\(\int \frac {-\csc ^4(a+b x)+\sec ^4(a+b x)}{\csc ^4(a+b x)+\sec ^4(a+b x)} \, dx\) [950]

Optimal result

Integrand size = 39, antiderivative size = 72 \[ \int \frac {-\csc ^4(a+b x)+\sec ^4(a+b x)}{\csc ^4(a+b x)+\sec ^4(a+b x)} \, dx=\frac {\log \left (1-\sqrt {2} \tan (a+b x)+\tan ^2(a+b x)\right )}{2 \sqrt {2} b}-\frac {\log \left (1+\sqrt {2} \tan (a+b x)+\tan ^2(a+b x)\right )}{2 \sqrt {2} b} \]

[Out]

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

Rubi [A] (verified)

Time = 1.59 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {1179, 642} \[ \int \frac {-\csc ^4(a+b x)+\sec ^4(a+b x)}{\csc ^4(a+b x)+\sec ^4(a+b x)} \, dx=\frac {\log \left (\tan ^2(a+b x)-\sqrt {2} \tan (a+b x)+1\right )}{2 \sqrt {2} b}-\frac {\log \left (\tan ^2(a+b x)+\sqrt {2} \tan (a+b x)+1\right )}{2 \sqrt {2} b} \]

[In]

Int[(-Csc[a + b*x]^4 + Sec[a + b*x]^4)/(Csc[a + b*x]^4 + Sec[a + b*x]^4),x]

[Out]

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

Rule 642

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 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {-1+x^2}{1+x^4} \, dx,x,\tan (a+b x)\right )}{b} \\ & = \frac {\text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\tan (a+b x)\right )}{2 \sqrt {2} b}+\frac {\text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\tan (a+b x)\right )}{2 \sqrt {2} b} \\ & = \frac {\log \left (1-\sqrt {2} \tan (a+b x)+\tan ^2(a+b x)\right )}{2 \sqrt {2} b}-\frac {\log \left (1+\sqrt {2} \tan (a+b x)+\tan ^2(a+b x)\right )}{2 \sqrt {2} b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.36 \[ \int \frac {-\csc ^4(a+b x)+\sec ^4(a+b x)}{\csc ^4(a+b x)+\sec ^4(a+b x)} \, dx=-\frac {\text {arctanh}\left (\frac {\sin (2 a+2 b x)}{\sqrt {2}}\right )}{\sqrt {2} b} \]

[In]

Integrate[(-Csc[a + b*x]^4 + Sec[a + b*x]^4)/(Csc[a + b*x]^4 + Sec[a + b*x]^4),x]

[Out]

-(ArcTanh[Sin[2*a + 2*b*x]/Sqrt[2]]/(Sqrt[2]*b))

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.02 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00

method	result	size
risch	\(\frac {\sqrt {2}\, \ln \left ({\mathrm e}^{4 i \left (x b +a \right )}-2 i \sqrt {2}\, {\mathrm e}^{2 i \left (x b +a \right )}-1\right )}{4 b}-\frac {\sqrt {2}\, \ln \left ({\mathrm e}^{4 i \left (x b +a \right )}+2 i \sqrt {2}\, {\mathrm e}^{2 i \left (x b +a \right )}-1\right )}{4 b}\)	\(72\)
derivativedivides	\(\frac {-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \tan \left (x b +a \right )+\tan \left (x b +a \right )^{2}}{1-\sqrt {2}\, \tan \left (x b +a \right )+\tan \left (x b +a \right )^{2}}\right )+2 \arctan \left (\sqrt {2}\, \tan \left (x b +a \right )+1\right )+2 \arctan \left (\sqrt {2}\, \tan \left (x b +a \right )-1\right )\right )}{8}+\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \tan \left (x b +a \right )+\tan \left (x b +a \right )^{2}}{1+\sqrt {2}\, \tan \left (x b +a \right )+\tan \left (x b +a \right )^{2}}\right )+2 \arctan \left (\sqrt {2}\, \tan \left (x b +a \right )+1\right )+2 \arctan \left (\sqrt {2}\, \tan \left (x b +a \right )-1\right )\right )}{8}}{b}\)	\(168\)
default	\(\frac {-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \tan \left (x b +a \right )+\tan \left (x b +a \right )^{2}}{1-\sqrt {2}\, \tan \left (x b +a \right )+\tan \left (x b +a \right )^{2}}\right )+2 \arctan \left (\sqrt {2}\, \tan \left (x b +a \right )+1\right )+2 \arctan \left (\sqrt {2}\, \tan \left (x b +a \right )-1\right )\right )}{8}+\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \tan \left (x b +a \right )+\tan \left (x b +a \right )^{2}}{1+\sqrt {2}\, \tan \left (x b +a \right )+\tan \left (x b +a \right )^{2}}\right )+2 \arctan \left (\sqrt {2}\, \tan \left (x b +a \right )+1\right )+2 \arctan \left (\sqrt {2}\, \tan \left (x b +a \right )-1\right )\right )}{8}}{b}\)	\(168\)

