\(\int \frac {\csc ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx\) [60]

Optimal result

Integrand size = 23, antiderivative size = 130 \[ \int \frac {\csc ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {(a-b)^{3/2} \sqrt {b} \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{a^3 f}-\frac {\left (3 a^2-12 a b+8 b^2\right ) \text {arctanh}(\cos (e+f x))}{8 a^3 f}-\frac {(5 a-4 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f} \]

[Out]

-1/8*(3*a^2-12*a*b+8*b^2)*arctanh(cos(f*x+e))/a^3/f-1/8*(5*a-4*b)*cot(f*x+e)*csc(f*x+e)/a^2/f-1/4*cot(f*x+e)^3
*csc(f*x+e)/a/f-(a-b)^(3/2)*arctan(sec(f*x+e)*b^(1/2)/(a-b)^(1/2))*b^(1/2)/a^3/f

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3745, 481, 541, 536, 213, 211} \[ \int \frac {\csc ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {\sqrt {b} (a-b)^{3/2} \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{a^3 f}-\frac {(5 a-4 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f}-\frac {\left (3 a^2-12 a b+8 b^2\right ) \text {arctanh}(\cos (e+f x))}{8 a^3 f}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f} \]

[In]

Int[Csc[e + f*x]^5/(a + b*Tan[e + f*x]^2),x]

[Out]

-(((a - b)^(3/2)*Sqrt[b]*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/(a^3*f)) - ((3*a^2 - 12*a*b + 8*b^2)*ArcT
anh[Cos[e + f*x]])/(8*a^3*f) - ((5*a - 4*b)*Cot[e + f*x]*Csc[e + f*x])/(8*a^2*f) - (Cot[e + f*x]^3*Csc[e + f*x
])/(4*a*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 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 481

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-a)*e^(
2*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(b*n*(b*c - a*d)*(p + 1))), x] + Dist[e^
(2*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1)
+ (a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0]
&& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 536

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 541

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[(
-(b*e - a*f))*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*n*(b*c - a*d)*(p + 1))), x] + Dist[1/(a*n*(b*c - a
*d)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*
f)*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 3745

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a - b + b*ff^2*x^2)^p/x^(m
+ 1)), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^4}{\left (-1+x^2\right )^3 \left (a-b+b x^2\right )} \, dx,x,\sec (e+f x)\right )}{f} \\ & = -\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f}-\frac {\text {Subst}\left (\int \frac {-a+b+(-4 a+3 b) x^2}{\left (-1+x^2\right )^2 \left (a-b+b x^2\right )} \, dx,x,\sec (e+f x)\right )}{4 a f} \\ & = -\frac {(5 a-4 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f}-\frac {\text {Subst}\left (\int \frac {-((3 a-4 b) (a-b))+(5 a-4 b) b x^2}{\left (-1+x^2\right ) \left (a-b+b x^2\right )} \, dx,x,\sec (e+f x)\right )}{8 a^2 f} \\ & = -\frac {(5 a-4 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f}-\frac {\left ((a-b)^2 b\right ) \text {Subst}\left (\int \frac {1}{a-b+b x^2} \, dx,x,\sec (e+f x)\right )}{a^3 f}+\frac {\left (3 a^2-12 a b+8 b^2\right ) \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sec (e+f x)\right )}{8 a^3 f} \\ & = -\frac {(a-b)^{3/2} \sqrt {b} \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{a^3 f}-\frac {\left (3 a^2-12 a b+8 b^2\right ) \text {arctanh}(\cos (e+f x))}{8 a^3 f}-\frac {(5 a-4 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(326\) vs. \(2(130)=260\).

Time = 6.71 (sec) , antiderivative size = 326, normalized size of antiderivative = 2.51 \[ \int \frac {\csc ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\frac {(a-b)^{3/2} \sqrt {b} \arctan \left (\frac {\sec \left (\frac {1}{2} (e+f x)\right ) \left (\sqrt {a-b} \cos \left (\frac {1}{2} (e+f x)\right )-\sqrt {a} \sin \left (\frac {1}{2} (e+f x)\right )\right )}{\sqrt {b}}\right )}{a^3 f}+\frac {(a-b)^{3/2} \sqrt {b} \arctan \left (\frac {\sec \left (\frac {1}{2} (e+f x)\right ) \left (\sqrt {a-b} \cos \left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sin \left (\frac {1}{2} (e+f x)\right )\right )}{\sqrt {b}}\right )}{a^3 f}+\frac {(-3 a+4 b) \csc ^2\left (\frac {1}{2} (e+f x)\right )}{32 a^2 f}-\frac {\csc ^4\left (\frac {1}{2} (e+f x)\right )}{64 a f}+\frac {\left (-3 a^2+12 a b-8 b^2\right ) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )}{8 a^3 f}+\frac {\left (3 a^2-12 a b+8 b^2\right ) \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )}{8 a^3 f}+\frac {(3 a-4 b) \sec ^2\left (\frac {1}{2} (e+f x)\right )}{32 a^2 f}+\frac {\sec ^4\left (\frac {1}{2} (e+f x)\right )}{64 a f} \]

