3.182 \(\int \csc ^5(a+b x) \sec ^3(a+b x) \, dx\)

Optimal. Leaf size=58 \[ \frac{\tan ^2(a+b x)}{2 b}-\frac{\cot ^4(a+b x)}{4 b}-\frac{3 \cot ^2(a+b x)}{2 b}+\frac{3 \log (\tan (a+b x))}{b} \]

[Out]

(-3*Cot[a + b*x]^2)/(2*b) - Cot[a + b*x]^4/(4*b) + (3*Log[Tan[a + b*x]])/b + Tan[a + b*x]^2/(2*b)

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Rubi [A]  time = 0.0398449, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {2620, 266, 43} \[ \frac{\tan ^2(a+b x)}{2 b}-\frac{\cot ^4(a+b x)}{4 b}-\frac{3 \cot ^2(a+b x)}{2 b}+\frac{3 \log (\tan (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^5*Sec[a + b*x]^3,x]

[Out]

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

Mathematica [A]  time = 0.340545, size = 54, normalized size = 0.93 \[ -\frac{\csc ^4(a+b x)+4 \csc ^2(a+b x)-2 \sec ^2(a+b x)-12 \log (\sin (a+b x))+12 \log (\cos (a+b x))}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^5*Sec[a + b*x]^3,x]

[Out]

-(4*Csc[a + b*x]^2 + Csc[a + b*x]^4 + 12*Log[Cos[a + b*x]] - 12*Log[Sin[a + b*x]] - 2*Sec[a + b*x]^2)/(4*b)

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Maple [A]  time = 0.026, size = 69, normalized size = 1.2 \begin{align*} -{\frac{1}{4\,b \left ( \sin \left ( bx+a \right ) \right ) ^{4} \left ( \cos \left ( bx+a \right ) \right ) ^{2}}}+{\frac{3}{4\,b \left ( \sin \left ( bx+a \right ) \right ) ^{2} \left ( \cos \left ( bx+a \right ) \right ) ^{2}}}-{\frac{3}{2\,b \left ( \sin \left ( bx+a \right ) \right ) ^{2}}}+3\,{\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)^5,x)

[Out]

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

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Maxima [A]  time = 0.985309, size = 100, normalized size = 1.72 \begin{align*} -\frac{\frac{6 \, \sin \left (b x + a\right )^{4} - 3 \, \sin \left (b x + a\right )^{2} - 1}{\sin \left (b x + a\right )^{6} - \sin \left (b x + a\right )^{4}} + 6 \, \log \left (\sin \left (b x + a\right )^{2} - 1\right ) - 6 \, \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)^3/sin(b*x+a)^5,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.92825, size = 366, normalized size = 6.31 \begin{align*} \frac{6 \, \cos \left (b x + a\right )^{4} - 9 \, \cos \left (b x + a\right )^{2} - 6 \,{\left (\cos \left (b x + a\right )^{6} - 2 \, \cos \left (b x + a\right )^{4} + \cos \left (b x + a\right )^{2}\right )} \log \left (\cos \left (b x + a\right )^{2}\right ) + 6 \,{\left (\cos \left (b x + a\right )^{6} - 2 \, \cos \left (b x + a\right )^{4} + \cos \left (b x + a\right )^{2}\right )} \log \left (-\frac{1}{4} \, \cos \left (b x + a\right )^{2} + \frac{1}{4}\right ) + 2}{4 \,{\left (b \cos \left (b x + a\right )^{6} - 2 \, b \cos \left (b x + a\right )^{4} + b \cos \left (b x + a\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^3/sin(b*x+a)^5,x, algorithm="fricas")

[Out]

1/4*(6*cos(b*x + a)^4 - 9*cos(b*x + a)^2 - 6*(cos(b*x + a)^6 - 2*cos(b*x + a)^4 + cos(b*x + a)^2)*log(cos(b*x
+ a)^2) + 6*(cos(b*x + a)^6 - 2*cos(b*x + a)^4 + cos(b*x + a)^2)*log(-1/4*cos(b*x + a)^2 + 1/4) + 2)/(b*cos(b*
x + a)^6 - 2*b*cos(b*x + a)^4 + b*cos(b*x + a)^2)

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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)**3/sin(b*x+a)**5,x)

[Out]

Timed out

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Giac [B]  time = 1.24591, size = 313, normalized size = 5.4 \begin{align*} \frac{\frac{20 \,{\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{\frac{18 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac{111 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac{36 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}} + \frac{72 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} - 1}{{\left (\frac{\cos \left (b x + a\right ) - 1}{\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}}\right )}^{2}} + 96 \, \log \left (\frac{{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right ) - 192 \, \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)^3/sin(b*x+a)^5,x, algorithm="giac")

[Out]

1/64*(20*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - (cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + (18*(cos(b*x + a)
- 1)/(cos(b*x + a) + 1) + 111*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 36*(cos(b*x + a) - 1)^3/(cos(b*x +
a) + 1)^3 + 72*(cos(b*x + a) - 1)^4/(cos(b*x + a) + 1)^4 - 1)/((cos(b*x + a) - 1)/(cos(b*x + a) + 1) + (cos(b*
x + a) - 1)^2/(cos(b*x + a) + 1)^2)^2 + 96*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)) - 192*log(abs(-(c
os(b*x + a) - 1)/(cos(b*x + a) + 1) - 1)))/b