3.949 \(\int \frac {-\csc ^3(a+b x)+\sec ^3(a+b x)}{\csc ^3(a+b x)+\sec ^3(a+b x)} \, dx\)

Optimal. Leaf size=54 \[ \frac {2 \log \left (\tan ^2(a+b x)-\tan (a+b x)+1\right )}{3 b}-\frac {\log (\tan (a+b x)+1)}{3 b}+\frac {\log (\cos (a+b x))}{b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.53, antiderivative size = 54, normalized size of antiderivative = 1.00, 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 = {6725, 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[(-Csc[a + b*x]^3 + Sec[a + b*x]^3)/(Csc[a + b*x]^3 + Sec[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 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]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {-\csc ^3(a+b x)+\sec ^3(a+b x)}{\csc ^3(a+b x)+\sec ^3(a+b x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {-1+x^3}{\left (1+x^2\right ) \left (1+x^3\right )} \, 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 = 0.23, size = 42, normalized size = 0.78 \[ \frac {2 \log (2-\sin (2 (a+b x)))}{3 b}-\frac {\log (\sin (a+b x)+\cos (a+b x))}{3 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(-Csc[a + b*x]^3 + Sec[a + b*x]^3)/(Csc[a + b*x]^3 + Sec[a + b*x]^3),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.94, size = 42, normalized size = 0.78 \[ -\frac {\log \left (2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right ) - 4 \, \log \left (-\cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right )}{6 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-csc(b*x+a)^3+sec(b*x+a)^3)/(csc(b*x+a)^3+sec(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

________________________________________________________________________________________

giac [A]  time = 0.39, size = 52, normalized size = 0.96 \[ \frac {4 \, \log \left (\tan \left (b x + a\right )^{2} - \tan \left (b x + a\right ) + 1\right ) - 3 \, \log \left (\tan \left (b x + a\right )^{2} + 1\right ) - 2 \, \log \left ({\left | \tan \left (b x + a\right ) + 1 \right |}\right )}{6 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-csc(b*x+a)^3+sec(b*x+a)^3)/(csc(b*x+a)^3+sec(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

________________________________________________________________________________________

maple [A]  time = 0.83, size = 56, normalized size = 1.04 \[ \frac {2 \ln \left (1-\tan \left (b x +a \right )+\tan ^{2}\left (b x +a \right )\right )}{3 b}-\frac {\ln \left (1+\tan ^{2}\left (b x +a \right )\right )}{2 b}-\frac {\ln \left (1+\tan \left (b x +a \right )\right )}{3 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-csc(b*x+a)^3+sec(b*x+a)^3)/(csc(b*x+a)^3+sec(b*x+a)^3),x)

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.44, size = 154, normalized size = 2.85 \[ \frac {2 \, \log \left (-\frac {2 \, \sin \left (b x + a\right )}{\cos \left (b x + a\right ) + 1} + \frac {2 \, \sin \left (b x + a\right )^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac {2 \, \sin \left (b x + a\right )^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}} + \frac {\sin \left (b x + a\right )^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} + 1\right ) - \log \left (-\frac {2 \, \sin \left (b x + a\right )}{\cos \left (b x + a\right ) + 1} + \frac {\sin \left (b x + a\right )^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} - 1\right ) - 3 \, \log \left (\frac {\sin \left (b x + a\right )^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + 1\right )}{3 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-csc(b*x+a)^3+sec(b*x+a)^3)/(csc(b*x+a)^3+sec(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

________________________________________________________________________________________

mupad [B]  time = 3.23, size = 106, normalized size = 1.96 \[ \frac {2\,\ln \left ({\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^3+2\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2-2\,\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )+1\right )}{3\,b}-\frac {\ln \left ({\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2-2\,\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )-1\right )}{3\,b}-\frac {\ln \left ({\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2+1\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/cos(a + b*x)^3 - 1/sin(a + b*x)^3)/(1/cos(a + b*x)^3 + 1/sin(a + b*x)^3),x)

[Out]

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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-csc(b*x+a)**3+sec(b*x+a)**3)/(csc(b*x+a)**3+sec(b*x+a)**3),x)

[Out]

Timed out

________________________________________________________________________________________