3.944 \(\int \frac{\cos ^3(a+b x)-\sin ^3(a+b x)}{\cos ^3(a+b x)+\sin ^3(a+b x)} \, dx\)

Optimal. Leaf size=55 \[ -\frac{2 \log \left (\tan ^2(a+b x)-\tan (a+b x)+1\right )}{3 b}+\frac{\log (\tan (a+b x)+1)}{3 b}-\frac{\log (\cos (a+b x))}{b} \]

[Out]

-(Log[Cos[a + b*x]]/b) + Log[1 + Tan[a + b*x]]/(3*b) - (2*Log[1 - Tan[a + b*x] + Tan[a + b*x]^2])/(3*b)

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Rubi [A]  time = 0.408613, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {2074, 260, 628} \[ -\frac{2 \log \left (\tan ^2(a+b x)-\tan (a+b x)+1\right )}{3 b}+\frac{\log (\tan (a+b x)+1)}{3 b}-\frac{\log (\cos (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[a + b*x]^3 - Sin[a + b*x]^3)/(Cos[a + b*x]^3 + Sin[a + b*x]^3),x]

[Out]

-(Log[Cos[a + b*x]]/b) + Log[1 + Tan[a + b*x]]/(3*b) - (2*Log[1 - Tan[a + b*x] + Tan[a + b*x]^2])/(3*b)

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rule 260

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

Rule 628

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

Rubi steps

\begin{align*} \int \frac{\cos ^3(a+b x)-\sin ^3(a+b x)}{\cos ^3(a+b x)+\sin ^3(a+b x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1-x^3}{1+x^2+x^3+x^5} \, dx,x,\tan (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{3 (1+x)}+\frac{x}{1+x^2}-\frac{2 (-1+2 x)}{3 \left (1-x+x^2\right )}\right ) \, dx,x,\tan (a+b x)\right )}{b}\\ &=\frac{\log (1+\tan (a+b x))}{3 b}-\frac{2 \operatorname{Subst}\left (\int \frac{-1+2 x}{1-x+x^2} \, dx,x,\tan (a+b x)\right )}{3 b}+\frac{\operatorname{Subst}\left (\int \frac{x}{1+x^2} \, dx,x,\tan (a+b x)\right )}{b}\\ &=-\frac{\log (\cos (a+b x))}{b}+\frac{\log (1+\tan (a+b x))}{3 b}-\frac{2 \log \left (1-\tan (a+b x)+\tan ^2(a+b x)\right )}{3 b}\\ \end{align*}

Mathematica [A]  time = 0.204279, size = 42, normalized size = 0.76 \[ \frac{\log (\sin (a+b x)+\cos (a+b x))}{3 b}-\frac{2 \log (2-\sin (2 (a+b x)))}{3 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[a + b*x]^3 - Sin[a + b*x]^3)/(Cos[a + b*x]^3 + Sin[a + b*x]^3),x]

[Out]

Log[Cos[a + b*x] + Sin[a + b*x]]/(3*b) - (2*Log[2 - Sin[2*(a + b*x)]])/(3*b)

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Maple [A]  time = 0.147, size = 56, normalized size = 1. \begin{align*}{\frac{\ln \left ( 1+\tan \left ( bx+a \right ) \right ) }{3\,b}}+{\frac{\ln \left ( 1+ \left ( \tan \left ( bx+a \right ) \right ) ^{2} \right ) }{2\,b}}-{\frac{2\,\ln \left ( 1-\tan \left ( bx+a \right ) + \left ( \tan \left ( bx+a \right ) \right ) ^{2} \right ) }{3\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(b*x+a)^3-sin(b*x+a)^3)/(cos(b*x+a)^3+sin(b*x+a)^3),x)

[Out]

1/3*ln(1+tan(b*x+a))/b+1/2/b*ln(1+tan(b*x+a)^2)-2/3*ln(1-tan(b*x+a)+tan(b*x+a)^2)/b

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Maxima [B]  time = 1.47065, size = 208, normalized size = 3.78 \begin{align*} -\frac{2 \, \log \left (-\frac{2 \, \sin \left (b x + a\right )}{\cos \left (b x + a\right ) + 1} + \frac{2 \, \sin \left (b x + a\right )^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac{2 \, \sin \left (b x + a\right )^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}} + \frac{\sin \left (b x + a\right )^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} + 1\right ) - \log \left (-\frac{2 \, \sin \left (b x + a\right )}{\cos \left (b x + a\right ) + 1} + \frac{\sin \left (b x + a\right )^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} - 1\right ) - 3 \, \log \left (\frac{\sin \left (b x + a\right )^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + 1\right )}{3 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((cos(b*x+a)^3-sin(b*x+a)^3)/(cos(b*x+a)^3+sin(b*x+a)^3),x, algorithm="maxima")

[Out]

-1/3*(2*log(-2*sin(b*x + a)/(cos(b*x + a) + 1) + 2*sin(b*x + a)^2/(cos(b*x + a) + 1)^2 + 2*sin(b*x + a)^3/(cos
(b*x + a) + 1)^3 + sin(b*x + a)^4/(cos(b*x + a) + 1)^4 + 1) - log(-2*sin(b*x + a)/(cos(b*x + a) + 1) + sin(b*x
+ a)^2/(cos(b*x + a) + 1)^2 - 1) - 3*log(sin(b*x + a)^2/(cos(b*x + a) + 1)^2 + 1))/b

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Fricas [A]  time = 2.23759, size = 116, normalized size = 2.11 \begin{align*} \frac{\log \left (2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right ) - 4 \, \log \left (-\cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right )}{6 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((cos(b*x+a)^3-sin(b*x+a)^3)/(cos(b*x+a)^3+sin(b*x+a)^3),x, algorithm="fricas")

[Out]

1/6*(log(2*cos(b*x + a)*sin(b*x + a) + 1) - 4*log(-cos(b*x + a)*sin(b*x + a) + 1))/b

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Sympy [A]  time = 1.74592, size = 76, normalized size = 1.38 \begin{align*} \begin{cases} \frac{\log{\left (\sin{\left (a + b x \right )} + \cos{\left (a + b x \right )} \right )}}{3 b} - \frac{2 \log{\left (\sin ^{2}{\left (a + b x \right )} - \sin{\left (a + b x \right )} \cos{\left (a + b x \right )} + \cos ^{2}{\left (a + b x \right )} \right )}}{3 b} & \text{for}\: b \neq 0 \\\frac{x \left (- \sin ^{3}{\left (a \right )} + \cos ^{3}{\left (a \right )}\right )}{\sin ^{3}{\left (a \right )} + \cos ^{3}{\left (a \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((cos(b*x+a)**3-sin(b*x+a)**3)/(cos(b*x+a)**3+sin(b*x+a)**3),x)

[Out]

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

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Giac [A]  time = 1.17244, size = 70, normalized size = 1.27 \begin{align*} -\frac{4 \, \log \left (\tan \left (b x + a\right )^{2} - \tan \left (b x + a\right ) + 1\right ) - 3 \, \log \left (\tan \left (b x + a\right )^{2} + 1\right ) - 2 \, \log \left ({\left | \tan \left (b x + a\right ) + 1 \right |}\right )}{6 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((cos(b*x+a)^3-sin(b*x+a)^3)/(cos(b*x+a)^3+sin(b*x+a)^3),x, algorithm="giac")

[Out]

-1/6*(4*log(tan(b*x + a)^2 - tan(b*x + a) + 1) - 3*log(tan(b*x + a)^2 + 1) - 2*log(abs(tan(b*x + a) + 1)))/b