3.66 \(\int \frac {\csc ^6(e+f x)}{(a+b \sec ^2(e+f x))^3} \, dx\)

Optimal. Leaf size=242 \[ -\frac {\sqrt {b} \left (15 a^2-40 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 f (a+b)^{11/2}}-\frac {b \left (35 a^2-40 a b+24 b^2\right ) \tan (e+f x)}{40 f (a+b)^5 \left (a+b \tan ^2(e+f x)+b\right )}-\frac {b \left (5 a^2+4 b^2\right ) \tan (e+f x)}{20 f (a+b)^4 \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {\left (5 a^2-20 a b+2 b^2\right ) \cot (e+f x)}{5 f (a+b)^5}-\frac {(10 a+b) \cot ^3(e+f x)}{15 f (a+b)^4}-\frac {\cot ^5(e+f x)}{5 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2} \]

[Out]

-1/5*(5*a^2-20*a*b+2*b^2)*cot(f*x+e)/(a+b)^5/f-1/15*(10*a+b)*cot(f*x+e)^3/(a+b)^4/f-1/8*(15*a^2-40*a*b+8*b^2)*
arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1/2))*b^(1/2)/(a+b)^(11/2)/f-1/5*cot(f*x+e)^5/(a+b)/f/(a+b+b*tan(f*x+e)^2)^2-
1/20*b*(5*a^2+4*b^2)*tan(f*x+e)/(a+b)^4/f/(a+b+b*tan(f*x+e)^2)^2-1/40*b*(35*a^2-40*a*b+24*b^2)*tan(f*x+e)/(a+b
)^5/f/(a+b+b*tan(f*x+e)^2)

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Rubi [A]  time = 0.37, antiderivative size = 242, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {4132, 462, 456, 1259, 1261, 205} \[ -\frac {\sqrt {b} \left (15 a^2-40 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 f (a+b)^{11/2}}-\frac {b \left (35 a^2-40 a b+24 b^2\right ) \tan (e+f x)}{40 f (a+b)^5 \left (a+b \tan ^2(e+f x)+b\right )}-\frac {b \left (5 a^2+4 b^2\right ) \tan (e+f x)}{20 f (a+b)^4 \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {\left (5 a^2-20 a b+2 b^2\right ) \cot (e+f x)}{5 f (a+b)^5}-\frac {(10 a+b) \cot ^3(e+f x)}{15 f (a+b)^4}-\frac {\cot ^5(e+f x)}{5 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^6/(a + b*Sec[e + f*x]^2)^3,x]

[Out]

-(Sqrt[b]*(15*a^2 - 40*a*b + 8*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(8*(a + b)^(11/2)*f) - ((5*a^2
- 20*a*b + 2*b^2)*Cot[e + f*x])/(5*(a + b)^5*f) - ((10*a + b)*Cot[e + f*x]^3)/(15*(a + b)^4*f) - Cot[e + f*x]
^5/(5*(a + b)*f*(a + b + b*Tan[e + f*x]^2)^2) - (b*(5*a^2 + 4*b^2)*Tan[e + f*x])/(20*(a + b)^4*f*(a + b + b*Ta
n[e + f*x]^2)^2) - (b*(35*a^2 - 40*a*b + 24*b^2)*Tan[e + f*x])/(40*(a + b)^5*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 1259

