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

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

[Out]

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

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Rubi [A]  time = 0.0382698, 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 = {2564, 270} \[ \frac{\sin ^7(a+b x)}{7 b}-\frac{2 \sin ^5(a+b x)}{5 b}+\frac{\sin ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^5*Sin[a + b*x]^2,x]

[Out]

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

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) \sin ^2(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int x^2 \left (1-x^2\right )^2 \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac{\sin ^3(a+b x)}{3 b}-\frac{2 \sin ^5(a+b x)}{5 b}+\frac{\sin ^7(a+b x)}{7 b}\\ \end{align*}

Mathematica [A]  time = 0.0922559, size = 37, normalized size = 0.8 \[ \frac{\sin ^3(a+b x) (108 \cos (2 (a+b x))+15 \cos (4 (a+b x))+157)}{840 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.034, size = 50, normalized size = 1.1 \begin{align*}{\frac{1}{b} \left ( -{\frac{\sin \left ( bx+a \right ) \left ( \cos \left ( bx+a \right ) \right ) ^{6}}{7}}+{\frac{\sin \left ( bx+a \right ) }{35} \left ({\frac{8}{3}}+ \left ( \cos \left ( bx+a \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}}{3}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(-1/7*sin(b*x+a)*cos(b*x+a)^6+1/35*(8/3+cos(b*x+a)^4+4/3*cos(b*x+a)^2)*sin(b*x+a))

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Maxima [A]  time = 0.985084, size = 49, normalized size = 1.07 \begin{align*} \frac{15 \, \sin \left (b x + a\right )^{7} - 42 \, \sin \left (b x + a\right )^{5} + 35 \, \sin \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)^5*sin(b*x+a)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.91543, size = 115, normalized size = 2.5 \begin{align*} -\frac{{\left (15 \, \cos \left (b x + a\right )^{6} - 3 \, \cos \left (b x + a\right )^{4} - 4 \, \cos \left (b x + a\right )^{2} - 8\right )} \sin \left (b x + a\right )}{105 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/105*(15*cos(b*x + a)^6 - 3*cos(b*x + a)^4 - 4*cos(b*x + a)^2 - 8)*sin(b*x + a)/b

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Sympy [A]  time = 7.79142, size = 66, normalized size = 1.43 \begin{align*} \begin{cases} \frac{8 \sin ^{7}{\left (a + b x \right )}}{105 b} + \frac{4 \sin ^{5}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{15 b} + \frac{\sin ^{3}{\left (a + b x \right )} \cos ^{4}{\left (a + b x \right )}}{3 b} & \text{for}\: b \neq 0 \\x \sin ^{2}{\left (a \right )} \cos ^{5}{\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)**5*sin(b*x+a)**2,x)

[Out]

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

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Giac [A]  time = 1.14921, size = 73, normalized size = 1.59 \begin{align*} -\frac{\sin \left (7 \, b x + 7 \, a\right )}{448 \, b} - \frac{3 \, \sin \left (5 \, b x + 5 \, a\right )}{320 \, b} - \frac{\sin \left (3 \, b x + 3 \, a\right )}{192 \, b} + \frac{5 \, \sin \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)^5*sin(b*x+a)^2,x, algorithm="giac")

[Out]

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