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

Optimal result

Integrand size = 8, antiderivative size = 46 \[ \int \sin ^4(a+b x) \, dx=\frac {3 x}{8}-\frac {3 \cos (a+b x) \sin (a+b x)}{8 b}-\frac {\cos (a+b x) \sin ^3(a+b x)}{4 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2715, 8} \[ \int \sin ^4(a+b x) \, dx=-\frac {\sin ^3(a+b x) \cos (a+b x)}{4 b}-\frac {3 \sin (a+b x) \cos (a+b x)}{8 b}+\frac {3 x}{8} \]

[In]

Int[Sin[a + b*x]^4,x]

[Out]

(3*x)/8 - (3*Cos[a + b*x]*Sin[a + b*x])/(8*b) - (Cos[a + b*x]*Sin[a + b*x]^3)/(4*b)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rubi steps \begin{align*} \text {integral}& = -\frac {\cos (a+b x) \sin ^3(a+b x)}{4 b}+\frac {3}{4} \int \sin ^2(a+b x) \, dx \\ & = -\frac {3 \cos (a+b x) \sin (a+b x)}{8 b}-\frac {\cos (a+b x) \sin ^3(a+b x)}{4 b}+\frac {3 \int 1 \, dx}{8} \\ & = \frac {3 x}{8}-\frac {3 \cos (a+b x) \sin (a+b x)}{8 b}-\frac {\cos (a+b x) \sin ^3(a+b x)}{4 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.72 \[ \int \sin ^4(a+b x) \, dx=\frac {12 (a+b x)-8 \sin (2 (a+b x))+\sin (4 (a+b x))}{32 b} \]

[In]

Integrate[Sin[a + b*x]^4,x]

[Out]

(12*(a + b*x) - 8*Sin[2*(a + b*x)] + Sin[4*(a + b*x)])/(32*b)

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.67

method	result	size
parallelrisch	\(\frac {12 b x +\sin \left (4 b x +4 a \right )-8 \sin \left (2 b x +2 a \right )}{32 b}\)	\(31\)
risch	\(\frac {3 x}{8}+\frac {\sin \left (4 b x +4 a \right )}{32 b}-\frac {\sin \left (2 b x +2 a \right )}{4 b}\)	\(33\)
derivativedivides	\(\frac {-\frac {\left (\sin ^{3}\left (b x +a \right )+\frac {3 \sin \left (b x +a \right )}{2}\right ) \cos \left (b x +a \right )}{4}+\frac {3 b x}{8}+\frac {3 a}{8}}{b}\)	\(38\)
default	\(\frac {-\frac {\left (\sin ^{3}\left (b x +a \right )+\frac {3 \sin \left (b x +a \right )}{2}\right ) \cos \left (b x +a \right )}{4}+\frac {3 b x}{8}+\frac {3 a}{8}}{b}\)	\(38\)
norman	\(\frac {\frac {3 x}{8}-\frac {3 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{4 b}-\frac {11 \left (\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{4 b}+\frac {11 \left (\tan ^{5}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{4 b}+\frac {3 \left (\tan ^{7}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{4 b}+\frac {3 x \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{2}+\frac {9 x \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{4}+\frac {3 x \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{2}+\frac {3 x \left (\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8}}{\left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )^{4}}\)	\(139\)

[In]

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

[Out]

1/32*(12*b*x+sin(4*b*x+4*a)-8*sin(2*b*x+2*a))/b

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \sin ^4(a+b x) \, dx=\frac {3 \, b x + {\left (2 \, \cos \left (b x + a\right )^{3} - 5 \, \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{8 \, b} \]

[In]

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

[Out]

1/8*(3*b*x + (2*cos(b*x + a)^3 - 5*cos(b*x + a))*sin(b*x + a))/b

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (41) = 82\).

Time = 0.15 (sec) , antiderivative size = 95, normalized size of antiderivative = 2.07 \[ \int \sin ^4(a+b x) \, dx=\begin {cases} \frac {3 x \sin ^{4}{\left (a + b x \right )}}{8} + \frac {3 x \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{4} + \frac {3 x \cos ^{4}{\left (a + b x \right )}}{8} - \frac {5 \sin ^{3}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{8 b} - \frac {3 \sin {\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{8 b} & \text {for}\: b \neq 0 \\x \sin ^{4}{\left (a \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.72 \[ \int \sin ^4(a+b x) \, dx=\frac {12 \, b x + 12 \, a + \sin \left (4 \, b x + 4 \, a\right ) - 8 \, \sin \left (2 \, b x + 2 \, a\right )}{32 \, b} \]

[In]

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

[Out]

1/32*(12*b*x + 12*a + sin(4*b*x + 4*a) - 8*sin(2*b*x + 2*a))/b

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.70 \[ \int \sin ^4(a+b x) \, dx=\frac {3}{8} \, x + \frac {\sin \left (4 \, b x + 4 \, a\right )}{32 \, b} - \frac {\sin \left (2 \, b x + 2 \, a\right )}{4 \, b} \]

[In]

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

[Out]

3/8*x + 1/32*sin(4*b*x + 4*a)/b - 1/4*sin(2*b*x + 2*a)/b

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.09 \[ \int \sin ^4(a+b x) \, dx=\frac {3\,x}{8}-\frac {\frac {5\,{\mathrm {tan}\left (a+b\,x\right )}^3}{8}+\frac {3\,\mathrm {tan}\left (a+b\,x\right )}{8}}{b\,\left ({\mathrm {tan}\left (a+b\,x\right )}^4+2\,{\mathrm {tan}\left (a+b\,x\right )}^2+1\right )} \]

[In]

int(sin(a + b*x)^4,x)

[Out]

(3*x)/8 - ((3*tan(a + b*x))/8 + (5*tan(a + b*x)^3)/8)/(b*(2*tan(a + b*x)^2 + tan(a + b*x)^4 + 1))