\(\int \frac {\csc ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\) [105]

Optimal result

Integrand size = 25, antiderivative size = 132 \[ \int \frac {\csc ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {\left (15 a^2+10 a b+3 b^2\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b)^3 f}-\frac {2 (5 a+3 b) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b)^2 f}-\frac {\cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 (a+b) f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {4217, 473, 464, 270} \[ \int \frac {\csc ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {\left (15 a^2+10 a b+3 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{15 f (a+b)^3}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{5 f (a+b)}-\frac {2 (5 a+3 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{15 f (a+b)^2} \]

[In]

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

[Out]

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

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*
c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(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 \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {\cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 (a+b) f}+\frac {\text {Subst}\left (\int \frac {2 (5 a+3 b)+5 (a+b) x^2}{x^4 \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{5 (a+b) f} \\ & = -\frac {2 (5 a+3 b) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b)^2 f}-\frac {\cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 (a+b) f}+\frac {\left (15 a^2+10 a b+3 b^2\right ) \text {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{15 (a+b)^2 f} \\ & = -\frac {\left (15 a^2+10 a b+3 b^2\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b)^3 f}-\frac {2 (5 a+3 b) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b)^2 f}-\frac {\cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 (a+b) f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.76 \[ \int \frac {\csc ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {(a+2 b+a \cos (2 (e+f x))) \left (8 a^2+8 a b+3 b^2-2 a (3 a+b) \cos (2 (e+f x))+a^2 \cos (4 (e+f x))\right ) \csc ^5(e+f x) \sec (e+f x)}{30 (a+b)^3 f \sqrt {a+b \sec ^2(e+f x)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 5.02 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.91

method	result	size
default	\(-\frac {\left (b +a \cos \left (f x +e \right )^{2}\right ) \left (8 \cos \left (f x +e \right )^{4} a^{2}-20 \cos \left (f x +e \right )^{2} a^{2}-4 \cos \left (f x +e \right )^{2} a b +15 a^{2}+10 a b +3 b^{2}\right ) \sec \left (f x +e \right ) \csc \left (f x +e \right )^{5}}{15 f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \sqrt {a +b \sec \left (f x +e \right )^{2}}}\)	\(120\)

[In]

int(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/15/f/(a^3+3*a^2*b+3*a*b^2+b^3)*(b+a*cos(f*x+e)^2)*(8*cos(f*x+e)^4*a^2-20*cos(f*x+e)^2*a^2-4*cos(f*x+e)^2*a*
b+15*a^2+10*a*b+3*b^2)/(a+b*sec(f*x+e)^2)^(1/2)*sec(f*x+e)*csc(f*x+e)^5

Fricas [A] (verification not implemented)

none

Time = 0.59 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.30 \[ \int \frac {\csc ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {{\left (8 \, a^{2} \cos \left (f x + e\right )^{5} - 4 \, {\left (5 \, a^{2} + a b\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{2} + 10 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \, {\left ({\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f\right )} \sin \left (f x + e\right )} \]

[In]

integrate(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

-1/15*(8*a^2*cos(f*x + e)^5 - 4*(5*a^2 + a*b)*cos(f*x + e)^3 + (15*a^2 + 10*a*b + 3*b^2)*cos(f*x + e))*sqrt((a
*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f*cos(f*x + e)^4 - 2*(a^3 + 3*a^2*b + 3
*a*b^2 + b^3)*f*cos(f*x + e)^2 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f)*sin(f*x + e))

Sympy [F]

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

[In]

integrate(csc(f*x+e)**6/(a+b*sec(f*x+e)**2)**(1/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.45 \[ \int \frac {\csc ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {\frac {15 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b}}{{\left (a + b\right )} \tan \left (f x + e\right )} - \frac {20 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} b}{{\left (a + b\right )}^{2} \tan \left (f x + e\right )} + \frac {8 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} b^{2}}{{\left (a + b\right )}^{3} \tan \left (f x + e\right )} + \frac {10 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b}}{{\left (a + b\right )} \tan \left (f x + e\right )^{3}} - \frac {4 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} b}{{\left (a + b\right )}^{2} \tan \left (f x + e\right )^{3}} + \frac {3 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b}}{{\left (a + b\right )} \tan \left (f x + e\right )^{5}}}{15 \, f} \]

[In]

integrate(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

-1/15*(15*sqrt(b*tan(f*x + e)^2 + a + b)/((a + b)*tan(f*x + e)) - 20*sqrt(b*tan(f*x + e)^2 + a + b)*b/((a + b)
^2*tan(f*x + e)) + 8*sqrt(b*tan(f*x + e)^2 + a + b)*b^2/((a + b)^3*tan(f*x + e)) + 10*sqrt(b*tan(f*x + e)^2 +
a + b)/((a + b)*tan(f*x + e)^3) - 4*sqrt(b*tan(f*x + e)^2 + a + b)*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]

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

[In]

integrate(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 30.64 (sec) , antiderivative size = 723, normalized size of antiderivative = 5.48 \[ \int \frac {\csc ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

(((32*a + 16*b)/(5*f*(6*a + 6*b)*(a*1i + b*1i)) + (32*a + 80*b)/(5*f*(6*a + 6*b)*(a*1i + b*1i)))*(a + b/(exp(-
e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2)^2)^(1/2)*(2*exp(e*2i + f*x*2i) + exp(e*4i + f*x*4i) + 1))/((exp(e*2i
+ f*x*2i) - 1)^3*(exp(e*2i + f*x*2i) + 1)) - ((a + b/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2)^2)^(1/2)
*((a*(2*a + b)*32i)/(15*f*(a*1i + b*1i)^2*(8*a + 8*b)) + (a*(2*a + 3*b)*32i)/(15*f*(a*1i + b*1i)^2*(8*a + 8*b)
))*(2*exp(e*2i + f*x*2i) + exp(e*4i + f*x*4i) + 1))/((exp(e*2i + f*x*2i) - 1)^2*(exp(e*2i + f*x*2i) + 1)) + ((
a + b/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2)^2)^(1/2)*(2*exp(e*2i + f*x*2i) + exp(e*4i + f*x*4i) + 1)
*((96*a + 32*b)/(5*f*(a*1i + b*1i)*(16*a + 16*b)) + (160*a + 160*b)/(5*f*(a*1i + b*1i)*(16*a + 16*b)) + (256*a
+ 320*b)/(5*f*(a*1i + b*1i)*(16*a + 16*b))))/((exp(e*2i + f*x*2i) - 1)^4*(exp(e*2i + f*x*2i) + 1)) - ((a + b/
(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2)^2)^(1/2)*(2*exp(e*2i + f*x*2i) + exp(e*4i + f*x*4i) + 1)*32i)/
(f*(exp(e*2i + f*x*2i) - 1)^5*(exp(e*2i + f*x*2i) + 1)*(10*a + 10*b)) - (8*a^2*(a + b/(exp(- e*1i - f*x*1i)/2
+ exp(e*1i + f*x*1i)/2)^2)^(1/2)*(2*exp(e*2i + f*x*2i) + exp(e*4i + f*x*4i) + 1))/(15*f*(exp(e*2i + f*x*2i) -
1)*(exp(e*2i + f*x*2i) + 1)*(a*1i + b*1i)^3)