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((-d)^(
m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*(d + e*x^2)^(q + 1))/(2*e^(2*p + m/2)*(q + 1)), x] + Dist[(-d)^(m/2 - 1)/
(2*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*(-d)^(-(m/2) + 1)*e^(2*p)*(q + 1)*
(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2)))/(d + e*x^2)], x], x]
, x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 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 )^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^2}{x^6 \left (a+b+b x^2\right )^3} \, 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 )^2}+\frac {\operatorname {Subst}\left (\int \frac {10 a+b+5 (a+b) x^2}{x^4 \left (a+b+b x^2\right )^3} \, 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 )^2}-\frac {b \left (5 a^2+4 b^2\right ) \tan (e+f x)}{20 (a+b)^4 f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b \operatorname {Subst}\left (\int \frac {-\frac {4 (10 a+b)}{b (a+b)}-\frac {4 \left (5 a^2+4 b^2\right ) x^2}{b (a+b)^2}+\frac {3 \left (5 a^2+4 b^2\right ) x^4}{(a+b)^3}}{x^4 \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{20 (a+b) f}\\ &=-\frac {\cot ^5(e+f x)}{5 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b \left (5 a^2+4 b^2\right ) \tan (e+f x)}{20 (a+b)^4 f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b \left (35 a^2-40 a b+24 b^2\right ) \tan (e+f x)}{40 (a+b)^5 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {-8 b (a+b) (10 a+b)-8 b \left (5 a^2-10 a b+3 b^2\right ) x^2+\frac {b^2 \left (35 a^2-40 a b+24 b^2\right ) x^4}{a+b}}{x^4 \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{40 b (a+b)^4 f}\\ &=-\frac {\cot ^5(e+f x)}{5 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b \left (5 a^2+4 b^2\right ) \tan (e+f x)}{20 (a+b)^4 f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b \left (35 a^2-40 a b+24 b^2\right ) \tan (e+f x)}{40 (a+b)^5 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {\operatorname {Subst}\left (\int \left (-\frac {8 b (10 a+b)}{x^4}-\frac {8 b \left (5 a^2-20 a b+2 b^2\right )}{(a+b) x^2}+\frac {5 b^2 \left (15 a^2-40 a b+8 b^2\right )}{(a+b) \left (a+b+b x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{40 b (a+b)^4 f}\\ &=-\frac {\left (5 a^2-20 a b+2 b^2\right ) \cot (e+f x)}{5 (a+b)^5 f}-\frac {(10 a+b) \cot ^3(e+f x)}{15 (a+b)^4 f}-\frac {\cot ^5(e+f x)}{5 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b \left (5 a^2+4 b^2\right ) \tan (e+f x)}{20 (a+b)^4 f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b \left (35 a^2-40 a b+24 b^2\right ) \tan (e+f x)}{40 (a+b)^5 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {\left (b \left (15 a^2-40 a b+8 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{8 (a+b)^5 f}\\ &=-\frac {\sqrt {b} \left (15 a^2-40 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 (a+b)^{11/2} f}-\frac {\left (5 a^2-20 a b+2 b^2\right ) \cot (e+f x)}{5 (a+b)^5 f}-\frac {(10 a+b) \cot ^3(e+f x)}{15 (a+b)^4 f}-\frac {\cot ^5(e+f x)}{5 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b \left (5 a^2+4 b^2\right ) \tan (e+f x)}{20 (a+b)^4 f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b \left (35 a^2-40 a b+24 b^2\right ) \tan (e+f x)}{40 (a+b)^5 f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end {align*}

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Mathematica [C]  time = 5.77, size = 479, normalized size = 1.98 \[ \frac {\sec ^6(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (8 \left (8 a^2-59 a b+23 b^2\right ) \csc (e) \sin (f x) \csc (e+f x) (a \cos (2 (e+f x))+a+2 b)^2+15 b \sec (2 e) \left (\left (9 a^2+16 a b-8 b^2\right ) \sin (2 e)+3 a (2 b-3 a) \sin (2 f x)\right ) (a \cos (2 (e+f x))+a+2 b)+\frac {15 b \left (15 a^2-40 a b+8 b^2\right ) (\cos (2 e)-i \sin (2 e)) (a \cos (2 (e+f x))+a+2 b)^2 \tan ^{-1}\left (\frac {(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right )}{\sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}-60 b^2 (a+b) \sec (2 e) ((a+2 b) \sin (2 e)-a \sin (2 f x))-24 (a+b)^2 \cot (e) \csc ^4(e+f x) (a \cos (2 (e+f x))+a+2 b)^2-8 (4 a-11 b) (a+b) \cot (e) \csc ^2(e+f x) (a \cos (2 (e+f x))+a+2 b)^2+24 (a+b)^2 \csc (e) \sin (f x) \csc ^5(e+f x) (a \cos (2 (e+f x))+a+2 b)^2+8 (4 a-11 b) (a+b) \csc (e) \sin (f x) \csc ^3(e+f x) (a \cos (2 (e+f x))+a+2 b)^2\right )}{960 f (a+b)^5 \left (a+b \sec ^2(e+f x)\right )^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Csc[e + f*x]^6/(a + b*Sec[e + f*x]^2)^3,x]

