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

Optimal. Leaf size=43 \[ \frac{\tan ^2(a+b x)}{2 b}-\frac{\cot ^2(a+b x)}{2 b}+\frac{2 \log (\tan (a+b x))}{b} \]

[Out]

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

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Rubi [A]  time = 0.0381614, antiderivative size = 43, 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} \[ \frac{\tan ^2(a+b x)}{2 b}-\frac{\cot ^2(a+b x)}{2 b}+\frac{2 \log (\tan (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^3*Sec[a + b*x]^3,x]

[Out]

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

Mathematica [A]  time = 0.0176275, size = 61, normalized size = 1.42 \[ 8 \left (-\frac{\csc ^2(a+b x)}{16 b}+\frac{\sec ^2(a+b x)}{16 b}+\frac{\log (\sin (a+b x))}{4 b}-\frac{\log (\cos (a+b x))}{4 b}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.02, size = 48, normalized size = 1.1 \begin{align*}{\frac{1}{2\,b \left ( \sin \left ( bx+a \right ) \right ) ^{2} \left ( \cos \left ( bx+a \right ) \right ) ^{2}}}-{\frac{1}{b \left ( \sin \left ( bx+a \right ) \right ) ^{2}}}+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)^3,x)

[Out]

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

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Maxima [A]  time = 1.00213, size = 86, normalized size = 2. \begin{align*} -\frac{\frac{2 \, \sin \left (b x + a\right )^{2} - 1}{\sin \left (b x + a\right )^{4} - \sin \left (b x + a\right )^{2}} + 2 \, \log \left (\sin \left (b x + a\right )^{2} - 1\right ) - 2 \, \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)^3,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.21115, size = 261, normalized size = 6.07 \begin{align*} \frac{2 \, \cos \left (b x + a\right )^{2} - 2 \,{\left (\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 ) + 2 \,{\left (\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 ) - 1}{2 \,{\left (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)^3,x, algorithm="fricas")

[Out]

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

[Out]

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

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Giac [B]  time = 1.24046, size = 254, normalized size = 5.91 \begin{align*} -\frac{\frac{{\left (\frac{8 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - 1\right )}{\left (\cos \left (b x + a\right ) + 1\right )}}{\cos \left (b x + a\right ) - 1} - \frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - \frac{8 \,{\left (\frac{4 \,{\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\right )}}{{\left (\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1\right )}^{2}} - 8 \, \log \left (\frac{{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right ) + 16 \, \log \left ({\left | -\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1 \right |}\right )}{8 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*((8*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 1)*(cos(b*x + a) + 1)/(cos(b*x + a) - 1) - (cos(b*x + a) - 1)
/(cos(b*x + a) + 1) - 8*(4*(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 - 8*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)) + 16
*log(abs(-(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 1)))/b