Integrand size = 23, antiderivative size = 130 \[ \int \frac {\csc ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {(a-b)^{3/2} \sqrt {b} \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{a^3 f}-\frac {\left (3 a^2-12 a b+8 b^2\right ) \text {arctanh}(\cos (e+f x))}{8 a^3 f}-\frac {(5 a-4 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f} \]
-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
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 \[ \int \frac {\csc ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\frac {(a-b)^{3/2} \sqrt {b} \arctan \left (\frac {\sec \left (\frac {1}{2} (e+f x)\right ) \left (\sqrt {a-b} \cos \left (\frac {1}{2} (e+f x)\right )-\sqrt {a} \sin \left (\frac {1}{2} (e+f x)\right )\right )}{\sqrt {b}}\right )}{a^3 f}+\frac {(a-b)^{3/2} \sqrt {b} \arctan \left (\frac {\sec \left (\frac {1}{2} (e+f x)\right ) \left (\sqrt {a-b} \cos \left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sin \left (\frac {1}{2} (e+f x)\right )\right )}{\sqrt {b}}\right )}{a^3 f}+\frac {(-3 a+4 b) \csc ^2\left (\frac {1}{2} (e+f x)\right )}{32 a^2 f}-\frac {\csc ^4\left (\frac {1}{2} (e+f x)\right )}{64 a f}+\frac {\left (-3 a^2+12 a b-8 b^2\right ) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )}{8 a^3 f}+\frac {\left (3 a^2-12 a b+8 b^2\right ) \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )}{8 a^3 f}+\frac {(3 a-4 b) \sec ^2\left (\frac {1}{2} (e+f x)\right )}{32 a^2 f}+\frac {\sec ^4\left (\frac {1}{2} (e+f x)\right )}{64 a f} \]
((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) + ((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)
Time = 0.37 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.15, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 4147, 25, 372, 402, 25, 397, 218, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\csc ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (e+f x)^5 \left (a+b \tan (e+f x)^2\right )}dx\) |
| \(\Big \downarrow \) 4147 |
| \(\displaystyle \frac {\int -\frac {\sec ^4(e+f x)}{\left (1-\sec ^2(e+f x)\right )^3 \left (b \sec ^2(e+f x)+a-b\right )}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \frac {\sec ^4(e+f x)}{\left (1-\sec ^2(e+f x)\right )^3 \left (b \sec ^2(e+f x)+a-b\right )}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 372 |
| \(\displaystyle \frac {\frac {\int \frac {(4 a-3 b) \sec ^2(e+f x)+a-b}{\left (1-\sec ^2(e+f x)\right )^2 \left (b \sec ^2(e+f x)+a-b\right )}d\sec (e+f x)}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\int -\frac {(3 a-4 b) (a-b)-(5 a-4 b) b \sec ^2(e+f x)}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )}d\sec (e+f x)}{2 a}+\frac {(5 a-4 b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right )}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {(5 a-4 b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right )}-\frac {\int \frac {(3 a-4 b) (a-b)-(5 a-4 b) b \sec ^2(e+f x)}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )}d\sec (e+f x)}{2 a}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\frac {(5 a-4 b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right )}-\frac {\frac {\left (3 a^2-12 a b+8 b^2\right ) \int \frac {1}{1-\sec ^2(e+f x)}d\sec (e+f x)}{a}+\frac {8 b (a-b)^2 \int \frac {1}{b \sec ^2(e+f x)+a-b}d\sec (e+f x)}{a}}{2 a}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {(5 a-4 b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right )}-\frac {\frac {\left (3 a^2-12 a b+8 b^2\right ) \int \frac {1}{1-\sec ^2(e+f x)}d\sec (e+f x)}{a}+\frac {8 \sqrt {b} (a-b)^{3/2} \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{a}}{2 a}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\frac {(5 a-4 b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right )}-\frac {\frac {\left (3 a^2-12 a b+8 b^2\right ) \text {arctanh}(\sec (e+f x))}{a}+\frac {8 \sqrt {b} (a-b)^{3/2} \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{a}}{2 a}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2}}{f}\) |
(-1/4*Sec[e + f*x]/(a*(1 - Sec[e + f*x]^2)^2) + (-1/2*((8*(a - b)^(3/2)*Sq rt[b]*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/a + ((3*a^2 - 12*a*b + 8 *b^2)*ArcTanh[Sec[e + f*x]])/a)/a + ((5*a - 4*b)*Sec[e + f*x])/(2*a*(1 - S ec[e + f*x]^2)))/(4*a))/f
3.1.60.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 )^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 )) Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 3] && IntBinomialQ[a , b, c, d, e, m, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Simp[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]
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 | \(\frac {3 a \,{\mathrm e}^{7 i \left (f x +e \right )}-4 b \,{\mathrm e}^{7 i \left (f x +e \right )}-11 a \,{\mathrm e}^{5 i \left (f x +e \right )}+4 b \,{\mathrm e}^{5 i \left (f x +e \right )}-11 a \,{\mathrm e}^{3 i \left (f x +e \right )}+4 b \,{\mathrm e}^{3 i \left (f x +e \right )}+3 a \,{\mathrm e}^{i \left (f x +e \right )}-4 b \,{\mathrm e}^{i \left (f x +e \right )}}{4 f \,a^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4}}-\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{8 a f}+\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) b}{2 a^{2} f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) b^{2}}{a^{3} f}+\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{8 a f}-\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) b}{2 a^{2} f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) b^{2}}{a^{3} f}+\frac {i \sqrt {a b -b^{2}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {a b -b^{2}}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right )}{2 f \,a^{2}}-\frac {i \sqrt {a b -b^{2}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {a b -b^{2}}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right ) b}{2 f \,a^{3}}-\frac {i \sqrt {a b -b^{2}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {a b -b^{2}}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right )}{2 f \,a^{2}}+\frac {i \sqrt {a b -b^{2}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {a b -b^{2}}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right ) b}{2 f \,a^{3}}\) | \(497\) |
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) )
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 ] \]
[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)*co s(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)]
\[ \int \frac {\csc ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\int \frac {\csc ^{5}{\left (e + f x \right )}}{a + b \tan ^{2}{\left (e + f x \right )}}\, dx \]
Exception generated. \[ \int \frac {\csc ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for more details)Is
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 \[ \int \frac {\csc ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {\frac {\frac {8 \, a {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {8 \, b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {a {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}}{a^{2}} - \frac {4 \, {\left (3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \log \left (\frac {{\left | -\cos \left (f x + e\right ) + 1 \right |}}{{\left | \cos \left (f x + e\right ) + 1 \right |}}\right )}{a^{3}} + \frac {64 \, {\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \arctan \left (-\frac {a \cos \left (f x + e\right ) - b \cos \left (f x + e\right ) - b}{\sqrt {a b - b^{2}} \cos \left (f x + e\right ) + \sqrt {a b - b^{2}}}\right )}{\sqrt {a b - b^{2}} a^{3}} + \frac {{\left (a^{2} - \frac {8 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {8 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {18 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {72 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {48 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{2}}{a^{3} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}}{64 \, f} \]
-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
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 )} \]
(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)/c os(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*co s(4*e + 4*f*x))