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

Optimal. Leaf size=122 \[ -\frac {a^2 \cot ^5(c+d x)}{5 d}+\frac {a^2 \cot ^3(c+d x)}{3 d}-\frac {a^2 \cot (c+d x)}{d}-a^2 x-\frac {2 a b \csc ^5(c+d x)}{5 d}+\frac {4 a b \csc ^3(c+d x)}{3 d}-\frac {2 a b \csc (c+d x)}{d}-\frac {b^2 \cot ^5(c+d x)}{5 d} \]

[Out]

-a^2*x-a^2*cot(d*x+c)/d+1/3*a^2*cot(d*x+c)^3/d-1/5*a^2*cot(d*x+c)^5/d-1/5*b^2*cot(d*x+c)^5/d-2*a*b*csc(d*x+c)/
d+4/3*a*b*csc(d*x+c)^3/d-2/5*a*b*csc(d*x+c)^5/d

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Rubi [A]  time = 0.13, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3886, 3473, 8, 2606, 194, 2607, 30} \[ -\frac {a^2 \cot ^5(c+d x)}{5 d}+\frac {a^2 \cot ^3(c+d x)}{3 d}-\frac {a^2 \cot (c+d x)}{d}-a^2 x-\frac {2 a b \csc ^5(c+d x)}{5 d}+\frac {4 a b \csc ^3(c+d x)}{3 d}-\frac {2 a b \csc (c+d x)}{d}-\frac {b^2 \cot ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*x) - (a^2*Cot[c + d*x])/d + (a^2*Cot[c + d*x]^3)/(3*d) - (a^2*Cot[c + d*x]^5)/(5*d) - (b^2*Cot[c + d*x]^
5)/(5*d) - (2*a*b*Csc[c + d*x])/d + (4*a*b*Csc[c + d*x]^3)/(3*d) - (2*a*b*Csc[c + d*x]^5)/(5*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3886

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[ExpandI
ntegrand[(e*Cot[c + d*x])^m, (a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0]

Rubi steps

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

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Mathematica [A]  time = 0.59, size = 198, normalized size = 1.62 \[ -\frac {\csc ^5(c+d x) \left (10 \left (5 a^2+3 b^2\right ) \cos (c+d x)+150 a^2 c \sin (c+d x)+150 a^2 d x \sin (c+d x)-75 a^2 c \sin (3 (c+d x))-75 a^2 d x \sin (3 (c+d x))+15 a^2 c \sin (5 (c+d x))+15 a^2 d x \sin (5 (c+d x))-25 a^2 \cos (3 (c+d x))+23 a^2 \cos (5 (c+d x))-80 a b \cos (2 (c+d x))+60 a b \cos (4 (c+d x))+116 a b+15 b^2 \cos (3 (c+d x))+3 b^2 \cos (5 (c+d x))\right )}{240 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/240*(Csc[c + d*x]^5*(116*a*b + 10*(5*a^2 + 3*b^2)*Cos[c + d*x] - 80*a*b*Cos[2*(c + d*x)] - 25*a^2*Cos[3*(c
+ d*x)] + 15*b^2*Cos[3*(c + d*x)] + 60*a*b*Cos[4*(c + d*x)] + 23*a^2*Cos[5*(c + d*x)] + 3*b^2*Cos[5*(c + d*x)]
+ 150*a^2*c*Sin[c + d*x] + 150*a^2*d*x*Sin[c + d*x] - 75*a^2*c*Sin[3*(c + d*x)] - 75*a^2*d*x*Sin[3*(c + d*x)]
+ 15*a^2*c*Sin[5*(c + d*x)] + 15*a^2*d*x*Sin[5*(c + d*x)]))/d

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fricas [A]  time = 0.48, size = 152, normalized size = 1.25 \[ -\frac {30 \, a b \cos \left (d x + c\right )^{4} + {\left (23 \, a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{5} - 35 \, a^{2} \cos \left (d x + c\right )^{3} - 40 \, a b \cos \left (d x + c\right )^{2} + 15 \, a^{2} \cos \left (d x + c\right ) + 16 \, a b + 15 \, {\left (a^{2} d x \cos \left (d x + c\right )^{4} - 2 \, a^{2} d x \cos \left (d x + c\right )^{2} + a^{2} d x\right )} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/15*(30*a*b*cos(d*x + c)^4 + (23*a^2 + 3*b^2)*cos(d*x + c)^5 - 35*a^2*cos(d*x + c)^3 - 40*a*b*cos(d*x + c)^2
+ 15*a^2*cos(d*x + c) + 16*a*b + 15*(a^2*d*x*cos(d*x + c)^4 - 2*a^2*d*x*cos(d*x + c)^2 + a^2*d*x)*sin(d*x + c
))/((d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)*sin(d*x + c))

