\(\int \frac {\csc ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx\) [39]

Optimal result

Integrand size = 23, antiderivative size = 76 \[ \int \frac {\csc ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {a \sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{(a+b)^{5/2} f}-\frac {a \cot (e+f x)}{(a+b)^2 f}-\frac {\cot ^3(e+f x)}{3 (a+b) f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {4217, 464, 331, 211} \[ \int \frac {\csc ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {a \sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{f (a+b)^{5/2}}-\frac {\cot ^3(e+f x)}{3 f (a+b)}-\frac {a \cot (e+f x)}{f (a+b)^2} \]

[In]

Int[Csc[e + f*x]^4/(a + b*Sec[e + f*x]^2),x]

[Out]

-((a*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/((a + b)^(5/2)*f)) - (a*Cot[e + f*x])/((a + b)^2*f) -
Cot[e + f*x]^3/(3*(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 331

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] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 1.86 (sec) , antiderivative size = 226, normalized size of antiderivative = 2.97 \[ \int \frac {\csc ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^2(e+f x) \left (3 a b \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (\cos (2 e)-i \sin (2 e))+\frac {1}{4} \sqrt {a+b} \csc (e) \csc ^3(e+f x) \sqrt {b (\cos (e)-i \sin (e))^4} (6 a \sin (f x)-3 b \sin (2 e+f x)+(-2 a+b) \sin (2 e+3 f x))\right )}{6 (a+b)^{5/2} f \left (a+b \sec ^2(e+f x)\right ) \sqrt {b (\cos (e)-i \sin (e))^4}} \]

[In]

Integrate[Csc[e + f*x]^4/(a + b*Sec[e + f*x]^2),x]

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^2*(3*a*b*ArcTan[(Sec[f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b)*Si
n[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]^3*Sqrt[b*(Cos[e] - I*Sin[e])^4]*(6*a*Sin[f*x] - 3*b*Sin[2*e + f*x] + (-2*a + b)*Sin[2
*e + 3*f*x]))/4))/(6*(a + b)^(5/2)*f*(a + b*Sec[e + f*x]^2)*Sqrt[b*(Cos[e] - I*Sin[e])^4])

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.91

method	result	size
derivativedivides	\(\frac {-\frac {a b \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{\left (a +b \right )^{2} \sqrt {\left (a +b \right ) b}}-\frac {1}{3 \left (a +b \right ) \tan \left (f x +e \right )^{3}}-\frac {a}{\left (a +b \right )^{2} \tan \left (f x +e \right )}}{f}\)	\(69\)
default	\(\frac {-\frac {a b \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{\left (a +b \right )^{2} \sqrt {\left (a +b \right ) b}}-\frac {1}{3 \left (a +b \right ) \tan \left (f x +e \right )^{3}}-\frac {a}{\left (a +b \right )^{2} \tan \left (f x +e \right )}}{f}\)	\(69\)
risch	\(\frac {2 i \left (3 b \,{\mathrm e}^{4 i \left (f x +e \right )}+6 a \,{\mathrm e}^{2 i \left (f x +e \right )}-2 a +b \right )}{3 f \left (a +b \right )^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{3}}+\frac {\sqrt {-\left (a +b \right ) b}\, a \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{2 \left (a +b \right )^{3} f}-\frac {\sqrt {-\left (a +b \right ) b}\, a \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{2 \left (a +b \right )^{3} f}\)	\(158\)

[In]

int(csc(f*x+e)^4/(a+b*sec(f*x+e)^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 155 vs. \(2 (66) = 132\).

Time = 0.30 (sec) , antiderivative size = 397, normalized size of antiderivative = 5.22 \[ \int \frac {\csc ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\left [-\frac {4 \, {\left (2 \, a - b\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left (a \cos \left (f x + e\right )^{2} - a\right )} \sqrt {-\frac {b}{a + b}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a + b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) \sin \left (f x + e\right ) - 12 \, a \cos \left (f x + e\right )}{12 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{2} + 2 \, a b + b^{2}\right )} f\right )} \sin \left (f x + e\right )}, -\frac {2 \, {\left (2 \, a - b\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left (a \cos \left (f x + e\right )^{2} - a\right )} \sqrt {\frac {b}{a + b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a + b}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 6 \, a \cos \left (f x + e\right )}{6 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{2} + 2 \, a b + b^{2}\right )} f\right )} \sin \left (f x + e\right )}\right ] \]

[In]

integrate(csc(f*x+e)^4/(a+b*sec(f*x+e)^2),x, algorithm="fricas")

[Out]

[-1/12*(4*(2*a - b)*cos(f*x + e)^3 - 3*(a*cos(f*x + e)^2 - a)*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) - 1
2*a*cos(f*x + e))/(((a^2 + 2*a*b + b^2)*f*cos(f*x + e)^2 - (a^2 + 2*a*b + b^2)*f)*sin(f*x + e)), -1/6*(2*(2*a
- b)*cos(f*x + e)^3 - 3*(a*cos(f*x + e)^2 - a)*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) - 6*a*cos(f*x + e))/(((a^2 + 2*a*b + b^2)*f*cos(f*x + e
)^2 - (a^2 + 2*a*b + b^2)*f)*sin(f*x + e))]

Sympy [F]

\[ \int \frac {\csc ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\int \frac {\csc ^{4}{\left (e + f x \right )}}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \]

[In]

integrate(csc(f*x+e)**4/(a+b*sec(f*x+e)**2),x)

[Out]

Integral(csc(e + f*x)**4/(a + b*sec(e + f*x)**2), x)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.08 \[ \int \frac {\csc ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {3 \, a b \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {{\left (a + b\right )} b}} + \frac {3 \, a \tan \left (f x + e\right )^{2} + a + b}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{3}}}{3 \, f} \]

[In]

integrate(csc(f*x+e)^4/(a+b*sec(f*x+e)^2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.36 \[ \int \frac {\csc ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {3 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} a b}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a b + b^{2}}} + \frac {3 \, a \tan \left (f x + e\right )^{2} + a + b}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{3}}}{3 \, f} \]

[In]

integrate(csc(f*x+e)^4/(a+b*sec(f*x+e)^2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 18.33 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.05 \[ \int \frac {\csc ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {1}{3\,\left (a+b\right )}+\frac {a\,{\mathrm {tan}\left (e+f\,x\right )}^2}{{\left (a+b\right )}^2}}{f\,{\mathrm {tan}\left (e+f\,x\right )}^3}-\frac {a\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )\,\left (a^2+2\,a\,b+b^2\right )}{{\left (a+b\right )}^{5/2}}\right )}{f\,{\left (a+b\right )}^{5/2}} \]

[In]

int(1/(sin(e + f*x)^4*(a + b/cos(e + f*x)^2)),x)

[Out]

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