\(\int \sin (\log (a+b x)) \, dx\) [25]

Optimal result

Integrand size = 7, antiderivative size = 39 \[ \int \sin (\log (a+b x)) \, dx=-\frac {(a+b x) \cos (\log (a+b x))}{2 b}+\frac {(a+b x) \sin (\log (a+b x))}{2 b} \]

[Out]

-1/2*(b*x+a)*cos(ln(b*x+a))/b+1/2*(b*x+a)*sin(ln(b*x+a))/b

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4563} \[ \int \sin (\log (a+b x)) \, dx=\frac {(a+b x) \sin (\log (a+b x))}{2 b}-\frac {(a+b x) \cos (\log (a+b x))}{2 b} \]

[In]

Int[Sin[Log[a + b*x]],x]

[Out]

-1/2*((a + b*x)*Cos[Log[a + b*x]])/b + ((a + b*x)*Sin[Log[a + b*x]])/(2*b)

Rule 4563

Int[Sin[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)], x_Symbol] :> Simp[x*(Sin[d*(a + b*Log[c*x^n])]/(b^2*d^2*
n^2 + 1)), x] - Simp[b*d*n*x*(Cos[d*(a + b*Log[c*x^n])]/(b^2*d^2*n^2 + 1)), x] /; FreeQ[{a, b, c, d, n}, x] &&
NeQ[b^2*d^2*n^2 + 1, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}(\int \sin (\log (x)) \, dx,x,a+b x)}{b} \\ & = -\frac {(a+b x) \cos (\log (a+b x))}{2 b}+\frac {(a+b x) \sin (\log (a+b x))}{2 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.74 \[ \int \sin (\log (a+b x)) \, dx=-\frac {(a+b x) (\cos (\log (a+b x))-\sin (\log (a+b x)))}{2 b} \]

[In]

Integrate[Sin[Log[a + b*x]],x]

[Out]

-1/2*((a + b*x)*(Cos[Log[a + b*x]] - Sin[Log[a + b*x]]))/b

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.87

method	result	size
default	\(\frac {-\frac {\left (x b +a \right ) \cos \left (\ln \left (x b +a \right )\right )}{2}+\frac {\left (x b +a \right ) \sin \left (\ln \left (x b +a \right )\right )}{2}}{b}\)	\(34\)
risch	\(\frac {\left (-\frac {1}{4}-\frac {i}{4}\right ) \left (x b +a \right ) \left (x b +a \right )^{i}}{b}+\frac {\left (-\frac {1}{4}+\frac {i}{4}\right ) \left (x b +a \right ) \left (x b +a \right )^{-i}}{b}\)	\(44\)
parallelrisch	\(\frac {\left (x b -a \right ) \tan \left (\ln \left (\sqrt {x b +a}\right )\right )^{2}+\left (2 x b +2 a \right ) \tan \left (\ln \left (\sqrt {x b +a}\right )\right )-x b -3 a}{2 b \left (1+\tan \left (\ln \left (\sqrt {x b +a}\right )\right )^{2}\right )}\)	\(66\)
norman	\(\frac {x \tan \left (\frac {\ln \left (x b +a \right )}{2}\right )+\frac {a \tan \left (\frac {\ln \left (x b +a \right )}{2}\right )}{b}+\frac {a \tan \left (\frac {\ln \left (x b +a \right )}{2}\right )^{2}}{b}-\frac {x}{2}+\frac {x \tan \left (\frac {\ln \left (x b +a \right )}{2}\right )^{2}}{2}}{1+\tan \left (\frac {\ln \left (x b +a \right )}{2}\right )^{2}}\)	\(76\)

[In]

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

[Out]

1/b*(-1/2*(b*x+a)*cos(ln(b*x+a))+1/2*(b*x+a)*sin(ln(b*x+a)))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.85 \[ \int \sin (\log (a+b x)) \, dx=-\frac {{\left (b x + a\right )} \cos \left (\log \left (b x + a\right )\right ) - {\left (b x + a\right )} \sin \left (\log \left (b x + a\right )\right )}{2 \, b} \]

[In]

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

[Out]

-1/2*((b*x + a)*cos(log(b*x + a)) - (b*x + a)*sin(log(b*x + a)))/b

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.44 \[ \int \sin (\log (a+b x)) \, dx=\begin {cases} \frac {a \sin {\left (\log {\left (a + b x \right )} \right )}}{2 b} - \frac {a \cos {\left (\log {\left (a + b x \right )} \right )}}{2 b} + \frac {x \sin {\left (\log {\left (a + b x \right )} \right )}}{2} - \frac {x \cos {\left (\log {\left (a + b x \right )} \right )}}{2} & \text {for}\: b \neq 0 \\x \sin {\left (\log {\left (a \right )} \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a*sin(log(a + b*x))/(2*b) - a*cos(log(a + b*x))/(2*b) + x*sin(log(a + b*x))/2 - x*cos(log(a + b*x))
/2, Ne(b, 0)), (x*sin(log(a)), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.69 \[ \int \sin (\log (a+b x)) \, dx=-\frac {{\left (b x + a\right )} {\left (\cos \left (\log \left (b x + a\right )\right ) - \sin \left (\log \left (b x + a\right )\right )\right )}}{2 \, b} \]

[In]

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

[Out]

-1/2*(b*x + a)*(cos(log(b*x + a)) - sin(log(b*x + a)))/b

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.90 \[ \int \sin (\log (a+b x)) \, dx=-\frac {{\left (b x + a\right )} \cos \left (\log \left (b x + a\right )\right )}{2 \, b} + \frac {{\left (b x + a\right )} \sin \left (\log \left (b x + a\right )\right )}{2 \, b} \]

[In]

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

[Out]

-1/2*(b*x + a)*cos(log(b*x + a))/b + 1/2*(b*x + a)*sin(log(b*x + a))/b

Mupad [B] (verification not implemented)

Time = 27.84 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.92 \[ \int \sin (\log (a+b x)) \, dx=\left \{\begin {array}{cl} x\,\sin \left (\ln \left (a\right )\right ) & \text {\ if\ \ }b=0\\ -\frac {\sqrt {2}\,\cos \left (\frac {\pi }{4}+\ln \left (a+b\,x\right )\right )\,\left (a+b\,x\right )}{2\,b} & \text {\ if\ \ }b\neq 0 \end {array}\right . \]

[In]

int(sin(log(a + b*x)),x)

[Out]

piecewise(b == 0, x*sin(log(a)), b ~= 0, -(2^(1/2)*cos(pi/4 + log(a + b*x))*(a + b*x))/(2*b))