3.132 \(\int \csc (a+b x) \sec ^7(a+b x) \, dx\)

Optimal. Leaf size=57 \[ \frac{\tan ^6(a+b x)}{6 b}+\frac{3 \tan ^4(a+b x)}{4 b}+\frac{3 \tan ^2(a+b x)}{2 b}+\frac{\log (\tan (a+b x))}{b} \]

[Out]

Log[Tan[a + b*x]]/b + (3*Tan[a + b*x]^2)/(2*b) + (3*Tan[a + b*x]^4)/(4*b) + Tan[a + b*x]^6/(6*b)

________________________________________________________________________________________

Rubi [A]  time = 0.0312009, antiderivative size = 57, 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 = {2620, 266, 43} \[ \frac{\tan ^6(a+b x)}{6 b}+\frac{3 \tan ^4(a+b x)}{4 b}+\frac{3 \tan ^2(a+b x)}{2 b}+\frac{\log (\tan (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]*Sec[a + b*x]^7,x]

[Out]

Log[Tan[a + b*x]]/b + (3*Tan[a + b*x]^2)/(2*b) + (3*Tan[a + b*x]^4)/(4*b) + Tan[a + b*x]^6/(6*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 (a+b x) \sec ^7(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^3}{x} \, dx,x,\tan (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(1+x)^3}{x} \, dx,x,\tan ^2(a+b x)\right )}{2 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (3+\frac{1}{x}+3 x+x^2\right ) \, dx,x,\tan ^2(a+b x)\right )}{2 b}\\ &=\frac{\log (\tan (a+b x))}{b}+\frac{3 \tan ^2(a+b x)}{2 b}+\frac{3 \tan ^4(a+b x)}{4 b}+\frac{\tan ^6(a+b x)}{6 b}\\ \end{align*}

Mathematica [A]  time = 0.145671, size = 56, normalized size = 0.98 \[ -\frac{-2 \sec ^6(a+b x)-3 \sec ^4(a+b x)-6 \sec ^2(a+b x)-12 \log (\sin (a+b x))+12 \log (\cos (a+b x))}{12 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]*Sec[a + b*x]^7,x]

[Out]

-(12*Log[Cos[a + b*x]] - 12*Log[Sin[a + b*x]] - 6*Sec[a + b*x]^2 - 3*Sec[a + b*x]^4 - 2*Sec[a + b*x]^6)/(12*b)

________________________________________________________________________________________

Maple [A]  time = 0.025, size = 52, normalized size = 0.9 \begin{align*}{\frac{1}{6\,b \left ( \cos \left ( bx+a \right ) \right ) ^{6}}}+{\frac{1}{4\,b \left ( \cos \left ( bx+a \right ) \right ) ^{4}}}+{\frac{1}{2\,b \left ( \cos \left ( bx+a \right ) \right ) ^{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)^7/sin(b*x+a),x)

[Out]

1/6/b/cos(b*x+a)^6+1/4/b/cos(b*x+a)^4+1/2/b/cos(b*x+a)^2+ln(tan(b*x+a))/b

________________________________________________________________________________________

Maxima [A]  time = 1.00839, size = 115, normalized size = 2.02 \begin{align*} -\frac{\frac{6 \, \sin \left (b x + a\right )^{4} - 15 \, \sin \left (b x + a\right )^{2} + 11}{\sin \left (b x + a\right )^{6} - 3 \, \sin \left (b x + a\right )^{4} + 3 \, \sin \left (b x + a\right )^{2} - 1} + 6 \, \log \left (\sin \left (b x + a\right )^{2} - 1\right ) - 6 \, \log \left (\sin \left (b x + a\right )^{2}\right )}{12 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^7/sin(b*x+a),x, algorithm="maxima")

[Out]

-1/12*((6*sin(b*x + a)^4 - 15*sin(b*x + a)^2 + 11)/(sin(b*x + a)^6 - 3*sin(b*x + a)^4 + 3*sin(b*x + a)^2 - 1)
+ 6*log(sin(b*x + a)^2 - 1) - 6*log(sin(b*x + a)^2))/b

________________________________________________________________________________________

Fricas [A]  time = 1.71733, size = 212, normalized size = 3.72 \begin{align*} -\frac{6 \, \cos \left (b x + a\right )^{6} \log \left (\cos \left (b x + a\right )^{2}\right ) - 6 \, \cos \left (b x + a\right )^{6} \log \left (-\frac{1}{4} \, \cos \left (b x + a\right )^{2} + \frac{1}{4}\right ) - 6 \, \cos \left (b x + a\right )^{4} - 3 \, \cos \left (b x + a\right )^{2} - 2}{12 \, b \cos \left (b x + a\right )^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^7/sin(b*x+a),x, algorithm="fricas")

[Out]

-1/12*(6*cos(b*x + a)^6*log(cos(b*x + a)^2) - 6*cos(b*x + a)^6*log(-1/4*cos(b*x + a)^2 + 1/4) - 6*cos(b*x + a)
^4 - 3*cos(b*x + a)^2 - 2)/(b*cos(b*x + a)^6)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.22457, size = 289, normalized size = 5.07 \begin{align*} \frac{\frac{\frac{522 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac{1485 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac{1580 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}} + \frac{1485 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} + \frac{522 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{5}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{5}} + \frac{147 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{6}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{6}} + 147}{{\left (\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1\right )}^{6}} + 30 \, \log \left (\frac{{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right ) - 60 \, \log \left ({\left | -\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1 \right |}\right )}{60 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^7/sin(b*x+a),x, algorithm="giac")

[Out]

1/60*((522*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 1485*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 1580*(cos(
b*x + a) - 1)^3/(cos(b*x + a) + 1)^3 + 1485*(cos(b*x + a) - 1)^4/(cos(b*x + a) + 1)^4 + 522*(cos(b*x + a) - 1)
^5/(cos(b*x + a) + 1)^5 + 147*(cos(b*x + a) - 1)^6/(cos(b*x + a) + 1)^6 + 147)/((cos(b*x + a) - 1)/(cos(b*x +
a) + 1) + 1)^6 + 30*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)) - 60*log(abs(-(cos(b*x + a) - 1)/(cos(b*
x + a) + 1) - 1)))/b