\(\int \cos ^3(a+b x) \sin ^4(a+b x) \, dx\) [82]

Optimal result

Integrand size = 17, antiderivative size = 31 \[ \int \cos ^3(a+b x) \sin ^4(a+b x) \, dx=\frac {\sin ^5(a+b x)}{5 b}-\frac {\sin ^7(a+b x)}{7 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2644, 14} \[ \int \cos ^3(a+b x) \sin ^4(a+b x) \, dx=\frac {\sin ^5(a+b x)}{5 b}-\frac {\sin ^7(a+b x)}{7 b} \]

[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 2644

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*} \text {integral}& = \frac {\text {Subst}\left (\int x^4 \left (1-x^2\right ) \, dx,x,\sin (a+b x)\right )}{b} \\ & = \frac {\text {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*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \cos ^3(a+b x) \sin ^4(a+b x) \, dx=\frac {(9+5 \cos (2 (a+b x))) \sin ^5(a+b x)}{70 b} \]

[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)

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(\frac {-\frac {\left (\sin ^{7}\left (b x +a \right )\right )}{7}+\frac {\left (\sin ^{5}\left (b x +a \right )\right )}{5}}{b}\)	\(26\)
default	\(\frac {-\frac {\left (\sin ^{7}\left (b x +a \right )\right )}{7}+\frac {\left (\sin ^{5}\left (b x +a \right )\right )}{5}}{b}\)	\(26\)
risch	\(\frac {3 \sin \left (b x +a \right )}{64 b}+\frac {\sin \left (7 b x +7 a \right )}{448 b}-\frac {\sin \left (5 b x +5 a \right )}{320 b}-\frac {\sin \left (3 b x +3 a \right )}{64 b}\)	\(55\)
norman	\(\frac {\frac {32 \left (\tan ^{5}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{5 b}-\frac {192 \left (\tan ^{7}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{35 b}+\frac {32 \left (\tan ^{9}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{5 b}}{\left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )^{7}}\)	\(66\)
parallelrisch	\(\frac {\left (\sin \left (\frac {5 b x}{2}+\frac {5 a}{2}\right )-5 \sin \left (\frac {3 b x}{2}+\frac {3 a}{2}\right )+10 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \left (9+5 \cos \left (2 b x +2 a \right )\right ) \left (\cos \left (\frac {5 b x}{2}+\frac {5 a}{2}\right )+5 \cos \left (\frac {3 b x}{2}+\frac {3 a}{2}\right )+10 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{560 b}\)	\(83\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.32 \[ \int \cos ^3(a+b x) \sin ^4(a+b x) \, dx=\frac {{\left (5 \, \cos \left (b x + a\right )^{6} - 8 \, \cos \left (b x + a\right )^{4} + \cos \left (b x + a\right )^{2} + 2\right )} \sin \left (b x + a\right )}{35 \, b} \]

[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

Sympy [A] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \[ \int \cos ^3(a+b x) \sin ^4(a+b x) \, dx=\begin {cases} \frac {2 \sin ^{7}{\left (a + b x \right )}}{35 b} + \frac {\sin ^{5}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{5 b} & \text {for}\: b \neq 0 \\x \sin ^{4}{\left (a \right )} \cos ^{3}{\left (a \right )} & \text {otherwise} \end {cases} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \cos ^3(a+b x) \sin ^4(a+b x) \, dx=-\frac {5 \, \sin \left (b x + a\right )^{7} - 7 \, \sin \left (b x + a\right )^{5}}{35 \, b} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \cos ^3(a+b x) \sin ^4(a+b x) \, dx=-\frac {5 \, \sin \left (b x + a\right )^{7} - 7 \, \sin \left (b x + a\right )^{5}}{35 \, b} \]

[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

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \cos ^3(a+b x) \sin ^4(a+b x) \, dx=\frac {7\,{\sin \left (a+b\,x\right )}^5-5\,{\sin \left (a+b\,x\right )}^7}{35\,b} \]

[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)