3.173 \(\int \csc ^6(c+d x) (a+b \sec (c+d x)) \, dx\)

Optimal. Leaf size=101 \[ -\frac{a \cot ^5(c+d x)}{5 d}-\frac{2 a \cot ^3(c+d x)}{3 d}-\frac{a \cot (c+d x)}{d}-\frac{b \csc ^5(c+d x)}{5 d}-\frac{b \csc ^3(c+d x)}{3 d}-\frac{b \csc (c+d x)}{d}+\frac{b \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

(b*ArcTanh[Sin[c + d*x]])/d - (a*Cot[c + d*x])/d - (2*a*Cot[c + d*x]^3)/(3*d) - (a*Cot[c + d*x]^5)/(5*d) - (b*
Csc[c + d*x])/d - (b*Csc[c + d*x]^3)/(3*d) - (b*Csc[c + d*x]^5)/(5*d)

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Rubi [A]  time = 0.110891, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {3872, 2838, 2621, 302, 207, 3767} \[ -\frac{a \cot ^5(c+d x)}{5 d}-\frac{2 a \cot ^3(c+d x)}{3 d}-\frac{a \cot (c+d x)}{d}-\frac{b \csc ^5(c+d x)}{5 d}-\frac{b \csc ^3(c+d x)}{3 d}-\frac{b \csc (c+d x)}{d}+\frac{b \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^6*(a + b*Sec[c + d*x]),x]

[Out]

(b*ArcTanh[Sin[c + d*x]])/d - (a*Cot[c + d*x])/d - (2*a*Cot[c + d*x]^3)/(3*d) - (a*Cot[c + d*x]^5)/(5*d) - (b*
Csc[c + d*x])/d - (b*Csc[c + d*x]^3)/(3*d) - (b*Csc[c + d*x]^5)/(5*d)

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2838

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]), x_Symbol] :> Dist[a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[(g*Cos[e + f*x
])^p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]

Rule 2621

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])

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])

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin{align*} \int \csc ^6(c+d x) (a+b \sec (c+d x)) \, dx &=-\int (-b-a \cos (c+d x)) \csc ^6(c+d x) \sec (c+d x) \, dx\\ &=a \int \csc ^6(c+d x) \, dx+b \int \csc ^6(c+d x) \sec (c+d x) \, dx\\ &=-\frac{a \operatorname{Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,\cot (c+d x)\right )}{d}-\frac{b \operatorname{Subst}\left (\int \frac{x^6}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{d}\\ &=-\frac{a \cot (c+d x)}{d}-\frac{2 a \cot ^3(c+d x)}{3 d}-\frac{a \cot ^5(c+d x)}{5 d}-\frac{b \operatorname{Subst}\left (\int \left (1+x^2+x^4+\frac{1}{-1+x^2}\right ) \, dx,x,\csc (c+d x)\right )}{d}\\ &=-\frac{a \cot (c+d x)}{d}-\frac{2 a \cot ^3(c+d x)}{3 d}-\frac{a \cot ^5(c+d x)}{5 d}-\frac{b \csc (c+d x)}{d}-\frac{b \csc ^3(c+d x)}{3 d}-\frac{b \csc ^5(c+d x)}{5 d}-\frac{b \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{d}\\ &=\frac{b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{a \cot (c+d x)}{d}-\frac{2 a \cot ^3(c+d x)}{3 d}-\frac{a \cot ^5(c+d x)}{5 d}-\frac{b \csc (c+d x)}{d}-\frac{b \csc ^3(c+d x)}{3 d}-\frac{b \csc ^5(c+d x)}{5 d}\\ \end{align*}

Mathematica [C]  time = 0.0257432, size = 91, normalized size = 0.9 \[ -\frac{b \csc ^5(c+d x) \text{Hypergeometric2F1}\left (-\frac{5}{2},1,-\frac{3}{2},\sin ^2(c+d x)\right )}{5 d}-\frac{8 a \cot (c+d x)}{15 d}-\frac{a \cot (c+d x) \csc ^4(c+d x)}{5 d}-\frac{4 a \cot (c+d x) \csc ^2(c+d x)}{15 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^6*(a + b*Sec[c + d*x]),x]

[Out]

(-8*a*Cot[c + d*x])/(15*d) - (4*a*Cot[c + d*x]*Csc[c + d*x]^2)/(15*d) - (a*Cot[c + d*x]*Csc[c + d*x]^4)/(5*d)
- (b*Csc[c + d*x]^5*Hypergeometric2F1[-5/2, 1, -3/2, Sin[c + d*x]^2])/(5*d)