[In]

int((-csc(b*x+a)^4+sec(b*x+a)^4)/(csc(b*x+a)^4+sec(b*x+a)^4),x,method=_RETURNVERBOSE)

[Out]

1/4*2^(1/2)/b*ln(exp(4*I*(b*x+a))-2*I*2^(1/2)*exp(2*I*(b*x+a))-1)-1/4*2^(1/2)/b*ln(exp(4*I*(b*x+a))+2*I*2^(1/2
)*exp(2*I*(b*x+a))-1)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.03 \[ \int \frac {-\csc ^4(a+b x)+\sec ^4(a+b x)}{\csc ^4(a+b x)+\sec ^4(a+b x)} \, dx=\frac {\sqrt {2} \log \left (-\frac {2 \, \cos \left (b x + a\right )^{4} + 2 \, \sqrt {2} \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 2 \, \cos \left (b x + a\right )^{2} - 1}{2 \, \cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1}\right )}{4 \, b} \]

[In]

integrate((-csc(b*x+a)^4+sec(b*x+a)^4)/(csc(b*x+a)^4+sec(b*x+a)^4),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {-\csc ^4(a+b x)+\sec ^4(a+b x)}{\csc ^4(a+b x)+\sec ^4(a+b x)} \, dx=- \int \frac {\csc ^{4}{\left (a + b x \right )}}{\csc ^{4}{\left (a + b x \right )} + \sec ^{4}{\left (a + b x \right )}}\, dx - \int \left (- \frac {\sec ^{4}{\left (a + b x \right )}}{\csc ^{4}{\left (a + b x \right )} + \sec ^{4}{\left (a + b x \right )}}\right )\, dx \]

[In]

integrate((-csc(b*x+a)**4+sec(b*x+a)**4)/(csc(b*x+a)**4+sec(b*x+a)**4),x)

[Out]

-Integral(csc(a + b*x)**4/(csc(a + b*x)**4 + sec(a + b*x)**4), x) - Integral(-sec(a + b*x)**4/(csc(a + b*x)**4
+ sec(a + b*x)**4), x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.81 \[ \int \frac {-\csc ^4(a+b x)+\sec ^4(a+b x)}{\csc ^4(a+b x)+\sec ^4(a+b x)} \, dx=-\frac {\sqrt {2} \log \left (\tan \left (b x + a\right )^{2} + \sqrt {2} \tan \left (b x + a\right ) + 1\right ) - \sqrt {2} \log \left (\tan \left (b x + a\right )^{2} - \sqrt {2} \tan \left (b x + a\right ) + 1\right )}{4 \, b} \]

[In]

integrate((-csc(b*x+a)^4+sec(b*x+a)^4)/(csc(b*x+a)^4+sec(b*x+a)^4),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.67 \[ \int \frac {-\csc ^4(a+b x)+\sec ^4(a+b x)}{\csc ^4(a+b x)+\sec ^4(a+b x)} \, dx=\frac {\sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 2 \, \sin \left (2 \, b x + 2 \, a\right ) \right |}}{{\left | 2 \, \sqrt {2} + 2 \, \sin \left (2 \, b x + 2 \, a\right ) \right |}}\right )}{4 \, b} \]

[In]

integrate((-csc(b*x+a)^4+sec(b*x+a)^4)/(csc(b*x+a)^4+sec(b*x+a)^4),x, algorithm="giac")

[Out]

1/4*sqrt(2)*log(abs(-2*sqrt(2) + 2*sin(2*b*x + 2*a))/abs(2*sqrt(2) + 2*sin(2*b*x + 2*a)))/b

Mupad [B] (verification not implemented)

Time = 26.87 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.32 \[ \int \frac {-\csc ^4(a+b x)+\sec ^4(a+b x)}{\csc ^4(a+b x)+\sec ^4(a+b x)} \, dx=-\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sin \left (2\,a+2\,b\,x\right )}{2}\right )}{2\,b} \]

[In]

int((1/cos(a + b*x)^4 - 1/sin(a + b*x)^4)/(1/cos(a + b*x)^4 + 1/sin(a + b*x)^4),x)

[Out]

-(2^(1/2)*atanh((2^(1/2)*sin(2*a + 2*b*x))/2))/(2*b)