\(\int \sin (\frac {x}{4}) \sin (x) \, dx\) [29]

Optimal result

Integrand size = 9, antiderivative size = 21 \[ \int \sin \left (\frac {x}{4}\right ) \sin (x) \, dx=\frac {2}{3} \sin \left (\frac {3 x}{4}\right )-\frac {2}{5} \sin \left (\frac {5 x}{4}\right ) \]

[Out]

2/3*sin(3/4*x)-2/5*sin(5/4*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4367} \[ \int \sin \left (\frac {x}{4}\right ) \sin (x) \, dx=\frac {2}{3} \sin \left (\frac {3 x}{4}\right )-\frac {2}{5} \sin \left (\frac {5 x}{4}\right ) \]

[In]

Int[Sin[x/4]*Sin[x],x]

[Out]

(2*Sin[(3*x)/4])/3 - (2*Sin[(5*x)/4])/5

Rule 4367

Int[sin[(a_.) + (b_.)*(x_)]*sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[a - c + (b - d)*x]/(2*(b - d)), x]
- Simp[Sin[a + c + (b + d)*x]/(2*(b + d)), x] /; FreeQ[{a, b, c, d}, x] && NeQ[b^2 - d^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {2}{3} \sin \left (\frac {3 x}{4}\right )-\frac {2}{5} \sin \left (\frac {5 x}{4}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \sin \left (\frac {x}{4}\right ) \sin (x) \, dx=\frac {2}{3} \sin \left (\frac {3 x}{4}\right )-\frac {2}{5} \sin \left (\frac {5 x}{4}\right ) \]

[In]

Integrate[Sin[x/4]*Sin[x],x]

[Out]

(2*Sin[(3*x)/4])/3 - (2*Sin[(5*x)/4])/5

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67

method	result	size
risch	\(\frac {2 \sin \left (\frac {3 x}{4}\right )}{3}-\frac {2 \sin \left (\frac {5 x}{4}\right )}{5}\)	\(14\)
parallelrisch	\(\frac {2 \sin \left (\frac {3 x}{4}\right )}{3}-\frac {2 \sin \left (\frac {5 x}{4}\right )}{5}\)	\(14\)
derivativedivides	\(-\frac {32 \left (\sin ^{5}\left (\frac {x}{4}\right )\right )}{5}+\frac {16 \left (\sin ^{3}\left (\frac {x}{4}\right )\right )}{3}\)	\(18\)
default	\(-\frac {32 \left (\sin ^{5}\left (\frac {x}{4}\right )\right )}{5}+\frac {16 \left (\sin ^{3}\left (\frac {x}{4}\right )\right )}{3}\)	\(18\)
norman	\(\frac {-\frac {8 \tan \left (\frac {x}{2}\right ) \left (\tan ^{2}\left (\frac {x}{8}\right )\right )}{15}+\frac {32 \left (\tan ^{2}\left (\frac {x}{2}\right )\right ) \tan \left (\frac {x}{8}\right )}{15}+\frac {8 \tan \left (\frac {x}{2}\right )}{15}-\frac {32 \tan \left (\frac {x}{8}\right )}{15}}{\left (1+\tan ^{2}\left (\frac {x}{8}\right )\right ) \left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )}\)	\(59\)

[In]

int(sin(1/4*x)*sin(x),x,method=_RETURNVERBOSE)

[Out]

2/3*sin(3/4*x)-2/5*sin(5/4*x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \sin \left (\frac {x}{4}\right ) \sin (x) \, dx=-\frac {16}{15} \, {\left (6 \, \cos \left (\frac {1}{4} \, x\right )^{4} - 7 \, \cos \left (\frac {1}{4} \, x\right )^{2} + 1\right )} \sin \left (\frac {1}{4} \, x\right ) \]

[In]

integrate(sin(1/4*x)*sin(x),x, algorithm="fricas")

[Out]

-16/15*(6*cos(1/4*x)^4 - 7*cos(1/4*x)^2 + 1)*sin(1/4*x)

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \sin \left (\frac {x}{4}\right ) \sin (x) \, dx=- \frac {16 \sin {\left (\frac {x}{4} \right )} \cos {\left (x \right )}}{15} + \frac {4 \sin {\left (x \right )} \cos {\left (\frac {x}{4} \right )}}{15} \]

[In]

integrate(sin(1/4*x)*sin(x),x)

[Out]

-16*sin(x/4)*cos(x)/15 + 4*sin(x)*cos(x/4)/15

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \sin \left (\frac {x}{4}\right ) \sin (x) \, dx=-\frac {2}{5} \, \sin \left (\frac {5}{4} \, x\right ) + \frac {2}{3} \, \sin \left (\frac {3}{4} \, x\right ) \]

[In]

integrate(sin(1/4*x)*sin(x),x, algorithm="maxima")

[Out]

-2/5*sin(5/4*x) + 2/3*sin(3/4*x)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \sin \left (\frac {x}{4}\right ) \sin (x) \, dx=-\frac {32}{5} \, \sin \left (\frac {1}{4} \, x\right )^{5} + \frac {16}{3} \, \sin \left (\frac {1}{4} \, x\right )^{3} \]

[In]

integrate(sin(1/4*x)*sin(x),x, algorithm="giac")

[Out]

-32/5*sin(1/4*x)^5 + 16/3*sin(1/4*x)^3

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \sin \left (\frac {x}{4}\right ) \sin (x) \, dx=\frac {2\,\sin \left (\frac {3\,x}{4}\right )}{3}-\frac {2\,\sin \left (\frac {5\,x}{4}\right )}{5} \]

[In]

int(sin(x/4)*sin(x),x)

[Out]

(2*sin((3*x)/4))/3 - (2*sin((5*x)/4))/5