3.220 \(\int \frac{\csc ^4(c+d x)}{(a+b \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=343 \[ \frac{a^3 b^2 \sin (c+d x)}{d \left (a^2-b^2\right )^3 (a \cos (c+d x)+b)}-\frac{4 a^2 b \left (a^2+b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}-\frac{2 a^2 b^3 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}-\frac{\sin (c+d x)}{12 d (a+b)^2 (1-\cos (c+d x))}-\frac{(a-b) \sin (c+d x)}{4 d (a+b)^3 (1-\cos (c+d x))}+\frac{(a+b) \sin (c+d x)}{4 d (a-b)^3 (\cos (c+d x)+1)}+\frac{\sin (c+d x)}{12 d (a-b)^2 (\cos (c+d x)+1)}-\frac{\sin (c+d x)}{12 d (a+b)^2 (1-\cos (c+d x))^2}+\frac{\sin (c+d x)}{12 d (a-b)^2 (\cos (c+d x)+1)^2} \]

[Out]

(-2*a^2*b^3*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(7/2)*(a + b)^(7/2)*d) - (4*a^2*b*(a
^2 + b^2)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(7/2)*(a + b)^(7/2)*d) - Sin[c + d*x]/
(12*(a + b)^2*d*(1 - Cos[c + d*x])^2) - ((a - b)*Sin[c + d*x])/(4*(a + b)^3*d*(1 - Cos[c + d*x])) - Sin[c + d*
x]/(12*(a + b)^2*d*(1 - Cos[c + d*x])) + Sin[c + d*x]/(12*(a - b)^2*d*(1 + Cos[c + d*x])^2) + Sin[c + d*x]/(12
*(a - b)^2*d*(1 + Cos[c + d*x])) + ((a + b)*Sin[c + d*x])/(4*(a - b)^3*d*(1 + Cos[c + d*x])) + (a^3*b^2*Sin[c
+ d*x])/((a^2 - b^2)^3*d*(b + a*Cos[c + d*x]))

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Rubi [A]  time = 0.547166, antiderivative size = 343, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {3872, 2897, 2650, 2648, 2664, 12, 2659, 208} \[ \frac{a^3 b^2 \sin (c+d x)}{d \left (a^2-b^2\right )^3 (a \cos (c+d x)+b)}-\frac{4 a^2 b \left (a^2+b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}-\frac{2 a^2 b^3 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}-\frac{\sin (c+d x)}{12 d (a+b)^2 (1-\cos (c+d x))}-\frac{(a-b) \sin (c+d x)}{4 d (a+b)^3 (1-\cos (c+d x))}+\frac{(a+b) \sin (c+d x)}{4 d (a-b)^3 (\cos (c+d x)+1)}+\frac{\sin (c+d x)}{12 d (a-b)^2 (\cos (c+d x)+1)}-\frac{\sin (c+d x)}{12 d (a+b)^2 (1-\cos (c+d x))^2}+\frac{\sin (c+d x)}{12 d (a-b)^2 (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^4/(a + b*Sec[c + d*x])^2,x]

[Out]

(-2*a^2*b^3*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(7/2)*(a + b)^(7/2)*d) - (4*a^2*b*(a
^2 + b^2)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(7/2)*(a + b)^(7/2)*d) - Sin[c + d*x]/
(12*(a + b)^2*d*(1 - Cos[c + d*x])^2) - ((a - b)*Sin[c + d*x])/(4*(a + b)^3*d*(1 - Cos[c + d*x])) - Sin[c + d*
x]/(12*(a + b)^2*d*(1 - Cos[c + d*x])) + Sin[c + d*x]/(12*(a - b)^2*d*(1 + Cos[c + d*x])^2) + Sin[c + d*x]/(12
*(a - b)^2*d*(1 + Cos[c + d*x])) + ((a + b)*Sin[c + d*x])/(4*(a - b)^3*d*(1 + Cos[c + d*x])) + (a^3*b^2*Sin[c
+ d*x])/((a^2 - b^2)^3*d*(b + a*Cos[c + d*x]))

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 2897

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Int[ExpandTrig[(d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m*(1 - sin[e + f*x]^2)^(p/2), x], x]
/; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[m, 2*n, p/2] && (LtQ[m, -1] || (EqQ[m, -1] && G
tQ[p, 0]))

Rule 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2664

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n +
1))/(d*(n + 1)*(a^2 - b^2)), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1
) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integer
Q[2*n]

