3.340 \(\int \cos ^5(a+b x) (c \sin (a+b x))^m \, dx\)

Optimal. Leaf size=74 \[ -\frac{2 (c \sin (a+b x))^{m+3}}{b c^3 (m+3)}+\frac{(c \sin (a+b x))^{m+5}}{b c^5 (m+5)}+\frac{(c \sin (a+b x))^{m+1}}{b c (m+1)} \]

[Out]

(c*Sin[a + b*x])^(1 + m)/(b*c*(1 + m)) - (2*(c*Sin[a + b*x])^(3 + m))/(b*c^3*(3 + m)) + (c*Sin[a + b*x])^(5 +
m)/(b*c^5*(5 + m))

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Rubi [A]  time = 0.0697001, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2564, 270} \[ -\frac{2 (c \sin (a+b x))^{m+3}}{b c^3 (m+3)}+\frac{(c \sin (a+b x))^{m+5}}{b c^5 (m+5)}+\frac{(c \sin (a+b x))^{m+1}}{b c (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^5*(c*Sin[a + b*x])^m,x]

[Out]

(c*Sin[a + b*x])^(1 + m)/(b*c*(1 + m)) - (2*(c*Sin[a + b*x])^(3 + m))/(b*c^3*(3 + m)) + (c*Sin[a + b*x])^(5 +
m)/(b*c^5*(5 + m))

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \cos ^5(a+b x) (c \sin (a+b x))^m \, dx &=\frac{\operatorname{Subst}\left (\int x^m \left (1-\frac{x^2}{c^2}\right )^2 \, dx,x,c \sin (a+b x)\right )}{b c}\\ &=\frac{\operatorname{Subst}\left (\int \left (x^m-\frac{2 x^{2+m}}{c^2}+\frac{x^{4+m}}{c^4}\right ) \, dx,x,c \sin (a+b x)\right )}{b c}\\ &=\frac{(c \sin (a+b x))^{1+m}}{b c (1+m)}-\frac{2 (c \sin (a+b x))^{3+m}}{b c^3 (3+m)}+\frac{(c \sin (a+b x))^{5+m}}{b c^5 (5+m)}\\ \end{align*}

Mathematica [A]  time = 0.315901, size = 55, normalized size = 0.74 \[ \frac{\sin (a+b x) \left (\frac{\sin ^4(a+b x)}{m+5}-\frac{2 \sin ^2(a+b x)}{m+3}+\frac{1}{m+1}\right ) (c \sin (a+b x))^m}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^5*(c*Sin[a + b*x])^m,x]

[Out]

(Sin[a + b*x]*(c*Sin[a + b*x])^m*((1 + m)^(-1) - (2*Sin[a + b*x]^2)/(3 + m) + Sin[a + b*x]^4/(5 + m)))/b

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Maple [F]  time = 1.115, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( bx+a \right ) \right ) ^{5} \left ( c\sin \left ( bx+a \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^5*(c*sin(b*x+a))^m,x)

[Out]

int(cos(b*x+a)^5*(c*sin(b*x+a))^m,x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^5*(c*sin(b*x+a))^m,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.60802, size = 177, normalized size = 2.39 \begin{align*} \frac{{\left ({\left (m^{2} + 4 \, m + 3\right )} \cos \left (b x + a\right )^{4} + 4 \,{\left (m + 1\right )} \cos \left (b x + a\right )^{2} + 8\right )} \left (c \sin \left (b x + a\right )\right )^{m} \sin \left (b x + a\right )}{b m^{3} + 9 \, b m^{2} + 23 \, b m + 15 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^5*(c*sin(b*x+a))^m,x, algorithm="fricas")

[Out]

((m^2 + 4*m + 3)*cos(b*x + a)^4 + 4*(m + 1)*cos(b*x + a)^2 + 8)*(c*sin(b*x + a))^m*sin(b*x + a)/(b*m^3 + 9*b*m
^2 + 23*b*m + 15*b)

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Sympy [A]  time = 79.3416, size = 2050, normalized size = 27.7 \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)**5*(c*sin(b*x+a))**m,x)

[Out]

