3.231 \(\int \frac{\csc ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=376 \[ \frac{3 b^4 \sin (c+d x)}{2 d \left (a^2-b^2\right )^3 (a \cos (c+d x)+b)}-\frac{b^3 \sin (c+d x)}{2 d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)^2}+\frac{b^2 \left (3 a^2-b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right )^3 (a \cos (c+d x)+b)}-\frac{2 b^3 \left (3 a^2-b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a d (a-b)^{7/2} (a+b)^{7/2}}-\frac{b^3 \left (a^2+2 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a d (a-b)^{7/2} (a+b)^{7/2}}-\frac{2 a b \left (3 a^2+b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}-\frac{\sin (c+d x)}{2 d (a+b)^3 (1-\cos (c+d x))}+\frac{\sin (c+d x)}{2 d (a-b)^3 (\cos (c+d x)+1)} \]

[Out]

(-2*b^3*(3*a^2 - b^2)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*(a - b)^(7/2)*(a + b)^(7/2)*d) -
(2*a*b*(3*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) - (
b^3*(a^2 + 2*b^2)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*(a - b)^(7/2)*(a + b)^(7/2)*d) - Sin
[c + d*x]/(2*(a + b)^3*d*(1 - Cos[c + d*x])) + Sin[c + d*x]/(2*(a - b)^3*d*(1 + Cos[c + d*x])) - (b^3*Sin[c +
d*x])/(2*(a^2 - b^2)^2*d*(b + a*Cos[c + d*x])^2) + (3*b^4*Sin[c + d*x])/(2*(a^2 - b^2)^3*d*(b + a*Cos[c + d*x]
)) + (b^2*(3*a^2 - 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 = 0.658311, antiderivative size = 376, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {3872, 2897, 2648, 2664, 12, 2659, 208, 2754} \[ \frac{3 b^4 \sin (c+d x)}{2 d \left (a^2-b^2\right )^3 (a \cos (c+d x)+b)}-\frac{b^3 \sin (c+d x)}{2 d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)^2}+\frac{b^2 \left (3 a^2-b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right )^3 (a \cos (c+d x)+b)}-\frac{2 b^3 \left (3 a^2-b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a d (a-b)^{7/2} (a+b)^{7/2}}-\frac{b^3 \left (a^2+2 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a d (a-b)^{7/2} (a+b)^{7/2}}-\frac{2 a b \left (3 a^2+b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}-\frac{\sin (c+d x)}{2 d (a+b)^3 (1-\cos (c+d x))}+\frac{\sin (c+d x)}{2 d (a-b)^3 (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^2/(a + b*Sec[c + d*x])^3,x]

[Out]