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 ^4(c+d x)}{(a+b \sec (c+d x))^2} \, dx &=\int \frac{\cot ^2(c+d x) \csc ^2(c+d x)}{(-b-a \cos (c+d x))^2} \, dx\\ &=\int \left (\frac{1}{4 (a-b)^2 (-1-\cos (c+d x))^2}+\frac{-a-b}{4 (a-b)^3 (-1-\cos (c+d x))}+\frac{1}{4 (a+b)^2 (1-\cos (c+d x))^2}+\frac{a-b}{4 (a+b)^3 (1-\cos (c+d x))}+\frac{a^2 b^2}{\left (a^2-b^2\right )^2 (-b-a \cos (c+d x))^2}+\frac{2 a^2 b \left (a^2+b^2\right )}{\left (a^2-b^2\right )^3 (-b-a \cos (c+d x))}\right ) \, dx\\ &=\frac{\int \frac{1}{(-1-\cos (c+d x))^2} \, dx}{4 (a-b)^2}+\frac{(a-b) \int \frac{1}{1-\cos (c+d x)} \, dx}{4 (a+b)^3}+\frac{\int \frac{1}{(1-\cos (c+d x))^2} \, dx}{4 (a+b)^2}-\frac{(a+b) \int \frac{1}{-1-\cos (c+d x)} \, dx}{4 (a-b)^3}+\frac{\left (a^2 b^2\right ) \int \frac{1}{(-b-a \cos (c+d x))^2} \, dx}{\left (a^2-b^2\right )^2}+\frac{\left (2 a^2 b \left (a^2+b^2\right )\right ) \int \frac{1}{-b-a \cos (c+d x)} \, dx}{\left (a^2-b^2\right )^3}\\ &=-\frac{\sin (c+d x)}{12 (a+b)^2 d (1-\cos (c+d x))^2}-\frac{(a-b) \sin (c+d x)}{4 (a+b)^3 d (1-\cos (c+d x))}+\frac{\sin (c+d x)}{12 (a-b)^2 d (1+\cos (c+d x))^2}+\frac{(a+b) \sin (c+d x)}{4 (a-b)^3 d (1+\cos (c+d x))}+\frac{a^3 b^2 \sin (c+d x)}{\left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}-\frac{\int \frac{1}{-1-\cos (c+d x)} \, dx}{12 (a-b)^2}+\frac{\int \frac{1}{1-\cos (c+d x)} \, dx}{12 (a+b)^2}+\frac{\left (a^2 b^2\right ) \int \frac{b}{-b-a \cos (c+d x)} \, dx}{\left (a^2-b^2\right )^3}+\frac{\left (4 a^2 b \left (a^2+b^2\right )\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{4 a^2 b \left (a^2+b^2\right ) \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{\sin (c+d x)}{12 (a+b)^2 d (1-\cos (c+d x))^2}-\frac{(a-b) \sin (c+d x)}{4 (a+b)^3 d (1-\cos (c+d x))}-\frac{\sin (c+d x)}{12 (a+b)^2 d (1-\cos (c+d x))}+\frac{\sin (c+d x)}{12 (a-b)^2 d (1+\cos (c+d x))^2}+\frac{\sin (c+d x)}{12 (a-b)^2 d (1+\cos (c+d x))}+\frac{(a+b) \sin (c+d x)}{4 (a-b)^3 d (1+\cos (c+d x))}+\frac{a^3 b^2 \sin (c+d x)}{\left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}+\frac{\left (a^2 b^3\right ) \int \frac{1}{-b-a \cos (c+d x)} \, dx}{\left (a^2-b^2\right )^3}\\ &=-\frac{4 a^2 b \left (a^2+b^2\right ) \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{\sin (c+d x)}{12 (a+b)^2 d (1-\cos (c+d x))^2}-\frac{(a-b) \sin (c+d x)}{4 (a+b)^3 d (1-\cos (c+d x))}-\frac{\sin (c+d x)}{12 (a+b)^2 d (1-\cos (c+d x))}+\frac{\sin (c+d x)}{12 (a-b)^2 d (1+\cos (c+d x))^2}+\frac{\sin (c+d x)}{12 (a-b)^2 d (1+\cos (c+d x))}+\frac{(a+b) \sin (c+d x)}{4 (a-b)^3 d (1+\cos (c+d x))}+\frac{a^3 b^2 \sin (c+d x)}{\left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}+\frac{\left (2 a^2 b^3\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^2 b^3 \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{4 a^2 b \left (a^2+b^2\right ) \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{\sin (c+d x)}{12 (a+b)^2 d (1-\cos (c+d x))^2}-\frac{(a-b) \sin (c+d x)}{4 (a+b)^3 d (1-\cos (c+d x))}-\frac{\sin (c+d x)}{12 (a+b)^2 d (1-\cos (c+d x))}+\frac{\sin (c+d x)}{12 (a-b)^2 d (1+\cos (c+d x))^2}+\frac{\sin (c+d x)}{12 (a-b)^2 d (1+\cos (c+d x))}+\frac{(a+b) \sin (c+d x)}{4 (a-b)^3 d (1+\cos (c+d x))}+\frac{a^3 b^2 \sin (c+d x)}{\left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.08097, size = 281, normalized size = 0.82 \[ \frac{\sec ^2(c+d x) (a \cos (c+d x)+b) \left (\frac{24 a^3 b^2 \sin (c+d x)}{(a-b)^3 (a+b)^3}+\frac{48 a^2 b \left (2 a^2+3 b^2\right ) (a \cos (c+d x)+b) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2}}+\frac{4 (2 a+b) \tan \left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{(a-b)^3}-\frac{4 (2 a-b) \cot \left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{(a+b)^3}-\frac{\cot \left (\frac{1}{2} (c+d x)\right ) \csc ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{(a+b)^2}+\frac{\tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{(a-b)^2}\right )}{24 d (a+b \sec (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^4/(a + b*Sec[c + d*x])^2,x]

