3.208 \(\int \frac{\csc ^6(c+d x)}{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=201 \[ \frac{\csc ^5(c+d x) (b-a \cos (c+d x))}{5 d \left (a^2-b^2\right )}+\frac{\csc ^3(c+d x) \left (5 a^2 b-a \left (4 a^2+b^2\right ) \cos (c+d x)\right )}{15 d \left (a^2-b^2\right )^2}+\frac{\csc (c+d x) \left (15 a^4 b-a \left (9 a^2 b^2+8 a^4-2 b^4\right ) \cos (c+d x)\right )}{15 d \left (a^2-b^2\right )^3}-\frac{2 a^5 b \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}} \]

[Out]

(-2*a^5*b*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(7/2)*(a + b)^(7/2)*d) + ((15*a^4*b -
a*(8*a^4 + 9*a^2*b^2 - 2*b^4)*Cos[c + d*x])*Csc[c + d*x])/(15*(a^2 - b^2)^3*d) + ((5*a^2*b - a*(4*a^2 + b^2)*C
os[c + d*x])*Csc[c + d*x]^3)/(15*(a^2 - b^2)^2*d) + ((b - a*Cos[c + d*x])*Csc[c + d*x]^5)/(5*(a^2 - b^2)*d)

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Rubi [A]  time = 0.519973, antiderivative size = 201, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3872, 2866, 12, 2659, 208} \[ \frac{\csc ^5(c+d x) (b-a \cos (c+d x))}{5 d \left (a^2-b^2\right )}+\frac{\csc ^3(c+d x) \left (5 a^2 b-a \left (4 a^2+b^2\right ) \cos (c+d x)\right )}{15 d \left (a^2-b^2\right )^2}+\frac{\csc (c+d x) \left (15 a^4 b-a \left (9 a^2 b^2+8 a^4-2 b^4\right ) \cos (c+d x)\right )}{15 d \left (a^2-b^2\right )^3}-\frac{2 a^5 b \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^5*b*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(7/2)*(a + b)^(7/2)*d) + ((15*a^4*b -
a*(8*a^4 + 9*a^2*b^2 - 2*b^4)*Cos[c + d*x])*Csc[c + d*x])/(15*(a^2 - b^2)^3*d) + ((5*a^2*b - a*(4*a^2 + b^2)*C
os[c + d*x])*Csc[c + d*x]^3)/(15*(a^2 - b^2)^2*d) + ((b - a*Cos[c + d*x])*Csc[c + d*x]^5)/(5*(a^2 - b^2)*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 2866

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[((g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c - a*d - (a*c -
b*d)*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p
+ 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegerQ[2*m]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{\csc ^6(c+d x)}{a+b \sec (c+d x)} \, dx &=-\int \frac{\cot (c+d x) \csc ^5(c+d x)}{-b-a \cos (c+d x)} \, dx\\ &=\frac{(b-a \cos (c+d x)) \csc ^5(c+d x)}{5 \left (a^2-b^2\right ) d}+\frac{\int \frac{\left (a b-4 a^2 \cos (c+d x)\right ) \csc ^4(c+d x)}{-b-a \cos (c+d x)} \, dx}{5 \left (a^2-b^2\right )}\\ &=\frac{\left (5 a^2 b-a \left (4 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^3(c+d x)}{15 \left (a^2-b^2\right )^2 d}+\frac{(b-a \cos (c+d x)) \csc ^5(c+d x)}{5 \left (a^2-b^2\right ) d}+\frac{\int \frac{\left (a b \left (7 a^2-2 b^2\right )-2 a^2 \left (4 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{-b-a \cos (c+d x)} \, dx}{15 \left (a^2-b^2\right )^2}\\ &=\frac{\left (15 a^4 b-a \left (8 a^4+9 a^2 b^2-2 b^4\right ) \cos (c+d x)\right ) \csc (c+d x)}{15 \left (a^2-b^2\right )^3 d}+\frac{\left (5 a^2 b-a \left (4 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^3(c+d x)}{15 \left (a^2-b^2\right )^2 d}+\frac{(b-a \cos (c+d x)) \csc ^5(c+d x)}{5 \left (a^2-b^2\right ) d}+\frac{\int \frac{15 a^5 b}{-b-a \cos (c+d x)} \, dx}{15 \left (a^2-b^2\right )^3}\\ &=\frac{\left (15 a^4 b-a \left (8 a^4+9 a^2 b^2-2 b^4\right ) \cos (c+d x)\right ) \csc (c+d x)}{15 \left (a^2-b^2\right )^3 d}+\frac{\left (5 a^2 b-a \left (4 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^3(c+d x)}{15 \left (a^2-b^2\right )^2 d}+\frac{(b-a \cos (c+d x)) \csc ^5(c+d x)}{5 \left (a^2-b^2\right ) d}+\frac{\left (a^5 b\right ) \int \frac{1}{-b-a \cos (c+d x)} \, dx}{\left (a^2-b^2\right )^3}\\ &=\frac{\left (15 a^4 b-a \left (8 a^4+9 a^2 b^2-2 b^4\right ) \cos (c+d x)\right ) \csc (c+d x)}{15 \left (a^2-b^2\right )^3 d}+\frac{\left (5 a^2 b-a \left (4 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^3(c+d x)}{15 \left (a^2-b^2\right )^2 d}+\frac{(b-a \cos (c+d x)) \csc ^5(c+d x)}{5 \left (a^2-b^2\right ) d}+\frac{\left (2 a^5 b\right ) \operatorname{Subst}\left (\int \frac{1}{-a-b+(a-b) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^3 d}\\ &=-\frac{2 a^5 b \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}+\frac{\left (15 a^4 b-a \left (8 a^4+9 a^2 b^2-2 b^4\right ) \cos (c+d x)\right ) \csc (c+d x)}{15 \left (a^2-b^2\right )^3 d}+\frac{\left (5 a^2 b-a \left (4 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^3(c+d x)}{15 \left (a^2-b^2\right )^2 d}+\frac{(b-a \cos (c+d x)) \csc ^5(c+d x)}{5 \left (a^2-b^2\right ) d}\\ \end{align*}

