\(\int \csc ^2(a+b x) \sec ^5(a+b x) \, dx\) [144]

Optimal result

Integrand size = 17, antiderivative size = 70 \[ \int \csc ^2(a+b x) \sec ^5(a+b x) \, dx=\frac {15 \text {arctanh}(\sin (a+b x))}{8 b}-\frac {15 \csc (a+b x)}{8 b}+\frac {5 \csc (a+b x) \sec ^2(a+b x)}{8 b}+\frac {\csc (a+b x) \sec ^4(a+b x)}{4 b} \]

[Out]

15/8*arctanh(sin(b*x+a))/b-15/8*csc(b*x+a)/b+5/8*csc(b*x+a)*sec(b*x+a)^2/b+1/4*csc(b*x+a)*sec(b*x+a)^4/b

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2701, 294, 327, 213} \[ \int \csc ^2(a+b x) \sec ^5(a+b x) \, dx=\frac {15 \text {arctanh}(\sin (a+b x))}{8 b}-\frac {15 \csc (a+b x)}{8 b}+\frac {\csc (a+b x) \sec ^4(a+b x)}{4 b}+\frac {5 \csc (a+b x) \sec ^2(a+b x)}{8 b} \]

[In]

Int[Csc[a + b*x]^2*Sec[a + b*x]^5,x]

[Out]

(15*ArcTanh[Sin[a + b*x]])/(8*b) - (15*Csc[a + b*x])/(8*b) + (5*Csc[a + b*x]*Sec[a + b*x]^2)/(8*b) + (Csc[a +
b*x]*Sec[a + b*x]^4)/(4*b)

Rule 213

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

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2701

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])

Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {x^6}{\left (-1+x^2\right )^3} \, dx,x,\csc (a+b x)\right )}{b} \\ & = \frac {\csc (a+b x) \sec ^4(a+b x)}{4 b}-\frac {5 \text {Subst}\left (\int \frac {x^4}{\left (-1+x^2\right )^2} \, dx,x,\csc (a+b x)\right )}{4 b} \\ & = \frac {5 \csc (a+b x) \sec ^2(a+b x)}{8 b}+\frac {\csc (a+b x) \sec ^4(a+b x)}{4 b}-\frac {15 \text {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{8 b} \\ & = -\frac {15 \csc (a+b x)}{8 b}+\frac {5 \csc (a+b x) \sec ^2(a+b x)}{8 b}+\frac {\csc (a+b x) \sec ^4(a+b x)}{4 b}-\frac {15 \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{8 b} \\ & = \frac {15 \text {arctanh}(\sin (a+b x))}{8 b}-\frac {15 \csc (a+b x)}{8 b}+\frac {5 \csc (a+b x) \sec ^2(a+b x)}{8 b}+\frac {\csc (a+b x) \sec ^4(a+b x)}{4 b} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.39 \[ \int \csc ^2(a+b x) \sec ^5(a+b x) \, dx=-\frac {\csc (a+b x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},3,\frac {1}{2},\sin ^2(a+b x)\right )}{b} \]

[In]

Integrate[Csc[a + b*x]^2*Sec[a + b*x]^5,x]

[Out]

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

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.97

method	result	size
derivativedivides	\(\frac {\frac {1}{4 \cos \left (b x +a \right )^{4} \sin \left (b x +a \right )}+\frac {5}{8 \cos \left (b x +a \right )^{2} \sin \left (b x +a \right )}-\frac {15}{8 \sin \left (b x +a \right )}+\frac {15 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{8}}{b}\)	\(68\)
default	\(\frac {\frac {1}{4 \cos \left (b x +a \right )^{4} \sin \left (b x +a \right )}+\frac {5}{8 \cos \left (b x +a \right )^{2} \sin \left (b x +a \right )}-\frac {15}{8 \sin \left (b x +a \right )}+\frac {15 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{8}}{b}\)	\(68\)
risch	\(-\frac {i \left (15 \,{\mathrm e}^{9 i \left (b x +a \right )}+40 \,{\mathrm e}^{7 i \left (b x +a \right )}+18 \,{\mathrm e}^{5 i \left (b x +a \right )}+40 \,{\mathrm e}^{3 i \left (b x +a \right )}+15 \,{\mathrm e}^{i \left (b x +a \right )}\right )}{4 b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{4} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}+\frac {15 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}{8 b}-\frac {15 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-i\right )}{8 b}\)	\(126\)
norman	\(\frac {-\frac {1}{2 b}+\frac {15 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{4 b}-\frac {5 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{4 b}-\frac {5 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{4 b}+\frac {15 \left (\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{4 b}-\frac {\tan ^{10}\left (\frac {b x}{2}+\frac {a}{2}\right )}{2 b}}{\left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{4} \tan \left (\frac {b x}{2}+\frac {a}{2}\right )}-\frac {15 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}{8 b}+\frac {15 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}{8 b}\)	\(149\)
parallelrisch	\(\frac {\left (-60 \cos \left (2 b x +2 a \right )-15 \cos \left (4 b x +4 a \right )-45\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )+\left (60 \cos \left (2 b x +2 a \right )+15 \cos \left (4 b x +4 a \right )+45\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )+\left (-170 \cos \left (b x +a \right )+60 \cos \left (2 b x +2 a \right )-30 \cos \left (3 b x +3 a \right )+140\right ) \cot \left (\frac {b x}{2}+\frac {a}{2}\right )-32 \sec \left (\frac {b x}{2}+\frac {a}{2}\right ) \csc \left (\frac {b x}{2}+\frac {a}{2}\right )}{8 b \left (\cos \left (4 b x +4 a \right )+4 \cos \left (2 b x +2 a \right )+3\right )}\)	\(167\)

