c) + b^2*d*sin(3*c) - 9*b*d^2*sin(3*c) - 9*d^3*sin(3
*c))*sin(-(b - d)*x - a - 2*c) + (b^3*sin(3*c) + 3*b^2*d*sin(3*c) - b*d^2*sin(3*c) - 3*d^3*sin(3*c))*sin(-(b -
3*d)*x - a + 6*c) - (b^3*sin(3*c) + 3*b^2*d*sin(3*c) - b*d^2*sin(3*c) - 3*d^3*sin(3*c))*sin(-(b - 3*d)*x - a)
)/(b^4*cos(3*c)^2 + b^4*sin(3*c)^2 + 9*(cos(3*c)^2 + sin(3*c)^2)*d^4 - 10*(b^2*cos(3*c)^2 + b^2*sin(3*c)^2)*d^
2)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.92 \[ \int \cos ^3(c+d x) \sin (a+b x) \, dx=-\frac {\cos \left (b x + 3 \, d x + a + 3 \, c\right )}{8 \, {\left (b + 3 \, d\right )}} - \frac {3 \, \cos \left (b x + d x + a + c\right )}{8 \, {\left (b + d\right )}} - \frac {3 \, \cos \left (b x - d x + a - c\right )}{8 \, {\left (b - d\right )}} - \frac {\cos \left (b x - 3 \, d x + a - 3 \, c\right )}{8 \, {\left (b - 3 \, d\right )}} \]

[In]

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

[Out]

-1/8*cos(b*x + 3*d*x + a + 3*c)/(b + 3*d) - 3/8*cos(b*x + d*x + a + c)/(b + d) - 3/8*cos(b*x - d*x + a - c)/(b
- d) - 1/8*cos(b*x - 3*d*x + a - 3*c)/(b - 3*d)

Mupad [B] (verification not implemented)

Time = 23.19 (sec) , antiderivative size = 297, normalized size of antiderivative = 3.26 \[ \int \cos ^3(c+d x) \sin (a+b x) \, dx=-{\mathrm {e}}^{a\,1{}\mathrm {i}-c\,3{}\mathrm {i}+b\,x\,1{}\mathrm {i}-d\,x\,3{}\mathrm {i}}\,\left (\frac {b+3\,d}{16\,b^2-144\,d^2}+\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,\left (b-3\,d\right )}{16\,b^2-144\,d^2}\right )-{\mathrm {e}}^{a\,1{}\mathrm {i}+c\,3{}\mathrm {i}+b\,x\,1{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,\left (\frac {b-3\,d}{16\,b^2-144\,d^2}+\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,\left (b+3\,d\right )}{16\,b^2-144\,d^2}\right )-{\mathrm {e}}^{a\,1{}\mathrm {i}-c\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\left (\frac {3\,b+3\,d}{16\,b^2-16\,d^2}+\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,\left (3\,b-3\,d\right )}{16\,b^2-16\,d^2}\right )-{\mathrm {e}}^{a\,1{}\mathrm {i}+c\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {3\,b-3\,d}{16\,b^2-16\,d^2}+\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,\left (3\,b+3\,d\right )}{16\,b^2-16\,d^2}\right ) \]

[In]

int(cos(c + d*x)^3*sin(a + b*x),x)

[Out]

- exp(a*1i - c*3i + b*x*1i - d*x*3i)*((b + 3*d)/(16*b^2 - 144*d^2) + (exp(- a*2i - b*x*2i)*(b - 3*d))/(16*b^2
- 144*d^2)) - exp(a*1i + c*3i + b*x*1i + d*x*3i)*((b - 3*d)/(16*b^2 - 144*d^2) + (exp(- a*2i - b*x*2i)*(b + 3*
d))/(16*b^2 - 144*d^2)) - exp(a*1i - c*1i + b*x*1i - d*x*1i)*((3*b + 3*d)/(16*b^2 - 16*d^2) + (exp(- a*2i - b*
x*2i)*(3*b - 3*d))/(16*b^2 - 16*d^2)) - exp(a*1i + c*1i + b*x*1i + d*x*1i)*((3*b - 3*d)/(16*b^2 - 16*d^2) + (e
xp(- a*2i - b*x*2i)*(3*b + 3*d))/(16*b^2 - 16*d^2))