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Maple [A]  time = 0.042, size = 115, normalized size = 1.1 \begin{align*} -{\frac{8\,a\cot \left ( dx+c \right ) }{15\,d}}-{\frac{a\cot \left ( dx+c \right ) \left ( \csc \left ( dx+c \right ) \right ) ^{4}}{5\,d}}-{\frac{4\,a\cot \left ( dx+c \right ) \left ( \csc \left ( dx+c \right ) \right ) ^{2}}{15\,d}}-{\frac{b}{5\,d \left ( \sin \left ( dx+c \right ) \right ) ^{5}}}-{\frac{b}{3\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{b}{d\sin \left ( dx+c \right ) }}+{\frac{b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^6*(a+b*sec(d*x+c)),x)

[Out]

-8/15*a*cot(d*x+c)/d-1/5/d*a*cot(d*x+c)*csc(d*x+c)^4-4/15/d*a*cot(d*x+c)*csc(d*x+c)^2-1/5/d*b/sin(d*x+c)^5-1/3
/d*b/sin(d*x+c)^3-1/d*b/sin(d*x+c)+1/d*b*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 0.969416, size = 130, normalized size = 1.29 \begin{align*} -\frac{b{\left (\frac{2 \,{\left (15 \, \sin \left (d x + c\right )^{4} + 5 \, \sin \left (d x + c\right )^{2} + 3\right )}}{\sin \left (d x + c\right )^{5}} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + \frac{2 \,{\left (15 \, \tan \left (d x + c\right )^{4} + 10 \, \tan \left (d x + c\right )^{2} + 3\right )} a}{\tan \left (d x + c\right )^{5}}}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^6*(a+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

-1/30*(b*(2*(15*sin(d*x + c)^4 + 5*sin(d*x + c)^2 + 3)/sin(d*x + c)^5 - 15*log(sin(d*x + c) + 1) + 15*log(sin(
d*x + c) - 1)) + 2*(15*tan(d*x + c)^4 + 10*tan(d*x + c)^2 + 3)*a/tan(d*x + c)^5)/d

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Fricas [A]  time = 1.7965, size = 473, normalized size = 4.68 \begin{align*} -\frac{16 \, a \cos \left (d x + c\right )^{5} + 30 \, b \cos \left (d x + c\right )^{4} - 40 \, a \cos \left (d x + c\right )^{3} - 70 \, b \cos \left (d x + c\right )^{2} - 15 \,{\left (b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 15 \,{\left (b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + b\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 30 \, a \cos \left (d x + c\right ) + 46 \, b}{30 \,{\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^6*(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

-1/30*(16*a*cos(d*x + c)^5 + 30*b*cos(d*x + c)^4 - 40*a*cos(d*x + c)^3 - 70*b*cos(d*x + c)^2 - 15*(b*cos(d*x +
c)^4 - 2*b*cos(d*x + c)^2 + b)*log(sin(d*x + c) + 1)*sin(d*x + c) + 15*(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2
+ b)*log(-sin(d*x + c) + 1)*sin(d*x + c) + 30*a*cos(d*x + c) + 46*b)/((d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2
+ d)*sin(d*x + c))

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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(csc(d*x+c)**6*(a+b*sec(d*x+c)),x)

[Out]

Timed out

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Giac [B]  time = 1.3389, size = 262, normalized size = 2.59 \begin{align*} \frac{3 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 25 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 35 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 480 \, b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 480 \, b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + 150 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 330 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{150 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 330 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 25 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 35 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, a + 3 \, b}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5}}}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^6*(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

1/480*(3*a*tan(1/2*d*x + 1/2*c)^5 - 3*b*tan(1/2*d*x + 1/2*c)^5 + 25*a*tan(1/2*d*x + 1/2*c)^3 - 35*b*tan(1/2*d*
x + 1/2*c)^3 + 480*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 480*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 150*a*tan
(1/2*d*x + 1/2*c) - 330*b*tan(1/2*d*x + 1/2*c) - (150*a*tan(1/2*d*x + 1/2*c)^4 + 330*b*tan(1/2*d*x + 1/2*c)^4
+ 25*a*tan(1/2*d*x + 1/2*c)^2 + 35*b*tan(1/2*d*x + 1/2*c)^2 + 3*a + 3*b)/tan(1/2*d*x + 1/2*c)^5)/d