[Out]

((b + a*Cos[c + d*x])*Sec[c + d*x]^2*((48*a^2*b*(2*a^2 + 3*b^2)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 -
b^2]]*(b + a*Cos[c + d*x]))/(a^2 - b^2)^(7/2) - (4*(2*a - b)*(b + a*Cos[c + d*x])*Cot[(c + d*x)/2])/(a + b)^3
- ((b + a*Cos[c + d*x])*Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2)/(a + b)^2 + (24*a^3*b^2*Sin[c + d*x])/((a - b)^3
*(a + b)^3) + (4*(2*a + b)*(b + a*Cos[c + d*x])*Tan[(c + d*x)/2])/(a - b)^3 + ((b + a*Cos[c + d*x])*Sec[(c + d
*x)/2]^2*Tan[(c + d*x)/2])/(a - b)^2))/(24*d*(a + b*Sec[c + d*x])^2)

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Maple [A]  time = 0.092, size = 242, normalized size = 0.7 \begin{align*}{\frac{1}{d} \left ({\frac{1}{ \left ( 8\,{a}^{2}-16\,ab+8\,{b}^{2} \right ) \left ( a-b \right ) } \left ({\frac{a}{3} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{b}{3} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+3\,a\tan \left ( 1/2\,dx+c/2 \right ) +b\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+2\,{\frac{{a}^{2}b}{ \left ( a-b \right ) ^{3} \left ( a+b \right ) ^{3}} \left ( -{\frac{\tan \left ( 1/2\,dx+c/2 \right ) ab}{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b-a-b}}-{\frac{2\,{a}^{2}+3\,{b}^{2}}{\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 ) } \right ) }-{\frac{1}{24\, \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}}-{\frac{3\,a-b}{8\, \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)^4/(a+b*sec(d*x+c))^2,x)

[Out]

1/d*(1/8/(a^2-2*a*b+b^2)/(a-b)*(1/3*tan(1/2*d*x+1/2*c)^3*a-1/3*b*tan(1/2*d*x+1/2*c)^3+3*a*tan(1/2*d*x+1/2*c)+b
*tan(1/2*d*x+1/2*c))+2*a^2*b/(a+b)^3/(a-b)^3*(-tan(1/2*d*x+1/2*c)*a*b/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*
c)^2*b-a-b)-(2*a^2+3*b^2)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2)))-1/24/(a+b
)^2/tan(1/2*d*x+1/2*c)^3-1/8*(3*a-b)/(a+b)^3/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)^4/(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.25787, size = 2268, normalized size = 6.61 \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)^4/(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