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^6*(-8*(4*a - 11*b)*(a + b)*(a + 2*b + a*Cos[2*(e + f*x)])^2*Cot[e
]*Csc[e + f*x]^2 - 24*(a + b)^2*(a + 2*b + a*Cos[2*(e + f*x)])^2*Cot[e]*Csc[e + f*x]^4 + (15*b*(15*a^2 - 40*a*
b + 8*b^2)*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)])^2*(Cos[2*e] - I*Sin[2*e]))/(Sqrt[a + b]*Sqrt[b
*(Cos[e] - I*Sin[e])^4]) + 8*(8*a^2 - 59*a*b + 23*b^2)*(a + 2*b + a*Cos[2*(e + f*x)])^2*Csc[e]*Csc[e + f*x]*Si
n[f*x] + 8*(4*a - 11*b)*(a + b)*(a + 2*b + a*Cos[2*(e + f*x)])^2*Csc[e]*Csc[e + f*x]^3*Sin[f*x] + 24*(a + b)^2
*(a + 2*b + a*Cos[2*(e + f*x)])^2*Csc[e]*Csc[e + f*x]^5*Sin[f*x] - 60*b^2*(a + b)*Sec[2*e]*((a + 2*b)*Sin[2*e]
- a*Sin[2*f*x]) + 15*b*(a + 2*b + a*Cos[2*(e + f*x)])*Sec[2*e]*((9*a^2 + 16*a*b - 8*b^2)*Sin[2*e] + 3*a*(-3*a
+ 2*b)*Sin[2*f*x])))/(960*(a + b)^5*f*(a + b*Sec[e + f*x]^2)^3)

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fricas [B]  time = 0.67, size = 1423, normalized size = 5.88 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")

[Out]

[-1/480*(4*(64*a^4 - 607*a^3*b + 274*a^2*b^2)*cos(f*x + e)^9 - 4*(160*a^4 - 1533*a^3*b + 1599*a^2*b^2 - 488*a*
b^3)*cos(f*x + e)^7 + 4*(120*a^4 - 1205*a^3*b + 2769*a^2*b^2 - 1392*a*b^3 + 184*b^4)*cos(f*x + e)^5 + 20*(75*a
^3*b - 305*a^2*b^2 + 320*a*b^3 - 56*b^4)*cos(f*x + e)^3 - 15*((15*a^4 - 40*a^3*b + 8*a^2*b^2)*cos(f*x + e)^8 -
2*(15*a^4 - 55*a^3*b + 48*a^2*b^2 - 8*a*b^3)*cos(f*x + e)^6 + (15*a^4 - 100*a^3*b + 183*a^2*b^2 - 72*a*b^3 +
8*b^4)*cos(f*x + e)^4 + 15*a^2*b^2 - 40*a*b^3 + 8*b^4 + 2*(15*a^3*b - 55*a^2*b^2 + 48*a*b^3 - 8*b^4)*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)*cos(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*(15*a^2*b^2 - 40*a*b^3 + 8*b^4)*cos(f*x + e))/(((a^7
+ 5*a^6*b + 10*a^5*b^2 + 10*a^4*b^3 + 5*a^3*b^4 + a^2*b^5)*f*cos(f*x + e)^8 - 2*(a^7 + 4*a^6*b + 5*a^5*b^2 - 5
*a^3*b^4 - 4*a^2*b^5 - a*b^6)*f*cos(f*x + e)^6 + (a^7 + a^6*b - 9*a^5*b^2 - 25*a^4*b^3 - 25*a^3*b^4 - 9*a^2*b^
5 + a*b^6 + b^7)*f*cos(f*x + e)^4 + 2*(a^6*b + 4*a^5*b^2 + 5*a^4*b^3 - 5*a^2*b^5 - 4*a*b^6 - b^7)*f*cos(f*x +
e)^2 + (a^5*b^2 + 5*a^4*b^3 + 10*a^3*b^4 + 10*a^2*b^5 + 5*a*b^6 + b^7)*f)*sin(f*x + e)), -1/240*(2*(64*a^4 - 6
07*a^3*b + 274*a^2*b^2)*cos(f*x + e)^9 - 2*(160*a^4 - 1533*a^3*b + 1599*a^2*b^2 - 488*a*b^3)*cos(f*x + e)^7 +
2*(120*a^4 - 1205*a^3*b + 2769*a^2*b^2 - 1392*a*b^3 + 184*b^4)*cos(f*x + e)^5 + 10*(75*a^3*b - 305*a^2*b^2 + 3
20*a*b^3 - 56*b^4)*cos(f*x + e)^3 - 15*((15*a^4 - 40*a^3*b + 8*a^2*b^2)*cos(f*x + e)^8 - 2*(15*a^4 - 55*a^3*b
+ 48*a^2*b^2 - 8*a*b^3)*cos(f*x + e)^6 + (15*a^4 - 100*a^3*b + 183*a^2*b^2 - 72*a*b^3 + 8*b^4)*cos(f*x + e)^4
+ 15*a^2*b^2 - 40*a*b^3 + 8*b^4 + 2*(15*a^3*b - 55*a^2*b^2 + 48*a*b^3 - 8*b^4)*cos(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*(1
5*a^2*b^2 - 40*a*b^3 + 8*b^4)*cos(f*x + e))/(((a^7 + 5*a^6*b + 10*a^5*b^2 + 10*a^4*b^3 + 5*a^3*b^4 + a^2*b^5)*
f*cos(f*x + e)^8 - 2*(a^7 + 4*a^6*b + 5*a^5*b^2 - 5*a^3*b^4 - 4*a^2*b^5 - a*b^6)*f*cos(f*x + e)^6 + (a^7 + a^6
*b - 9*a^5*b^2 - 25*a^4*b^3 - 25*a^3*b^4 - 9*a^2*b^5 + a*b^6 + b^7)*f*cos(f*x + e)^4 + 2*(a^6*b + 4*a^5*b^2 +
5*a^4*b^3 - 5*a^2*b^5 - 4*a*b^6 - b^7)*f*cos(f*x + e)^2 + (a^5*b^2 + 5*a^4*b^3 + 10*a^3*b^4 + 10*a^2*b^5 + 5*a
*b^6 + b^7)*f)*sin(f*x + e))]

