\(\int \frac {\csc ^6(e+f x)}{(a+b \sec ^2(e+f x))^{3/2}} \, dx\) [118]

Optimal result

Integrand size = 25, antiderivative size = 183 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\left (15 a^2-10 a b-b^2\right ) \cot (e+f x)}{15 (a+b)^3 f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {2 (5 a+2 b) \cot ^3(e+f x)}{15 (a+b)^2 f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {\cot ^5(e+f x)}{5 (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {2 b \left (15 a^2-10 a b-b^2\right ) \tan (e+f x)}{15 (a+b)^4 f \sqrt {a+b+b \tan ^2(e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4217, 473, 464, 277, 197} \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {2 b \left (15 a^2-10 a b-b^2\right ) \tan (e+f x)}{15 f (a+b)^4 \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\left (15 a^2-10 a b-b^2\right ) \cot (e+f x)}{15 f (a+b)^3 \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\cot ^5(e+f x)}{5 f (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {2 (5 a+2 b) \cot ^3(e+f x)}{15 f (a+b)^2 \sqrt {a+b \tan ^2(e+f x)+b}} \]

[In]

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

[Out]

-1/15*((15*a^2 - 10*a*b - b^2)*Cot[e + f*x])/((a + b)^3*f*Sqrt[a + b + b*Tan[e + f*x]^2]) - (2*(5*a + 2*b)*Cot
[e + f*x]^3)/(15*(a + b)^2*f*Sqrt[a + b + b*Tan[e + f*x]^2]) - Cot[e + f*x]^5/(5*(a + b)*f*Sqrt[a + b + b*Tan[
e + f*x]^2]) - (2*b*(15*a^2 - 10*a*b - b^2)*Tan[e + f*x])/(15*(a + b)^4*f*Sqrt[a + b + b*Tan[e + f*x]^2])

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 277

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^n)^(p + 1)/(a*(m + 1))), x]
- Dist[b*((m + n*(p + 1) + 1)/(a*(m + 1))), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 464

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

Rule 473

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 4217

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

Mathematica [A] (verified)

Time = 0.74 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.69 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {(a+2 b+a \cos (2 (e+f x))) \left (-8 a^2 (a-5 b)+4 a \left (a^2-4 a b-5 b^2\right ) \csc ^2(e+f x)+(a-5 b) (a+b)^2 \csc ^4(e+f x)+3 (a+b)^3 \csc ^6(e+f x)\right ) \sec ^2(e+f x) \tan (e+f x)}{30 (a+b)^4 f \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]

[In]

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

[Out]

-1/30*((a + 2*b + a*Cos[2*(e + f*x)])*(-8*a^2*(a - 5*b) + 4*a*(a^2 - 4*a*b - 5*b^2)*Csc[e + f*x]^2 + (a - 5*b)
*(a + b)^2*Csc[e + f*x]^4 + 3*(a + b)^3*Csc[e + f*x]^6)*Sec[e + f*x]^2*Tan[e + f*x])/((a + b)^4*f*(a + b*Sec[e
+ f*x]^2)^(3/2))

Maple [A] (verified)

Time = 7.08 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.19

method	result	size
default	\(-\frac {\left (b +a \cos \left (f x +e \right )^{2}\right ) \left (8 a^{3} \cos \left (f x +e \right )^{6}-40 \cos \left (f x +e \right )^{6} a^{2} b -20 \cos \left (f x +e \right )^{4} a^{3}+104 \cos \left (f x +e \right )^{4} a^{2} b -20 \cos \left (f x +e \right )^{4} a \,b^{2}+15 \cos \left (f x +e \right )^{2} a^{3}-85 \cos \left (f x +e \right )^{2} a^{2} b +49 \cos \left (f x +e \right )^{2} a \,b^{2}+5 \cos \left (f x +e \right )^{2} b^{3}+30 a^{2} b -20 a \,b^{2}-2 b^{3}\right ) \sec \left (f x +e \right )^{3} \csc \left (f x +e \right )^{5}}{15 f \left (a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right ) \left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}\)	\(217\)

[In]

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

[Out]