[In]

Integrate[Csc[e + f*x]^5/(a + b*Tan[e + f*x]^2),x]

[Out]

((a - b)^(3/2)*Sqrt[b]*ArcTan[(Sec[(e + f*x)/2]*(Sqrt[a - b]*Cos[(e + f*x)/2] - Sqrt[a]*Sin[(e + f*x)/2]))/Sqr
t[b]])/(a^3*f) + ((a - b)^(3/2)*Sqrt[b]*ArcTan[(Sec[(e + f*x)/2]*(Sqrt[a - b]*Cos[(e + f*x)/2] + Sqrt[a]*Sin[(
e + f*x)/2]))/Sqrt[b]])/(a^3*f) + ((-3*a + 4*b)*Csc[(e + f*x)/2]^2)/(32*a^2*f) - Csc[(e + f*x)/2]^4/(64*a*f) +
((-3*a^2 + 12*a*b - 8*b^2)*Log[Cos[(e + f*x)/2]])/(8*a^3*f) + ((3*a^2 - 12*a*b + 8*b^2)*Log[Sin[(e + f*x)/2]]
)/(8*a^3*f) + ((3*a - 4*b)*Sec[(e + f*x)/2]^2)/(32*a^2*f) + Sec[(e + f*x)/2]^4/(64*a*f)

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.42

method	result	size
derivativedivides	\(\frac {\frac {b \left (a^{2}-2 a b +b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {b \left (a -b \right )}}\right )}{a^{3} \sqrt {b \left (a -b \right )}}+\frac {1}{16 a \left (\cos \left (f x +e \right )+1\right )^{2}}-\frac {-3 a +4 b}{16 a^{2} \left (\cos \left (f x +e \right )+1\right )}+\frac {\left (-3 a^{2}+12 a b -8 b^{2}\right ) \ln \left (\cos \left (f x +e \right )+1\right )}{16 a^{3}}-\frac {1}{16 a \left (\cos \left (f x +e \right )-1\right )^{2}}-\frac {-3 a +4 b}{16 a^{2} \left (\cos \left (f x +e \right )-1\right )}+\frac {\left (3 a^{2}-12 a b +8 b^{2}\right ) \ln \left (\cos \left (f x +e \right )-1\right )}{16 a^{3}}}{f}\)	\(185\)
default	\(\frac {\frac {b \left (a^{2}-2 a b +b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {b \left (a -b \right )}}\right )}{a^{3} \sqrt {b \left (a -b \right )}}+\frac {1}{16 a \left (\cos \left (f x +e \right )+1\right )^{2}}-\frac {-3 a +4 b}{16 a^{2} \left (\cos \left (f x +e \right )+1\right )}+\frac {\left (-3 a^{2}+12 a b -8 b^{2}\right ) \ln \left (\cos \left (f x +e \right )+1\right )}{16 a^{3}}-\frac {1}{16 a \left (\cos \left (f x +e \right )-1\right )^{2}}-\frac {-3 a +4 b}{16 a^{2} \left (\cos \left (f x +e \right )-1\right )}+\frac {\left (3 a^{2}-12 a b +8 b^{2}\right ) \ln \left (\cos \left (f x +e \right )-1\right )}{16 a^{3}}}{f}\)	\(185\)
risch	\(\frac {3 a \,{\mathrm e}^{7 i \left (f x +e \right )}-4 b \,{\mathrm e}^{7 i \left (f x +e \right )}-11 a \,{\mathrm e}^{5 i \left (f x +e \right )}+4 b \,{\mathrm e}^{5 i \left (f x +e \right )}-11 a \,{\mathrm e}^{3 i \left (f x +e \right )}+4 b \,{\mathrm e}^{3 i \left (f x +e \right )}+3 a \,{\mathrm e}^{i \left (f x +e \right )}-4 b \,{\mathrm e}^{i \left (f x +e \right )}}{4 f \,a^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4}}-\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{8 a f}+\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) b}{2 a^{2} f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) b^{2}}{a^{3} f}+\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{8 a f}-\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) b}{2 a^{2} f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) b^{2}}{a^{3} f}+\frac {i \sqrt {a b -b^{2}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {a b -b^{2}}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right )}{2 f \,a^{2}}-\frac {i \sqrt {a b -b^{2}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {a b -b^{2}}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right ) b}{2 f \,a^{3}}-\frac {i \sqrt {a b -b^{2}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {a b -b^{2}}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right )}{2 f \,a^{2}}+\frac {i \sqrt {a b -b^{2}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {a b -b^{2}}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right ) b}{2 f \,a^{3}}\)	\(497\)

