3.3 \(\int \frac{A+B \sin (x)}{1-\cos (x)} \, dx\)

Optimal. Leaf size=23 \[ B \log (1-\cos (x))-\frac{A \sin (x)}{1-\cos (x)} \]

[Out]

B*Log[1 - Cos[x]] - (A*Sin[x])/(1 - Cos[x])

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Rubi [A]  time = 0.0783222, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {4401, 2648, 2667, 31} \[ B \log (1-\cos (x))-\frac{A \sin (x)}{1-\cos (x)} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sin[x])/(1 - Cos[x]),x]

[Out]

B*Log[1 - Cos[x]] - (A*Sin[x])/(1 - Cos[x])

Rule 4401

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

Rule 2648

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

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{A+B \sin (x)}{1-\cos (x)} \, dx &=\int \left (-\frac{A}{-1+\cos (x)}-\frac{B \sin (x)}{-1+\cos (x)}\right ) \, dx\\ &=-\left (A \int \frac{1}{-1+\cos (x)} \, dx\right )-B \int \frac{\sin (x)}{-1+\cos (x)} \, dx\\ &=-\frac{A \sin (x)}{1-\cos (x)}+B \operatorname{Subst}\left (\int \frac{1}{-1+x} \, dx,x,\cos (x)\right )\\ &=B \log (1-\cos (x))-\frac{A \sin (x)}{1-\cos (x)}\\ \end{align*}

Mathematica [A]  time = 0.0465049, size = 20, normalized size = 0.87 \[ 2 B \log \left (\sin \left (\frac{x}{2}\right )\right )-A \cot \left (\frac{x}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sin[x])/(1 - Cos[x]),x]

[Out]

-(A*Cot[x/2]) + 2*B*Log[Sin[x/2]]

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Maple [A]  time = 0.02, size = 31, normalized size = 1.4 \begin{align*} -B\ln \left ( \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) -{A \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{-1}}+2\,B\ln \left ( \tan \left ( x/2 \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sin(x))/(1-cos(x)),x)

[Out]

-B*ln(tan(1/2*x)^2+1)-A/tan(1/2*x)+2*B*ln(tan(1/2*x))

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Maxima [A]  time = 0.95204, size = 26, normalized size = 1.13 \begin{align*} B \log \left (\cos \left (x\right ) - 1\right ) - \frac{A{\left (\cos \left (x\right ) + 1\right )}}{\sin \left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(x))/(1-cos(x)),x, algorithm="maxima")

[Out]

B*log(cos(x) - 1) - A*(cos(x) + 1)/sin(x)

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Fricas [A]  time = 2.11904, size = 77, normalized size = 3.35 \begin{align*} \frac{B \log \left (-\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) \sin \left (x\right ) - A \cos \left (x\right ) - A}{\sin \left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(x))/(1-cos(x)),x, algorithm="fricas")

[Out]

(B*log(-1/2*cos(x) + 1/2)*sin(x) - A*cos(x) - A)/sin(x)

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Sympy [A]  time = 0.680834, size = 27, normalized size = 1.17 \begin{align*} - \frac{A}{\tan{\left (\frac{x}{2} \right )}} - B \log{\left (\tan ^{2}{\left (\frac{x}{2} \right )} + 1 \right )} + 2 B \log{\left (\tan{\left (\frac{x}{2} \right )} \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(x))/(1-cos(x)),x)

[Out]

-A/tan(x/2) - B*log(tan(x/2)**2 + 1) + 2*B*log(tan(x/2))

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Giac [A]  time = 1.17906, size = 53, normalized size = 2.3 \begin{align*} -B \log \left (\tan \left (\frac{1}{2} \, x\right )^{2} + 1\right ) + 2 \, B \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) \right |}\right ) - \frac{2 \, B \tan \left (\frac{1}{2} \, x\right ) + A}{\tan \left (\frac{1}{2} \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(x))/(1-cos(x)),x, algorithm="giac")

[Out]

-B*log(tan(1/2*x)^2 + 1) + 2*B*log(abs(tan(1/2*x))) - (2*B*tan(1/2*x) + A)/tan(1/2*x)