[-1/6*(22*a^5*b^2 - 14*a^3*b^4 - 8*a*b^6 + 2*(2*a^7 + 10*a^5*b^2 - 11*a^3*b^4 - a*b^6)*cos(d*x + c)^4 - 2*(4*a
^6*b - 7*a^4*b^3 + 2*a^2*b^5 + b^7)*cos(d*x + c)^3 - 3*(2*a^4*b^2 + 3*a^2*b^4 - (2*a^5*b + 3*a^3*b^3)*cos(d*x
+ c)^3 - (2*a^4*b^2 + 3*a^2*b^4)*cos(d*x + c)^2 + (2*a^5*b + 3*a^3*b^3)*cos(d*x + c))*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) - 6*(a^7 + 6*a^5*b^2 - 5*a^3*b^4 - 2*a*b^6
)*cos(d*x + c)^2 + 10*(a^6*b - 2*a^4*b^3 + a^2*b^5)*cos(d*x + c))/(((a^9 - 4*a^7*b^2 + 6*a^5*b^4 - 4*a^3*b^6 +
a*b^8)*d*cos(d*x + c)^3 + (a^8*b - 4*a^6*b^3 + 6*a^4*b^5 - 4*a^2*b^7 + b^9)*d*cos(d*x + c)^2 - (a^9 - 4*a^7*b
^2 + 6*a^5*b^4 - 4*a^3*b^6 + a*b^8)*d*cos(d*x + c) - (a^8*b - 4*a^6*b^3 + 6*a^4*b^5 - 4*a^2*b^7 + b^9)*d)*sin(
d*x + c)), -1/3*(11*a^5*b^2 - 7*a^3*b^4 - 4*a*b^6 + (2*a^7 + 10*a^5*b^2 - 11*a^3*b^4 - a*b^6)*cos(d*x + c)^4 -
(4*a^6*b - 7*a^4*b^3 + 2*a^2*b^5 + b^7)*cos(d*x + c)^3 - 3*(2*a^4*b^2 + 3*a^2*b^4 - (2*a^5*b + 3*a^3*b^3)*cos
(d*x + c)^3 - (2*a^4*b^2 + 3*a^2*b^4)*cos(d*x + c)^2 + (2*a^5*b + 3*a^3*b^3)*cos(d*x + c))*sqrt(-a^2 + b^2)*ar
ctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c)))*sin(d*x + c) - 3*(a^7 + 6*a^5*b^2 - 5*
a^3*b^4 - 2*a*b^6)*cos(d*x + c)^2 + 5*(a^6*b - 2*a^4*b^3 + a^2*b^5)*cos(d*x + c))/(((a^9 - 4*a^7*b^2 + 6*a^5*b
^4 - 4*a^3*b^6 + a*b^8)*d*cos(d*x + c)^3 + (a^8*b - 4*a^6*b^3 + 6*a^4*b^5 - 4*a^2*b^7 + b^9)*d*cos(d*x + c)^2
- (a^9 - 4*a^7*b^2 + 6*a^5*b^4 - 4*a^3*b^6 + a*b^8)*d*cos(d*x + c) - (a^8*b - 4*a^6*b^3 + 6*a^4*b^5 - 4*a^2*b^
7 + b^9)*d)*sin(d*x + c))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc ^{4}{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**4/(a+b*sec(d*x+c))**2,x)

[Out]

Integral(csc(c + d*x)**4/(a + b*sec(c + d*x))**2, x)

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Giac [A]  time = 1.37596, size = 617, normalized size = 1.8 \begin{align*} -\frac{\frac{48 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a - b\right )}} - \frac{48 \,{\left (2 \, a^{4} b + 3 \, a^{2} b^{3}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt{-a^{2} + b^{2}}} - \frac{a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 24 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 18 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{6} - 6 \, a^{5} b + 15 \, a^{4} b^{2} - 20 \, a^{3} b^{3} + 15 \, a^{2} b^{4} - 6 \, a b^{5} + b^{6}} + \frac{9 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a + b}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4/(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

-1/24*(48*a^3*b^2*tan(1/2*d*x + 1/2*c)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(
1/2*d*x + 1/2*c)^2 - a - b)) - 48*(2*a^4*b + 3*a^2*b^3)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(2*a - 2*b) + arc
tan((a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*
sqrt(-a^2 + b^2)) - (a^4*tan(1/2*d*x + 1/2*c)^3 - 4*a^3*b*tan(1/2*d*x + 1/2*c)^3 + 6*a^2*b^2*tan(1/2*d*x + 1/2
*c)^3 - 4*a*b^3*tan(1/2*d*x + 1/2*c)^3 + b^4*tan(1/2*d*x + 1/2*c)^3 + 9*a^4*tan(1/2*d*x + 1/2*c) - 24*a^3*b*ta
n(1/2*d*x + 1/2*c) + 18*a^2*b^2*tan(1/2*d*x + 1/2*c) - 3*b^4*tan(1/2*d*x + 1/2*c))/(a^6 - 6*a^5*b + 15*a^4*b^2
- 20*a^3*b^3 + 15*a^2*b^4 - 6*a*b^5 + b^6) + (9*a*tan(1/2*d*x + 1/2*c)^2 - 3*b*tan(1/2*d*x + 1/2*c)^2 + a + b
)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*tan(1/2*d*x + 1/2*c)^3))/d