[In]

int(csc(f*x+e)^5/(a+b*tan(f*x+e)^2),x,method=_RETURNVERBOSE)

[Out]

1/f*(b*(a^2-2*a*b+b^2)/a^3/(b*(a-b))^(1/2)*arctan((a-b)*cos(f*x+e)/(b*(a-b))^(1/2))+1/16/a/(cos(f*x+e)+1)^2-1/
16*(-3*a+4*b)/a^2/(cos(f*x+e)+1)+1/16/a^3*(-3*a^2+12*a*b-8*b^2)*ln(cos(f*x+e)+1)-1/16/a/(cos(f*x+e)-1)^2-1/16*
(-3*a+4*b)/a^2/(cos(f*x+e)-1)+1/16*(3*a^2-12*a*b+8*b^2)/a^3*ln(cos(f*x+e)-1))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 298 vs. \(2 (116) = 232\).

Time = 0.36 (sec) , antiderivative size = 630, normalized size of antiderivative = 4.85 \[ \int \frac {\csc ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\left [\frac {2 \, {\left (3 \, a^{2} - 4 \, a b\right )} \cos \left (f x + e\right )^{3} - 8 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (a - b\right )} \cos \left (f x + e\right )^{2} + a - b\right )} \sqrt {-a b + b^{2}} \log \left (\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-a b + b^{2}} \cos \left (f x + e\right ) - b}{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}\right ) - 2 \, {\left (5 \, a^{2} - 4 \, a b\right )} \cos \left (f x + e\right ) - {\left ({\left (3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + {\left ({\left (3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right )}{16 \, {\left (a^{3} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} f \cos \left (f x + e\right )^{2} + a^{3} f\right )}}, \frac {2 \, {\left (3 \, a^{2} - 4 \, a b\right )} \cos \left (f x + e\right )^{3} + 16 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (a - b\right )} \cos \left (f x + e\right )^{2} + a - b\right )} \sqrt {a b - b^{2}} \arctan \left (\frac {\sqrt {a b - b^{2}} \cos \left (f x + e\right )}{b}\right ) - 2 \, {\left (5 \, a^{2} - 4 \, a b\right )} \cos \left (f x + e\right ) - {\left ({\left (3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + {\left ({\left (3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right )}{16 \, {\left (a^{3} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} f \cos \left (f x + e\right )^{2} + a^{3} f\right )}}\right ] \]

[In]

integrate(csc(f*x+e)^5/(a+b*tan(f*x+e)^2),x, algorithm="fricas")

[Out]

[1/16*(2*(3*a^2 - 4*a*b)*cos(f*x + e)^3 - 8*((a - b)*cos(f*x + e)^4 - 2*(a - b)*cos(f*x + e)^2 + a - b)*sqrt(-
a*b + b^2)*log(((a - b)*cos(f*x + e)^2 - 2*sqrt(-a*b + b^2)*cos(f*x + e) - b)/((a - b)*cos(f*x + e)^2 + b)) -
2*(5*a^2 - 4*a*b)*cos(f*x + e) - ((3*a^2 - 12*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a^2 - 12*a*b + 8*b^2)*cos(f*x
+ e)^2 + 3*a^2 - 12*a*b + 8*b^2)*log(1/2*cos(f*x + e) + 1/2) + ((3*a^2 - 12*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(
3*a^2 - 12*a*b + 8*b^2)*cos(f*x + e)^2 + 3*a^2 - 12*a*b + 8*b^2)*log(-1/2*cos(f*x + e) + 1/2))/(a^3*f*cos(f*x
+ e)^4 - 2*a^3*f*cos(f*x + e)^2 + a^3*f), 1/16*(2*(3*a^2 - 4*a*b)*cos(f*x + e)^3 + 16*((a - b)*cos(f*x + e)^4
- 2*(a - b)*cos(f*x + e)^2 + a - b)*sqrt(a*b - b^2)*arctan(sqrt(a*b - b^2)*cos(f*x + e)/b) - 2*(5*a^2 - 4*a*b)
*cos(f*x + e) - ((3*a^2 - 12*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a^2 - 12*a*b + 8*b^2)*cos(f*x + e)^2 + 3*a^2 -
12*a*b + 8*b^2)*log(1/2*cos(f*x + e) + 1/2) + ((3*a^2 - 12*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a^2 - 12*a*b +
8*b^2)*cos(f*x + e)^2 + 3*a^2 - 12*a*b + 8*b^2)*log(-1/2*cos(f*x + e) + 1/2))/(a^3*f*cos(f*x + e)^4 - 2*a^3*f*
cos(f*x + e)^2 + a^3*f)]