Mathematica [A]  time = 1.24476, size = 277, normalized size = 1.38 \[ \frac{\sec (c+d x) (a \cos (c+d x)+b) \left (\frac{2 \left (64 a^2-43 a b+9 b^2\right ) \tan \left (\frac{1}{2} (c+d x)\right )}{(a-b)^3}-\frac{2 \left (64 a^2+43 a b+9 b^2\right ) \cot \left (\frac{1}{2} (c+d x)\right )}{(a+b)^3}+\frac{960 a^5 b \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2}}+\frac{96 \sin ^6\left (\frac{1}{2} (c+d x)\right ) \csc ^5(c+d x)}{a-b}+\frac{8 (19 a-9 b) \sin ^4\left (\frac{1}{2} (c+d x)\right ) \csc ^3(c+d x)}{(a-b)^2}-\frac{3 \sin (c+d x) \csc ^6\left (\frac{1}{2} (c+d x)\right )}{2 (a+b)}-\frac{(19 a+9 b) \sin (c+d x) \csc ^4\left (\frac{1}{2} (c+d x)\right )}{2 (a+b)^2}\right )}{480 d (a+b \sec (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b + a*Cos[c + d*x])*Sec[c + d*x]*((960*a^5*b*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^
2)^(7/2) - (2*(64*a^2 + 43*a*b + 9*b^2)*Cot[(c + d*x)/2])/(a + b)^3 + (8*(19*a - 9*b)*Csc[c + d*x]^3*Sin[(c +
d*x)/2]^4)/(a - b)^2 + (96*Csc[c + d*x]^5*Sin[(c + d*x)/2]^6)/(a - b) - ((19*a + 9*b)*Csc[(c + d*x)/2]^4*Sin[c
+ d*x])/(2*(a + b)^2) - (3*Csc[(c + d*x)/2]^6*Sin[c + d*x])/(2*(a + b)) + (2*(64*a^2 - 43*a*b + 9*b^2)*Tan[(c
+ d*x)/2])/(a - b)^3))/(480*d*(a + b*Sec[c + d*x]))

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Maple [A]  time = 0.074, size = 282, normalized size = 1.4 \begin{align*}{\frac{1}{d} \left ({\frac{1}{32\, \left ( a-b \right ) ^{3}} \left ({\frac{{a}^{2}}{5} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{2\,ab}{5} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{{b}^{2}}{5} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{5\,{a}^{2}}{3} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{8\,ab}{3} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}{b}^{2}+10\,{a}^{2}\tan \left ( 1/2\,dx+c/2 \right ) -8\,\tan \left ( 1/2\,dx+c/2 \right ) ab+2\,{b}^{2}\tan \left ( 1/2\,dx+c/2 \right ) \right ) }-2\,{\frac{b{a}^{5}}{ \left ( a-b \right ) ^{3} \left ( a+b \right ) ^{3}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-{\frac{1}{160\,a+160\,b} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-5}}-{\frac{5\,a+3\,b}{96\, \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}}-{\frac{10\,{a}^{2}+8\,ab+2\,{b}^{2}}{32\, \left ( a+b \right ) ^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}} \right ) } \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]

