3.53 \(\int \frac{\csc ^6(e+f x)}{(a+b \sec ^2(e+f x))^2} \, dx\)

Optimal. Leaf size=188 \[ -\frac{b \left (5 a^2+2 b^2\right ) \tan (e+f x)}{10 f (a+b)^4 \left (a+b \tan ^2(e+f x)+b\right )}-\frac{\left (5 a^2-10 a b-b^2\right ) \cot (e+f x)}{5 f (a+b)^4}-\frac{a \sqrt{b} (3 a-4 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 f (a+b)^{9/2}}-\frac{(10 a+3 b) \cot ^3(e+f x)}{15 f (a+b)^3}-\frac{\cot ^5(e+f x)}{5 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )} \]

[Out]

-(a*(3*a - 4*b)*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(2*(a + b)^(9/2)*f) - ((5*a^2 - 10*a*b - b
^2)*Cot[e + f*x])/(5*(a + b)^4*f) - ((10*a + 3*b)*Cot[e + f*x]^3)/(15*(a + b)^3*f) - Cot[e + f*x]^5/(5*(a + b)
*f*(a + b + b*Tan[e + f*x]^2)) - (b*(5*a^2 + 2*b^2)*Tan[e + f*x])/(10*(a + b)^4*f*(a + b + b*Tan[e + f*x]^2))

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Rubi [A]  time = 0.26434, antiderivative size = 188, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {4132, 462, 456, 1261, 205} \[ -\frac{b \left (5 a^2+2 b^2\right ) \tan (e+f x)}{10 f (a+b)^4 \left (a+b \tan ^2(e+f x)+b\right )}-\frac{\left (5 a^2-10 a b-b^2\right ) \cot (e+f x)}{5 f (a+b)^4}-\frac{a \sqrt{b} (3 a-4 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 f (a+b)^{9/2}}-\frac{(10 a+3 b) \cot ^3(e+f x)}{15 f (a+b)^3}-\frac{\cot ^5(e+f x)}{5 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^6/(a + b*Sec[e + f*x]^2)^2,x]

[Out]

-(a*(3*a - 4*b)*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(2*(a + b)^(9/2)*f) - ((5*a^2 - 10*a*b - b
^2)*Cot[e + f*x])/(5*(a + b)^4*f) - ((10*a + 3*b)*Cot[e + f*x]^3)/(15*(a + b)^3*f) - Cot[e + f*x]^5/(5*(a + b)
*f*(a + b + b*Tan[e + f*x]^2)) - (b*(5*a^2 + 2*b^2)*Tan[e + f*x])/(10*(a + b)^4*f*(a + b + b*Tan[e + f*x]^2))

Rule 4132

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = Fr
eeFactors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p)/(
1 + ff^2*x^2)^(m/2 + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integer
Q[n/2]

Rule 462

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[(c^2*(e*x)^(
m + 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 0]

Rule 456

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)]
- ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &
& ILtQ[m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1261

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rule 205

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

Rubi steps

\begin{align*} \int \frac{\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^2}{x^6 \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cot ^5(e+f x)}{5 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{10 a+3 b+5 (a+b) x^2}{x^4 \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{5 (a+b) f}\\ &=-\frac{\cot ^5(e+f x)}{5 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{b \left (5 a^2+2 b^2\right ) \tan (e+f x)}{10 (a+b)^4 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{b \operatorname{Subst}\left (\int \frac{-\frac{2 (10 a+3 b)}{b (a+b)}-\frac{2 \left (5 a^2+2 b^2\right ) x^2}{b (a+b)^2}+\frac{\left (5 a^2+2 b^2\right ) x^4}{(a+b)^3}}{x^4 \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{10 (a+b) f}\\ &=-\frac{\cot ^5(e+f x)}{5 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{b \left (5 a^2+2 b^2\right ) \tan (e+f x)}{10 (a+b)^4 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{b \operatorname{Subst}\left (\int \left (-\frac{2 (10 a+3 b)}{b (a+b)^2 x^4}+\frac{2 \left (-5 a^2+10 a b+b^2\right )}{b (a+b)^3 x^2}+\frac{5 a (3 a-4 b)}{(a+b)^3 \left (a+b+b x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{10 (a+b) f}\\ &=-\frac{\left (5 a^2-10 a b-b^2\right ) \cot (e+f x)}{5 (a+b)^4 f}-\frac{(10 a+3 b) \cot ^3(e+f x)}{15 (a+b)^3 f}-\frac{\cot ^5(e+f x)}{5 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{b \left (5 a^2+2 b^2\right ) \tan (e+f x)}{10 (a+b)^4 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{(a (3 a-4 b) b) \operatorname{Subst}\left (\int \frac{1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{2 (a+b)^4 f}\\ &=-\frac{a (3 a-4 b) \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 (a+b)^{9/2} f}-\frac{\left (5 a^2-10 a b-b^2\right ) \cot (e+f x)}{5 (a+b)^4 f}-\frac{(10 a+3 b) \cot ^3(e+f x)}{15 (a+b)^3 f}-\frac{\cot ^5(e+f x)}{5 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{b \left (5 a^2+2 b^2\right ) \tan (e+f x)}{10 (a+b)^4 f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end{align*}

