\(\int \frac {\csc ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx\) [40]

Optimal result

Integrand size = 23, antiderivative size = 105 \[ \int \frac {\csc ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {a^2 \sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{(a+b)^{7/2} f}-\frac {a^2 \cot (e+f x)}{(a+b)^3 f}-\frac {(2 a+b) \cot ^3(e+f x)}{3 (a+b)^2 f}-\frac {\cot ^5(e+f x)}{5 (a+b) f} \]

[Out]

-a^2*cot(f*x+e)/(a+b)^3/f-1/3*(2*a+b)*cot(f*x+e)^3/(a+b)^2/f-1/5*cot(f*x+e)^5/(a+b)/f-a^2*arctan(b^(1/2)*tan(f
*x+e)/(a+b)^(1/2))*b^(1/2)/(a+b)^(7/2)/f

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4217, 472, 211} \[ \int \frac {\csc ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {a^2 \sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{f (a+b)^{7/2}}-\frac {a^2 \cot (e+f x)}{f (a+b)^3}-\frac {\cot ^5(e+f x)}{5 f (a+b)}-\frac {(2 a+b) \cot ^3(e+f x)}{3 f (a+b)^2} \]

[In]

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

[Out]

-((a^2*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/((a + b)^(7/2)*f)) - (a^2*Cot[e + f*x])/((a + b)^3*
f) - ((2*a + b)*Cot[e + f*x]^3)/(3*(a + b)^2*f) - Cot[e + f*x]^5/(5*(a + b)*f)

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 472

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[(e*x)^m*((a + b*x^n)^p/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

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 )} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \left (\frac {1}{(a+b) x^6}+\frac {2 a+b}{(a+b)^2 x^4}+\frac {a^2}{(a+b)^3 x^2}-\frac {a^2 b}{(a+b)^3 \left (a+b+b x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {a^2 \cot (e+f x)}{(a+b)^3 f}-\frac {(2 a+b) \cot ^3(e+f x)}{3 (a+b)^2 f}-\frac {\cot ^5(e+f x)}{5 (a+b) f}-\frac {\left (a^2 b\right ) \text {Subst}\left (\int \frac {1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{(a+b)^3 f} \\ & = -\frac {a^2 \sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{(a+b)^{7/2} f}-\frac {a^2 \cot (e+f x)}{(a+b)^3 f}-\frac {(2 a+b) \cot ^3(e+f x)}{3 (a+b)^2 f}-\frac {\cot ^5(e+f x)}{5 (a+b) f} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 1.78 (sec) , antiderivative size = 318, normalized size of antiderivative = 3.03 \[ \int \frac {\csc ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^2(e+f x) \left (240 a^2 b \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (\cos (2 e)-i \sin (2 e))+\sqrt {a+b} \csc (e) \csc ^5(e+f x) \sqrt {b (\cos (e)-i \sin (e))^4} \left (10 \left (8 a^2+b^2\right ) \sin (f x)-30 b (3 a+b) \sin (2 e+f x)-40 a^2 \sin (2 e+3 f x)+30 a b \sin (2 e+3 f x)+10 b^2 \sin (2 e+3 f x)+15 a b \sin (4 e+3 f x)+8 a^2 \sin (4 e+5 f x)-9 a b \sin (4 e+5 f x)-2 b^2 \sin (4 e+5 f x)\right )\right )}{480 (a+b)^{7/2} f \left (a+b \sec ^2(e+f x)\right ) \sqrt {b (\cos (e)-i \sin (e))^4}} \]

[In]

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

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^2*(240*a^2*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])]*(Cos[2*e] - I*Sin[2*e]) + Sqrt
[a + b]*Csc[e]*Csc[e + f*x]^5*Sqrt[b*(Cos[e] - I*Sin[e])^4]*(10*(8*a^2 + b^2)*Sin[f*x] - 30*b*(3*a + b)*Sin[2*
e + f*x] - 40*a^2*Sin[2*e + 3*f*x] + 30*a*b*Sin[2*e + 3*f*x] + 10*b^2*Sin[2*e + 3*f*x] + 15*a*b*Sin[4*e + 3*f*
x] + 8*a^2*Sin[4*e + 5*f*x] - 9*a*b*Sin[4*e + 5*f*x] - 2*b^2*Sin[4*e + 5*f*x])))/(480*(a + b)^(7/2)*f*(a + b*S
ec[e + f*x]^2)*Sqrt[b*(Cos[e] - I*Sin[e])^4])

