Integrand size = 23, antiderivative size = 147 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=-\frac {(3 a-4 b) \sqrt {b} \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{2 a^3 \sqrt {a-b} f}-\frac {(a-4 b) \text {arctanh}(\cos (e+f x))}{2 a^3 f}-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {b \sec (e+f x)}{a^2 f \left (a-b+b \sec ^2(e+f x)\right )} \]
-1/2*(a-4*b)*arctanh(cos(f*x+e))/a^3/f-1/2*cot(f*x+e)*csc(f*x+e)/a/f/(a-b+ b*sec(f*x+e)^2)-b*sec(f*x+e)/a^2/f/(a-b+b*sec(f*x+e)^2)-1/2*(3*a-4*b)*arct an(sec(f*x+e)*b^(1/2)/(a-b)^(1/2))*b^(1/2)/a^3/f/(a-b)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(325\) vs. \(2(147)=294\).
Time = 6.94 (sec) , antiderivative size = 325, normalized size of antiderivative = 2.21 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=-\frac {(3 a-4 b) \sqrt {a-b} \sqrt {b} \arctan \left (\frac {\sec \left (\frac {1}{2} (e+f x)\right ) \left (\sqrt {a-b} \cos \left (\frac {1}{2} (e+f x)\right )-\sqrt {a} \sin \left (\frac {1}{2} (e+f x)\right )\right )}{\sqrt {b}}\right )}{2 a^3 (-a+b) f}-\frac {(3 a-4 b) \sqrt {a-b} \sqrt {b} \arctan \left (\frac {\sec \left (\frac {1}{2} (e+f x)\right ) \left (\sqrt {a-b} \cos \left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sin \left (\frac {1}{2} (e+f x)\right )\right )}{\sqrt {b}}\right )}{2 a^3 (-a+b) f}-\frac {b \cos (e+f x)}{a^2 f (a+b+a \cos (2 (e+f x))-b \cos (2 (e+f x)))}-\frac {\csc ^2\left (\frac {1}{2} (e+f x)\right )}{8 a^2 f}+\frac {(-a+4 b) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )}{2 a^3 f}+\frac {(a-4 b) \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )}{2 a^3 f}+\frac {\sec ^2\left (\frac {1}{2} (e+f x)\right )}{8 a^2 f} \]
-1/2*((3*a - 4*b)*Sqrt[a - b]*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*(-a + b)*f) - ((3*a - 4*b)*Sqrt[a - b]*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]])/(2*a^3*(-a + b)*f) - (b*Cos[e + f*x])/(a^2*f*(a + b + a*Cos[2*(e + f*x)] - b*Cos[2*(e + f*x) ])) - Csc[(e + f*x)/2]^2/(8*a^2*f) + ((-a + 4*b)*Log[Cos[(e + f*x)/2]])/(2 *a^3*f) + ((a - 4*b)*Log[Sin[(e + f*x)/2]])/(2*a^3*f) + Sec[(e + f*x)/2]^2 /(8*a^2*f)
Time = 0.35 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 4147, 373, 402, 27, 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.