[In]

int(sec(b*x+a)^5/sin(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

1/b*(1/4/cos(b*x+a)^4/sin(b*x+a)+5/8/cos(b*x+a)^2/sin(b*x+a)-15/8/sin(b*x+a)+15/8*ln(sec(b*x+a)+tan(b*x+a)))

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.36 \[ \int \csc ^2(a+b x) \sec ^5(a+b x) \, dx=\frac {15 \, \cos \left (b x + a\right )^{4} \log \left (\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 15 \, \cos \left (b x + a\right )^{4} \log \left (-\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 30 \, \cos \left (b x + a\right )^{4} + 10 \, \cos \left (b x + a\right )^{2} + 4}{16 \, b \cos \left (b x + a\right )^{4} \sin \left (b x + a\right )} \]

[In]

integrate(sec(b*x+a)^5/sin(b*x+a)^2,x, algorithm="fricas")

[Out]

1/16*(15*cos(b*x + a)^4*log(sin(b*x + a) + 1)*sin(b*x + a) - 15*cos(b*x + a)^4*log(-sin(b*x + a) + 1)*sin(b*x
+ a) - 30*cos(b*x + a)^4 + 10*cos(b*x + a)^2 + 4)/(b*cos(b*x + a)^4*sin(b*x + a))

Sympy [F]

\[ \int \csc ^2(a+b x) \sec ^5(a+b x) \, dx=\int \frac {\sec ^{5}{\left (a + b x \right )}}{\sin ^{2}{\left (a + b x \right )}}\, dx \]

[In]

integrate(sec(b*x+a)**5/sin(b*x+a)**2,x)

[Out]

Integral(sec(a + b*x)**5/sin(a + b*x)**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.13 \[ \int \csc ^2(a+b x) \sec ^5(a+b x) \, dx=-\frac {\frac {2 \, {\left (15 \, \sin \left (b x + a\right )^{4} - 25 \, \sin \left (b x + a\right )^{2} + 8\right )}}{\sin \left (b x + a\right )^{5} - 2 \, \sin \left (b x + a\right )^{3} + \sin \left (b x + a\right )} - 15 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 15 \, \log \left (\sin \left (b x + a\right ) - 1\right )}{16 \, b} \]

[In]

integrate(sec(b*x+a)^5/sin(b*x+a)^2,x, algorithm="maxima")

[Out]

-1/16*(2*(15*sin(b*x + a)^4 - 25*sin(b*x + a)^2 + 8)/(sin(b*x + a)^5 - 2*sin(b*x + a)^3 + sin(b*x + a)) - 15*l
og(sin(b*x + a) + 1) + 15*log(sin(b*x + a) - 1))/b

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.04 \[ \int \csc ^2(a+b x) \sec ^5(a+b x) \, dx=-\frac {\frac {2 \, {\left (7 \, \sin \left (b x + a\right )^{3} - 9 \, \sin \left (b x + a\right )\right )}}{{\left (\sin \left (b x + a\right )^{2} - 1\right )}^{2}} + \frac {16}{\sin \left (b x + a\right )} - 15 \, \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right ) + 15 \, \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right )}{16 \, b} \]

[In]

integrate(sec(b*x+a)^5/sin(b*x+a)^2,x, algorithm="giac")

[Out]

-1/16*(2*(7*sin(b*x + a)^3 - 9*sin(b*x + a))/(sin(b*x + a)^2 - 1)^2 + 16/sin(b*x + a) - 15*log(abs(sin(b*x + a
) + 1)) + 15*log(abs(sin(b*x + a) - 1)))/b

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.96 \[ \int \csc ^2(a+b x) \sec ^5(a+b x) \, dx=\frac {15\,\mathrm {atanh}\left (\sin \left (a+b\,x\right )\right )}{8\,b}-\frac {\frac {15\,{\sin \left (a+b\,x\right )}^4}{8}-\frac {25\,{\sin \left (a+b\,x\right )}^2}{8}+1}{b\,\left ({\sin \left (a+b\,x\right )}^5-2\,{\sin \left (a+b\,x\right )}^3+\sin \left (a+b\,x\right )\right )} \]

[In]

int(1/(cos(a + b*x)^5*sin(a + b*x)^2),x)

[Out]

(15*atanh(sin(a + b*x)))/(8*b) - ((15*sin(a + b*x)^4)/8 - (25*sin(a + b*x)^2)/8 + 1)/(b*(sin(a + b*x) - 2*sin(
a + b*x)^3 + sin(a + b*x)^5))