Mathematica [C]  time = 3.53815, size = 777, normalized size = 4.13 \[ \frac{\sec ^4(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (-\csc (e) \sec (2 e) \csc ^5(e+f x) \left (10 a \left (16 a^2+34 a b+123 b^2\right ) \sin (f x)-a \left (16 a^2-223 a b+1336 b^2\right ) \sin (3 f x)+640 a^2 b \sin (2 e-f x)-715 a^2 b \sin (2 e+f x)+415 a^2 b \sin (4 e+f x)+165 a^2 b \sin (2 e+3 f x)+208 a^2 b \sin (4 e+3 f x)+180 a^2 b \sin (6 e+3 f x)-268 a^2 b \sin (2 e+5 f x)-223 a^2 b \sin (6 e+5 f x)-45 a^2 b \sin (8 e+5 f x)+83 a^2 b \sin (4 e+7 f x)-15 a^2 b \sin (6 e+7 f x)+68 a^2 b \sin (8 e+7 f x)+240 a^3 \sin (2 e-f x)-240 a^3 \sin (2 e+f x)+160 a^3 \sin (4 e+f x)-16 a^3 \sin (4 e+3 f x)+48 a^3 \sin (2 e+5 f x)+48 a^3 \sin (6 e+5 f x)-16 a^3 \sin (4 e+7 f x)-16 a^3 \sin (8 e+7 f x)-1460 a b^2 \sin (2 e-f x)+860 a b^2 \sin (2 e+f x)+1830 a b^2 \sin (4 e+f x)-30 a b^2 \sin (2 e+3 f x)-1036 a b^2 \sin (4 e+3 f x)-330 a b^2 \sin (6 e+3 f x)+290 a b^2 \sin (2 e+5 f x)+230 a b^2 \sin (6 e+5 f x)+60 a b^2 \sin (8 e+5 f x)-6 a b^2 \sin (4 e+7 f x)-6 a b^2 \sin (8 e+7 f x)+240 b^3 \sin (2 e-f x)-240 b^3 \sin (2 e+f x)+120 b^3 \sin (2 e+3 f x)+120 b^3 \sin (6 e+3 f x)-24 b^3 \sin (2 e+5 f x)-24 b^3 \sin (6 e+5 f x)\right )+\frac{960 a b (3 a-4 b) (\cos (2 e)-i \sin (2 e)) (a \cos (2 (e+f x))+a+2 b) \tan ^{-1}\left (\frac{(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4}}\right )}{\sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4}}\right )}{7680 f (a+b)^4 \left (a+b \sec ^2(e+f x)\right )^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Csc[e + f*x]^6/(a + b*Sec[e + f*x]^2)^2,x]

