3.25 \(\int \sin (\log (a+b x)) \, dx\)

Optimal. Leaf size=39 \[ \frac {(a+b x) \sin (\log (a+b x))}{2 b}-\frac {(a+b x) \cos (\log (a+b x))}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4475} \[ \frac {(a+b x) \sin (\log (a+b x))}{2 b}-\frac {(a+b x) \cos (\log (a+b x))}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[Log[a + b*x]],x]

[Out]

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

Rule 4475

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*} \int \sin (\log (a+b x)) \, dx &=\frac {\operatorname {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*}

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Mathematica [A]  time = 0.01, size = 29, normalized size = 0.74 \[ -\frac {(a+b x) (\cos (\log (a+b x))-\sin (\log (a+b x)))}{2 b} \]

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.79, size = 33, normalized size = 0.85 \[ -\frac {{\left (b x + a\right )} \cos \left (\log \left (b x + a\right )\right ) - {\left (b x + a\right )} \sin \left (\log \left (b x + a\right )\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.16, size = 35, normalized size = 0.90 \[ -\frac {{\left (b x + a\right )} \cos \left (\log \left (b x + a\right )\right )}{2 \, b} + \frac {{\left (b x + a\right )} \sin \left (\log \left (b x + a\right )\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [B]  time = 0.02, size = 76, normalized size = 1.95 \[ \frac {x \tan \left (\frac {\ln \left (b x +a \right )}{2}\right )+\frac {a \tan \left (\frac {\ln \left (b x +a \right )}{2}\right )}{b}+\frac {a \left (\tan ^{2}\left (\frac {\ln \left (b x +a \right )}{2}\right )\right )}{b}-\frac {x}{2}+\frac {x \left (\tan ^{2}\left (\frac {\ln \left (b x +a \right )}{2}\right )\right )}{2}}{1+\tan ^{2}\left (\frac {\ln \left (b x +a \right )}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*tan(1/2*ln(b*x+a))+a/b*tan(1/2*ln(b*x+a))+a/b*tan(1/2*ln(b*x+a))^2-1/2*x+1/2*x*tan(1/2*ln(b*x+a))^2)/(1+tan
(1/2*ln(b*x+a))^2)

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maxima [A]  time = 0.32, size = 27, normalized size = 0.69 \[ -\frac {{\left (b x + a\right )} {\left (\cos \left (\log \left (b x + a\right )\right ) - \sin \left (\log \left (b x + a\right )\right )\right )}}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 2.16, size = 36, normalized size = 0.92 \[ \left \{\begin {array}{cl} x\,\sin \left (\ln \relax (a)\right ) & \text {\ if\ \ }b=0\\ -\frac {\sqrt {2}\,\cos \left (\frac {\pi }{4}+\ln \left (a+b\,x\right )\right )\,\left (a+b\,x\right )}{2\,b} & \text {\ if\ \ }b\neq 0 \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.70, size = 56, normalized size = 1.44 \[ \begin {cases} \frac {a \sin {\left (\log {\left (a + b x \right )} \right )}}{2 b} - \frac {a \cos {\left (\log {\left (a + b x \right )} \right )}}{2 b} + \frac {x \sin {\left (\log {\left (a + b x \right )} \right )}}{2} - \frac {x \cos {\left (\log {\left (a + b x \right )} \right )}}{2} & \text {for}\: b \neq 0 \\x \sin {\left (\log {\relax (a )} \right )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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