3.131 \(\int \csc (a+b x) \sec ^6(a+b x) \, dx\)

Optimal. Leaf size=53 \[ \frac{\sec ^5(a+b x)}{5 b}+\frac{\sec ^3(a+b x)}{3 b}+\frac{\sec (a+b x)}{b}-\frac{\tanh ^{-1}(\cos (a+b x))}{b} \]

[Out]

-(ArcTanh[Cos[a + b*x]]/b) + Sec[a + b*x]/b + Sec[a + b*x]^3/(3*b) + Sec[a + b*x]^5/(5*b)

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Rubi [A]  time = 0.0279177, antiderivative size = 53, 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 = {2622, 302, 207} \[ \frac{\sec ^5(a+b x)}{5 b}+\frac{\sec ^3(a+b x)}{3 b}+\frac{\sec (a+b x)}{b}-\frac{\tanh ^{-1}(\cos (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]*Sec[a + b*x]^6,x]

[Out]

-(ArcTanh[Cos[a + b*x]]/b) + Sec[a + b*x]/b + Sec[a + b*x]^3/(3*b) + Sec[a + b*x]^5/(5*b)

Rule 2622

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(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 (a+b x) \sec ^6(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^6}{-1+x^2} \, dx,x,\sec (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \left (1+x^2+x^4+\frac{1}{-1+x^2}\right ) \, dx,x,\sec (a+b x)\right )}{b}\\ &=\frac{\sec (a+b x)}{b}+\frac{\sec ^3(a+b x)}{3 b}+\frac{\sec ^5(a+b x)}{5 b}+\frac{\operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\sec (a+b x)\right )}{b}\\ &=-\frac{\tanh ^{-1}(\cos (a+b x))}{b}+\frac{\sec (a+b x)}{b}+\frac{\sec ^3(a+b x)}{3 b}+\frac{\sec ^5(a+b x)}{5 b}\\ \end{align*}

Mathematica [A]  time = 0.022692, size = 72, normalized size = 1.36 \[ \frac{\sec ^5(a+b x)}{5 b}+\frac{\sec ^3(a+b x)}{3 b}+\frac{\sec (a+b x)}{b}+\frac{\log \left (\sin \left (\frac{1}{2} (a+b x)\right )\right )}{b}-\frac{\log \left (\cos \left (\frac{1}{2} (a+b x)\right )\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]*Sec[a + b*x]^6,x]

[Out]

-(Log[Cos[(a + b*x)/2]]/b) + Log[Sin[(a + b*x)/2]]/b + Sec[a + b*x]/b + Sec[a + b*x]^3/(3*b) + Sec[a + b*x]^5/
(5*b)

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Maple [A]  time = 0.026, size = 60, normalized size = 1.1 \begin{align*}{\frac{1}{5\,b \left ( \cos \left ( bx+a \right ) \right ) ^{5}}}+{\frac{1}{3\,b \left ( \cos \left ( bx+a \right ) \right ) ^{3}}}+{\frac{1}{b\cos \left ( bx+a \right ) }}+{\frac{\ln \left ( \csc \left ( bx+a \right ) -\cot \left ( bx+a \right ) \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^6/sin(b*x+a),x)

[Out]

1/5/b/cos(b*x+a)^5+1/3/b/cos(b*x+a)^3+1/b/cos(b*x+a)+1/b*ln(csc(b*x+a)-cot(b*x+a))

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Maxima [A]  time = 0.96447, size = 81, normalized size = 1.53 \begin{align*} \frac{\frac{2 \,{\left (15 \, \cos \left (b x + a\right )^{4} + 5 \, \cos \left (b x + a\right )^{2} + 3\right )}}{\cos \left (b x + a\right )^{5}} - 15 \, \log \left (\cos \left (b x + a\right ) + 1\right ) + 15 \, \log \left (\cos \left (b x + a\right ) - 1\right )}{30 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^6/sin(b*x+a),x, algorithm="maxima")

[Out]

1/30*(2*(15*cos(b*x + a)^4 + 5*cos(b*x + a)^2 + 3)/cos(b*x + a)^5 - 15*log(cos(b*x + a) + 1) + 15*log(cos(b*x
+ a) - 1))/b

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Fricas [A]  time = 1.69654, size = 225, normalized size = 4.25 \begin{align*} -\frac{15 \, \cos \left (b x + a\right )^{5} \log \left (\frac{1}{2} \, \cos \left (b x + a\right ) + \frac{1}{2}\right ) - 15 \, \cos \left (b x + a\right )^{5} \log \left (-\frac{1}{2} \, \cos \left (b x + a\right ) + \frac{1}{2}\right ) - 30 \, \cos \left (b x + a\right )^{4} - 10 \, \cos \left (b x + a\right )^{2} - 6}{30 \, b \cos \left (b x + a\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^6/sin(b*x+a),x, algorithm="fricas")

[Out]

-1/30*(15*cos(b*x + a)^5*log(1/2*cos(b*x + a) + 1/2) - 15*cos(b*x + a)^5*log(-1/2*cos(b*x + a) + 1/2) - 30*cos
(b*x + a)^4 - 10*cos(b*x + a)^2 - 6)/(b*cos(b*x + a)^5)

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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)**6/sin(b*x+a),x)

[Out]

Timed out

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Giac [B]  time = 1.19563, size = 196, normalized size = 3.7 \begin{align*} \frac{\frac{4 \,{\left (\frac{70 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac{140 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac{90 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}} + \frac{45 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} + 23\right )}}{{\left (\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1\right )}^{5}} + 15 \, \log \left (\frac{{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right )}{30 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^6/sin(b*x+a),x, algorithm="giac")

[Out]

1/30*(4*(70*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 140*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 90*(cos(b*
x + a) - 1)^3/(cos(b*x + a) + 1)^3 + 45*(cos(b*x + a) - 1)^4/(cos(b*x + a) + 1)^4 + 23)/((cos(b*x + a) - 1)/(c
os(b*x + a) + 1) + 1)^5 + 15*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)))/b