\(\int \csc ^3(a+b x) \sec ^5(a+b x) \, dx\) [156]

Optimal result

Integrand size = 17, antiderivative size = 58 \[ \int \csc ^3(a+b x) \sec ^5(a+b x) \, dx=-\frac {\cot ^2(a+b x)}{2 b}+\frac {3 \log (\tan (a+b x))}{b}+\frac {3 \tan ^2(a+b x)}{2 b}+\frac {\tan ^4(a+b x)}{4 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2700, 272, 45} \[ \int \csc ^3(a+b x) \sec ^5(a+b x) \, dx=\frac {\tan ^4(a+b x)}{4 b}+\frac {3 \tan ^2(a+b x)}{2 b}-\frac {\cot ^2(a+b x)}{2 b}+\frac {3 \log (\tan (a+b x))}{b} \]

[In]

Int[Csc[a + b*x]^3*Sec[a + b*x]^5,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.16 \[ \int \csc ^3(a+b x) \sec ^5(a+b x) \, dx=-\frac {\csc ^2(a+b x)}{2 b}-\frac {3 \log (\cos (a+b x))}{b}+\frac {3 \log (\sin (a+b x))}{b}+\frac {\sec ^2(a+b x)}{b}+\frac {\sec ^4(a+b x)}{4 b} \]

[In]

Integrate[Csc[a + b*x]^3*Sec[a + b*x]^5,x]

[Out]

-1/2*Csc[a + b*x]^2/b - (3*Log[Cos[a + b*x]])/b + (3*Log[Sin[a + b*x]])/b + Sec[a + b*x]^2/b + Sec[a + b*x]^4/
(4*b)

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.05

method	result	size
derivativedivides	\(\frac {\frac {1}{4 \cos \left (b x +a \right )^{4} \sin \left (b x +a \right )^{2}}+\frac {3}{4 \cos \left (b x +a \right )^{2} \sin \left (b x +a \right )^{2}}-\frac {3}{2 \sin \left (b x +a \right )^{2}}+3 \ln \left (\tan \left (b x +a \right )\right )}{b}\)	\(61\)
default	\(\frac {\frac {1}{4 \cos \left (b x +a \right )^{4} \sin \left (b x +a \right )^{2}}+\frac {3}{4 \cos \left (b x +a \right )^{2} \sin \left (b x +a \right )^{2}}-\frac {3}{2 \sin \left (b x +a \right )^{2}}+3 \ln \left (\tan \left (b x +a \right )\right )}{b}\)	\(61\)
risch	\(\frac {6 \,{\mathrm e}^{10 i \left (b x +a \right )}+12 \,{\mathrm e}^{8 i \left (b x +a \right )}-4 \,{\mathrm e}^{6 i \left (b x +a \right )}+12 \,{\mathrm e}^{4 i \left (b x +a \right )}+6 \,{\mathrm e}^{2 i \left (b x +a \right )}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{4} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{b}+\frac {3 \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}{b}\)	\(123\)
norman	\(\frac {-\frac {1}{8 b}-\frac {\tan ^{12}\left (\frac {b x}{2}+\frac {a}{2}\right )}{8 b}-\frac {10 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}+\frac {57 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8 b}+\frac {57 \left (\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8 b}}{\left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{4} \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}+\frac {3 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}-\frac {3 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}{b}-\frac {3 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}{b}\)	\(148\)
parallelrisch	\(\frac {\left (-24 \cos \left (2 b x +2 a \right )-6 \cos \left (4 b x +4 a \right )-18\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )+\left (-24 \cos \left (2 b x +2 a \right )-6 \cos \left (4 b x +4 a \right )-18\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )+\left (24 \cos \left (2 b x +2 a \right )+6 \cos \left (4 b x +4 a \right )+18\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\left (-36 \cos \left (b x +a \right )+24 \cos \left (2 b x +2 a \right )-12 \cos \left (3 b x +3 a \right )+3 \cos \left (4 b x +4 a \right )+9\right ) \left (\cot ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-2 \left (\csc ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \left (\sec ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-6\right )}{2 b \left (\cos \left (4 b x +4 a \right )+4 \cos \left (2 b x +2 a \right )+3\right )}\)	\(221\)

[In]

int(sec(b*x+a)^5/sin(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 112 vs. \(2 (52) = 104\).