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giac [A]  time = 1.09, size = 382, normalized size = 1.58 \[ -\frac {\frac {15 \, {\left (15 \, a^{2} b - 40 \, a b^{2} + 8 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )}}{{\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} \sqrt {a b + b^{2}}} + \frac {15 \, {\left (7 \, a^{2} b^{2} \tan \left (f x + e\right )^{3} - 8 \, a b^{3} \tan \left (f x + e\right )^{3} + 9 \, a^{3} b \tan \left (f x + e\right ) + a^{2} b^{2} \tan \left (f x + e\right ) - 8 \, a b^{3} \tan \left (f x + e\right )\right )}}{{\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2}} + \frac {8 \, {\left (15 \, a^{2} \tan \left (f x + e\right )^{4} - 60 \, a b \tan \left (f x + e\right )^{4} + 15 \, b^{2} \tan \left (f x + e\right )^{4} + 10 \, a^{2} \tan \left (f x + e\right )^{2} + 5 \, a b \tan \left (f x + e\right )^{2} - 5 \, b^{2} \tan \left (f x + e\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}\right )}}{{\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} \tan \left (f x + e\right )^{5}}}{120 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^3,x, algorithm="giac")

[Out]

-1/120*(15*(15*a^2*b - 40*a*b^2 + 8*b^3)*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b
+ b^2)))/((a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*sqrt(a*b + b^2)) + 15*(7*a^2*b^2*tan(f*x
+ e)^3 - 8*a*b^3*tan(f*x + e)^3 + 9*a^3*b*tan(f*x + e) + a^2*b^2*tan(f*x + e) - 8*a*b^3*tan(f*x + e))/((a^5 +
5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*(b*tan(f*x + e)^2 + a + b)^2) + 8*(15*a^2*tan(f*x + e)^4 -
60*a*b*tan(f*x + e)^4 + 15*b^2*tan(f*x + e)^4 + 10*a^2*tan(f*x + e)^2 + 5*a*b*tan(f*x + e)^2 - 5*b^2*tan(f*x +
e)^2 + 3*a^2 + 6*a*b + 3*b^2)/((a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*tan(f*x + e)^5))/f