1/d*(1/32/(a-b)^3*(1/5*tan(1/2*d*x+1/2*c)^5*a^2-2/5*tan(1/2*d*x+1/2*c)^5*a*b+1/5*b^2*tan(1/2*d*x+1/2*c)^5+5/3*
tan(1/2*d*x+1/2*c)^3*a^2-8/3*tan(1/2*d*x+1/2*c)^3*a*b+tan(1/2*d*x+1/2*c)^3*b^2+10*a^2*tan(1/2*d*x+1/2*c)-8*tan
(1/2*d*x+1/2*c)*a*b+2*b^2*tan(1/2*d*x+1/2*c))-2/(a-b)^3/(a+b)^3*b*a^5/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/
2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))-1/160/(a+b)/tan(1/2*d*x+1/2*c)^5-1/96*(5*a+3*b)/(a+b)^2/tan(1/2*d*x+1/2*c)^3
-1/32/(a+b)^3*(10*a^2+8*a*b+2*b^2)/tan(1/2*d*x+1/2*c))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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]

Exception raised: ValueError

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Fricas [B]  time = 2.09674, size = 1917, normalized size = 9.54 \begin{align*} \text{result too large to display} \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*(46*a^6*b - 68*a^4*b^3 + 28*a^2*b^5 - 6*b^7 - 2*(8*a^7 + a^5*b^2 - 11*a^3*b^4 + 2*a*b^6)*cos(d*x + c)^5
+ 30*(a^6*b - a^4*b^3)*cos(d*x + c)^4 + 10*(4*a^7 - a^5*b^2 - 4*a^3*b^4 + a*b^6)*cos(d*x + c)^3 - 15*(a^5*b*co
s(d*x + c)^4 - 2*a^5*b*cos(d*x + c)^2 + a^5*b)*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x
+ c)^2 + 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d
*x + c) + b^2))*sin(d*x + c) - 10*(7*a^6*b - 8*a^4*b^3 + a^2*b^5)*cos(d*x + c)^2 - 30*(a^7 - a^5*b^2)*cos(d*x
+ c))/(((a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*d*cos(d*x + c)^4 - 2*(a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*
a^2*b^6 + b^8)*d*cos(d*x + c)^2 + (a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*d)*sin(d*x + c)), 1/15*(23*a
^6*b - 34*a^4*b^3 + 14*a^2*b^5 - 3*b^7 - (8*a^7 + a^5*b^2 - 11*a^3*b^4 + 2*a*b^6)*cos(d*x + c)^5 + 15*(a^6*b -
a^4*b^3)*cos(d*x + c)^4 + 5*(4*a^7 - a^5*b^2 - 4*a^3*b^4 + a*b^6)*cos(d*x + c)^3 - 15*(a^5*b*cos(d*x + c)^4 -
2*a^5*b*cos(d*x + c)^2 + a^5*b)*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*s
in(d*x + c)))*sin(d*x + c) - 5*(7*a^6*b - 8*a^4*b^3 + a^2*b^5)*cos(d*x + c)^2 - 15*(a^7 - a^5*b^2)*cos(d*x + c
))/(((a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*d*cos(d*x + c)^4 - 2*(a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2
*b^6 + b^8)*d*cos(d*x + c)^2 + (a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*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 = 1.32361, size = 730, normalized size = 3.63 \begin{align*} \text{result too large to display} \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*(960*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*
x + 1/2*c))/sqrt(-a^2 + b^2)))*a^5*b/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sqrt(-a^2 + b^2)) - (3*a^4*tan(1/2*d
*x + 1/2*c)^5 - 12*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 18*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 - 12*a*b^3*tan(1/2*d*x + 1
/2*c)^5 + 3*b^4*tan(1/2*d*x + 1/2*c)^5 + 25*a^4*tan(1/2*d*x + 1/2*c)^3 - 90*a^3*b*tan(1/2*d*x + 1/2*c)^3 + 120
*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 70*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 15*b^4*tan(1/2*d*x + 1/2*c)^3 + 150*a^4*ta
n(1/2*d*x + 1/2*c) - 420*a^3*b*tan(1/2*d*x + 1/2*c) + 420*a^2*b^2*tan(1/2*d*x + 1/2*c) - 180*a*b^3*tan(1/2*d*x
+ 1/2*c) + 30*b^4*tan(1/2*d*x + 1/2*c))/(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5) + (150*a^2*
tan(1/2*d*x + 1/2*c)^4 + 120*a*b*tan(1/2*d*x + 1/2*c)^4 + 30*b^2*tan(1/2*d*x + 1/2*c)^4 + 25*a^2*tan(1/2*d*x +
1/2*c)^2 + 40*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)/((a^3 + 3*a
^2*b + 3*a*b^2 + b^3)*tan(1/2*d*x + 1/2*c)^5))/d