3.3.31 \(\int \frac {\csc ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx\) [231]

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

[Out]

-2*b^3*(3*a^2-b^2)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a/(a-b)^(7/2)/(a+b)^(7/2)/d-2*a*b*(3*a^
2+b^2)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(7/2)/(a+b)^(7/2)/d-b^3*(a^2+2*b^2)*arctanh((
a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a/(a-b)^(7/2)/(a+b)^(7/2)/d-1/2*sin(d*x+c)/(a+b)^3/d/(1-cos(d*x+c))
+1/2*sin(d*x+c)/(a-b)^3/d/(1+cos(d*x+c))-1/2*b^3*sin(d*x+c)/(a^2-b^2)^2/d/(b+a*cos(d*x+c))^2+3/2*b^4*sin(d*x+c
)/(a^2-b^2)^3/d/(b+a*cos(d*x+c))+b^2*(3*a^2-b^2)*sin(d*x+c)/(a^2-b^2)^3/d/(b+a*cos(d*x+c))

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Rubi [A]
time = 0.49, antiderivative size = 376, normalized size of antiderivative = 1.00, 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 = {3957, 2976, 2727, 2743, 12, 2738, 214, 2833} \begin {gather*} \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 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 {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 {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 {\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)} \end {gather*}

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 12

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

Rule 214

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

Rule 2727

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 2738

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 2743

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] && Integ
erQ[2*n]

Rule 2833

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]

Rule 2976

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 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[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]

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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.71, size = 231, normalized size = 0.61 \begin {gather*} \frac {(b+a \cos (c+d x)) \sec ^3(c+d x) \left (\frac {6 a b \left (2 a^2+3 b^2\right ) \tanh ^{-1}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right ) (b+a \cos (c+d x))^2}{\left (a^2-b^2\right )^{7/2}}-\frac {(b+a \cos (c+d x))^2 \cot \left (\frac {1}{2} (c+d x)\right )}{(a+b)^3}-\frac {b^3 \sin (c+d x)}{(a-b)^2 (a+b)^2}+\frac {b^2 \left (6 a^2+b^2\right ) (b+a \cos (c+d x)) \sin (c+d x)}{(a-b)^3 (a+b)^3}+\frac {(b+a \cos (c+d x))^2 \tan \left (\frac {1}{2} (c+d x)\right )}{(a-b)^3}\right )}{2 d (a+b \sec (c+d x))^3} \end {gather*}

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 = 0.25, size = 234, normalized size = 0.62

method	result	size
derivativedivides	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{3}-6 b \,a^{2}+6 b^{2} a -2 b^{3}}-\frac {1}{2 \left (a +b \right )^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {2 b \left (\frac {\left (-3 b \,a^{3}+\frac {5}{2} b^{2} a^{2}-\frac {1}{2} b^{3} a +b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 b \,a^{3}+\frac {5}{2} b^{2} a^{2}+\frac {1}{2} b^{3} a +b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b \right )^{2}}-\frac {3 \left (2 a^{2}+3 b^{2}\right ) a \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{2 \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3}}}{d}\)	\(234\)
default	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{3}-6 b \,a^{2}+6 b^{2} a -2 b^{3}}-\frac {1}{2 \left (a +b \right )^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {2 b \left (\frac {\left (-3 b \,a^{3}+\frac {5}{2} b^{2} a^{2}-\frac {1}{2} b^{3} a +b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 b \,a^{3}+\frac {5}{2} b^{2} a^{2}+\frac {1}{2} b^{3} a +b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b \right )^{2}}-\frac {3 \left (2 a^{2}+3 b^{2}\right ) a \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{2 \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3}}}{d}\)	\(234\)
risch	\(-\frac {i \left (6 a^{5} b \,{\mathrm e}^{5 i \left (d x +c \right )}+9 a^{3} b^{3} {\mathrm e}^{5 i \left (d x +c \right )}-2 a^{6} {\mathrm e}^{4 i \left (d x +c \right )}+24 a^{4} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+21 a^{2} b^{4} {\mathrm e}^{4 i \left (d x +c \right )}+2 b^{6} {\mathrm e}^{4 i \left (d x +c \right )}+4 a^{5} b \,{\mathrm e}^{3 i \left (d x +c \right )}+14 a^{3} b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+12 a \,b^{5} {\mathrm e}^{3 i \left (d x +c \right )}-4 a^{6} {\mathrm e}^{2 i \left (d x +c \right )}+4 a^{4} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-28 a^{2} b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-2 b^{6} {\mathrm e}^{2 i \left (d x +c \right )}-2 a^{5} b \,{\mathrm e}^{i \left (d x +c \right )}-39 a^{3} b^{3} {\mathrm e}^{i \left (d x +c \right )}-4 a \,b^{5} {\mathrm e}^{i \left (d x +c \right )}-2 a^{6}-12 a^{4} b^{2}-a^{2} b^{4}\right )}{a \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left (-a^{2}+b^{2}\right )^{3} d}-\frac {3 a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}-\frac {9 a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}+\frac {3 a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}+\frac {9 a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}\)	\(670\)