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.89

method	result	size
derivativedivides	\(\frac {-\frac {a^{2} b \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{\left (a +b \right )^{3} \sqrt {\left (a +b \right ) b}}-\frac {1}{5 \left (a +b \right ) \tan \left (f x +e \right )^{5}}-\frac {a^{2}}{\left (a +b \right )^{3} \tan \left (f x +e \right )}-\frac {2 a +b}{3 \left (a +b \right )^{2} \tan \left (f x +e \right )^{3}}}{f}\)	\(93\)
default	\(\frac {-\frac {a^{2} b \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{\left (a +b \right )^{3} \sqrt {\left (a +b \right ) b}}-\frac {1}{5 \left (a +b \right ) \tan \left (f x +e \right )^{5}}-\frac {a^{2}}{\left (a +b \right )^{3} \tan \left (f x +e \right )}-\frac {2 a +b}{3 \left (a +b \right )^{2} \tan \left (f x +e \right )^{3}}}{f}\)	\(93\)
risch	\(\frac {2 i \left (15 a b \,{\mathrm e}^{8 i \left (f x +e \right )}-90 a b \,{\mathrm e}^{6 i \left (f x +e \right )}-30 b^{2} {\mathrm e}^{6 i \left (f x +e \right )}-80 a^{2} {\mathrm e}^{4 i \left (f x +e \right )}-10 b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+40 a^{2} {\mathrm e}^{2 i \left (f x +e \right )}-30 a b \,{\mathrm e}^{2 i \left (f x +e \right )}-10 b^{2} {\mathrm e}^{2 i \left (f x +e \right )}-8 a^{2}+9 a b +2 b^{2}\right )}{15 f \left (a +b \right )^{3} \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{5}}-\frac {\sqrt {-\left (a +b \right ) b}\, a^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{2 \left (a +b \right )^{4} f}+\frac {\sqrt {-\left (a +b \right ) b}\, a^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{2 \left (a +b \right )^{4} f}\)	\(257\)

[In]

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

[Out]

1/f*(-a^2*b/(a+b)^3/((a+b)*b)^(1/2)*arctan(b*tan(f*x+e)/((a+b)*b)^(1/2))-1/5/(a+b)/tan(f*x+e)^5-a^2/(a+b)^3/ta
n(f*x+e)-1/3*(2*a+b)/(a+b)^2/tan(f*x+e)^3)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (93) = 186\).

Time = 0.29 (sec) , antiderivative size = 587, normalized size of antiderivative = 5.59 \[ \int \frac {\csc ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\left [-\frac {4 \, {\left (8 \, a^{2} - 9 \, a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - 20 \, {\left (4 \, a^{2} - 3 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left (a^{2} \cos \left (f x + e\right )^{4} - 2 \, a^{2} \cos \left (f x + e\right )^{2} + a^{2}\right )} \sqrt {-\frac {b}{a + b}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a + b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) \sin \left (f x + e\right ) + 60 \, a^{2} \cos \left (f x + e\right )}{60 \, {\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 )}, -\frac {2 \, {\left (8 \, a^{2} - 9 \, a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - 10 \, {\left (4 \, a^{2} - 3 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left (a^{2} \cos \left (f x + e\right )^{4} - 2 \, a^{2} \cos \left (f x + e\right )^{2} + a^{2}\right )} \sqrt {\frac {b}{a + b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a + b}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 30 \, a^{2} \cos \left (f x + e\right )}{30 \, {\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 )}\right ] \]

[In]

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

[Out]

[-1/60*(4*(8*a^2 - 9*a*b - 2*b^2)*cos(f*x + e)^5 - 20*(4*a^2 - 3*a*b - b^2)*cos(f*x + e)^3 - 15*(a^2*cos(f*x +
e)^4 - 2*a^2*cos(f*x + e)^2 + a^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*a^2*cos(f*x + e))/(((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)), -1/30*(2*(8*a^2 - 9*a*b - 2*b^2)*cos(f*x + e)^5 - 10*(4*a^2 - 3*a*b -
b^2)*cos(f*x + e)^3 - 15*(a^2*cos(f*x + e)^4 - 2*a^2*cos(f*x + e)^2 + a^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*a^2*cos(f*x + e))/(((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]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.30 \[ \int \frac {\csc ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {15 \, a^{2} b \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {{\left (a + b\right )} b}} + \frac {15 \, a^{2} \tan \left (f x + e\right )^{4} + 5 \, {\left (2 \, a^{2} + 3 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \tan \left (f x + e\right )^{5}}}{15 \, f} \]

[In]

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

[Out]

-1/15*(15*a^2*b*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sqrt((a + b)*b)) + (15
*a^2*tan(f*x + e)^4 + 5*(2*a^2 + 3*a*b + b^2)*tan(f*x + e)^2 + 3*a^2 + 6*a*b + 3*b^2)/((a^3 + 3*a^2*b + 3*a*b^
2 + b^3)*tan(f*x + e)^5))/f

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.65 \[ \int \frac {\csc ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {15 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} a^{2} b}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {a b + b^{2}}} + \frac {15 \, a^{2} \tan \left (f x + e\right )^{4} + 10 \, a^{2} \tan \left (f x + e\right )^{2} + 15 \, a b \tan \left (f x + e\right )^{2} + 5 \, b^{2} \tan \left (f x + e\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \tan \left (f x + e\right )^{5}}}{15 \, f} \]

[In]

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

[Out]

-1/15*(15*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))*a^2*b/((a^3 + 3*a^2*b
+ 3*a*b^2 + b^3)*sqrt(a*b + b^2)) + (15*a^2*tan(f*x + e)^4 + 10*a^2*tan(f*x + e)^2 + 15*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^3 + 3*a^2*b + 3*a*b^2 + b^3)*tan(f*x + e)^5))/f

Mupad [B] (verification not implemented)

Time = 19.03 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.07 \[ \int \frac {\csc ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {1}{5\,\left (a+b\right )}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (2\,a+b\right )}{3\,{\left (a+b\right )}^2}+\frac {a^2\,{\mathrm {tan}\left (e+f\,x\right )}^4}{{\left (a+b\right )}^3}}{f\,{\mathrm {tan}\left (e+f\,x\right )}^5}-\frac {a^2\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}{{\left (a+b\right )}^{7/2}}\right )}{f\,{\left (a+b\right )}^{7/2}} \]

[In]

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

[Out]

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