3.128 \(\int \csc (a+b x) \sec ^3(a+b x) \, dx\)

Optimal. Leaf size=27 \[ \frac{\tan ^2(a+b x)}{2 b}+\frac{\log (\tan (a+b x))}{b} \]

[Out]

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

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Rubi [A]  time = 0.0202023, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2620, 14} \[ \frac{\tan ^2(a+b x)}{2 b}+\frac{\log (\tan (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]*Sec[a + b*x]^3,x]

[Out]

Log[Tan[a + b*x]]/b + Tan[a + b*x]^2/(2*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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

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

Mathematica [A]  time = 0.0309752, size = 36, normalized size = 1.33 \[ -\frac{-\sec ^2(a+b x)-2 \log (\sin (a+b x))+2 \log (\cos (a+b x))}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]*Sec[a + b*x]^3,x]

[Out]

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

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Maple [A]  time = 0.022, size = 26, normalized size = 1. \begin{align*}{\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)^3/sin(b*x+a),x)

[Out]

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

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Maxima [A]  time = 1.04927, size = 54, normalized size = 2. \begin{align*} -\frac{\frac{1}{\sin \left (b x + a\right )^{2} - 1} + \log \left (\sin \left (b x + a\right )^{2} - 1\right ) - \log \left (\sin \left (b x + a\right )^{2}\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^3/sin(b*x+a),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.68809, size = 154, normalized size = 5.7 \begin{align*} -\frac{\cos \left (b x + a\right )^{2} \log \left (\cos \left (b x + a\right )^{2}\right ) - \cos \left (b x + a\right )^{2} \log \left (-\frac{1}{4} \, \cos \left (b x + a\right )^{2} + \frac{1}{4}\right ) - 1}{2 \, b \cos \left (b x + a\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^3/sin(b*x+a),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{3}{\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)**3/sin(b*x+a),x)

[Out]

Integral(sec(a + b*x)**3/sin(a + b*x), x)

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Giac [B]  time = 1.22026, size = 167, normalized size = 6.19 \begin{align*} \frac{\frac{\frac{2 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac{3 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + 3}{{\left (\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1\right )}^{2}} + \log \left (\frac{{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right ) - 2 \, \log \left ({\left | -\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1 \right |}\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^3/sin(b*x+a),x, algorithm="giac")

[Out]

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