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

Optimal. Leaf size=46 \[ \frac{\sin ^{10}(a+b x)}{10 b}-\frac{\sin ^8(a+b x)}{4 b}+\frac{\sin ^6(a+b x)}{6 b} \]

[Out]

Sin[a + b*x]^6/(6*b) - Sin[a + b*x]^8/(4*b) + Sin[a + b*x]^10/(10*b)

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Rubi [A]  time = 0.0402625, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {2564, 266, 43} \[ \frac{\sin ^{10}(a+b x)}{10 b}-\frac{\sin ^8(a+b x)}{4 b}+\frac{\sin ^6(a+b x)}{6 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^5*Sin[a + b*x]^5,x]

[Out]

Sin[a + b*x]^6/(6*b) - Sin[a + b*x]^8/(4*b) + Sin[a + b*x]^10/(10*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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \cos ^5(a+b x) \sin ^5(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int x^5 \left (1-x^2\right )^2 \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int (1-x)^2 x^2 \, dx,x,\sin ^2(a+b x)\right )}{2 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (x^2-2 x^3+x^4\right ) \, dx,x,\sin ^2(a+b x)\right )}{2 b}\\ &=\frac{\sin ^6(a+b x)}{6 b}-\frac{\sin ^8(a+b x)}{4 b}+\frac{\sin ^{10}(a+b x)}{10 b}\\ \end{align*}

Mathematica [A]  time = 0.0267049, size = 50, normalized size = 1.09 \[ \frac{1}{32} \left (-\frac{5 \cos (2 (a+b x))}{16 b}+\frac{5 \cos (6 (a+b x))}{96 b}-\frac{\cos (10 (a+b x))}{160 b}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-5*Cos[2*(a + b*x)])/(16*b) + (5*Cos[6*(a + b*x)])/(96*b) - Cos[10*(a + b*x)]/(160*b))/32

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Maple [A]  time = 0.013, size = 52, normalized size = 1.1 \begin{align*}{\frac{1}{b} \left ( -{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{6} \left ( \sin \left ( bx+a \right ) \right ) ^{4}}{10}}-{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{6} \left ( \sin \left ( bx+a \right ) \right ) ^{2}}{20}}-{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{6}}{60}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(-1/10*cos(b*x+a)^6*sin(b*x+a)^4-1/20*cos(b*x+a)^6*sin(b*x+a)^2-1/60*cos(b*x+a)^6)

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Maxima [A]  time = 0.996927, size = 49, normalized size = 1.07 \begin{align*} \frac{6 \, \sin \left (b x + a\right )^{10} - 15 \, \sin \left (b x + a\right )^{8} + 10 \, \sin \left (b x + a\right )^{6}}{60 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(6*sin(b*x + a)^10 - 15*sin(b*x + a)^8 + 10*sin(b*x + a)^6)/b

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Fricas [A]  time = 1.67699, size = 93, normalized size = 2.02 \begin{align*} -\frac{6 \, \cos \left (b x + a\right )^{10} - 15 \, \cos \left (b x + a\right )^{8} + 10 \, \cos \left (b x + a\right )^{6}}{60 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(6*cos(b*x + a)^10 - 15*cos(b*x + a)^8 + 10*cos(b*x + a)^6)/b

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Sympy [A]  time = 31.496, size = 63, normalized size = 1.37 \begin{align*} \begin{cases} \frac{\sin ^{10}{\left (a + b x \right )}}{60 b} + \frac{\sin ^{8}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{12 b} + \frac{\sin ^{6}{\left (a + b x \right )} \cos ^{4}{\left (a + b x \right )}}{6 b} & \text{for}\: b \neq 0 \\x \sin ^{5}{\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)**5,x)

[Out]

Piecewise((sin(a + b*x)**10/(60*b) + sin(a + b*x)**8*cos(a + b*x)**2/(12*b) + sin(a + b*x)**6*cos(a + b*x)**4/
(6*b), Ne(b, 0)), (x*sin(a)**5*cos(a)**5, True))

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Giac [A]  time = 1.11881, size = 58, normalized size = 1.26 \begin{align*} -\frac{\cos \left (10 \, b x + 10 \, a\right )}{5120 \, b} + \frac{5 \, \cos \left (6 \, b x + 6 \, a\right )}{3072 \, b} - \frac{5 \, \cos \left (2 \, b x + 2 \, a\right )}{512 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5120*cos(10*b*x + 10*a)/b + 5/3072*cos(6*b*x + 6*a)/b - 5/512*cos(2*b*x + 2*a)/b