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]

-1/5*(5*a^2-10*a*b-b^2)*cot(f*x+e)/(a+b)^4/f-1/15*(10*a+3*b)*cot(f*x+e)^3/(a+b)^3/f-1/2*a*(3*a-4*b)*arctan(b^(
1/2)*tan(f*x+e)/(a+b)^(1/2))*b^(1/2)/(a+b)^(9/2)/f-1/5*cot(f*x+e)^5/(a+b)/f/(a+b+b*tan(f*x+e)^2)-1/10*b*(5*a^2
+2*b^2)*tan(f*x+e)/(a+b)^4/f/(a+b+b*tan(f*x+e)^2)

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Rubi [A]  time = 0.26, antiderivative size = 188, normalized size of antiderivative = 1.00, 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 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]

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 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 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 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]

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*}

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Mathematica [C]  time = 3.22, size = 777, normalized size = 4.13 \[ \frac {\sec ^4(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (-\csc (e) \sec (2 e) \csc ^5(e+f x) \left (240 a^3 \sin (2 e-f x)-240 a^3 \sin (2 e+f x)+160 a^3 \sin (4 e+f x)-16 a^3 \sin (4 e+3 f x)+48 a^3 \sin (2 e+5 f x)+48 a^3 \sin (6 e+5 f x)-16 a^3 \sin (4 e+7 f x)-16 a^3 \sin (8 e+7 f x)+10 a \left (16 a^2+34 a b+123 b^2\right ) \sin (f x)-a \left (16 a^2-223 a b+1336 b^2\right ) \sin (3 f x)+640 a^2 b \sin (2 e-f x)-715 a^2 b \sin (2 e+f x)+415 a^2 b \sin (4 e+f x)+165 a^2 b \sin (2 e+3 f x)+208 a^2 b \sin (4 e+3 f x)+180 a^2 b \sin (6 e+3 f x)-268 a^2 b \sin (2 e+5 f x)-223 a^2 b \sin (6 e+5 f x)-45 a^2 b \sin (8 e+5 f x)+83 a^2 b \sin (4 e+7 f x)-15 a^2 b \sin (6 e+7 f x)+68 a^2 b \sin (8 e+7 f x)-1460 a b^2 \sin (2 e-f x)+860 a b^2 \sin (2 e+f x)+1830 a b^2 \sin (4 e+f x)-30 a b^2 \sin (2 e+3 f x)-1036 a b^2 \sin (4 e+3 f x)-330 a b^2 \sin (6 e+3 f x)+290 a b^2 \sin (2 e+5 f x)+230 a b^2 \sin (6 e+5 f x)+60 a b^2 \sin (8 e+5 f x)-6 a b^2 \sin (4 e+7 f x)-6 a b^2 \sin (8 e+7 f x)+240 b^3 \sin (2 e-f x)-240 b^3 \sin (2 e+f x)+120 b^3 \sin (2 e+3 f x)+120 b^3 \sin (6 e+3 f x)-24 b^3 \sin (2 e+5 f x)-24 b^3 \sin (6 e+5 f x)\right )+\frac {960 a b (3 a-4 b) (\cos (2 e)-i \sin (2 e)) (a \cos (2 (e+f x))+a+2 b) \tan ^{-1}\left (\frac {(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right )}{\sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right )}{7680 f (a+b)^4 \left (a+b \sec ^2(e+f x)\right )^2} \]

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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fricas [B]  time = 0.75, size = 987, normalized size = 5.25 \[ \text {result too large to display} \]

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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giac [A]  time = 0.51, size = 263, normalized size = 1.40 \[ -\frac {\frac {15 \, a^{2} b \tan \left (f x + e\right )}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}} + \frac {15 \, {\left (3 \, a^{2} b - 4 \, a b^{2}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )}}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \sqrt {a b + b^{2}}} + \frac {2 \, {\left (15 \, a^{2} \tan \left (f x + e\right )^{4} - 30 \, a b \tan \left (f x + e\right )^{4} + 10 \, a^{2} \tan \left (f x + e\right )^{2} + 10 \, a b \tan \left (f x + e\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}\right )}}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \tan \left (f x + e\right )^{5}}}{30 \, f} \]

