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]

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

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Rubi [A]  time = 0.04, antiderivative size = 43, normalized size of antiderivative = 1.00, 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 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])

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 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]

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*}

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Mathematica [A]  time = 0.01, 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*(-1/16*Csc[a + b*x]^2/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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fricas [B]  time = 0.43, size = 102, normalized size = 2.37 \[ \frac {2 \, \cos \left (b x + a\right )^{2} - 2 \, {\left (\cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2}\right )} \log \left (\cos \left (b x + a\right )^{2}\right ) + 2 \, {\left (\cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2}\right )} \log \left (-\frac {1}{4} \, \cos \left (b x + a\right )^{2} + \frac {1}{4}\right ) - 1}{2 \, {\left (b \cos \left (b x + a\right )^{4} - b \cos \left (b x + a\right )^{2}\right )}} \]

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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giac [B]  time = 0.25, size = 188, normalized size = 4.37 \[ -\frac {\frac {{\left (\frac {8 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - 1\right )} {\left (\cos \left (b x + a\right ) + 1\right )}}{\cos \left (b x + a\right ) - 1} - \frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - \frac {8 \, {\left (\frac {4 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac {3 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + 3\right )}}{{\left (\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1\right )}^{2}} - 8 \, \log \left (\frac {{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right ) + 16 \, \log \left ({\left | -\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1 \right |}\right )}{8 \, b} \]

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

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maple [A]  time = 0.04, size = 48, normalized size = 1.12 \[ \frac {1}{2 b \sin \left (b x +a \right )^{2} \cos \left (b x +a \right )^{2}}-\frac {1}{\sin \left (b x +a \right )^{2} b}+\frac {2 \ln \left (\tan \left (b x +a \right )\right )}{b} \]

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 = 0.30, size = 64, normalized size = 1.49 \[ -\frac {\frac {2 \, \sin \left (b x + a\right )^{2} - 1}{\sin \left (b x + a\right )^{4} - \sin \left (b x + a\right )^{2}} + 2 \, \log \left (\sin \left (b x + a\right )^{2} - 1\right ) - 2 \, \log \left (\sin \left (b x + a\right )^{2}\right )}{2 \, b} \]

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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mupad [B]  time = 0.38, size = 39, normalized size = 0.91 \[ \frac {{\mathrm {tan}\left (a+b\,x\right )}^2}{2\,b}-\frac {1}{2\,b\,{\mathrm {tan}\left (a+b\,x\right )}^2}+\frac {2\,\ln \left (\mathrm {tan}\left (a+b\,x\right )\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(a + b*x)^3*sin(a + b*x)^3),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{3}{\left (a + b x \right )}}{\sin ^{3}{\left (a + b x \right )}}\, dx \]

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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