Piecewise((x*(c*sin(a))**m*cos(a)**5, Eq(b, 0)), ((log(sin(a + b*x))/b + cos(a + b*x)**2/(2*b*sin(a + b*x)**2)
- cos(a + b*x)**4/(4*b*sin(a + b*x)**4))/c**5, Eq(m, -5)), ((16*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)
**6/(8*b*tan(a/2 + b*x/2)**6 + 16*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) + 32*log(tan(a/2 + b*x/2)**
2 + 1)*tan(a/2 + b*x/2)**4/(8*b*tan(a/2 + b*x/2)**6 + 16*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) + 16
*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**2/(8*b*tan(a/2 + b*x/2)**6 + 16*b*tan(a/2 + b*x/2)**4 + 8*b*ta
n(a/2 + b*x/2)**2) - 16*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**6/(8*b*tan(a/2 + b*x/2)**6 + 16*b*tan(a/2 + b*
x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) - 32*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**4/(8*b*tan(a/2 + b*x/2)**6 + 1
6*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) - 16*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**2/(8*b*tan(a/2
+ b*x/2)**6 + 16*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) - tan(a/2 + b*x/2)**8/(8*b*tan(a/2 + b*x/2)
**6 + 16*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) + 18*tan(a/2 + b*x/2)**4/(8*b*tan(a/2 + b*x/2)**6 +
16*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) - 1/(8*b*tan(a/2 + b*x/2)**6 + 16*b*tan(a/2 + b*x/2)**4 +
8*b*tan(a/2 + b*x/2)**2))/c**3, Eq(m, -3)), ((-log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**8/(b*tan(a/2 + b
*x/2)**8 + 4*b*tan(a/2 + b*x/2)**6 + 6*b*tan(a/2 + b*x/2)**4 + 4*b*tan(a/2 + b*x/2)**2 + b) - 4*log(tan(a/2 +
b*x/2)**2 + 1)*tan(a/2 + b*x/2)**6/(b*tan(a/2 + b*x/2)**8 + 4*b*tan(a/2 + b*x/2)**6 + 6*b*tan(a/2 + b*x/2)**4
+ 4*b*tan(a/2 + b*x/2)**2 + b) - 6*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**4/(b*tan(a/2 + b*x/2)**8 + 4
*b*tan(a/2 + b*x/2)**6 + 6*b*tan(a/2 + b*x/2)**4 + 4*b*tan(a/2 + b*x/2)**2 + b) - 4*log(tan(a/2 + b*x/2)**2 +
1)*tan(a/2 + b*x/2)**2/(b*tan(a/2 + b*x/2)**8 + 4*b*tan(a/2 + b*x/2)**6 + 6*b*tan(a/2 + b*x/2)**4 + 4*b*tan(a/
2 + b*x/2)**2 + b) - log(tan(a/2 + b*x/2)**2 + 1)/(b*tan(a/2 + b*x/2)**8 + 4*b*tan(a/2 + b*x/2)**6 + 6*b*tan(a
/2 + b*x/2)**4 + 4*b*tan(a/2 + b*x/2)**2 + b) + log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**8/(b*tan(a/2 + b*x/2)*
*8 + 4*b*tan(a/2 + b*x/2)**6 + 6*b*tan(a/2 + b*x/2)**4 + 4*b*tan(a/2 + b*x/2)**2 + b) + 4*log(tan(a/2 + b*x/2)
)*tan(a/2 + b*x/2)**6/(b*tan(a/2 + b*x/2)**8 + 4*b*tan(a/2 + b*x/2)**6 + 6*b*tan(a/2 + b*x/2)**4 + 4*b*tan(a/2
+ b*x/2)**2 + b) + 6*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**4/(b*tan(a/2 + b*x/2)**8 + 4*b*tan(a/2 + b*x/2)*
*6 + 6*b*tan(a/2 + b*x/2)**4 + 4*b*tan(a/2 + b*x/2)**2 + b) + 4*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**2/(b*t
an(a/2 + b*x/2)**8 + 4*b*tan(a/2 + b*x/2)**6 + 6*b*tan(a/2 + b*x/2)**4 + 4*b*tan(a/2 + b*x/2)**2 + b) + log(ta
n(a/2 + b*x/2))/(b*tan(a/2 + b*x/2)**8 + 4*b*tan(a/2 + b*x/2)**6 + 6*b*tan(a/2 + b*x/2)**4 + 4*b*tan(a/2 + b*x
/2)**2 + b) - 4*tan(a/2 + b*x/2)**6/(b*tan(a/2 + b*x/2)**8 + 4*b*tan(a/2 + b*x/2)**6 + 6*b*tan(a/2 + b*x/2)**4
+ 4*b*tan(a/2 + b*x/2)**2 + b) - 4*tan(a/2 + b*x/2)**4/(b*tan(a/2 + b*x/2)**8 + 4*b*tan(a/2 + b*x/2)**6 + 6*b
*tan(a/2 + b*x/2)**4 + 4*b*tan(a/2 + b*x/2)**2 + b) - 4*tan(a/2 + b*x/2)**2/(b*tan(a/2 + b*x/2)**8 + 4*b*tan(a
/2 + b*x/2)**6 + 6*b*tan(a/2 + b*x/2)**4 + 4*b*tan(a/2 + b*x/2)**2 + b))/c, Eq(m, -1)), (c**m*m**2*sin(a + b*x
)*sin(a + b*x)**m*cos(a + b*x)**4/(b*m**3 + 9*b*m**2 + 23*b*m + 15*b) + 4*c**m*m*sin(a + b*x)**3*sin(a + b*x)*
*m*cos(a + b*x)**2/(b*m**3 + 9*b*m**2 + 23*b*m + 15*b) + 8*c**m*m*sin(a + b*x)*sin(a + b*x)**m*cos(a + b*x)**4
/(b*m**3 + 9*b*m**2 + 23*b*m + 15*b) + 8*c**m*sin(a + b*x)**5*sin(a + b*x)**m/(b*m**3 + 9*b*m**2 + 23*b*m + 15
*b) + 20*c**m*sin(a + b*x)**3*sin(a + b*x)**m*cos(a + b*x)**2/(b*m**3 + 9*b*m**2 + 23*b*m + 15*b) + 15*c**m*si
n(a + b*x)*sin(a + b*x)**m*cos(a + b*x)**4/(b*m**3 + 9*b*m**2 + 23*b*m + 15*b), True))