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

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maple [A]  time = 1.25, size = 189, normalized size = 1.01 \[ -\frac {b \,a^{2} \tan \left (f x +e \right )}{2 f \left (a +b \right )^{4} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}-\frac {3 b \,a^{2} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{2 f \left (a +b \right )^{4} \sqrt {\left (a +b \right ) b}}+\frac {2 b^{2} a \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f \left (a +b \right )^{4} \sqrt {\left (a +b \right ) b}}-\frac {1}{5 f \left (a +b \right )^{2} \tan \left (f x +e \right )^{5}}-\frac {a^{2}}{f \left (a +b \right )^{4} \tan \left (f x +e \right )}+\frac {2 a b}{f \left (a +b \right )^{4} \tan \left (f x +e \right )}-\frac {2 a}{3 f \left (a +b \right )^{3} \tan \left (f x +e \right )^{3}} \]

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]

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

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maxima [A]  time = 0.44, size = 268, normalized size = 1.43 \[ -\frac {\frac {15 \, {\left (3 \, a^{2} b - 4 \, a b^{2}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \sqrt {{\left (a + b\right )} b}} + \frac {15 \, {\left (3 \, a^{2} b - 4 \, a b^{2}\right )} \tan \left (f x + e\right )^{6} + 10 \, {\left (3 \, a^{3} - a^{2} b - 4 \, a b^{2}\right )} \tan \left (f x + e\right )^{4} + 6 \, a^{3} + 18 \, a^{2} b + 18 \, a b^{2} + 6 \, b^{3} + 2 \, {\left (10 \, a^{3} + 23 \, a^{2} b + 16 \, a b^{2} + 3 \, b^{3}\right )} \tan \left (f x + e\right )^{2}}{{\left (a^{4} b + 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} + 4 \, a b^{4} + b^{5}\right )} \tan \left (f x + e\right )^{7} + {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} \tan \left (f x + e\right )^{5}}}{30 \, f} \]

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]

-1/30*(15*(3*a^2*b - 4*a*b^2)*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b
^4)*sqrt((a + b)*b)) + (15*(3*a^2*b - 4*a*b^2)*tan(f*x + e)^6 + 10*(3*a^3 - a^2*b - 4*a*b^2)*tan(f*x + e)^4 +
6*a^3 + 18*a^2*b + 18*a*b^2 + 6*b^3 + 2*(10*a^3 + 23*a^2*b + 16*a*b^2 + 3*b^3)*tan(f*x + e)^2)/((a^4*b + 4*a^3
*b^2 + 6*a^2*b^3 + 4*a*b^4 + b^5)*tan(f*x + e)^7 + (a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*t
an(f*x + e)^5))/f

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mupad [B]  time = 6.31, size = 198, normalized size = 1.05 \[ \frac {a\,\sqrt {b}\,\mathrm {atan}\left (\frac {a\,\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )\,\left (3\,a-4\,b\right )\,\left (a^4+4\,a^3\,b+6\,a^2\,b^2+4\,a\,b^3+b^4\right )}{{\left (a+b\right )}^{9/2}\,\left (4\,a\,b-3\,a^2\right )}\right )\,\left (3\,a-4\,b\right )}{2\,f\,{\left (a+b\right )}^{9/2}}-\frac {\frac {1}{5\,\left (a+b\right )}-\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (4\,a\,b-3\,a^2\right )}{3\,{\left (a+b\right )}^3}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (10\,a+3\,b\right )}{15\,{\left (a+b\right )}^2}-\frac {b\,{\mathrm {tan}\left (e+f\,x\right )}^6\,\left (4\,a\,b-3\,a^2\right )}{2\,{\left (a+b\right )}^4}}{f\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^7+\left (a+b\right )\,{\mathrm {tan}\left (e+f\,x\right )}^5\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(e + f*x)^6*(a + b/cos(e + f*x)^2)^2),x)

[Out]

(a*b^(1/2)*atan((a*b^(1/2)*tan(e + f*x)*(3*a - 4*b)*(4*a*b^3 + 4*a^3*b + a^4 + b^4 + 6*a^2*b^2))/((a + b)^(9/2
)*(4*a*b - 3*a^2)))*(3*a - 4*b))/(2*f*(a + b)^(9/2)) - (1/(5*(a + b)) - (tan(e + f*x)^4*(4*a*b - 3*a^2))/(3*(a
+ b)^3) + (tan(e + f*x)^2*(10*a + 3*b))/(15*(a + b)^2) - (b*tan(e + f*x)^6*(4*a*b - 3*a^2))/(2*(a + b)^4))/(f
*(tan(e + f*x)^5*(a + b) + b*tan(e + f*x)^7))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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