3.98 \(\int \cos ^7(a+b x) \sin ^5(a+b x) \, dx\)

Optimal. Leaf size=46 \[ -\frac{\cos ^{12}(a+b x)}{12 b}+\frac{\cos ^{10}(a+b x)}{5 b}-\frac{\cos ^8(a+b x)}{8 b} \]

[Out]

-Cos[a + b*x]^8/(8*b) + Cos[a + b*x]^10/(5*b) - Cos[a + b*x]^12/(12*b)

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Rubi [A]  time = 0.0410329, 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 = {2565, 266, 43} \[ -\frac{\cos ^{12}(a+b x)}{12 b}+\frac{\cos ^{10}(a+b x)}{5 b}-\frac{\cos ^8(a+b x)}{8 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^7*Sin[a + b*x]^5,x]

[Out]

-Cos[a + b*x]^8/(8*b) + Cos[a + b*x]^10/(5*b) - Cos[a + b*x]^12/(12*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 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 ^7(a+b x) \sin ^5(a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int x^7 \left (1-x^2\right )^2 \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac{\operatorname{Subst}\left (\int (1-x)^2 x^3 \, dx,x,\cos ^2(a+b x)\right )}{2 b}\\ &=-\frac{\operatorname{Subst}\left (\int \left (x^3-2 x^4+x^5\right ) \, dx,x,\cos ^2(a+b x)\right )}{2 b}\\ &=-\frac{\cos ^8(a+b x)}{8 b}+\frac{\cos ^{10}(a+b x)}{5 b}-\frac{\cos ^{12}(a+b x)}{12 b}\\ \end{align*}

Mathematica [A]  time = 0.378623, size = 68, normalized size = 1.48 \[ -\frac{600 \cos (2 (a+b x))+75 \cos (4 (a+b x))-100 \cos (6 (a+b x))-30 \cos (8 (a+b x))+12 \cos (10 (a+b x))+5 \cos (12 (a+b x))}{122880 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^7*Sin[a + b*x]^5,x]

[Out]

-(600*Cos[2*(a + b*x)] + 75*Cos[4*(a + b*x)] - 100*Cos[6*(a + b*x)] - 30*Cos[8*(a + b*x)] + 12*Cos[10*(a + b*x
)] + 5*Cos[12*(a + b*x)])/(122880*b)

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Maple [A]  time = 0.013, size = 52, normalized size = 1.1 \begin{align*}{\frac{1}{b} \left ( -{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{4} \left ( \cos \left ( bx+a \right ) \right ) ^{8}}{12}}-{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{2} \left ( \cos \left ( bx+a \right ) \right ) ^{8}}{30}}-{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{8}}{120}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^7*sin(b*x+a)^5,x)

[Out]

1/b*(-1/12*sin(b*x+a)^4*cos(b*x+a)^8-1/30*sin(b*x+a)^2*cos(b*x+a)^8-1/120*cos(b*x+a)^8)

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Maxima [A]  time = 1.02676, size = 62, normalized size = 1.35 \begin{align*} -\frac{10 \, \sin \left (b x + a\right )^{12} - 36 \, \sin \left (b x + a\right )^{10} + 45 \, \sin \left (b x + a\right )^{8} - 20 \, \sin \left (b x + a\right )^{6}}{120 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^7*sin(b*x+a)^5,x, algorithm="maxima")

[Out]

-1/120*(10*sin(b*x + a)^12 - 36*sin(b*x + a)^10 + 45*sin(b*x + a)^8 - 20*sin(b*x + a)^6)/b

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Fricas [A]  time = 1.70107, size = 97, normalized size = 2.11 \begin{align*} -\frac{10 \, \cos \left (b x + a\right )^{12} - 24 \, \cos \left (b x + a\right )^{10} + 15 \, \cos \left (b x + a\right )^{8}}{120 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^7*sin(b*x+a)^5,x, algorithm="fricas")

[Out]

-1/120*(10*cos(b*x + a)^12 - 24*cos(b*x + a)^10 + 15*cos(b*x + a)^8)/b

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Sympy [A]  time = 81.0591, size = 83, normalized size = 1.8 \begin{align*} \begin{cases} \frac{\sin ^{12}{\left (a + b x \right )}}{120 b} + \frac{\sin ^{10}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{20 b} + \frac{\sin ^{8}{\left (a + b x \right )} \cos ^{4}{\left (a + b x \right )}}{8 b} + \frac{\sin ^{6}{\left (a + b x \right )} \cos ^{6}{\left (a + b x \right )}}{6 b} & \text{for}\: b \neq 0 \\x \sin ^{5}{\left (a \right )} \cos ^{7}{\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)**7*sin(b*x+a)**5,x)

[Out]

Piecewise((sin(a + b*x)**12/(120*b) + sin(a + b*x)**10*cos(a + b*x)**2/(20*b) + sin(a + b*x)**8*cos(a + b*x)**
4/(8*b) + sin(a + b*x)**6*cos(a + b*x)**6/(6*b), Ne(b, 0)), (x*sin(a)**5*cos(a)**7, True))

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Giac [B]  time = 1.17461, size = 115, normalized size = 2.5 \begin{align*} -\frac{\cos \left (12 \, b x + 12 \, a\right )}{24576 \, b} - \frac{\cos \left (10 \, b x + 10 \, a\right )}{10240 \, b} + \frac{\cos \left (8 \, b x + 8 \, a\right )}{4096 \, b} + \frac{5 \, \cos \left (6 \, b x + 6 \, a\right )}{6144 \, b} - \frac{5 \, \cos \left (4 \, b x + 4 \, a\right )}{8192 \, b} - \frac{5 \, \cos \left (2 \, b x + 2 \, a\right )}{1024 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^7*sin(b*x+a)^5,x, algorithm="giac")

[Out]

-1/24576*cos(12*b*x + 12*a)/b - 1/10240*cos(10*b*x + 10*a)/b + 1/4096*cos(8*b*x + 8*a)/b + 5/6144*cos(6*b*x +
6*a)/b - 5/8192*cos(4*b*x + 4*a)/b - 5/1024*cos(2*b*x + 2*a)/b