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,method=_RETURNVERBOSE)

[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)-1/2/(a+b)^3/tan(1/2*d*x+1/2*c)+2*b/(a-b)^3/(a+b)^3*(((-3
*b*a^3+5/2*b^2*a^2-1/2*b^3*a+b^4)*tan(1/2*d*x+1/2*c)^3+(3*b*a^3+5/2*b^2*a^2+1/2*b^3*a+b^4)*tan(1/2*d*x+1/2*c))
/(a*tan(1/2*d*x+1/2*c)^2-b*tan(1/2*d*x+1/2*c)^2-a-b)^2-3/2*(2*a^2+3*b^2)*a/((a+b)*(a-b))^(1/2)*arctanh((a-b)*t
an(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more de

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Fricas [A]
time = 4.49, size = 841, normalized size = 2.24 \begin {gather*} \left [\frac {22 \, a^{4} b^{3} - 14 \, a^{2} b^{5} - 8 \, b^{7} - 2 \, {\left (2 \, a^{7} + 10 \, a^{5} b^{2} - 11 \, a^{3} b^{4} - a b^{6}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (2 \, a^{3} b^{3} + 3 \, a b^{5} + {\left (2 \, a^{5} b + 3 \, a^{3} b^{3}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) \sin \left (d x + c\right ) + 2 \, {\left (2 \, a^{6} b - 17 \, a^{4} b^{3} + 13 \, a^{2} b^{5} + 2 \, b^{7}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (16 \, a^{5} b^{2} - 17 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )}{4 \, {\left ({\left (a^{10} - 4 \, a^{8} b^{2} + 6 \, a^{6} b^{4} - 4 \, a^{4} b^{6} + a^{2} b^{8}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{9} b - 4 \, a^{7} b^{3} + 6 \, a^{5} b^{5} - 4 \, a^{3} b^{7} + a b^{9}\right )} d \cos \left (d x + c\right ) + {\left (a^{8} b^{2} - 4 \, a^{6} b^{4} + 6 \, a^{4} b^{6} - 4 \, a^{2} b^{8} + b^{10}\right )} d\right )} \sin \left (d x + c\right )}, \frac {11 \, a^{4} b^{3} - 7 \, a^{2} b^{5} - 4 \, b^{7} - {\left (2 \, a^{7} + 10 \, a^{5} b^{2} - 11 \, a^{3} b^{4} - a b^{6}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (2 \, a^{3} b^{3} + 3 \, a b^{5} + {\left (2 \, a^{5} b + 3 \, a^{3} b^{3}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + {\left (2 \, a^{6} b - 17 \, a^{4} b^{3} + 13 \, a^{2} b^{5} + 2 \, b^{7}\right )} \cos \left (d x + c\right )^{2} + {\left (16 \, a^{5} b^{2} - 17 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )}{2 \, {\left ({\left (a^{10} - 4 \, a^{8} b^{2} + 6 \, a^{6} b^{4} - 4 \, a^{4} b^{6} + a^{2} b^{8}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{9} b - 4 \, a^{7} b^{3} + 6 \, a^{5} b^{5} - 4 \, a^{3} b^{7} + a b^{9}\right )} d \cos \left (d x + c\right ) + {\left (a^{8} b^{2} - 4 \, a^{6} b^{4} + 6 \, a^{4} b^{6} - 4 \, a^{2} b^{8} + b^{10}\right )} d\right )} \sin \left (d x + c\right )}\right ] \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\csc ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{3}}\, dx \end {gather*}

