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

Optimal. Leaf size=38 \[ -\frac{\csc ^3(a+b x)}{3 b}-\frac{\csc (a+b x)}{b}+\frac{\tanh ^{-1}(\sin (a+b x))}{b} \]

[Out]

ArcTanh[Sin[a + b*x]]/b - Csc[a + b*x]/b - Csc[a + b*x]^3/(3*b)

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Rubi [A]  time = 0.0258361, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2621, 302, 207} \[ -\frac{\csc ^3(a+b x)}{3 b}-\frac{\csc (a+b x)}{b}+\frac{\tanh ^{-1}(\sin (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^4*Sec[a + b*x],x]

[Out]

ArcTanh[Sin[a + b*x]]/b - Csc[a + b*x]/b - Csc[a + b*x]^3/(3*b)

Rule 2621

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(f*a^n)^(-1), Subst
[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && Integer
Q[(n + 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [C]  time = 0.012873, size = 31, normalized size = 0.82 \[ -\frac{\csc ^3(a+b x) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\sin ^2(a+b x)\right )}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Csc[a + b*x]^3*Hypergeometric2F1[-3/2, 1, -1/2, Sin[a + b*x]^2])/(3*b)

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Maple [A]  time = 0.021, size = 46, normalized size = 1.2 \begin{align*} -{\frac{1}{3\, \left ( \sin \left ( bx+a \right ) \right ) ^{3}b}}-{\frac{1}{b\sin \left ( bx+a \right ) }}+{\frac{\ln \left ( \sec \left ( bx+a \right ) +\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)/sin(b*x+a)^4,x)

[Out]

-1/3/sin(b*x+a)^3/b-1/b/sin(b*x+a)+1/b*ln(sec(b*x+a)+tan(b*x+a))

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Maxima [A]  time = 0.980149, size = 68, normalized size = 1.79 \begin{align*} -\frac{\frac{2 \,{\left (3 \, \sin \left (b x + a\right )^{2} + 1\right )}}{\sin \left (b x + a\right )^{3}} - 3 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 3 \, \log \left (\sin \left (b x + a\right ) - 1\right )}{6 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.24198, size = 252, normalized size = 6.63 \begin{align*} \frac{3 \,{\left (\cos \left (b x + a\right )^{2} - 1\right )} \log \left (\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 3 \,{\left (\cos \left (b x + a\right )^{2} - 1\right )} \log \left (-\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 6 \, \cos \left (b x + a\right )^{2} + 8}{6 \,{\left (b \cos \left (b x + a\right )^{2} - b\right )} \sin \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(cos(b*x + a)^2 - 1)*log(sin(b*x + a) + 1)*sin(b*x + a) - 3*(cos(b*x + a)^2 - 1)*log(-sin(b*x + a) + 1)
*sin(b*x + a) - 6*cos(b*x + a)^2 + 8)/((b*cos(b*x + a)^2 - b)*sin(b*x + a))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.17546, size = 70, normalized size = 1.84 \begin{align*} -\frac{\frac{2 \,{\left (3 \, \sin \left (b x + a\right )^{2} + 1\right )}}{\sin \left (b x + a\right )^{3}} - 3 \, \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right ) + 3 \, \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right )}{6 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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