3.143 \(\int \csc ^2(a+b x) \sec ^4(a+b x) \, dx\)

Optimal. Leaf size=38 \[ \frac{\tan ^3(a+b x)}{3 b}+\frac{2 \tan (a+b x)}{b}-\frac{\cot (a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.0367211, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2620, 270} \[ \frac{\tan ^3(a+b x)}{3 b}+\frac{2 \tan (a+b x)}{b}-\frac{\cot (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^2*Sec[a + b*x]^4,x]

[Out]

-(Cot[a + b*x]/b) + (2*Tan[a + b*x])/b + Tan[a + b*x]^3/(3*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 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \csc ^2(a+b x) \sec ^4(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^2}{x^2} \, dx,x,\tan (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \left (2+\frac{1}{x^2}+x^2\right ) \, dx,x,\tan (a+b x)\right )}{b}\\ &=-\frac{\cot (a+b x)}{b}+\frac{2 \tan (a+b x)}{b}+\frac{\tan ^3(a+b x)}{3 b}\\ \end{align*}

Mathematica [A]  time = 0.0330181, size = 46, normalized size = 1.21 \[ \frac{5 \tan (a+b x)}{3 b}-\frac{\cot (a+b x)}{b}+\frac{\tan (a+b x) \sec ^2(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^2*Sec[a + b*x]^4,x]

[Out]

-(Cot[a + b*x]/b) + (5*Tan[a + b*x])/(3*b) + (Sec[a + b*x]^2*Tan[a + b*x])/(3*b)

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Maple [A]  time = 0.022, size = 50, normalized size = 1.3 \begin{align*}{\frac{1}{b} \left ({\frac{1}{3\, \left ( \cos \left ( bx+a \right ) \right ) ^{3}\sin \left ( bx+a \right ) }}+{\frac{4}{3\,\cos \left ( bx+a \right ) \sin \left ( bx+a \right ) }}-{\frac{8\,\cot \left ( bx+a \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^4/sin(b*x+a)^2,x)

[Out]

1/b*(1/3/sin(b*x+a)/cos(b*x+a)^3+4/3/sin(b*x+a)/cos(b*x+a)-8/3*cot(b*x+a))

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Maxima [A]  time = 0.985613, size = 43, normalized size = 1.13 \begin{align*} \frac{\tan \left (b x + a\right )^{3} - \frac{3}{\tan \left (b x + a\right )} + 6 \, \tan \left (b x + a\right )}{3 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^4/sin(b*x+a)^2,x, algorithm="maxima")

[Out]

1/3*(tan(b*x + a)^3 - 3/tan(b*x + a) + 6*tan(b*x + a))/b

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Fricas [A]  time = 1.78534, size = 108, normalized size = 2.84 \begin{align*} -\frac{8 \, \cos \left (b x + a\right )^{4} - 4 \, \cos \left (b x + a\right )^{2} - 1}{3 \, b \cos \left (b x + a\right )^{3} \sin \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^4/sin(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/3*(8*cos(b*x + a)^4 - 4*cos(b*x + a)^2 - 1)/(b*cos(b*x + a)^3*sin(b*x + a))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{4}{\left (a + b x \right )}}{\sin ^{2}{\left (a + b x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**4/sin(b*x+a)**2,x)

[Out]

Integral(sec(a + b*x)**4/sin(a + b*x)**2, x)

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Giac [A]  time = 1.19946, size = 43, normalized size = 1.13 \begin{align*} \frac{\tan \left (b x + a\right )^{3} - \frac{3}{\tan \left (b x + a\right )} + 6 \, \tan \left (b x + a\right )}{3 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^4/sin(b*x+a)^2,x, algorithm="giac")

[Out]

1/3*(tan(b*x + a)^3 - 3/tan(b*x + a) + 6*tan(b*x + a))/b