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

Rule 2754

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

Rubi steps

\begin{align*} \int \frac{\csc ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx &=-\int \frac{\cos (c+d x) \cot ^2(c+d x)}{(-b-a \cos (c+d x))^3} \, dx\\ &=\int \left (-\frac{1}{2 (a-b)^3 (-1-\cos (c+d x))}+\frac{1}{2 (a+b)^3 (1-\cos (c+d x))}+\frac{3 a^2 b^2-b^4}{a \left (a^2-b^2\right )^2 (-b-a \cos (c+d x))^2}+\frac{a \left (3 a^2 b+b^3\right )}{\left (a^2-b^2\right )^3 (-b-a \cos (c+d x))}+\frac{b^3}{a \left (-a^2+b^2\right ) (b+a \cos (c+d x))^3}\right ) \, dx\\ &=-\frac{\int \frac{1}{-1-\cos (c+d x)} \, dx}{2 (a-b)^3}+\frac{\int \frac{1}{1-\cos (c+d x)} \, dx}{2 (a+b)^3}-\frac{b^3 \int \frac{1}{(b+a \cos (c+d x))^3} \, dx}{a \left (a^2-b^2\right )}+\frac{\left (b^2 \left (3 a^2-b^2\right )\right ) \int \frac{1}{(-b-a \cos (c+d x))^2} \, dx}{a \left (a^2-b^2\right )^2}+\frac{\left (a b \left (3 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)}{2 (a+b)^3 d (1-\cos (c+d x))}+\frac{\sin (c+d x)}{2 (a-b)^3 d (1+\cos (c+d x))}-\frac{b^3 \sin (c+d x)}{2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))^2}+\frac{b^2 \left (3 a^2-b^2\right ) \sin (c+d x)}{\left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}-\frac{b^3 \int \frac{-2 b+a \cos (c+d x)}{(b+a \cos (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )^2}+\frac{\left (b^2 \left (3 a^2-b^2\right )\right ) \int \frac{b}{-b-a \cos (c+d x)} \, dx}{a \left (a^2-b^2\right )^3}+\frac{\left (2 a b \left (3 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{2 a b \left (3 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)}{2 (a+b)^3 d (1-\cos (c+d x))}+\frac{\sin (c+d x)}{2 (a-b)^3 d (1+\cos (c+d x))}-\frac{b^3 \sin (c+d x)}{2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))^2}+\frac{3 b^4 \sin (c+d x)}{2 \left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}+\frac{b^2 \left (3 a^2-b^2\right ) \sin (c+d x)}{\left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}-\frac{b^3 \int \frac{a^2+2 b^2}{b+a \cos (c+d x)} \, dx}{2 a \left (a^2-b^2\right )^3}+\frac{\left (b^3 \left (3 a^2-b^2\right )\right ) \int \frac{1}{-b-a \cos (c+d x)} \, dx}{a \left (a^2-b^2\right )^3}\\ &=-\frac{2 a b \left (3 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)}{2 (a+b)^3 d (1-\cos (c+d x))}+\frac{\sin (c+d x)}{2 (a-b)^3 d (1+\cos (c+d x))}-\frac{b^3 \sin (c+d x)}{2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))^2}+\frac{3 b^4 \sin (c+d x)}{2 \left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}+\frac{b^2 \left (3 a^2-b^2\right ) \sin (c+d x)}{\left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}-\frac{\left (b^3 \left (a^2+2 b^2\right )\right ) \int \frac{1}{b+a \cos (c+d x)} \, dx}{2 a \left (a^2-b^2\right )^3}+\frac{\left (2 b^3 \left (3 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 )}{a \left (a^2-b^2\right )^3 d}\\ &=-\frac{2 b^3 \left (3 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 (a-b)^{7/2} (a+b)^{7/2} d}-\frac{2 a b \left (3 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)}{2 (a+b)^3 d (1-\cos (c+d x))}+\frac{\sin (c+d x)}{2 (a-b)^3 d (1+\cos (c+d x))}-\frac{b^3 \sin (c+d x)}{2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))^2}+\frac{3 b^4 \sin (c+d x)}{2 \left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}+\frac{b^2 \left (3 a^2-b^2\right ) \sin (c+d x)}{\left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}-\frac{\left (b^3 \left (a^2+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 )}{a \left (a^2-b^2\right )^3 d}\\ &=-\frac{2 b^3 \left (3 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 (a-b)^{7/2} (a+b)^{7/2} d}-\frac{2 a b \left (3 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{b^3 \left (a^2+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 (a-b)^{7/2} (a+b)^{7/2} d}-\frac{\sin (c+d x)}{2 (a+b)^3 d (1-\cos (c+d x))}+\frac{\sin (c+d x)}{2 (a-b)^3 d (1+\cos (c+d x))}-\frac{b^3 \sin (c+d x)}{2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))^2}+\frac{3 b^4 \sin (c+d x)}{2 \left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}+\frac{b^2 \left (3 a^2-b^2\right ) \sin (c+d x)}{\left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.919154, size = 231, normalized size = 0.61 \[ \frac{\sec ^3(c+d x) (a \cos (c+d x)+b) \left (\frac{b^2 \left (6 a^2+b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)}{(a-b)^3 (a+b)^3}+\frac{6 a b \left (2 a^2+3 b^2\right ) (a \cos (c+d x)+b)^2 \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2}}-\frac{b^3 \sin (c+d x)}{(a-b)^2 (a+b)^2}+\frac{\tan \left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)^2}{(a-b)^3}-\frac{\cot \left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)^2}{(a+b)^3}\right )}{2 d (a+b \sec (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^2/(a + b*Sec[c + d*x])^3,x]

[Out]

((b + a*Cos[c + d*x])*Sec[c + d*x]^3*((6*a*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])^2)/(a^2 - b^2)^(7/2) - ((b + a*Cos[c + d*x])^2*Cot[(c + d*x)/2])/(a + b)^3 - (b^3*Sin
[c + d*x])/((a - b)^2*(a + b)^2) + (b^2*(6*a^2 + b^2)*(b + a*Cos[c + d*x])*Sin[c + d*x])/((a - b)^3*(a + b)^3)
+ ((b + a*Cos[c + d*x])^2*Tan[(c + d*x)/2])/(a - b)^3))/(2*d*(a + b*Sec[c + d*x])^3)

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Maple [A]  time = 0.091, size = 234, normalized size = 0.6 \begin{align*}{\frac{1}{d} \left ({\frac{1}{2\,{a}^{3}-6\,{a}^{2}b+6\,a{b}^{2}-2\,{b}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{b}{ \left ( a-b \right ) ^{3} \left ( a+b \right ) ^{3}} \left ({\frac{ \left ( -3\,{a}^{3}b+5/2\,{a}^{2}{b}^{2}-1/2\,a{b}^{3}+{b}^{4} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}+ \left ( 3\,{a}^{3}b+5/2\,{a}^{2}{b}^{2}+1/2\,a{b}^{3}+{b}^{4} \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{ \left ( \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 \right ) ^{2}}}-3/2\,{\frac{ \left ( 2\,{a}^{2}+3\,{b}^{2} \right ) a}{\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}{2\, \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)^2/(a+b*sec(d*x+c))^3,x)