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Giac [B]  time = 1.12502, size = 335, normalized size = 4.53 \begin{align*} \frac{\left (c \sin \left (b x + a\right )\right )^{m} c^{5} m^{2} \sin \left (b x + a\right )^{5} + 4 \, \left (c \sin \left (b x + a\right )\right )^{m} c^{5} m \sin \left (b x + a\right )^{5} - 2 \, \left (c \sin \left (b x + a\right )\right )^{m} c^{5} m^{2} \sin \left (b x + a\right )^{3} + 3 \, \left (c \sin \left (b x + a\right )\right )^{m} c^{5} \sin \left (b x + a\right )^{5} - 12 \, \left (c \sin \left (b x + a\right )\right )^{m} c^{5} m \sin \left (b x + a\right )^{3} + \left (c \sin \left (b x + a\right )\right )^{m} c^{5} m^{2} \sin \left (b x + a\right ) - 10 \, \left (c \sin \left (b x + a\right )\right )^{m} c^{5} \sin \left (b x + a\right )^{3} + 8 \, \left (c \sin \left (b x + a\right )\right )^{m} c^{5} m \sin \left (b x + a\right ) + 15 \, \left (c \sin \left (b x + a\right )\right )^{m} c^{5} \sin \left (b x + a\right )}{{\left (c^{4} m^{3} + 9 \, c^{4} m^{2} + 23 \, c^{4} m + 15 \, c^{4}\right )} b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^5*(c*sin(b*x+a))^m,x, algorithm="giac")

[Out]

((c*sin(b*x + a))^m*c^5*m^2*sin(b*x + a)^5 + 4*(c*sin(b*x + a))^m*c^5*m*sin(b*x + a)^5 - 2*(c*sin(b*x + a))^m*
c^5*m^2*sin(b*x + a)^3 + 3*(c*sin(b*x + a))^m*c^5*sin(b*x + a)^5 - 12*(c*sin(b*x + a))^m*c^5*m*sin(b*x + a)^3
+ (c*sin(b*x + a))^m*c^5*m^2*sin(b*x + a) - 10*(c*sin(b*x + a))^m*c^5*sin(b*x + a)^3 + 8*(c*sin(b*x + a))^m*c^
5*m*sin(b*x + a) + 15*(c*sin(b*x + a))^m*c^5*sin(b*x + a))/((c^4*m^3 + 9*c^4*m^2 + 23*c^4*m + 15*c^4)*b*c)