| \(\displaystyle \int \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (e+f x)^3 \left (a+b \tan (e+f x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 4147 |
| \(\displaystyle \frac {\int \frac {\sec ^2(e+f x)}{\left (1-\sec ^2(e+f x)\right )^2 \left (b \sec ^2(e+f x)+a-b\right )^2}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 373 |
| \(\displaystyle \frac {\frac {\sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )}-\frac {\int \frac {-3 b \sec ^2(e+f x)+a-b}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )^2}d\sec (e+f x)}{2 a}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )}-\frac {\frac {2 b \sec (e+f x)}{a \left (a+b \sec ^2(e+f x)-b\right )}-\frac {\int -\frac {2 (a-b) \left (-2 b \sec ^2(e+f x)+a-2 b\right )}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )}d\sec (e+f x)}{2 a (a-b)}}{2 a}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )}-\frac {\frac {\int \frac {-2 b \sec ^2(e+f x)+a-2 b}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )}d\sec (e+f x)}{a}+\frac {2 b \sec (e+f x)}{a \left (a+b \sec ^2(e+f x)-b\right )}}{2 a}}{f}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )}-\frac {\frac {\frac {(a-4 b) \int \frac {1}{1-\sec ^2(e+f x)}d\sec (e+f x)}{a}+\frac {b (3 a-4 b) \int \frac {1}{b \sec ^2(e+f x)+a-b}d\sec (e+f x)}{a}}{a}+\frac {2 b \sec (e+f x)}{a \left (a+b \sec ^2(e+f x)-b\right )}}{2 a}}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )}-\frac {\frac {\frac {(a-4 b) \int \frac {1}{1-\sec ^2(e+f x)}d\sec (e+f x)}{a}+\frac {\sqrt {b} (3 a-4 b) \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{a \sqrt {a-b}}}{a}+\frac {2 b \sec (e+f x)}{a \left (a+b \sec ^2(e+f x)-b\right )}}{2 a}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )}-\frac {\frac {\frac {\sqrt {b} (3 a-4 b) \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{a \sqrt {a-b}}+\frac {(a-4 b) \text {arctanh}(\sec (e+f x))}{a}}{a}+\frac {2 b \sec (e+f x)}{a \left (a+b \sec ^2(e+f x)-b\right )}}{2 a}}{f}\) |
(Sec[e + f*x]/(2*a*(1 - Sec[e + f*x]^2)*(a - b + b*Sec[e + f*x]^2)) - (((( 3*a - 4*b)*Sqrt[b]*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/(a*Sqrt[a - b]) + ((a - 4*b)*ArcTanh[Sec[e + f*x]])/a)/a + (2*b*Sec[e + f*x])/(a*(a - b + b*Sec[e + f*x]^2)))/(2*a))/f
3.1.72.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*(b*c - a*d)*(p + 1))), x] - Simp[e^2/(2*(b*c - a*d)*(p + 1)) Int[(e *x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(m - 1) + d*(m + 2*p + 2*q + 3)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 1] && LeQ[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 = 0.61 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.06
| method | result | size |
| derivativedivides | \(\frac {\frac {b \left (-\frac {a \cos \left (f x +e \right )}{2 \left (a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b \right )}+\frac {\left (3 a -4 b \right ) \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {b \left (a -b \right )}}\right )}{2 \sqrt {b \left (a -b \right )}}\right )}{a^{3}}+\frac {1}{4 a^{2} \left (\cos \left (f x +e \right )+1\right )}+\frac {\left (-a +4 b \right ) \ln \left (\cos \left (f x +e \right )+1\right )}{4 a^{3}}+\frac {1}{4 a^{2} \left (\cos \left (f x +e \right )-1\right )}+\frac {\left (a -4 b \right ) \ln \left (\cos \left (f x +e \right )-1\right )}{4 a^{3}}}{f}\) | \(156\) |
