3.103 \(\int \cos ^2(a+b x) \sin ^5(a+b x) \, dx\)

Optimal. Leaf size=46 \[ -\frac{\cos ^7(a+b x)}{7 b}+\frac{2 \cos ^5(a+b x)}{5 b}-\frac{\cos ^3(a+b x)}{3 b} \]

[Out]

-Cos[a + b*x]^3/(3*b) + (2*Cos[a + b*x]^5)/(5*b) - Cos[a + b*x]^7/(7*b)

________________________________________________________________________________________

Rubi [A]  time = 0.0354446, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2565, 270} \[ -\frac{\cos ^7(a+b x)}{7 b}+\frac{2 \cos ^5(a+b x)}{5 b}-\frac{\cos ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^2*Sin[a + b*x]^5,x]

[Out]

-Cos[a + b*x]^3/(3*b) + (2*Cos[a + b*x]^5)/(5*b) - Cos[a + b*x]^7/(7*b)

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[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 ^2(a+b x) \sin ^5(a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int x^2 \left (1-x^2\right )^2 \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac{\operatorname{Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac{\cos ^3(a+b x)}{3 b}+\frac{2 \cos ^5(a+b x)}{5 b}-\frac{\cos ^7(a+b x)}{7 b}\\ \end{align*}

Mathematica [A]  time = 0.0895744, size = 37, normalized size = 0.8 \[ \frac{\cos ^3(a+b x) (108 \cos (2 (a+b x))-15 \cos (4 (a+b x))-157)}{840 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^2*Sin[a + b*x]^5,x]

[Out]

(Cos[a + b*x]^3*(-157 + 108*Cos[2*(a + b*x)] - 15*Cos[4*(a + b*x)]))/(840*b)

________________________________________________________________________________________

Maple [A]  time = 0.012, size = 52, normalized size = 1.1 \begin{align*}{\frac{1}{b} \left ( -{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{3} \left ( \sin \left ( bx+a \right ) \right ) ^{4}}{7}}-{\frac{4\, \left ( \cos \left ( bx+a \right ) \right ) ^{3} \left ( \sin \left ( bx+a \right ) \right ) ^{2}}{35}}-{\frac{8\, \left ( \cos \left ( bx+a \right ) \right ) ^{3}}{105}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^2*sin(b*x+a)^5,x)

[Out]

1/b*(-1/7*cos(b*x+a)^3*sin(b*x+a)^4-4/35*cos(b*x+a)^3*sin(b*x+a)^2-8/105*cos(b*x+a)^3)

________________________________________________________________________________________

Maxima [A]  time = 0.981742, size = 49, normalized size = 1.07 \begin{align*} -\frac{15 \, \cos \left (b x + a\right )^{7} - 42 \, \cos \left (b x + a\right )^{5} + 35 \, \cos \left (b x + a\right )^{3}}{105 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2*sin(b*x+a)^5,x, algorithm="maxima")

[Out]

-1/105*(15*cos(b*x + a)^7 - 42*cos(b*x + a)^5 + 35*cos(b*x + a)^3)/b

________________________________________________________________________________________

Fricas [A]  time = 1.59817, size = 95, normalized size = 2.07 \begin{align*} -\frac{15 \, \cos \left (b x + a\right )^{7} - 42 \, \cos \left (b x + a\right )^{5} + 35 \, \cos \left (b x + a\right )^{3}}{105 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2*sin(b*x+a)^5,x, algorithm="fricas")

[Out]

-1/105*(15*cos(b*x + a)^7 - 42*cos(b*x + a)^5 + 35*cos(b*x + a)^3)/b

________________________________________________________________________________________

Sympy [A]  time = 6.94841, size = 68, normalized size = 1.48 \begin{align*} \begin{cases} - \frac{\sin ^{4}{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{3 b} - \frac{4 \sin ^{2}{\left (a + b x \right )} \cos ^{5}{\left (a + b x \right )}}{15 b} - \frac{8 \cos ^{7}{\left (a + b x \right )}}{105 b} & \text{for}\: b \neq 0 \\x \sin ^{5}{\left (a \right )} \cos ^{2}{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**2*sin(b*x+a)**5,x)

[Out]

Piecewise((-sin(a + b*x)**4*cos(a + b*x)**3/(3*b) - 4*sin(a + b*x)**2*cos(a + b*x)**5/(15*b) - 8*cos(a + b*x)*
*7/(105*b), Ne(b, 0)), (x*sin(a)**5*cos(a)**2, True))

________________________________________________________________________________________

Giac [A]  time = 1.16678, size = 73, normalized size = 1.59 \begin{align*} -\frac{\cos \left (7 \, b x + 7 \, a\right )}{448 \, b} + \frac{3 \, \cos \left (5 \, b x + 5 \, a\right )}{320 \, b} - \frac{\cos \left (3 \, b x + 3 \, a\right )}{192 \, b} - \frac{5 \, \cos \left (b x + a\right )}{64 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2*sin(b*x+a)^5,x, algorithm="giac")

[Out]

-1/448*cos(7*b*x + 7*a)/b + 3/320*cos(5*b*x + 5*a)/b - 1/192*cos(3*b*x + 3*a)/b - 5/64*cos(b*x + a)/b