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maple [A]  time = 1.18, size = 411, normalized size = 1.70 \[ -\frac {7 b^{2} \left (\tan ^{3}\left (f x +e \right )\right ) a^{2}}{8 f \left (a +b \right )^{5} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}+\frac {b^{3} \left (\tan ^{3}\left (f x +e \right )\right ) a}{f \left (a +b \right )^{5} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {9 b \tan \left (f x +e \right ) a^{3}}{8 f \left (a +b \right )^{5} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {b^{2} \tan \left (f x +e \right ) a^{2}}{8 f \left (a +b \right )^{5} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}+\frac {b^{3} \tan \left (f x +e \right ) a}{f \left (a +b \right )^{5} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {15 b \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right ) a^{2}}{8 f \left (a +b \right )^{5} \sqrt {\left (a +b \right ) b}}+\frac {5 b^{2} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right ) a}{f \left (a +b \right )^{5} \sqrt {\left (a +b \right ) b}}-\frac {b^{3} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f \left (a +b \right )^{5} \sqrt {\left (a +b \right ) b}}-\frac {1}{5 f \left (a +b \right )^{3} \tan \left (f x +e \right )^{5}}-\frac {2 a}{3 f \left (a +b \right )^{4} \tan \left (f x +e \right )^{3}}+\frac {b}{3 f \left (a +b \right )^{4} \tan \left (f x +e \right )^{3}}-\frac {a^{2}}{f \left (a +b \right )^{5} \tan \left (f x +e \right )}+\frac {4 a b}{f \left (a +b \right )^{5} \tan \left (f x +e \right )}-\frac {b^{2}}{f \left (a +b \right )^{5} \tan \left (f x +e \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^3,x)

[Out]

-7/8/f*b^2/(a+b)^5/(a+b+b*tan(f*x+e)^2)^2*tan(f*x+e)^3*a^2+1/f*b^3/(a+b)^5/(a+b+b*tan(f*x+e)^2)^2*tan(f*x+e)^3
*a-9/8/f*b/(a+b)^5/(a+b+b*tan(f*x+e)^2)^2*tan(f*x+e)*a^3-1/8/f*b^2/(a+b)^5/(a+b+b*tan(f*x+e)^2)^2*tan(f*x+e)*a
^2+1/f*b^3/(a+b)^5/(a+b+b*tan(f*x+e)^2)^2*tan(f*x+e)*a-15/8/f*b/(a+b)^5/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((
a+b)*b)^(1/2))*a^2+5/f*b^2/(a+b)^5/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))*a-1/f*b^3/(a+b)^5/((a+
b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))-1/5/f/(a+b)^3/tan(f*x+e)^5-2/3/f/(a+b)^4/tan(f*x+e)^3*a+1/3/f
/(a+b)^4/tan(f*x+e)^3*b-1/f/(a+b)^5/tan(f*x+e)*a^2+4/f/(a+b)^5/tan(f*x+e)*a*b-1/f/(a+b)^5/tan(f*x+e)*b^2

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maxima [A]  time = 0.45, size = 434, normalized size = 1.79 \[ -\frac {\frac {15 \, {\left (15 \, a^{2} b - 40 \, a b^{2} + 8 \, b^{3}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} \sqrt {{\left (a + b\right )} b}} + \frac {15 \, {\left (15 \, a^{2} b^{2} - 40 \, a b^{3} + 8 \, b^{4}\right )} \tan \left (f x + e\right )^{8} + 25 \, {\left (15 \, a^{3} b - 25 \, a^{2} b^{2} - 32 \, a b^{3} + 8 \, b^{4}\right )} \tan \left (f x + e\right )^{6} + 8 \, {\left (15 \, a^{4} - 10 \, a^{3} b - 57 \, a^{2} b^{2} - 24 \, a b^{3} + 8 \, b^{4}\right )} \tan \left (f x + e\right )^{4} + 24 \, a^{4} + 96 \, a^{3} b + 144 \, a^{2} b^{2} + 96 \, a b^{3} + 24 \, b^{4} + 8 \, {\left (10 \, a^{4} + 31 \, a^{3} b + 33 \, a^{2} b^{2} + 13 \, a b^{3} + b^{4}\right )} \tan \left (f x + e\right )^{2}}{{\left (a^{5} b^{2} + 5 \, a^{4} b^{3} + 10 \, a^{3} b^{4} + 10 \, a^{2} b^{5} + 5 \, a b^{6} + b^{7}\right )} \tan \left (f x + e\right )^{9} + 2 \, {\left (a^{6} b + 6 \, a^{5} b^{2} + 15 \, a^{4} b^{3} + 20 \, a^{3} b^{4} + 15 \, a^{2} b^{5} + 6 \, a b^{6} + b^{7}\right )} \tan \left (f x + e\right )^{7} + {\left (a^{7} + 7 \, a^{6} b + 21 \, a^{5} b^{2} + 35 \, a^{4} b^{3} + 35 \, a^{3} b^{4} + 21 \, a^{2} b^{5} + 7 \, a b^{6} + b^{7}\right )} \tan \left (f x + e\right )^{5}}}{120 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")