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giac [B]  time = 1.21, size = 273, normalized size = 2.24 \[ \frac {3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 35 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 50 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 480 \, {\left (d x + c\right )} a^{2} + 330 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 300 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 30 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {330 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 300 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 30 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 35 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 50 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(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 - 35*a^2*tan
(1/2*d*x + 1/2*c)^3 + 50*a*b*tan(1/2*d*x + 1/2*c)^3 - 15*b^2*tan(1/2*d*x + 1/2*c)^3 - 480*(d*x + c)*a^2 + 330*
a^2*tan(1/2*d*x + 1/2*c) - 300*a*b*tan(1/2*d*x + 1/2*c) + 30*b^2*tan(1/2*d*x + 1/2*c) - (330*a^2*tan(1/2*d*x +
1/2*c)^4 + 300*a*b*tan(1/2*d*x + 1/2*c)^4 + 30*b^2*tan(1/2*d*x + 1/2*c)^4 - 35*a^2*tan(1/2*d*x + 1/2*c)^2 - 5
0*a*b*tan(1/2*d*x + 1/2*c)^2 - 15*b^2*tan(1/2*d*x + 1/2*c)^2 + 3*a^2 + 6*a*b + 3*b^2)/tan(1/2*d*x + 1/2*c)^5)/
d

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maple [A]  time = 0.81, size = 154, normalized size = 1.26 \[ \frac {a^{2} \left (-\frac {\left (\cot ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\cot ^{3}\left (d x +c \right )\right )}{3}-\cot \left (d x +c \right )-d x -c \right )+2 a b \left (-\frac {\cos ^{6}\left (d x +c \right )}{5 \sin \left (d x +c \right )^{5}}+\frac {\cos ^{6}\left (d x +c \right )}{15 \sin \left (d x +c \right )^{3}}-\frac {\cos ^{6}\left (d x +c \right )}{5 \sin \left (d x +c \right )}-\frac {\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}\right )-\frac {b^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{5 \sin \left (d x +c \right )^{5}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^6*(a+b*sec(d*x+c))^2,x)

[Out]

1/d*(a^2*(-1/5*cot(d*x+c)^5+1/3*cot(d*x+c)^3-cot(d*x+c)-d*x-c)+2*a*b*(-1/5/sin(d*x+c)^5*cos(d*x+c)^6+1/15/sin(
d*x+c)^3*cos(d*x+c)^6-1/5/sin(d*x+c)*cos(d*x+c)^6-1/5*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))-1/5*b^2/
sin(d*x+c)^5*cos(d*x+c)^5)

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maxima [A]  time = 0.66, size = 96, normalized size = 0.79 \[ -\frac {{\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} - 5 \, \tan \left (d x + c\right )^{2} + 3}{\tan \left (d x + c\right )^{5}}\right )} a^{2} + \frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{4} - 10 \, \sin \left (d x + c\right )^{2} + 3\right )} a b}{\sin \left (d x + c\right )^{5}} + \frac {3 \, b^{2}}{\tan \left (d x + c\right )^{5}}}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.50, size = 191, normalized size = 1.57 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\left (a-b\right )}^2}{160\,d}-a^2\,x-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {a^2}{16}-\frac {a\,b}{12}+\frac {b^2}{48}+\frac {{\left (a-b\right )}^2}{96}\right )}{d}-\frac {\frac {2\,a\,b}{5}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (22\,a^2+20\,a\,b+2\,b^2\right )+\frac {a^2}{5}+\frac {b^2}{5}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {7\,a^2}{3}+\frac {10\,a\,b}{3}+b^2\right )}{32\,d\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {21\,a^2}{32}-\frac {9\,a\,b}{16}+\frac {b^2}{32}+\frac {{\left (a-b\right )}^2}{32}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^6*(a + b/cos(c + d*x))^2,x)

[Out]

(tan(c/2 + (d*x)/2)^5*(a - b)^2)/(160*d) - a^2*x - (tan(c/2 + (d*x)/2)^3*(a^2/16 - (a*b)/12 + b^2/48 + (a - b)
^2/96))/d - ((2*a*b)/5 + tan(c/2 + (d*x)/2)^4*(20*a*b + 22*a^2 + 2*b^2) + a^2/5 + b^2/5 - tan(c/2 + (d*x)/2)^2
*((10*a*b)/3 + (7*a^2)/3 + b^2))/(32*d*tan(c/2 + (d*x)/2)^5) + (tan(c/2 + (d*x)/2)*((21*a^2)/32 - (9*a*b)/16 +
b^2/32 + (a - b)^2/32))/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sec {\left (c + d x \right )}\right )^{2} \cot ^{6}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**6*(a+b*sec(d*x+c))**2,x)

[Out]

Integral((a + b*sec(c + d*x))**2*cot(c + d*x)**6, x)

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