\(\int \csc (a+b x) \sec ^7(a+b x) \, dx\) [132]

Optimal result

Integrand size = 15, antiderivative size = 57 \[ \int \csc (a+b x) \sec ^7(a+b x) \, dx=\frac {\log (\tan (a+b x))}{b}+\frac {3 \tan ^2(a+b x)}{2 b}+\frac {3 \tan ^4(a+b x)}{4 b}+\frac {\tan ^6(a+b x)}{6 b} \]

[Out]

ln(tan(b*x+a))/b+3/2*tan(b*x+a)^2/b+3/4*tan(b*x+a)^4/b+1/6*tan(b*x+a)^6/b

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2700, 272, 45} \[ \int \csc (a+b x) \sec ^7(a+b x) \, dx=\frac {\tan ^6(a+b x)}{6 b}+\frac {3 \tan ^4(a+b x)}{4 b}+\frac {3 \tan ^2(a+b x)}{2 b}+\frac {\log (\tan (a+b x))}{b} \]

[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 45

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 272

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 2700

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*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (1+x^2\right )^3}{x} \, dx,x,\tan (a+b x)\right )}{b} \\ & = \frac {\text {Subst}\left (\int \frac {(1+x)^3}{x} \, dx,x,\tan ^2(a+b x)\right )}{2 b} \\ & = \frac {\text {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] (verified)

Time = 0.02 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.21 \[ \int \csc (a+b x) \sec ^7(a+b x) \, dx=-\frac {\log (\cos (a+b x))}{b}+\frac {\log (\sin (a+b x))}{b}+\frac {\sec ^2(a+b x)}{2 b}+\frac {\sec ^4(a+b x)}{4 b}+\frac {\sec ^6(a+b x)}{6 b} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.75

method	result	size
derivativedivides	\(\frac {\frac {1}{6 \cos \left (b x +a \right )^{6}}+\frac {1}{4 \cos \left (b x +a \right )^{4}}+\frac {1}{2 \cos \left (b x +a \right )^{2}}+\ln \left (\tan \left (b x +a \right )\right )}{b}\)	\(43\)
default	\(\frac {\frac {1}{6 \cos \left (b x +a \right )^{6}}+\frac {1}{4 \cos \left (b x +a \right )^{4}}+\frac {1}{2 \cos \left (b x +a \right )^{2}}+\ln \left (\tan \left (b x +a \right )\right )}{b}\)	\(43\)
risch	\(\frac {2 \,{\mathrm e}^{10 i \left (b x +a \right )}+12 \,{\mathrm e}^{8 i \left (b x +a \right )}+\frac {92 \,{\mathrm e}^{6 i \left (b x +a \right )}}{3}+12 \,{\mathrm e}^{4 i \left (b x +a \right )}+2 \,{\mathrm e}^{2 i \left (b x +a \right )}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{6}}-\frac {\ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{b}+\frac {\ln \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}{b}\)	\(109\)
norman	\(\frac {\frac {6 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}+\frac {6 \left (\tan ^{10}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}-\frac {12 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}-\frac {12 \left (\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}+\frac {68 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}}{\left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{6}}+\frac {\ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}-\frac {\ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}{b}-\frac {\ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}{b}\)	\(147\)
parallelrisch	\(\frac {\left (-180 \cos \left (2 b x +2 a \right )-72 \cos \left (4 b x +4 a \right )-12 \cos \left (6 b x +6 a \right )-120\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )+\left (-180 \cos \left (2 b x +2 a \right )-72 \cos \left (4 b x +4 a \right )-12 \cos \left (6 b x +6 a \right )-120\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )+\left (180 \cos \left (2 b x +2 a \right )+72 \cos \left (4 b x +4 a \right )+12 \cos \left (6 b x +6 a \right )+120\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )-21 \cos \left (2 b x +2 a \right )-42 \cos \left (4 b x +4 a \right )-11 \cos \left (6 b x +6 a \right )+74}{12 b \left (\cos \left (6 b x +6 a \right )+6 \cos \left (4 b x +4 a \right )+15 \cos \left (2 b x +2 a \right )+10\right )}\)	\(218\)

[In]

int(sec(b*x+a)^7/sin(b*x+a),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.35 \[ \int \csc (a+b x) \sec ^7(a+b x) \, dx=-\frac {6 \, \cos \left (b x + a\right )^{6} \log \left (\cos \left (b x + a\right )^{2}\right ) - 6 \, \cos \left (b x + a\right )^{6} \log \left (-\frac {1}{4} \, \cos \left (b x + a\right )^{2} + \frac {1}{4}\right ) - 6 \, \cos \left (b x + a\right )^{4} - 3 \, \cos \left (b x + a\right )^{2} - 2}{12 \, b \cos \left (b x + a\right )^{6}} \]

[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]

\[ \int \csc (a+b x) \sec ^7(a+b x) \, dx=\int \frac {\sec ^{7}{\left (a + b x \right )}}{\sin {\left (a + b x \right )}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.49 \[ \int \csc (a+b x) \sec ^7(a+b x) \, dx=-\frac {\frac {6 \, \sin \left (b x + a\right )^{4} - 15 \, \sin \left (b x + a\right )^{2} + 11}{\sin \left (b x + a\right )^{6} - 3 \, \sin \left (b x + a\right )^{4} + 3 \, \sin \left (b x + a\right )^{2} - 1} + 6 \, \log \left (\sin \left (b x + a\right )^{2} - 1\right ) - 6 \, \log \left (\sin \left (b x + a\right )^{2}\right )}{12 \, b} \]

[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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 214 vs. \(2 (51) = 102\).

Time = 0.34 (sec) , antiderivative size = 214, normalized size of antiderivative = 3.75 \[ \int \csc (a+b x) \sec ^7(a+b x) \, dx=\frac {\frac {\frac {522 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac {1485 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac {1580 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}} + \frac {1485 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} + \frac {522 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{5}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{5}} + \frac {147 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{6}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{6}} + 147}{{\left (\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1\right )}^{6}} + 30 \, \log \left (\frac {{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right ) - 60 \, \log \left ({\left | -\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1 \right |}\right )}{60 \, b} \]

[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

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.98 \[ \int \csc (a+b x) \sec ^7(a+b x) \, dx=\frac {\frac {\ln \left ({\sin \left (a+b\,x\right )}^2\right )}{2}-\ln \left (\cos \left (a+b\,x\right )\right )+\frac {\frac {{\cos \left (a+b\,x\right )}^4}{2}+\frac {{\cos \left (a+b\,x\right )}^2}{4}+\frac {1}{6}}{{\cos \left (a+b\,x\right )}^6}}{b} \]

[In]

int(1/(cos(a + b*x)^7*sin(a + b*x)),x)

[Out]

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