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]

-(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))

________________________________________________________________________________________

Rubi [A]  time = 0.369588, antiderivative size = 242, normalized size of antiderivative = 1., 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 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 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 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 )^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*}

Mathematica [C]  time = 6.06661, 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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Maple [A]  time = 0.136, size = 411, normalized size = 1.7 \begin{align*} -{\frac{1}{5\,f \left ( a+b \right ) ^{3} \left ( \tan \left ( fx+e \right ) \right ) ^{5}}}-{\frac{2\,a}{3\,f \left ( a+b \right ) ^{4} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}}+{\frac{b}{3\,f \left ( a+b \right ) ^{4} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}}-{\frac{{a}^{2}}{f \left ( a+b \right ) ^{5}\tan \left ( fx+e \right ) }}+4\,{\frac{ab}{f \left ( a+b \right ) ^{5}\tan \left ( fx+e \right ) }}-{\frac{{b}^{2}}{f \left ( a+b \right ) ^{5}\tan \left ( fx+e \right ) }}-{\frac{7\,{b}^{2} \left ( \tan \left ( fx+e \right ) \right ) ^{3}{a}^{2}}{8\,f \left ( a+b \right ) ^{5} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{{b}^{3} \left ( \tan \left ( fx+e \right ) \right ) ^{3}a}{f \left ( a+b \right ) ^{5} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{9\,b\tan \left ( fx+e \right ){a}^{3}}{8\,f \left ( a+b \right ) ^{5} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{{b}^{2}\tan \left ( fx+e \right ){a}^{2}}{8\,f \left ( a+b \right ) ^{5} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{{b}^{3}\tan \left ( fx+e \right ) a}{f \left ( a+b \right ) ^{5} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{15\,{a}^{2}b}{8\,f \left ( a+b \right ) ^{5}}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}}+5\,{\frac{a{b}^{2}}{f \left ( a+b \right ) ^{5}\sqrt{ \left ( a+b \right ) b}}\arctan \left ({\frac{b\tan \left ( fx+e \right ) }{\sqrt{ \left ( a+b \right ) b}}} \right ) }-{\frac{{b}^{3}}{f \left ( a+b \right ) ^{5}}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \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)^3,x)

[Out]

-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^2-7/8/f/(a+b)^5*b^2/(a+b+b*tan(f*x+e)^2)^2*tan(f*x+e)^3
*a^2+1/f/(a+b)^5*b^3/(a+b+b*tan(f*x+e)^2)^2*tan(f*x+e)^3*a-9/8/f/(a+b)^5*b/(a+b+b*tan(f*x+e)^2)^2*tan(f*x+e)*a
^3-1/8/f/(a+b)^5*b^2/(a+b+b*tan(f*x+e)^2)^2*tan(f*x+e)*a^2+1/f/(a+b)^5*b^3/(a+b+b*tan(f*x+e)^2)^2*tan(f*x+e)*a
-15/8/f/(a+b)^5*b/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))*a^2+5/f/(a+b)^5*b^2/((a+b)*b)^(1/2)*arc
tan(tan(f*x+e)*b/((a+b)*b)^(1/2))*a-1/f/(a+b)^5*b^3/((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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.898391, size = 3312, normalized size = 13.69 \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)^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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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)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.29214, size = 516, normalized size = 2.13 \begin{align*} -\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}\left (b\right ) + \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} \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)^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