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

Optimal. Leaf size=39 \[ \frac{\tan ^4(a+b x)}{4 b}+\frac{\tan ^2(a+b x)}{b}+\frac{\log (\tan (a+b x))}{b} \]

[Out]

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

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Rubi [A]  time = 0.0262298, antiderivative size = 39, 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} \[ \frac{\tan ^4(a+b x)}{4 b}+\frac{\tan ^2(a+b x)}{b}+\frac{\log (\tan (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]*Sec[a + b*x]^5,x]

[Out]

Log[Tan[a + b*x]]/b + Tan[a + b*x]^2/b + Tan[a + b*x]^4/(4*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 (a+b x) \sec ^5(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^2}{x} \, dx,x,\tan (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(1+x)^2}{x} \, dx,x,\tan ^2(a+b x)\right )}{2 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (2+\frac{1}{x}+x\right ) \, dx,x,\tan ^2(a+b x)\right )}{2 b}\\ &=\frac{\log (\tan (a+b x))}{b}+\frac{\tan ^2(a+b x)}{b}+\frac{\tan ^4(a+b x)}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.0907424, size = 46, normalized size = 1.18 \[ -\frac{-\sec ^4(a+b x)-2 \sec ^2(a+b x)-4 \log (\sin (a+b x))+4 \log (\cos (a+b x))}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.023, size = 39, normalized size = 1. \begin{align*}{\frac{1}{4\,b \left ( \cos \left ( bx+a \right ) \right ) ^{4}}}+{\frac{1}{2\,b \left ( \cos \left ( bx+a \right ) \right ) ^{2}}}+{\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)^5/sin(b*x+a),x)

[Out]

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

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Maxima [A]  time = 0.974171, size = 88, normalized size = 2.26 \begin{align*} -\frac{\frac{2 \, \sin \left (b x + a\right )^{2} - 3}{\sin \left (b x + a\right )^{4} - 2 \, \sin \left (b x + a\right )^{2} + 1} + 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)^5/sin(b*x+a),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.74276, size = 185, normalized size = 4.74 \begin{align*} -\frac{2 \, \cos \left (b x + a\right )^{4} \log \left (\cos \left (b x + a\right )^{2}\right ) - 2 \, \cos \left (b x + a\right )^{4} \log \left (-\frac{1}{4} \, \cos \left (b x + a\right )^{2} + \frac{1}{4}\right ) - 2 \, \cos \left (b x + a\right )^{2} - 1}{4 \, b \cos \left (b x + a\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{5}{\left (a + b x \right )}}{\sin{\left (a + b x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.19017, size = 230, normalized size = 5.9 \begin{align*} \frac{\frac{\frac{52 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac{102 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac{52 \,{\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}{{\left (\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1\right )}^{4}} + 6 \, \log \left (\frac{{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right ) - 12 \, \log \left ({\left | -\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1 \right |}\right )}{12 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*((52*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 102*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 52*(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 + 6*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)) - 12*log(abs(-(cos(b*x + a) - 1)/(
cos(b*x + a) + 1) - 1)))/b