\(\int \csc ^5(a+b x) \sec ^5(a+b x) \, dx\) [184]

Optimal result

Integrand size = 17, antiderivative size = 69 \[ \int \csc ^5(a+b x) \sec ^5(a+b x) \, dx=-\frac {2 \cot ^2(a+b x)}{b}-\frac {\cot ^4(a+b x)}{4 b}+\frac {6 \log (\tan (a+b x))}{b}+\frac {2 \tan ^2(a+b x)}{b}+\frac {\tan ^4(a+b x)}{4 b} \]

[Out]

-2*cot(b*x+a)^2/b-1/4*cot(b*x+a)^4/b+6*ln(tan(b*x+a))/b+2*tan(b*x+a)^2/b+1/4*tan(b*x+a)^4/b

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2700, 272, 45} \[ \int \csc ^5(a+b x) \sec ^5(a+b x) \, dx=\frac {\tan ^4(a+b x)}{4 b}+\frac {2 \tan ^2(a+b x)}{b}-\frac {\cot ^4(a+b x)}{4 b}-\frac {2 \cot ^2(a+b x)}{b}+\frac {6 \log (\tan (a+b x))}{b} \]

[In]

Int[Csc[a + b*x]^5*Sec[a + b*x]^5,x]

[Out]

(-2*Cot[a + b*x]^2)/b - Cot[a + b*x]^4/(4*b) + (6*Log[Tan[a + b*x]])/b + (2*Tan[a + b*x]^2)/b + Tan[a + b*x]^4
/(4*b)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2700

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (1+x^2\right )^4}{x^5} \, dx,x,\tan (a+b x)\right )}{b} \\ & = \frac {\text {Subst}\left (\int \frac {(1+x)^4}{x^3} \, dx,x,\tan ^2(a+b x)\right )}{2 b} \\ & = \frac {\text {Subst}\left (\int \left (4+\frac {1}{x^3}+\frac {4}{x^2}+\frac {6}{x}+x\right ) \, dx,x,\tan ^2(a+b x)\right )}{2 b} \\ & = -\frac {2 \cot ^2(a+b x)}{b}-\frac {\cot ^4(a+b x)}{4 b}+\frac {6 \log (\tan (a+b x))}{b}+\frac {2 \tan ^2(a+b x)}{b}+\frac {\tan ^4(a+b x)}{4 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.32 \[ \int \csc ^5(a+b x) \sec ^5(a+b x) \, dx=32 \left (-\frac {3 \csc ^2(a+b x)}{64 b}-\frac {\csc ^4(a+b x)}{128 b}-\frac {3 \log (\cos (a+b x))}{16 b}+\frac {3 \log (\sin (a+b x))}{16 b}+\frac {3 \sec ^2(a+b x)}{64 b}+\frac {\sec ^4(a+b x)}{128 b}\right ) \]

[In]

Integrate[Csc[a + b*x]^5*Sec[a + b*x]^5,x]

[Out]

32*((-3*Csc[a + b*x]^2)/(64*b) - Csc[a + b*x]^4/(128*b) - (3*Log[Cos[a + b*x]])/(16*b) + (3*Log[Sin[a + b*x]])
/(16*b) + (3*Sec[a + b*x]^2)/(64*b) + Sec[a + b*x]^4/(128*b))

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.14

method	result	size
derivativedivides	\(\frac {\frac {1}{4 \sin \left (b x +a \right )^{4} \cos \left (b x +a \right )^{4}}-\frac {1}{2 \cos \left (b x +a \right )^{2} \sin \left (b x +a \right )^{4}}+\frac {3}{2 \cos \left (b x +a \right )^{2} \sin \left (b x +a \right )^{2}}-\frac {3}{\sin \left (b x +a \right )^{2}}+6 \ln \left (\tan \left (b x +a \right )\right )}{b}\)	\(79\)
default	\(\frac {\frac {1}{4 \sin \left (b x +a \right )^{4} \cos \left (b x +a \right )^{4}}-\frac {1}{2 \cos \left (b x +a \right )^{2} \sin \left (b x +a \right )^{4}}+\frac {3}{2 \cos \left (b x +a \right )^{2} \sin \left (b x +a \right )^{2}}-\frac {3}{\sin \left (b x +a \right )^{2}}+6 \ln \left (\tan \left (b x +a \right )\right )}{b}\)	\(79\)
risch	\(\frac {12 \,{\mathrm e}^{14 i \left (b x +a \right )}-44 \,{\mathrm e}^{10 i \left (b x +a \right )}-44 \,{\mathrm e}^{6 i \left (b x +a \right )}+12 \,{\mathrm e}^{2 i \left (b x +a \right )}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{4} \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{4}}+\frac {6 \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}{b}-\frac {6 \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{b}\)	\(112\)
parallelrisch	\(\frac {\left (-384 \cos \left (2 b x +2 a \right )-96 \cos \left (4 b x +4 a \right )-288\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )+\left (-384 \cos \left (2 b x +2 a \right )-96 \cos \left (4 b x +4 a \right )-288\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )+\left (384 \cos \left (2 b x +2 a \right )+96 \cos \left (4 b x +4 a \right )+288\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\left (3384 \cos \left (b x +a \right )-1284 \cos \left (2 b x +2 a \right )+328 \cos \left (3 b x +3 a \right )-41 \cos \left (4 b x +4 a \right )-3227\right ) \left (\cot ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+592 \left (\cot ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \left (\csc ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-2 \left (\csc ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \left (\sec ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )+16 \left (\sec ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-140\right )}{16 b \left (\cos \left (4 b x +4 a \right )+4 \cos \left (2 b x +2 a \right )+3\right )}\)	\(258\)

[In]

int(sec(b*x+a)^5/sin(b*x+a)^5,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 148 vs. \(2 (65) = 130\).