[Out]

1/d*(1/2/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c)+2*b/(a-b)^3/(a+b)^3*(((-3*a^3*b+5/2*a^2*b^2-1/2*a*b^3+b^
4)*tan(1/2*d*x+1/2*c)^3+(3*a^3*b+5/2*a^2*b^2+1/2*a*b^3+b^4)*tan(1/2*d*x+1/2*c))/(tan(1/2*d*x+1/2*c)^2*a-tan(1/
2*d*x+1/2*c)^2*b-a-b)^2-3/2*(2*a^2+3*b^2)*a/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))
^(1/2)))-1/2/(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)^2/(a+b*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.2484, size = 1854, normalized size = 4.93 \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)^2/(a+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

[1/4*(22*a^4*b^3 - 14*a^2*b^5 - 8*b^7 - 2*(2*a^7 + 10*a^5*b^2 - 11*a^3*b^4 - a*b^6)*cos(d*x + c)^3 - 3*(2*a^3*
b^3 + 3*a*b^5 + (2*a^5*b + 3*a^3*b^3)*cos(d*x + c)^2 + 2*(2*a^4*b^2 + 3*a^2*b^4)*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) + 2*(2*a^6*b - 17*a^4*b^3 + 13*a^
2*b^5 + 2*b^7)*cos(d*x + c)^2 + 2*(16*a^5*b^2 - 17*a^3*b^4 + a*b^6)*cos(d*x + c))/(((a^10 - 4*a^8*b^2 + 6*a^6*
b^4 - 4*a^4*b^6 + a^2*b^8)*d*cos(d*x + c)^2 + 2*(a^9*b - 4*a^7*b^3 + 6*a^5*b^5 - 4*a^3*b^7 + a*b^9)*d*cos(d*x
+ c) + (a^8*b^2 - 4*a^6*b^4 + 6*a^4*b^6 - 4*a^2*b^8 + b^10)*d)*sin(d*x + c)), 1/2*(11*a^4*b^3 - 7*a^2*b^5 - 4*
b^7 - (2*a^7 + 10*a^5*b^2 - 11*a^3*b^4 - a*b^6)*cos(d*x + c)^3 - 3*(2*a^3*b^3 + 3*a*b^5 + (2*a^5*b + 3*a^3*b^3
)*cos(d*x + c)^2 + 2*(2*a^4*b^2 + 3*a^2*b^4)*cos(d*x + c))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*
x + c) + a)/((a^2 - b^2)*sin(d*x + c)))*sin(d*x + c) + (2*a^6*b - 17*a^4*b^3 + 13*a^2*b^5 + 2*b^7)*cos(d*x + c
)^2 + (16*a^5*b^2 - 17*a^3*b^4 + a*b^6)*cos(d*x + c))/(((a^10 - 4*a^8*b^2 + 6*a^6*b^4 - 4*a^4*b^6 + a^2*b^8)*d
*cos(d*x + c)^2 + 2*(a^9*b - 4*a^7*b^3 + 6*a^5*b^5 - 4*a^3*b^7 + a*b^9)*d*cos(d*x + c) + (a^8*b^2 - 4*a^6*b^4
+ 6*a^4*b^6 - 4*a^2*b^8 + b^10)*d)*sin(d*x + c))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc ^{2}{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**2/(a+b*sec(d*x+c))**3,x)

[Out]

Integral(csc(c + d*x)**2/(a + b*sec(c + d*x))**3, x)

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Giac [A]  time = 1.35506, size = 521, normalized size = 1.39 \begin{align*} \frac{\frac{6 \,{\left (2 \, a^{3} b + 3 \, a 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{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac{2 \,{\left (6 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 5 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 6 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 5 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\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 )}^{2}} - \frac{1}{{\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 )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2/(a+b*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/2*(6*(2*a^3*b + 3*a*b^3)*(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^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sqrt(-a^2 + b^2)) + tan(1/2*d
*x + 1/2*c)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - 2*(6*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 - 5*a^2*b^3*tan(1/2*d*x + 1/
2*c)^3 + a*b^4*tan(1/2*d*x + 1/2*c)^3 - 2*b^5*tan(1/2*d*x + 1/2*c)^3 - 6*a^3*b^2*tan(1/2*d*x + 1/2*c) - 5*a^2*
b^3*tan(1/2*d*x + 1/2*c) - a*b^4*tan(1/2*d*x + 1/2*c) - 2*b^5*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)^2) - 1/((a^3 + 3*a^2*b + 3*a*b^2 + b^
3)*tan(1/2*d*x + 1/2*c)))/d