Sympy [F]

\[ \int \frac {\csc ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\int \frac {\csc ^{5}{\left (e + f x \right )}}{a + b \tan ^{2}{\left (e + f x \right )}}\, dx \]

[In]

integrate(csc(f*x+e)**5/(a+b*tan(f*x+e)**2),x)

[Out]

Integral(csc(e + f*x)**5/(a + b*tan(e + f*x)**2), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\csc ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(csc(f*x+e)^5/(a+b*tan(f*x+e)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for mor
e details)Is

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 354 vs. \(2 (116) = 232\).

Time = 0.47 (sec) , antiderivative size = 354, normalized size of antiderivative = 2.72 \[ \int \frac {\csc ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {\frac {\frac {8 \, a {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {8 \, b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {a {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}}{a^{2}} - \frac {4 \, {\left (3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \log \left (\frac {{\left | -\cos \left (f x + e\right ) + 1 \right |}}{{\left | \cos \left (f x + e\right ) + 1 \right |}}\right )}{a^{3}} + \frac {64 \, {\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \arctan \left (-\frac {a \cos \left (f x + e\right ) - b \cos \left (f x + e\right ) - b}{\sqrt {a b - b^{2}} \cos \left (f x + e\right ) + \sqrt {a b - b^{2}}}\right )}{\sqrt {a b - b^{2}} a^{3}} + \frac {{\left (a^{2} - \frac {8 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {8 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {18 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {72 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {48 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{2}}{a^{3} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}}{64 \, f} \]

[In]

integrate(csc(f*x+e)^5/(a+b*tan(f*x+e)^2),x, algorithm="giac")

[Out]

-1/64*((8*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 8*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - a*(cos(f*x + e
) - 1)^2/(cos(f*x + e) + 1)^2)/a^2 - 4*(3*a^2 - 12*a*b + 8*b^2)*log(abs(-cos(f*x + e) + 1)/abs(cos(f*x + e) +
1))/a^3 + 64*(a^2*b - 2*a*b^2 + b^3)*arctan(-(a*cos(f*x + e) - b*cos(f*x + e) - b)/(sqrt(a*b - b^2)*cos(f*x +
e) + sqrt(a*b - b^2)))/(sqrt(a*b - b^2)*a^3) + (a^2 - 8*a^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 8*a*b*(cos
(f*x + e) - 1)/(cos(f*x + e) + 1) + 18*a^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 72*a*b*(cos(f*x + e) -
1)^2/(cos(f*x + e) + 1)^2 + 48*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)*(cos(f*x + e) + 1)^2/(a^3*(cos(f
*x + e) - 1)^2))/f

Mupad [B] (verification not implemented)