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^4*((960*a*(3*a - 4*b)*b*ArcTan[(Sec[f*x]*(Cos[2*e] - I*Sin[2*e])*
(-((a + 2*b)*Sin[f*x]) + a*Sin[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])]*(a + 2*b + a*Cos[2*
(e + f*x)])*(Cos[2*e] - I*Sin[2*e]))/(Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4]) - Csc[e]*Csc[e + f*x]^5*Sec[2
*e]*(10*a*(16*a^2 + 34*a*b + 123*b^2)*Sin[f*x] - a*(16*a^2 - 223*a*b + 1336*b^2)*Sin[3*f*x] + 240*a^3*Sin[2*e
- f*x] + 640*a^2*b*Sin[2*e - f*x] - 1460*a*b^2*Sin[2*e - f*x] + 240*b^3*Sin[2*e - f*x] - 240*a^3*Sin[2*e + f*x
] - 715*a^2*b*Sin[2*e + f*x] + 860*a*b^2*Sin[2*e + f*x] - 240*b^3*Sin[2*e + f*x] + 160*a^3*Sin[4*e + f*x] + 41
5*a^2*b*Sin[4*e + f*x] + 1830*a*b^2*Sin[4*e + f*x] + 165*a^2*b*Sin[2*e + 3*f*x] - 30*a*b^2*Sin[2*e + 3*f*x] +
120*b^3*Sin[2*e + 3*f*x] - 16*a^3*Sin[4*e + 3*f*x] + 208*a^2*b*Sin[4*e + 3*f*x] - 1036*a*b^2*Sin[4*e + 3*f*x]
+ 180*a^2*b*Sin[6*e + 3*f*x] - 330*a*b^2*Sin[6*e + 3*f*x] + 120*b^3*Sin[6*e + 3*f*x] + 48*a^3*Sin[2*e + 5*f*x]
- 268*a^2*b*Sin[2*e + 5*f*x] + 290*a*b^2*Sin[2*e + 5*f*x] - 24*b^3*Sin[2*e + 5*f*x] + 48*a^3*Sin[6*e + 5*f*x]
- 223*a^2*b*Sin[6*e + 5*f*x] + 230*a*b^2*Sin[6*e + 5*f*x] - 24*b^3*Sin[6*e + 5*f*x] - 45*a^2*b*Sin[8*e + 5*f*
x] + 60*a*b^2*Sin[8*e + 5*f*x] - 16*a^3*Sin[4*e + 7*f*x] + 83*a^2*b*Sin[4*e + 7*f*x] - 6*a*b^2*Sin[4*e + 7*f*x
] - 15*a^2*b*Sin[6*e + 7*f*x] - 16*a^3*Sin[8*e + 7*f*x] + 68*a^2*b*Sin[8*e + 7*f*x] - 6*a*b^2*Sin[8*e + 7*f*x]
)))/(7680*(a + b)^4*f*(a + b*Sec[e + f*x]^2)^2)

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Maple [A]  time = 0.118, size = 189, normalized size = 1. \begin{align*} -{\frac{1}{5\,f \left ( a+b \right ) ^{2} \left ( \tan \left ( fx+e \right ) \right ) ^{5}}}-{\frac{{a}^{2}}{f \left ( a+b \right ) ^{4}\tan \left ( fx+e \right ) }}+2\,{\frac{ab}{f \left ( a+b \right ) ^{4}\tan \left ( fx+e \right ) }}-{\frac{2\,a}{3\,f \left ( a+b \right ) ^{3} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}}-{\frac{{a}^{2}b\tan \left ( fx+e \right ) }{2\,f \left ( a+b \right ) ^{4} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{3\,{a}^{2}b}{2\,f \left ( a+b \right ) ^{4}}\arctan \left ({\tan \left ( fx+e \right ) b{\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}}+2\,{\frac{a{b}^{2}}{f \left ( a+b \right ) ^{4}\sqrt{ \left ( a+b \right ) b}}\arctan \left ({\frac{\tan \left ( fx+e \right ) b}{\sqrt{ \left ( a+b \right ) b}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^2,x)

[Out]

-1/5/f/(a+b)^2/tan(f*x+e)^5-1/f*a^2/(a+b)^4/tan(f*x+e)+2/f*a/(a+b)^4/tan(f*x+e)*b-2/3/f*a/(a+b)^3/tan(f*x+e)^3
-1/2/f*a^2/(a+b)^4*b*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)-3/2/f*a^2/(a+b)^4*b/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/(
(a+b)*b)^(1/2))+2/f*a/(a+b)^4*b^2/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))

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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(f*x+e)^6/(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.746461, size = 2260, normalized size = 12.02 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

