3.184 \(\int \csc ^6(c+d x) (a+b \sec (c+d x))^2 \, dx\)

Optimal. Leaf size=143 \[ -\frac{\left (a^2+b^2\right ) \cot ^5(c+d x)}{5 d}-\frac{\left (2 a^2+3 b^2\right ) \cot ^3(c+d x)}{3 d}-\frac{\left (a^2+3 b^2\right ) \cot (c+d x)}{d}-\frac{2 a b \csc ^5(c+d x)}{5 d}-\frac{2 a b \csc ^3(c+d x)}{3 d}-\frac{2 a b \csc (c+d x)}{d}+\frac{2 a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac{b^2 \tan (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.407985, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3872, 2911, 2621, 302, 207, 448} \[ -\frac{\left (a^2+b^2\right ) \cot ^5(c+d x)}{5 d}-\frac{\left (2 a^2+3 b^2\right ) \cot ^3(c+d x)}{3 d}-\frac{\left (a^2+3 b^2\right ) \cot (c+d x)}{d}-\frac{2 a b \csc ^5(c+d x)}{5 d}-\frac{2 a b \csc ^3(c+d x)}{3 d}-\frac{2 a b \csc (c+d x)}{d}+\frac{2 a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac{b^2 \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^6*(a + b*Sec[c + d*x])^2,x]

[Out]

(2*a*b*ArcTanh[Sin[c + d*x]])/d - ((a^2 + 3*b^2)*Cot[c + d*x])/d - ((2*a^2 + 3*b^2)*Cot[c + d*x]^3)/(3*d) - ((
a^2 + b^2)*Cot[c + d*x]^5)/(5*d) - (2*a*b*Csc[c + d*x])/d - (2*a*b*Csc[c + d*x]^3)/(3*d) - (2*a*b*Csc[c + d*x]
^5)/(5*d) + (b^2*Tan[c + d*x])/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 2911

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^2, x_Symbol] :> Dist[(2*a*b)/d, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n + 1), x], x] + Int[(g*Cos[e
+ f*x])^p*(d*Sin[e + f*x])^n*(a^2 + b^2*Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && NeQ[a^2 -
b^2, 0]

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 448

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 0]

Rubi steps

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

Mathematica [B]  time = 0.721495, size = 368, normalized size = 2.57 \[ -\frac{\csc ^7\left (\frac{1}{2} (c+d x)\right ) \sec ^5\left (\frac{1}{2} (c+d x)\right ) \left (20 \left (a^2+6 b^2\right ) \cos (2 (c+d x))-16 a^2 \cos (4 (c+d x))+4 a^2 \cos (6 (c+d x))+40 a^2+196 a b \cos (c+d x)-130 a b \cos (3 (c+d x))+30 a b \cos (5 (c+d x))+75 a b \sin (2 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-75 a b \sin (2 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-60 a b \sin (4 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+60 a b \sin (4 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+15 a b \sin (6 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-15 a b \sin (6 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-96 b^2 \cos (4 (c+d x))+24 b^2 \cos (6 (c+d x))\right )}{7680 d \left (\cot ^2\left (\frac{1}{2} (c+d x)\right )-1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^6*(a + b*Sec[c + d*x])^2,x]

[Out]

-(Csc[(c + d*x)/2]^7*Sec[(c + d*x)/2]^5*(40*a^2 + 196*a*b*Cos[c + d*x] + 20*(a^2 + 6*b^2)*Cos[2*(c + d*x)] - 1
30*a*b*Cos[3*(c + d*x)] - 16*a^2*Cos[4*(c + d*x)] - 96*b^2*Cos[4*(c + d*x)] + 30*a*b*Cos[5*(c + d*x)] + 4*a^2*
Cos[6*(c + d*x)] + 24*b^2*Cos[6*(c + d*x)] + 75*a*b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[2*(c + d*x)]
- 75*a*b*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[2*(c + d*x)] - 60*a*b*Log[Cos[(c + d*x)/2] - Sin[(c + d*
x)/2]]*Sin[4*(c + d*x)] + 60*a*b*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[4*(c + d*x)] + 15*a*b*Log[Cos[(c
+ d*x)/2] - Sin[(c + d*x)/2]]*Sin[6*(c + d*x)] - 15*a*b*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[6*(c + d
*x)]))/(7680*d*(-1 + Cot[(c + d*x)/2]^2))

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Maple [A]  time = 0.048, size = 212, normalized size = 1.5 \begin{align*} -{\frac{8\,{a}^{2}\cot \left ( dx+c \right ) }{15\,d}}-{\frac{{a}^{2}\cot \left ( dx+c \right ) \left ( \csc \left ( dx+c \right ) \right ) ^{4}}{5\,d}}-{\frac{4\,{a}^{2}\cot \left ( dx+c \right ) \left ( \csc \left ( dx+c \right ) \right ) ^{2}}{15\,d}}-{\frac{2\,ab}{5\,d \left ( \sin \left ( dx+c \right ) \right ) ^{5}}}-{\frac{2\,ab}{3\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}-2\,{\frac{ab}{d\sin \left ( dx+c \right ) }}+2\,{\frac{ab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-{\frac{{b}^{2}}{5\,d \left ( \sin \left ( dx+c \right ) \right ) ^{5}\cos \left ( dx+c \right ) }}-{\frac{2\,{b}^{2}}{5\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}\cos \left ( dx+c \right ) }}+{\frac{8\,{b}^{2}}{5\,d\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }}-{\frac{16\,{b}^{2}\cot \left ( dx+c \right ) }{5\,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))^2,x)