| default | \(\frac {\frac {b \left (-\frac {a \cos \left (f x +e \right )}{2 \left (a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b \right )}+\frac {\left (3 a -4 b \right ) \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {b \left (a -b \right )}}\right )}{2 \sqrt {b \left (a -b \right )}}\right )}{a^{3}}+\frac {1}{4 a^{2} \left (\cos \left (f x +e \right )+1\right )}+\frac {\left (-a +4 b \right ) \ln \left (\cos \left (f x +e \right )+1\right )}{4 a^{3}}+\frac {1}{4 a^{2} \left (\cos \left (f x +e \right )-1\right )}+\frac {\left (a -4 b \right ) \ln \left (\cos \left (f x +e \right )-1\right )}{4 a^{3}}}{f}\) | \(156\) |
| risch | \(\frac {a \,{\mathrm e}^{7 i \left (f x +e \right )}-2 b \,{\mathrm e}^{7 i \left (f x +e \right )}+3 a \,{\mathrm e}^{5 i \left (f x +e \right )}+2 b \,{\mathrm e}^{5 i \left (f x +e \right )}+3 a \,{\mathrm e}^{3 i \left (f x +e \right )}+2 b \,{\mathrm e}^{3 i \left (f x +e \right )}+a \,{\mathrm e}^{i \left (f x +e \right )}-2 b \,{\mathrm e}^{i \left (f x +e \right )}}{f \,a^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2} \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}-b \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+2 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a -b \right )}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{2 a^{2} f}+\frac {2 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) b}{a^{3} f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{2 a^{2} f}-\frac {2 \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) b}{a^{3} f}-\frac {3 i \sqrt {b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right )}{4 \left (a -b \right ) f \,a^{2}}+\frac {i \sqrt {b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right ) b}{\left (a -b \right ) f \,a^{3}}+\frac {3 i \sqrt {b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right )}{4 \left (a -b \right ) f \,a^{2}}-\frac {i \sqrt {b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right ) b}{\left (a -b \right ) f \,a^{3}}\) | \(515\) |
1/f*(b/a^3*(-1/2*a*cos(f*x+e)/(a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)+1/2*(3*a-4 *b)/(b*(a-b))^(1/2)*arctan((a-b)*cos(f*x+e)/(b*(a-b))^(1/2)))+1/4/a^2/(cos (f*x+e)+1)+1/4/a^3*(-a+4*b)*ln(cos(f*x+e)+1)+1/4/a^2/(cos(f*x+e)-1)+1/4*(a -4*b)/a^3*ln(cos(f*x+e)-1))
Leaf count of result is larger than twice the leaf count of optimal. 317 vs. \(2 (133) = 266\).
Time = 0.38 (sec) , antiderivative size = 672, normalized size of antiderivative = 4.57 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\left [\frac {2 \, {\left (a^{2} - 2 \, a b\right )} \cos \left (f x + e\right )^{3} + 4 \, a b \cos \left (f x + e\right ) - {\left ({\left (3 \, a^{2} - 7 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (3 \, a^{2} - 10 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 3 \, a b + 4 \, b^{2}\right )} \sqrt {-\frac {b}{a - b}} \log \left (-\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (a - b\right )} \sqrt {-\frac {b}{a - b}} \cos \left (f x + e\right ) - b}{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}\right ) - {\left ({\left (a^{2} - 5 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (a^{2} - 6 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - a b + 4 \, b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + {\left ({\left (a^{2} - 5 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (a^{2} - 6 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - a b + 4 \, b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right )}{4 \, {\left ({\left (a^{4} - a^{3} b\right )} f \cos \left (f x + e\right )^{4} - a^{3} b f - {\left (a^{4} - 2 \, a^{3} b\right )} f \cos \left (f x + e\right )^{2}\right )}}, \frac {2 \, {\left (a^{2} - 2 \, a b\right )} \cos \left (f x + e\right )^{3} + 4 \, a b \cos \left (f x + e\right ) - 2 \, {\left ({\left (3 \, a^{2} - 7 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (3 \, a^{2} - 10 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 3 \, a b + 4 \, b^{2}\right )} \sqrt {\frac {b}{a - b}} \arctan \left (-\frac {{\left (a - b\right )} \sqrt {\frac {b}{a - b}} \cos \left (f x + e\right )}{b}\right ) - {\left ({\left (a^{2} - 5 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (a^{2} - 6 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - a b + 4 \, b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + {\left ({\left (a^{2} - 5 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (a^{2} - 6 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - a b + 4 \, b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right )}{4 \, {\left ({\left (a^{4} - a^{3} b\right )} f \cos \left (f x + e\right )^{4} - a^{3} b f - {\left (a^{4} - 2 \, a^{3} b\right )} f \cos \left (f x + e\right )^{2}\right )}}\right ] \]
[1/4*(2*(a^2 - 2*a*b)*cos(f*x + e)^3 + 4*a*b*cos(f*x + e) - ((3*a^2 - 7*a* b + 4*b^2)*cos(f*x + e)^4 - (3*a^2 - 10*a*b + 8*b^2)*cos(f*x + e)^2 - 3*a* b + 4*b^2)*sqrt(-b/(a - b))*log(-((a - b)*cos(f*x + e)^2 - 2*(a - b)*sqrt( -b/(a - b))*cos(f*x + e) - b)/((a - b)*cos(f*x + e)^2 + b)) - ((a^2 - 5*a* b + 4*b^2)*cos(f*x + e)^4 - (a^2 - 6*a*b + 8*b^2)*cos(f*x + e)^2 - a*b + 4 *b^2)*log(1/2*cos(f*x + e) + 1/2) + ((a^2 - 5*a*b + 4*b^2)*cos(f*x + e)^4 - (a^2 - 6*a*b + 8*b^2)*cos(f*x + e)^2 - a*b + 4*b^2)*log(-1/2*cos(f*x + e ) + 1/2))/((a^4 - a^3*b)*f*cos(f*x + e)^4 - a^3*b*f - (a^4 - 2*a^3*b)*f*co s(f*x + e)^2), 1/4*(2*(a^2 - 2*a*b)*cos(f*x + e)^3 + 4*a*b*cos(f*x + e) - 2*((3*a^2 - 7*a*b + 4*b^2)*cos(f*x + e)^4 - (3*a^2 - 10*a*b + 8*b^2)*cos(f *x + e)^2 - 3*a*b + 4*b^2)*sqrt(b/(a - b))*arctan(-(a - b)*sqrt(b/(a - b)) *cos(f*x + e)/b) - ((a^2 - 5*a*b + 4*b^2)*cos(f*x + e)^4 - (a^2 - 6*a*b + 8*b^2)*cos(f*x + e)^2 - a*b + 4*b^2)*log(1/2*cos(f*x + e) + 1/2) + ((a^2 - 5*a*b + 4*b^2)*cos(f*x + e)^4 - (a^2 - 6*a*b + 8*b^2)*cos(f*x + e)^2 - a* b + 4*b^2)*log(-1/2*cos(f*x + e) + 1/2))/((a^4 - a^3*b)*f*cos(f*x + e)^4 - a^3*b*f - (a^4 - 2*a^3*b)*f*cos(f*x + e)^2)]
\[ \int \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\int \frac {\csc ^{3}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\text {Exception raised: ValueError} \]