[-1/120*(4*(16*a^3 - 83*a^2*b + 6*a*b^2)*cos(f*x + e)^7 - 4*(40*a^3 - 201*a^2*b + 68*a*b^2 - 6*b^3)*cos(f*x +
e)^5 + 20*(6*a^3 - 29*a^2*b + 28*a*b^2)*cos(f*x + e)^3 + 15*((3*a^3 - 4*a^2*b)*cos(f*x + e)^6 - (6*a^3 - 11*a^
2*b + 4*a*b^2)*cos(f*x + e)^4 + 3*a^2*b - 4*a*b^2 + (3*a^3 - 10*a^2*b + 8*a*b^2)*cos(f*x + e)^2)*sqrt(-b/(a +
b))*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a*b + 4*b^2)*cos(f*x + e)^2 - 4*((a^2 + 3*a*b + 2*b^2)*co
s(f*x + e)^3 - (a*b + b^2)*cos(f*x + e))*sqrt(-b/(a + b))*sin(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(
f*x + e)^2 + b^2))*sin(f*x + e) + 60*(3*a^2*b - 4*a*b^2)*cos(f*x + e))/(((a^5 + 4*a^4*b + 6*a^3*b^2 + 4*a^2*b^
3 + a*b^4)*f*cos(f*x + e)^6 - (2*a^5 + 7*a^4*b + 8*a^3*b^2 + 2*a^2*b^3 - 2*a*b^4 - b^5)*f*cos(f*x + e)^4 + (a^
5 + 2*a^4*b - 2*a^3*b^2 - 8*a^2*b^3 - 7*a*b^4 - 2*b^5)*f*cos(f*x + e)^2 + (a^4*b + 4*a^3*b^2 + 6*a^2*b^3 + 4*a
*b^4 + b^5)*f)*sin(f*x + e)), -1/60*(2*(16*a^3 - 83*a^2*b + 6*a*b^2)*cos(f*x + e)^7 - 2*(40*a^3 - 201*a^2*b +
68*a*b^2 - 6*b^3)*cos(f*x + e)^5 + 10*(6*a^3 - 29*a^2*b + 28*a*b^2)*cos(f*x + e)^3 - 15*((3*a^3 - 4*a^2*b)*cos
(f*x + e)^6 - (6*a^3 - 11*a^2*b + 4*a*b^2)*cos(f*x + e)^4 + 3*a^2*b - 4*a*b^2 + (3*a^3 - 10*a^2*b + 8*a*b^2)*c
os(f*x + e)^2)*sqrt(b/(a + b))*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)*sqrt(b/(a + b))/(b*cos(f*x + e)*sin(f
*x + e)))*sin(f*x + e) + 30*(3*a^2*b - 4*a*b^2)*cos(f*x + e))/(((a^5 + 4*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + a*b^4
)*f*cos(f*x + e)^6 - (2*a^5 + 7*a^4*b + 8*a^3*b^2 + 2*a^2*b^3 - 2*a*b^4 - b^5)*f*cos(f*x + e)^4 + (a^5 + 2*a^4
*b - 2*a^3*b^2 - 8*a^2*b^3 - 7*a*b^4 - 2*b^5)*f*cos(f*x + e)^2 + (a^4*b + 4*a^3*b^2 + 6*a^2*b^3 + 4*a*b^4 + b^
5)*f)*sin(f*x + e))]

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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(f*x+e)**6/(a+b*sec(f*x+e)**2)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.3106, size = 355, normalized size = 1.89 \begin{align*} -\frac{\frac{15 \, a^{2} b \tan \left (f x + e\right )}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )}{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}} + \frac{15 \,{\left (3 \, a^{2} b - 4 \, a b^{2}\right )}{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (f x + e\right )}{\sqrt{a b + b^{2}}}\right )\right )}}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \sqrt{a b + b^{2}}} + \frac{2 \,{\left (15 \, a^{2} \tan \left (f x + e\right )^{4} - 30 \, a b \tan \left (f x + e\right )^{4} + 10 \, a^{2} \tan \left (f x + e\right )^{2} + 10 \, a b \tan \left (f x + e\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}\right )}}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \tan \left (f x + e\right )^{5}}}{30 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")

[Out]

-1/30*(15*a^2*b*tan(f*x + e)/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*(b*tan(f*x + e)^2 + a + b)) + 15*(3*
a^2*b - 4*a*b^2)*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))/((a^4 + 4*a^3*
b + 6*a^2*b^2 + 4*a*b^3 + b^4)*sqrt(a*b + b^2)) + 2*(15*a^2*tan(f*x + e)^4 - 30*a*b*tan(f*x + e)^4 + 10*a^2*ta
n(f*x + e)^2 + 10*a*b*tan(f*x + e)^2 + 3*a^2 + 6*a*b + 3*b^2)/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*tan
(f*x + e)^5))/f