[Out]

-1/120*(15*(15*a^2*b - 40*a*b^2 + 8*b^3)*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/((a^5 + 5*a^4*b + 10*a^3*b^2 +
10*a^2*b^3 + 5*a*b^4 + b^5)*sqrt((a + b)*b)) + (15*(15*a^2*b^2 - 40*a*b^3 + 8*b^4)*tan(f*x + e)^8 + 25*(15*a^
3*b - 25*a^2*b^2 - 32*a*b^3 + 8*b^4)*tan(f*x + e)^6 + 8*(15*a^4 - 10*a^3*b - 57*a^2*b^2 - 24*a*b^3 + 8*b^4)*ta
n(f*x + e)^4 + 24*a^4 + 96*a^3*b + 144*a^2*b^2 + 96*a*b^3 + 24*b^4 + 8*(10*a^4 + 31*a^3*b + 33*a^2*b^2 + 13*a*
b^3 + b^4)*tan(f*x + e)^2)/((a^5*b^2 + 5*a^4*b^3 + 10*a^3*b^4 + 10*a^2*b^5 + 5*a*b^6 + b^7)*tan(f*x + e)^9 + 2
*(a^6*b + 6*a^5*b^2 + 15*a^4*b^3 + 20*a^3*b^4 + 15*a^2*b^5 + 6*a*b^6 + b^7)*tan(f*x + e)^7 + (a^7 + 7*a^6*b +
21*a^5*b^2 + 35*a^4*b^3 + 35*a^3*b^4 + 21*a^2*b^5 + 7*a*b^6 + b^7)*tan(f*x + e)^5))/f

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mupad [B]  time = 7.49, size = 267, normalized size = 1.10 \[ -\frac {\frac {1}{5\,\left (a+b\right )}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (10\,a+b\right )}{15\,{\left (a+b\right )}^2}+\frac {5\,{\mathrm {tan}\left (e+f\,x\right )}^6\,\left (15\,a^2\,b-40\,a\,b^2+8\,b^3\right )}{24\,{\left (a+b\right )}^4}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (15\,a^2-40\,a\,b+8\,b^2\right )}{15\,{\left (a+b\right )}^3}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^8\,\left (15\,a^2\,b^2-40\,a\,b^3+8\,b^4\right )}{8\,{\left (a+b\right )}^5}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^5\,\left (a^2+2\,a\,b+b^2\right )+{\mathrm {tan}\left (e+f\,x\right )}^7\,\left (2\,b^2+2\,a\,b\right )+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^9\right )}-\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )\,\left (a^5+5\,a^4\,b+10\,a^3\,b^2+10\,a^2\,b^3+5\,a\,b^4+b^5\right )}{{\left (a+b\right )}^{11/2}}\right )\,\left (15\,a^2-40\,a\,b+8\,b^2\right )}{8\,f\,{\left (a+b\right )}^{11/2}} \]

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)^3),x)

[Out]

- (1/(5*(a + b)) + (tan(e + f*x)^2*(10*a + b))/(15*(a + b)^2) + (5*tan(e + f*x)^6*(15*a^2*b - 40*a*b^2 + 8*b^3
))/(24*(a + b)^4) + (tan(e + f*x)^4*(15*a^2 - 40*a*b + 8*b^2))/(15*(a + b)^3) + (tan(e + f*x)^8*(8*b^4 - 40*a*
b^3 + 15*a^2*b^2))/(8*(a + b)^5))/(f*(tan(e + f*x)^5*(2*a*b + a^2 + b^2) + tan(e + f*x)^7*(2*a*b + 2*b^2) + b^
2*tan(e + f*x)^9)) - (b^(1/2)*atan((b^(1/2)*tan(e + f*x)*(5*a*b^4 + 5*a^4*b + a^5 + b^5 + 10*a^2*b^3 + 10*a^3*
b^2))/(a + b)^(11/2))*(15*a^2 - 40*a*b + 8*b^2))/(8*f*(a + b)^(11/2))

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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)**3,x)

[Out]

Timed out

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