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 = 0.56, size = 386, normalized size = 1.03 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 2.86, size = 423, normalized size = 1.12 \begin {gather*} \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d\,{\left (a-b\right )}^3}-\frac {\frac {a^3-3\,a^2\,b+3\,a\,b^2-b^3}{a+b}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^5-5\,a^4\,b+22\,a^3\,b^2-20\,a^2\,b^3+7\,a\,b^4-5\,b^5\right )}{{\left (a+b\right )}^3}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^4-4\,a^3\,b+12\,a^2\,b^2-5\,a\,b^3+3\,b^4\right )}{{\left (a+b\right )}^2}}{d\,\left (\left (2\,a^5-10\,a^4\,b+20\,a^3\,b^2-20\,a^2\,b^3+10\,a\,b^4-2\,b^5\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-4\,a^5+12\,a^4\,b-8\,a^3\,b^2-8\,a^2\,b^3+12\,a\,b^4-4\,b^5\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,a^5-2\,a^4\,b-4\,a^3\,b^2+4\,a^2\,b^3+2\,a\,b^4-2\,b^5\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {a\,b\,\mathrm {atan}\left (\frac {1{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^6-3{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,b^2+3{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^4-1{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^6}{{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{5/2}}\right )\,\left (2\,a^2+3\,b^2\right )\,3{}\mathrm {i}}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(c + d*x)^2*(a + b/cos(c + d*x))^3),x)

[Out]

tan(c/2 + (d*x)/2)/(2*d*(a - b)^3) - ((3*a*b^2 - 3*a^2*b + a^3 - b^3)/(a + b) + (tan(c/2 + (d*x)/2)^4*(7*a*b^4
- 5*a^4*b + a^5 - 5*b^5 - 20*a^2*b^3 + 22*a^3*b^2))/(a + b)^3 - (2*tan(c/2 + (d*x)/2)^2*(a^4 - 4*a^3*b - 5*a*
b^3 + 3*b^4 + 12*a^2*b^2))/(a + b)^2)/(d*(tan(c/2 + (d*x)/2)*(2*a*b^4 - 2*a^4*b + 2*a^5 - 2*b^5 + 4*a^2*b^3 -
4*a^3*b^2) - tan(c/2 + (d*x)/2)^3*(4*a^5 - 12*a^4*b - 12*a*b^4 + 4*b^5 + 8*a^2*b^3 + 8*a^3*b^2) + tan(c/2 + (d
*x)/2)^5*(10*a*b^4 - 10*a^4*b + 2*a^5 - 2*b^5 - 20*a^2*b^3 + 20*a^3*b^2))) + (a*b*atan((a^6*tan(c/2 + (d*x)/2)
*1i - b^6*tan(c/2 + (d*x)/2)*1i + a^2*b^4*tan(c/2 + (d*x)/2)*3i - a^4*b^2*tan(c/2 + (d*x)/2)*3i)/((a + b)^(7/2
)*(a - b)^(5/2)))*(2*a^2 + 3*b^2)*3i)/(d*(a + b)^(7/2)*(a - b)^(7/2))

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