Time = 12.90 (sec) , antiderivative size = 740, normalized size of antiderivative = 5.69 \[ \int \frac {\csc ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\frac {a^2\,\left (\frac {3\,\cos \left (3\,e+3\,f\,x\right )}{4}-\frac {11\,\cos \left (e+f\,x\right )}{4}+\frac {9\,\ln \left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{8}-\frac {3\,\cos \left (2\,e+2\,f\,x\right )\,\ln \left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{2}+\frac {3\,\cos \left (4\,e+4\,f\,x\right )\,\ln \left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{8}\right )+3\,b^2\,\ln \left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )-a\,\left (b\,\cos \left (3\,e+3\,f\,x\right )-b\,\cos \left (e+f\,x\right )+\frac {9\,b\,\ln \left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{2}-6\,b\,\cos \left (2\,e+2\,f\,x\right )\,\ln \left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )+\frac {3\,b\,\cos \left (4\,e+4\,f\,x\right )\,\ln \left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{2}\right )-4\,b^2\,\cos \left (2\,e+2\,f\,x\right )\,\ln \left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )+b^2\,\cos \left (4\,e+4\,f\,x\right )\,\ln \left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )+3\,\sqrt {b}\,\mathrm {atan}\left (\frac {a^4\,\cos \left (e+f\,x\right )-a^3\,b-3\,a\,b^3+b^4\,\cos \left (e+f\,x\right )+b^4+3\,a^2\,b^2+6\,a^2\,b^2\,\cos \left (e+f\,x\right )-4\,a\,b^3\,\cos \left (e+f\,x\right )-4\,a^3\,b\,\cos \left (e+f\,x\right )}{2\,\sqrt {b}\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,{\left (a-b\right )}^{7/2}}\right )\,{\left (a-b\right )}^{3/2}-4\,\sqrt {b}\,\mathrm {atan}\left (\frac {a^4\,\cos \left (e+f\,x\right )-a^3\,b-3\,a\,b^3+b^4\,\cos \left (e+f\,x\right )+b^4+3\,a^2\,b^2+6\,a^2\,b^2\,\cos \left (e+f\,x\right )-4\,a\,b^3\,\cos \left (e+f\,x\right )-4\,a^3\,b\,\cos \left (e+f\,x\right )}{2\,\sqrt {b}\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,{\left (a-b\right )}^{7/2}}\right )\,\cos \left (2\,e+2\,f\,x\right )\,{\left (a-b\right )}^{3/2}+\sqrt {b}\,\mathrm {atan}\left (\frac {a^4\,\cos \left (e+f\,x\right )-a^3\,b-3\,a\,b^3+b^4\,\cos \left (e+f\,x\right )+b^4+3\,a^2\,b^2+6\,a^2\,b^2\,\cos \left (e+f\,x\right )-4\,a\,b^3\,\cos \left (e+f\,x\right )-4\,a^3\,b\,\cos \left (e+f\,x\right )}{2\,\sqrt {b}\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,{\left (a-b\right )}^{7/2}}\right )\,\cos \left (4\,e+4\,f\,x\right )\,{\left (a-b\right )}^{3/2}}{3\,a^3\,f-4\,a^3\,f\,\cos \left (2\,e+2\,f\,x\right )+a^3\,f\,\cos \left (4\,e+4\,f\,x\right )} \]

[In]

int(1/(sin(e + f*x)^5*(a + b*tan(e + f*x)^2)),x)

[Out]

(a^2*((3*cos(3*e + 3*f*x))/4 - (11*cos(e + f*x))/4 + (9*log(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)))/8 - (3*cos
(2*e + 2*f*x)*log(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)))/2 + (3*cos(4*e + 4*f*x)*log(sin(e/2 + (f*x)/2)/cos(e
/2 + (f*x)/2)))/8) + 3*b^2*log(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)) - a*(b*cos(3*e + 3*f*x) - b*cos(e + f*x)
+ (9*b*log(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)))/2 - 6*b*cos(2*e + 2*f*x)*log(sin(e/2 + (f*x)/2)/cos(e/2 +
(f*x)/2)) + (3*b*cos(4*e + 4*f*x)*log(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)))/2) - 4*b^2*cos(2*e + 2*f*x)*log(
sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)) + b^2*cos(4*e + 4*f*x)*log(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)) + 3*b
^(1/2)*atan((a^4*cos(e + f*x) - a^3*b - 3*a*b^3 + b^4*cos(e + f*x) + b^4 + 3*a^2*b^2 + 6*a^2*b^2*cos(e + f*x)
- 4*a*b^3*cos(e + f*x) - 4*a^3*b*cos(e + f*x))/(2*b^(1/2)*cos(e/2 + (f*x)/2)^2*(a - b)^(7/2)))*(a - b)^(3/2) -
4*b^(1/2)*atan((a^4*cos(e + f*x) - a^3*b - 3*a*b^3 + b^4*cos(e + f*x) + b^4 + 3*a^2*b^2 + 6*a^2*b^2*cos(e + f
*x) - 4*a*b^3*cos(e + f*x) - 4*a^3*b*cos(e + f*x))/(2*b^(1/2)*cos(e/2 + (f*x)/2)^2*(a - b)^(7/2)))*cos(2*e + 2
*f*x)*(a - b)^(3/2) + b^(1/2)*atan((a^4*cos(e + f*x) - a^3*b - 3*a*b^3 + b^4*cos(e + f*x) + b^4 + 3*a^2*b^2 +
6*a^2*b^2*cos(e + f*x) - 4*a*b^3*cos(e + f*x) - 4*a^3*b*cos(e + f*x))/(2*b^(1/2)*cos(e/2 + (f*x)/2)^2*(a - b)^
(7/2)))*cos(4*e + 4*f*x)*(a - b)^(3/2))/(3*a^3*f - 4*a^3*f*cos(2*e + 2*f*x) + a^3*f*cos(4*e + 4*f*x))