-1/15/f/(a^4+4*a^3*b+6*a^2*b^2+4*a*b^3+b^4)*(b+a*cos(f*x+e)^2)*(8*a^3*cos(f*x+e)^6-40*cos(f*x+e)^6*a^2*b-20*co
s(f*x+e)^4*a^3+104*cos(f*x+e)^4*a^2*b-20*cos(f*x+e)^4*a*b^2+15*cos(f*x+e)^2*a^3-85*cos(f*x+e)^2*a^2*b+49*cos(f
*x+e)^2*a*b^2+5*cos(f*x+e)^2*b^3+30*a^2*b-20*a*b^2-2*b^3)/(a+b*sec(f*x+e)^2)^(3/2)*sec(f*x+e)^3*csc(f*x+e)^5

Fricas [A] (verification not implemented)

none

Time = 2.34 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.72 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {{\left (8 \, {\left (a^{3} - 5 \, a^{2} b\right )} \cos \left (f x + e\right )^{7} - 4 \, {\left (5 \, a^{3} - 26 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (f x + e\right )^{5} + {\left (15 \, a^{3} - 85 \, a^{2} b + 49 \, a b^{2} + 5 \, b^{3}\right )} \cos \left (f x + e\right )^{3} + 2 \, {\left (15 \, a^{2} b - 10 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \, {\left ({\left (a^{5} + 4 \, a^{4} b + 6 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + a b^{4}\right )} f \cos \left (f x + e\right )^{6} - {\left (2 \, a^{5} + 7 \, a^{4} b + 8 \, a^{3} b^{2} + 2 \, a^{2} b^{3} - 2 \, a b^{4} - b^{5}\right )} f \cos \left (f x + e\right )^{4} + {\left (a^{5} + 2 \, a^{4} b - 2 \, a^{3} b^{2} - 8 \, a^{2} b^{3} - 7 \, a b^{4} - 2 \, b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} b + 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} + 4 \, a b^{4} + b^{5}\right )} f\right )} \sin \left (f x + e\right )} \]

[In]

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

[Out]

-1/15*(8*(a^3 - 5*a^2*b)*cos(f*x + e)^7 - 4*(5*a^3 - 26*a^2*b + 5*a*b^2)*cos(f*x + e)^5 + (15*a^3 - 85*a^2*b +
49*a*b^2 + 5*b^3)*cos(f*x + e)^3 + 2*(15*a^2*b - 10*a*b^2 - b^3)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/co
s(f*x + e)^2)/(((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))

Sympy [F]

\[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\csc ^{6}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.54 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\frac {30 \, b \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{2}} - \frac {80 \, b^{2} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{3}} + \frac {48 \, b^{3} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{4}} + \frac {15}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \tan \left (f x + e\right )} - \frac {40 \, b}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{2} \tan \left (f x + e\right )} + \frac {24 \, b^{2}}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{3} \tan \left (f x + e\right )} + \frac {10}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \tan \left (f x + e\right )^{3}} - \frac {6 \, b}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{2} \tan \left (f x + e\right )^{3}} + \frac {3}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \tan \left (f x + e\right )^{5}}}{15 \, f} \]

[In]

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

[Out]

-1/15*(30*b*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)^2) - 80*b^2*tan(f*x + e)/(sqrt(b*tan(f*x + e)
^2 + a + b)*(a + b)^3) + 48*b^3*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)^4) + 15/(sqrt(b*tan(f*x +
e)^2 + a + b)*(a + b)*tan(f*x + e)) - 40*b/(sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)^2*tan(f*x + e)) + 24*b^2/(
sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)^3*tan(f*x + e)) + 10/(sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)*tan(f*x +
e)^3) - 6*b/(sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)^2*tan(f*x + e)^3) + 3/(sqrt(b*tan(f*x + e)^2 + a + b)*(a +
b)*tan(f*x + e)^5))/f

Giac [F]

\[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\csc \left (f x + e\right )^{6}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\text {Hanged} \]

[In]

int(1/(sin(e + f*x)^6*(a + b/cos(e + f*x)^2)^(3/2)),x)

[Out]

\text{Hanged}