[Out]

-8/15*a^2*cot(d*x+c)/d-1/5/d*a^2*cot(d*x+c)*csc(d*x+c)^4-4/15/d*a^2*cot(d*x+c)*csc(d*x+c)^2-2/5/d*a*b/sin(d*x+
c)^5-2/3/d*a*b/sin(d*x+c)^3-2/d*a*b/sin(d*x+c)+2/d*a*b*ln(sec(d*x+c)+tan(d*x+c))-1/5/d*b^2/sin(d*x+c)^5/cos(d*
x+c)-2/5/d*b^2/sin(d*x+c)^3/cos(d*x+c)+8/5/d*b^2/sin(d*x+c)/cos(d*x+c)-16/5/d*b^2*cot(d*x+c)

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Maxima [A]  time = 1.04026, size = 193, normalized size = 1.35 \begin{align*} -\frac{a 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 )} + 3 \, b^{2}{\left (\frac{15 \, \tan \left (d x + c\right )^{4} + 5 \, \tan \left (d x + c\right )^{2} + 1}{\tan \left (d x + c\right )^{5}} - 5 \, \tan \left (d x + c\right )\right )} + \frac{{\left (15 \, \tan \left (d x + c\right )^{4} + 10 \, \tan \left (d x + c\right )^{2} + 3\right )} a^{2}}{\tan \left (d x + c\right )^{5}}}{15 \, 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))^2,x, algorithm="maxima")

[Out]

-1/15*(a*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(si
n(d*x + c) - 1)) + 3*b^2*((15*tan(d*x + c)^4 + 5*tan(d*x + c)^2 + 1)/tan(d*x + c)^5 - 5*tan(d*x + c)) + (15*ta
n(d*x + c)^4 + 10*tan(d*x + c)^2 + 3)*a^2/tan(d*x + c)^5)/d

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Fricas [A]  time = 1.86416, size = 629, normalized size = 4.4 \begin{align*} -\frac{30 \, a b \cos \left (d x + c\right )^{5} + 8 \,{\left (a^{2} + 6 \, b^{2}\right )} \cos \left (d x + c\right )^{6} - 70 \, a b \cos \left (d x + c\right )^{3} - 20 \,{\left (a^{2} + 6 \, b^{2}\right )} \cos \left (d x + c\right )^{4} + 46 \, a b \cos \left (d x + c\right ) + 15 \,{\left (a^{2} + 6 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 15 \,{\left (a b \cos \left (d x + c\right )^{5} - 2 \, a b \cos \left (d x + c\right )^{3} + a b \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 15 \,{\left (a b \cos \left (d x + c\right )^{5} - 2 \, a b \cos \left (d x + c\right )^{3} + a b \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 15 \, b^{2}}{15 \,{\left (d \cos \left (d x + c\right )^{5} - 2 \, d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )\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))^2,x, algorithm="fricas")

[Out]

-1/15*(30*a*b*cos(d*x + c)^5 + 8*(a^2 + 6*b^2)*cos(d*x + c)^6 - 70*a*b*cos(d*x + c)^3 - 20*(a^2 + 6*b^2)*cos(d
*x + c)^4 + 46*a*b*cos(d*x + c) + 15*(a^2 + 6*b^2)*cos(d*x + c)^2 - 15*(a*b*cos(d*x + c)^5 - 2*a*b*cos(d*x + c
)^3 + a*b*cos(d*x + c))*log(sin(d*x + c) + 1)*sin(d*x + c) + 15*(a*b*cos(d*x + c)^5 - 2*a*b*cos(d*x + c)^3 + a
*b*cos(d*x + c))*log(-sin(d*x + c) + 1)*sin(d*x + c) - 15*b^2)/((d*cos(d*x + c)^5 - 2*d*cos(d*x + c)^3 + d*cos
(d*x + c))*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))**2,x)

[Out]

Timed out

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Giac [B]  time = 1.36358, size = 440, normalized size = 3.08 \begin{align*} \frac{3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 25 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 70 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 45 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 960 \, a b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 960 \, a b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + 150 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 660 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 570 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{960 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1} - \frac{150 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 660 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 570 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 25 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 70 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 45 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}}{\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))^2,x, algorithm="giac")

[Out]

1/480*(3*a^2*tan(1/2*d*x + 1/2*c)^5 - 6*a*b*tan(1/2*d*x + 1/2*c)^5 + 3*b^2*tan(1/2*d*x + 1/2*c)^5 + 25*a^2*tan
(1/2*d*x + 1/2*c)^3 - 70*a*b*tan(1/2*d*x + 1/2*c)^3 + 45*b^2*tan(1/2*d*x + 1/2*c)^3 + 960*a*b*log(abs(tan(1/2*
d*x + 1/2*c) + 1)) - 960*a*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 150*a^2*tan(1/2*d*x + 1/2*c) - 660*a*b*tan(1
/2*d*x + 1/2*c) + 570*b^2*tan(1/2*d*x + 1/2*c) - 960*b^2*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) - (
150*a^2*tan(1/2*d*x + 1/2*c)^4 + 660*a*b*tan(1/2*d*x + 1/2*c)^4 + 570*b^2*tan(1/2*d*x + 1/2*c)^4 + 25*a^2*tan(
1/2*d*x + 1/2*c)^2 + 70*a*b*tan(1/2*d*x + 1/2*c)^2 + 45*b^2*tan(1/2*d*x + 1/2*c)^2 + 3*a^2 + 6*a*b + 3*b^2)/ta
n(1/2*d*x + 1/2*c)^5)/d