Time = 0.36 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.93 \[ \int \csc ^3(a+b x) \sec ^5(a+b x) \, dx=\frac {6 \, \cos \left (b x + a\right )^{4} - 3 \, \cos \left (b x + a\right )^{2} - 6 \, {\left (\cos \left (b x + a\right )^{6} - \cos \left (b x + a\right )^{4}\right )} \log \left (\cos \left (b x + a\right )^{2}\right ) + 6 \, {\left (\cos \left (b x + a\right )^{6} - \cos \left (b x + a\right )^{4}\right )} \log \left (-\frac {1}{4} \, \cos \left (b x + a\right )^{2} + \frac {1}{4}\right ) - 1}{4 \, {\left (b \cos \left (b x + a\right )^{6} - b \cos \left (b x + a\right )^{4}\right )}} \]

[In]

integrate(sec(b*x+a)^5/sin(b*x+a)^3,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \csc ^3(a+b x) \sec ^5(a+b x) \, dx=\int \frac {\sec ^{5}{\left (a + b x \right )}}{\sin ^{3}{\left (a + b x \right )}}\, dx \]

[In]

integrate(sec(b*x+a)**5/sin(b*x+a)**3,x)

[Out]

Integral(sec(a + b*x)**5/sin(a + b*x)**3, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.41 \[ \int \csc ^3(a+b x) \sec ^5(a+b x) \, dx=-\frac {\frac {6 \, \sin \left (b x + a\right )^{4} - 9 \, \sin \left (b x + a\right )^{2} + 2}{\sin \left (b x + a\right )^{6} - 2 \, \sin \left (b x + a\right )^{4} + \sin \left (b x + a\right )^{2}} + 6 \, \log \left (\sin \left (b x + a\right )^{2} - 1\right ) - 6 \, \log \left (\sin \left (b x + a\right )^{2}\right )}{4 \, b} \]

[In]

integrate(sec(b*x+a)^5/sin(b*x+a)^3,x, algorithm="maxima")

[Out]

-1/4*((6*sin(b*x + a)^4 - 9*sin(b*x + a)^2 + 2)/(sin(b*x + a)^6 - 2*sin(b*x + a)^4 + sin(b*x + a)^2) + 6*log(s
in(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. 232 vs. \(2 (52) = 104\).

Time = 0.36 (sec) , antiderivative size = 232, normalized size of antiderivative = 4.00 \[ \int \csc ^3(a+b x) \sec ^5(a+b x) \, dx=-\frac {\frac {{\left (\frac {12 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - 1\right )} {\left (\cos \left (b x + a\right ) + 1\right )}}{\cos \left (b x + a\right ) - 1} - \frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - \frac {2 \, {\left (\frac {76 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac {118 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac {76 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}} + \frac {25 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} + 25\right )}}{{\left (\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1\right )}^{4}} - 12 \, \log \left (\frac {{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right ) + 24 \, \log \left ({\left | -\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1 \right |}\right )}{8 \, b} \]

[In]

integrate(sec(b*x+a)^5/sin(b*x+a)^3,x, algorithm="giac")

[Out]

-1/8*((12*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 1)*(cos(b*x + a) + 1)/(cos(b*x + a) - 1) - (cos(b*x + a) - 1
)/(cos(b*x + a) + 1) - 2*(76*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 118*(cos(b*x + a) - 1)^2/(cos(b*x + a) +
1)^2 + 76*(cos(b*x + a) - 1)^3/(cos(b*x + a) + 1)^3 + 25*(cos(b*x + a) - 1)^4/(cos(b*x + a) + 1)^4 + 25)/((cos
(b*x + a) - 1)/(cos(b*x + a) + 1) + 1)^4 - 12*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)) + 24*log(abs(-
(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 1)))/b

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.28 \[ \int \csc ^3(a+b x) \sec ^5(a+b x) \, dx=\frac {3\,\ln \left ({\sin \left (a+b\,x\right )}^2\right )}{2\,b}-\frac {3\,\ln \left (\cos \left (a+b\,x\right )\right )}{b}+\frac {-\frac {3\,{\cos \left (a+b\,x\right )}^4}{2}+\frac {3\,{\cos \left (a+b\,x\right )}^2}{4}+\frac {1}{4}}{b\,\left ({\cos \left (a+b\,x\right )}^4-{\cos \left (a+b\,x\right )}^6\right )} \]

[In]

int(1/(cos(a + b*x)^5*sin(a + b*x)^3),x)

[Out]

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