3.180 \(\int \csc ^5(a+b x) \sec (a+b x) \, dx\)

Optimal. Leaf size=40 \[ -\frac{\cot ^4(a+b x)}{4 b}-\frac{\cot ^2(a+b x)}{b}+\frac{\log (\tan (a+b x))}{b} \]

[Out]

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

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Rubi [A]  time = 0.0271886, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2620, 266, 43} \[ -\frac{\cot ^4(a+b x)}{4 b}-\frac{\cot ^2(a+b x)}{b}+\frac{\log (\tan (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^5*Sec[a + b*x],x]

[Out]

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

Rule 2620

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]

Rule 266

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 43

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])

Rubi steps

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

Mathematica [A]  time = 0.111257, size = 44, normalized size = 1.1 \[ -\frac{\csc ^4(a+b x)+2 \csc ^2(a+b x)-4 \log (\sin (a+b x))+4 \log (\cos (a+b x))}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(2*Csc[a + b*x]^2 + Csc[a + b*x]^4 + 4*Log[Cos[a + b*x]] - 4*Log[Sin[a + b*x]])/(4*b)

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Maple [A]  time = 0.02, size = 39, normalized size = 1. \begin{align*} -{\frac{1}{4\, \left ( \sin \left ( bx+a \right ) \right ) ^{4}b}}-{\frac{1}{2\, \left ( \sin \left ( bx+a \right ) \right ) ^{2}b}}+{\frac{\ln \left ( \tan \left ( bx+a \right ) \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.971075, size = 69, normalized size = 1.72 \begin{align*} -\frac{\frac{2 \, \sin \left (b x + a\right )^{2} + 1}{\sin \left (b x + a\right )^{4}} + 2 \, \log \left (\sin \left (b x + a\right )^{2} - 1\right ) - 2 \, \log \left (\sin \left (b x + a\right )^{2}\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.86571, size = 285, normalized size = 7.12 \begin{align*} \frac{2 \, \cos \left (b x + a\right )^{2} - 2 \,{\left (\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1\right )} \log \left (\cos \left (b x + a\right )^{2}\right ) + 2 \,{\left (\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1\right )} \log \left (-\frac{1}{4} \, \cos \left (b x + a\right )^{2} + \frac{1}{4}\right ) - 3}{4 \,{\left (b \cos \left (b x + a\right )^{4} - 2 \, b \cos \left (b x + a\right )^{2} + b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.33617, size = 223, normalized size = 5.58 \begin{align*} \frac{\frac{{\left (\frac{12 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac{48 \,{\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{12 \,{\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}} + 32 \, \log \left (\frac{{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right ) - 64 \, \log \left ({\left | -\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1 \right |}\right )}{64 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*((12*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 48*(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 + 12*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - (cos(b*x + a) - 1)^2/(cos(b*x + a
) + 1)^2 + 32*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)) - 64*log(abs(-(cos(b*x + a) - 1)/(cos(b*x + a)
+ 1) - 1)))/b