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. 390 vs. \(2 (133) = 266\).
Time = 0.57 (sec) , antiderivative size = 390, normalized size of antiderivative = 2.65 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\frac {\frac {6 \, {\left (a - 4 \, b\right )} \log \left (\frac {{\left | -\cos \left (f x + e\right ) + 1 \right |}}{{\left | \cos \left (f x + e\right ) + 1 \right |}}\right )}{a^{3}} - \frac {12 \, {\left (3 \, a b - 4 \, b^{2}\right )} \arctan \left (-\frac {a \cos \left (f x + e\right ) - b \cos \left (f x + e\right ) - b}{\sqrt {a b - b^{2}} \cos \left (f x + e\right ) + \sqrt {a b - b^{2}}}\right )}{\sqrt {a b - b^{2}} a^{3}} - \frac {3 \, {\left (\cos \left (f x + e\right ) - 1\right )}}{a^{2} {\left (\cos \left (f x + e\right ) + 1\right )}} + \frac {3 \, a^{2} + \frac {4 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {28 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {16 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {2 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {8 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}{a^{3} {\left (\frac {a {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {2 \, a {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {4 \, b {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}}{24 \, f} \]
1/24*(6*(a - 4*b)*log(abs(-cos(f*x + e) + 1)/abs(cos(f*x + e) + 1))/a^3 - 12*(3*a*b - 4*b^2)*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) - 3*(cos(f* x + e) - 1)/(a^2*(cos(f*x + e) + 1)) + (3*a^2 + 4*a^2*(cos(f*x + e) - 1)/( cos(f*x + e) + 1) - 28*a*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - a^2*(co s(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 16*b^2*(cos(f*x + e) - 1)^2/(cos( f*x + e) + 1)^2 - 2*a^2*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 8*a*b* (cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3)/(a^3*(a*(cos(f*x + e) - 1)/(cos (f*x + e) + 1) + 2*a*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 4*b*(cos( f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + a*(cos(f*x + e) - 1)^3/(cos(f*x + e ) + 1)^3)))/f
Time = 11.39 (sec) , antiderivative size = 917, normalized size of antiderivative = 6.24 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2}{8\,a^2\,f}-\frac {\frac {a}{2}-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (a-6\,b\right )+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (a^2-8\,a\,b+16\,b^2\right )}{2\,a}}{f\,\left (4\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+4\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (16\,a^2\,b-8\,a^3\right )\right )}+\frac {\ln \left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )\,\left (a-4\,b\right )}{2\,a^3\,f}+\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {2\,a^2\,\left (\frac {\sqrt {b}\,\left (3\,a-4\,b\right )\,\left (12\,a^6\,b-106\,a^5\,b^2+240\,a^4\,b^3-160\,a^3\,b^4\right )}{2\,a^9\,\sqrt {a-b}}+\frac {b^{3/2}\,{\left (3\,a-4\,b\right )}^3\,\left (8\,a^{11}-32\,a^{10}\,b+32\,a^9\,b^2\right )}{32\,a^{15}\,{\left (a-b\right )}^{3/2}}\right )\,\left (a-b\right )\,\left (15\,a^4-182\,a^3\,b+648\,a^2\,b^2-864\,a\,b^3+384\,b^4\right )}{\left (9\,a^2\,b-24\,a\,b^2+16\,b^3\right )\,\left (4\,a^3-27\,a^2\,b+72\,a\,b^2-48\,b^3\right )}-\frac {4\,a^7\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,{\left (a-b\right )}^{3/2}\,\left (\frac {\left (\frac {4\,\left (9\,a^2\,b^2-24\,a\,b^3+16\,b^4\right )}{a^5}-\frac {b\,{\left (3\,a-4\,b\right )}^2\,\left (2\,a^8-46\,a^7\,b+344\,a^6\,b^2-672\,a^5\,b^3+384\,a^4\,b^4\right )}{4\,a^{11}\,\left (a-b\right )}\right )\,\left (a^4-31\,a^3\,b+180\,a^2\,b^2-336\,a\,b^3+192\,b^4\right )}{\sqrt {b}\,\left (b\,\left (27\,a^7+b\,\left (48\,a^5\,b-72\,a^6\right )\right )-4\,a^8\right )}+\frac {\left (\frac {\sqrt {b}\,\left (3\,a-4\,b\right )\,\left (9\,a^5\,b-78\,a^4\,b^2+268\,a^3\,b^3-384\,a^2\,b^4+192\,a\,b^5\right )}{a^8\,\sqrt {a-b}}-\frac {b^{3/2}\,{\left (3\,a-4\,b\right )}^3\,\left (-4\,a^{10}+104\,a^9\,b-288\,a^8\,b^2+192\,a^7\,b^3\right )}{16\,a^{14}\,{\left (a-b\right )}^{3/2}}\right )\,\left (15\,a^4-182\,a^3\,b+648\,a^2\,b^2-864\,a\,b^3+384\,b^4\right )}{2\,a^5\,\sqrt {a-b}\,\left (4\,a^3-27\,a^2\,b+72\,a\,b^2-48\,b^3\right )}\right )}{9\,a^2\,b-24\,a\,b^2+16\,b^3}+\frac {4\,a^7\,{\left (a-b\right )}^{3/2}\,\left (\frac {2\,\left (9\,a^3\,b^2-60\,a^2\,b^3+112\,a\,b^4-64\,b^5\right )}{a^6}+\frac {b\,{\left (3\,a-4\,b\right )}^2\,\left (-4\,a^9+56\,a^8\,b-160\,a^7\,b^2+128\,a^6\,b^3\right )}{8\,a^{12}\,\left (a-b\right )}\right )\,\left (a^4-31\,a^3\,b+180\,a^2\,b^2-336\,a\,b^3+192\,b^4\right )}{\sqrt {b}\,\left (b\,\left (27\,a^7+b\,\left (48\,a^5\,b-72\,a^6\right )\right )-4\,a^8\right )\,\left (9\,a^2\,b-24\,a\,b^2+16\,b^3\right )}\right )\,\left (3\,a-4\,b\right )}{2\,a^3\,f\,\sqrt {a-b}} \]
tan(e/2 + (f*x)/2)^2/(8*a^2*f) - (a/2 - tan(e/2 + (f*x)/2)^2*(a - 6*b) + ( tan(e/2 + (f*x)/2)^4*(a^2 - 8*a*b + 16*b^2))/(2*a))/(f*(4*a^3*tan(e/2 + (f *x)/2)^2 + 4*a^3*tan(e/2 + (f*x)/2)^6 + tan(e/2 + (f*x)/2)^4*(16*a^2*b - 8 *a^3))) + (log(tan(e/2 + (f*x)/2))*(a - 4*b))/(2*a^3*f) + (b^(1/2)*atan((2 *a^2*((b^(1/2)*(3*a - 4*b)*(12*a^6*b - 160*a^3*b^4 + 240*a^4*b^3 - 106*a^5 *b^2))/(2*a^9*(a - b)^(1/2)) + (b^(3/2)*(3*a - 4*b)^3*(8*a^11 - 32*a^10*b + 32*a^9*b^2))/(32*a^15*(a - b)^(3/2)))*(a - b)*(15*a^4 - 182*a^3*b - 864* a*b^3 + 384*b^4 + 648*a^2*b^2))/((9*a^2*b - 24*a*b^2 + 16*b^3)*(72*a*b^2 - 27*a^2*b + 4*a^3 - 48*b^3)) - (4*a^7*tan(e/2 + (f*x)/2)^2*(a - b)^(3/2)*( (((4*(16*b^4 - 24*a*b^3 + 9*a^2*b^2))/a^5 - (b*(3*a - 4*b)^2*(2*a^8 - 46*a ^7*b + 384*a^4*b^4 - 672*a^5*b^3 + 344*a^6*b^2))/(4*a^11*(a - b)))*(a^4 - 31*a^3*b - 336*a*b^3 + 192*b^4 + 180*a^2*b^2))/(b^(1/2)*(b*(27*a^7 + b*(48 *a^5*b - 72*a^6)) - 4*a^8)) + (((b^(1/2)*(3*a - 4*b)*(192*a*b^5 + 9*a^5*b - 384*a^2*b^4 + 268*a^3*b^3 - 78*a^4*b^2))/(a^8*(a - b)^(1/2)) - (b^(3/2)* (3*a - 4*b)^3*(104*a^9*b - 4*a^10 + 192*a^7*b^3 - 288*a^8*b^2))/(16*a^14*( a - b)^(3/2)))*(15*a^4 - 182*a^3*b - 864*a*b^3 + 384*b^4 + 648*a^2*b^2))/( 2*a^5*(a - b)^(1/2)*(72*a*b^2 - 27*a^2*b + 4*a^3 - 48*b^3))))/(9*a^2*b - 2 4*a*b^2 + 16*b^3) + (4*a^7*(a - b)^(3/2)*((2*(112*a*b^4 - 64*b^5 - 60*a^2* b^3 + 9*a^3*b^2))/a^6 + (b*(3*a - 4*b)^2*(56*a^8*b - 4*a^9 + 128*a^6*b^3 - 160*a^7*b^2))/(8*a^12*(a - b)))*(a^4 - 31*a^3*b - 336*a*b^3 + 192*b^4 ...