Time = 0.33 (sec) , antiderivative size = 148, normalized size of antiderivative = 2.14 \[ \int \csc ^5(a+b x) \sec ^5(a+b x) \, dx=\frac {12 \, \cos \left (b x + a\right )^{6} - 18 \, \cos \left (b x + a\right )^{4} + 4 \, \cos \left (b x + a\right )^{2} - 12 \, {\left (\cos \left (b x + a\right )^{8} - 2 \, \cos \left (b x + a\right )^{6} + \cos \left (b x + a\right )^{4}\right )} \log \left (\cos \left (b x + a\right )^{2}\right ) + 12 \, {\left (\cos \left (b x + a\right )^{8} - 2 \, \cos \left (b x + a\right )^{6} + \cos \left (b x + a\right )^{4}\right )} \log \left (-\frac {1}{4} \, \cos \left (b x + a\right )^{2} + \frac {1}{4}\right ) + 1}{4 \, {\left (b \cos \left (b x + a\right )^{8} - 2 \, b \cos \left (b x + a\right )^{6} + b \cos \left (b x + a\right )^{4}\right )}} \]

[In]

integrate(sec(b*x+a)^5/sin(b*x+a)^5,x, algorithm="fricas")

[Out]

1/4*(12*cos(b*x + a)^6 - 18*cos(b*x + a)^4 + 4*cos(b*x + a)^2 - 12*(cos(b*x + a)^8 - 2*cos(b*x + a)^6 + cos(b*
x + a)^4)*log(cos(b*x + a)^2) + 12*(cos(b*x + a)^8 - 2*cos(b*x + a)^6 + cos(b*x + a)^4)*log(-1/4*cos(b*x + a)^
2 + 1/4) + 1)/(b*cos(b*x + a)^8 - 2*b*cos(b*x + a)^6 + b*cos(b*x + a)^4)

Sympy [F]

\[ \int \csc ^5(a+b x) \sec ^5(a+b x) \, dx=\int \frac {\sec ^{5}{\left (a + b x \right )}}{\sin ^{5}{\left (a + b x \right )}}\, dx \]

[In]

integrate(sec(b*x+a)**5/sin(b*x+a)**5,x)

[Out]

Integral(sec(a + b*x)**5/sin(a + b*x)**5, x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.33 \[ \int \csc ^5(a+b x) \sec ^5(a+b x) \, dx=-\frac {\frac {12 \, \sin \left (b x + a\right )^{6} - 18 \, \sin \left (b x + a\right )^{4} + 4 \, \sin \left (b x + a\right )^{2} + 1}{\sin \left (b x + a\right )^{8} - 2 \, \sin \left (b x + a\right )^{6} + \sin \left (b x + a\right )^{4}} + 12 \, \log \left (\sin \left (b x + a\right )^{2} - 1\right ) - 12 \, \log \left (\sin \left (b x + a\right )^{2}\right )}{4 \, b} \]

[In]

integrate(sec(b*x+a)^5/sin(b*x+a)^5,x, algorithm="maxima")

[Out]

-1/4*((12*sin(b*x + a)^6 - 18*sin(b*x + a)^4 + 4*sin(b*x + a)^2 + 1)/(sin(b*x + a)^8 - 2*sin(b*x + a)^6 + sin(
b*x + a)^4) + 12*log(sin(b*x + a)^2 - 1) - 12*log(sin(b*x + a)^2))/b

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 278 vs. \(2 (65) = 130\).

Time = 0.32 (sec) , antiderivative size = 278, normalized size of antiderivative = 4.03 \[ \int \csc ^5(a+b x) \sec ^5(a+b x) \, dx=\frac {\frac {{\left (\frac {28 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac {288 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} - 1\right )} {\left (\cos \left (b x + a\right ) + 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) - 1\right )}^{2}} + \frac {28 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac {{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac {32 \, {\left (\frac {84 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac {126 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac {84 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}} + \frac {25 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} + 25\right )}}{{\left (\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1\right )}^{4}} + 192 \, \log \left (\frac {{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right ) - 384 \, \log \left ({\left | -\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1 \right |}\right )}{64 \, b} \]

[In]

integrate(sec(b*x+a)^5/sin(b*x+a)^5,x, algorithm="giac")

[Out]

1/64*((28*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 288*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 - 1)*(cos(b*x
+ a) + 1)^2/(cos(b*x + a) - 1)^2 + 28*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - (cos(b*x + a) - 1)^2/(cos(b*x +
a) + 1)^2 + 32*(84*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 126*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 84*
(cos(b*x + a) - 1)^3/(cos(b*x + a) + 1)^3 + 25*(cos(b*x + a) - 1)^4/(cos(b*x + a) + 1)^4 + 25)/((cos(b*x + a)
- 1)/(cos(b*x + a) + 1) + 1)^4 + 192*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)) - 384*log(abs(-(cos(b*x
+ a) - 1)/(cos(b*x + a) + 1) - 1)))/b

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.93 \[ \int \csc ^5(a+b x) \sec ^5(a+b x) \, dx=\frac {2\,{\mathrm {tan}\left (a+b\,x\right )}^2}{b}+\frac {{\mathrm {tan}\left (a+b\,x\right )}^4}{4\,b}+\frac {6\,\ln \left (\mathrm {tan}\left (a+b\,x\right )\right )}{b}-\frac {{\mathrm {cot}\left (a+b\,x\right )}^4\,\left (2\,{\mathrm {tan}\left (a+b\,x\right )}^2+\frac {1}{4}\right )}{b} \]

[In]

int(1/(cos(a + b*x)^5*sin(a + b*x)^5),x)

[Out]

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