3.82 \(\int \cos ^3(a+b x) \sin ^4(a+b x) \, dx\)

Optimal. Leaf size=31 \[ \frac {\sin ^5(a+b x)}{5 b}-\frac {\sin ^7(a+b x)}{7 b} \]

[Out]

1/5*sin(b*x+a)^5/b-1/7*sin(b*x+a)^7/b

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Rubi [A]  time = 0.03, antiderivative size = 31, normalized size of antiderivative = 1.00, 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, 14} \[ \frac {\sin ^5(a+b x)}{5 b}-\frac {\sin ^7(a+b x)}{7 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^3*Sin[a + b*x]^4,x]

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
 &&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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])

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 27, normalized size = 0.87 \[ \frac {\sin ^5(a+b x) (5 \cos (2 (a+b x))+9)}{70 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^3*Sin[a + b*x]^4,x]

[Out]

((9 + 5*Cos[2*(a + b*x)])*Sin[a + b*x]^5)/(70*b)

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fricas [A]  time = 0.42, size = 41, normalized size = 1.32 \[ \frac {{\left (5 \, \cos \left (b x + a\right )^{6} - 8 \, \cos \left (b x + a\right )^{4} + \cos \left (b x + a\right )^{2} + 2\right )} \sin \left (b x + a\right )}{35 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3*sin(b*x+a)^4,x, algorithm="fricas")

[Out]

1/35*(5*cos(b*x + a)^6 - 8*cos(b*x + a)^4 + cos(b*x + a)^2 + 2)*sin(b*x + a)/b

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giac [A]  time = 0.24, size = 26, normalized size = 0.84 \[ -\frac {5 \, \sin \left (b x + a\right )^{7} - 7 \, \sin \left (b x + a\right )^{5}}{35 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3*sin(b*x+a)^4,x, algorithm="giac")

[Out]

-1/35*(5*sin(b*x + a)^7 - 7*sin(b*x + a)^5)/b

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maple [B]  time = 0.02, size = 58, normalized size = 1.87 \[ \frac {-\frac {\left (\cos ^{4}\left (b x +a \right )\right ) \left (\sin ^{3}\left (b x +a \right )\right )}{7}-\frac {3 \sin \left (b x +a \right ) \left (\cos ^{4}\left (b x +a \right )\right )}{35}+\frac {\left (2+\cos ^{2}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{35}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^3*sin(b*x+a)^4,x)

[Out]

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

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maxima [A]  time = 0.32, size = 26, normalized size = 0.84 \[ -\frac {5 \, \sin \left (b x + a\right )^{7} - 7 \, \sin \left (b x + a\right )^{5}}{35 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3*sin(b*x+a)^4,x, algorithm="maxima")

[Out]

-1/35*(5*sin(b*x + a)^7 - 7*sin(b*x + a)^5)/b

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mupad [B]  time = 0.03, size = 26, normalized size = 0.84 \[ \frac {7\,{\sin \left (a+b\,x\right )}^5-5\,{\sin \left (a+b\,x\right )}^7}{35\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^3*sin(a + b*x)^4,x)

[Out]

(7*sin(a + b*x)^5 - 5*sin(a + b*x)^7)/(35*b)

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sympy [A]  time = 7.36, size = 44, normalized size = 1.42 \[ \begin {cases} \frac {2 \sin ^{7}{\left (a + b x \right )}}{35 b} + \frac {\sin ^{5}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{5 b} & \text {for}\: b \neq 0 \\x \sin ^{4}{\relax (a )} \cos ^{3}{\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**3*sin(b*x+a)**4,x)

[Out]

Piecewise((2*sin(a + b*x)**7/(35*b) + sin(a + b*x)**5*cos(a + b*x)**2/(5*b), Ne(b, 0)), (x*sin(a)**4*cos(a)**3
, True))

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