Integrand size = 23, antiderivative size = 76 \[ \int \frac {\csc ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {(a-b) \sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{a^{5/2} f}-\frac {(a-b) \cot (e+f x)}{a^2 f}-\frac {\cot ^3(e+f x)}{3 a f} \]
-(a-b)*cot(f*x+e)/a^2/f-1/3*cot(f*x+e)^3/a/f-(a-b)*arctan(b^(1/2)*tan(f*x+ e)/a^(1/2))*b^(1/2)/a^(5/2)/f
Time = 0.65 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.96 \[ \int \frac {\csc ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\frac {3 \sqrt {b} (-a+b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )-\sqrt {a} \cot (e+f x) \left (2 a-3 b+a \csc ^2(e+f x)\right )}{3 a^{5/2} f} \]
(3*Sqrt[b]*(-a + b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]] - Sqrt[a]*Cot[e + f*x]*(2*a - 3*b + a*Csc[e + f*x]^2))/(3*a^(5/2)*f)
Time = 0.26 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4146, 359, 264, 218}
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.
\(\displaystyle \int \frac {\csc ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (e+f x)^4 \left (a+b \tan (e+f x)^2\right )}dx\) |
\(\Big \downarrow \) 4146 |
\(\displaystyle \frac {\int \frac {\cot ^4(e+f x) \left (\tan ^2(e+f x)+1\right )}{b \tan ^2(e+f x)+a}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 359 |
\(\displaystyle \frac {\frac {(a-b) \int \frac {\cot ^2(e+f x)}{b \tan ^2(e+f x)+a}d\tan (e+f x)}{a}-\frac {\cot ^3(e+f x)}{3 a}}{f}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {\frac {(a-b) \left (-\frac {b \int \frac {1}{b \tan ^2(e+f x)+a}d\tan (e+f x)}{a}-\frac {\cot (e+f x)}{a}\right )}{a}-\frac {\cot ^3(e+f x)}{3 a}}{f}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {(a-b) \left (-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\cot (e+f x)}{a}\right )}{a}-\frac {\cot ^3(e+f x)}{3 a}}{f}\) |
(-1/3*Cot[e + f*x]^3/a + ((a - b)*(-((Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x] )/Sqrt[a]])/a^(3/2)) - Cot[e + f*x]/a))/a)/f
3.1.66.3.1 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[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_ )])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[c*(ff^(m + 1)/f) Subst[Int[x^m*((a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2)^(m/ 2 + 1)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x ] && IntegerQ[m/2]
Time = 0.43 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.88
method | result | size |
derivativedivides | \(\frac {-\frac {1}{3 a \tan \left (f x +e \right )^{3}}-\frac {a -b}{a^{2} \tan \left (f x +e \right )}-\frac {b \left (a -b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{a^{2} \sqrt {a b}}}{f}\) | \(67\) |
default | \(\frac {-\frac {1}{3 a \tan \left (f x +e \right )^{3}}-\frac {a -b}{a^{2} \tan \left (f x +e \right )}-\frac {b \left (a -b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{a^{2} \sqrt {a b}}}{f}\) | \(67\) |
risch | \(\frac {2 i \left (3 b \,{\mathrm e}^{4 i \left (f x +e \right )}+6 a \,{\mathrm e}^{2 i \left (f x +e \right )}-6 b \,{\mathrm e}^{2 i \left (f x +e \right )}-2 a +3 b \right )}{3 f \,a^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{3}}+\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right )}{2 a^{2} f}-\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right ) b}{2 a^{3} f}-\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right )}{2 a^{2} f}+\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right ) b}{2 a^{3} f}\) | \(259\) |
1/f*(-1/3/a/tan(f*x+e)^3-(a-b)/a^2/tan(f*x+e)-b*(a-b)/a^2/(a*b)^(1/2)*arct an(b*tan(f*x+e)/(a*b)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 142 vs. \(2 (66) = 132\).
Time = 0.30 (sec) , antiderivative size = 373, normalized size of antiderivative = 4.91 \[ \int \frac {\csc ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\left [-\frac {4 \, {\left (2 \, a - 3 \, b\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - a + b\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{3} - a b \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a}} \sin \left (f x + e\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}}\right ) \sin \left (f x + e\right ) - 12 \, {\left (a - b\right )} \cos \left (f x + e\right )}{12 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f\right )} \sin \left (f x + e\right )}, -\frac {2 \, {\left (2 \, a - 3 \, b\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - a + b\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {{\left ({\left (a + b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 6 \, {\left (a - b\right )} \cos \left (f x + e\right )}{6 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f\right )} \sin \left (f x + e\right )}\right ] \]
[-1/12*(4*(2*a - 3*b)*cos(f*x + e)^3 + 3*((a - b)*cos(f*x + e)^2 - a + b)* sqrt(-b/a)*log(((a^2 + 6*a*b + b^2)*cos(f*x + e)^4 - 2*(3*a*b + b^2)*cos(f *x + e)^2 - 4*((a^2 + a*b)*cos(f*x + e)^3 - a*b*cos(f*x + e))*sqrt(-b/a)*s in(f*x + e) + b^2)/((a^2 - 2*a*b + b^2)*cos(f*x + e)^4 + 2*(a*b - b^2)*cos (f*x + e)^2 + b^2))*sin(f*x + e) - 12*(a - b)*cos(f*x + e))/((a^2*f*cos(f* x + e)^2 - a^2*f)*sin(f*x + e)), -1/6*(2*(2*a - 3*b)*cos(f*x + e)^3 - 3*(( a - b)*cos(f*x + e)^2 - a + b)*sqrt(b/a)*arctan(1/2*((a + b)*cos(f*x + e)^ 2 - b)*sqrt(b/a)/(b*cos(f*x + e)*sin(f*x + e)))*sin(f*x + e) - 6*(a - b)*c os(f*x + e))/((a^2*f*cos(f*x + e)^2 - a^2*f)*sin(f*x + e))]
\[ \int \frac {\csc ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\int \frac {\csc ^{4}{\left (e + f x \right )}}{a + b \tan ^{2}{\left (e + f x \right )}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.89 \[ \int \frac {\csc ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {\frac {3 \, {\left (a b - b^{2}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2}} + \frac {3 \, {\left (a - b\right )} \tan \left (f x + e\right )^{2} + a}{a^{2} \tan \left (f x + e\right )^{3}}}{3 \, f} \]
-1/3*(3*(a*b - b^2)*arctan(b*tan(f*x + e)/sqrt(a*b))/(sqrt(a*b)*a^2) + (3* (a - b)*tan(f*x + e)^2 + a)/(a^2*tan(f*x + e)^3))/f
Time = 0.48 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.21 \[ \int \frac {\csc ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {\frac {3 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )\right )} {\left (a b - b^{2}\right )}}{\sqrt {a b} a^{2}} + \frac {3 \, a \tan \left (f x + e\right )^{2} - 3 \, b \tan \left (f x + e\right )^{2} + a}{a^{2} \tan \left (f x + e\right )^{3}}}{3 \, f} \]
-1/3*(3*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt( a*b)))*(a*b - b^2)/(sqrt(a*b)*a^2) + (3*a*tan(f*x + e)^2 - 3*b*tan(f*x + e )^2 + a)/(a^2*tan(f*x + e)^3))/f
Time = 10.07 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.88 \[ \int \frac {\csc ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {\frac {1}{3\,a}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (a-b\right )}{a^2}}{f\,{\mathrm {tan}\left (e+f\,x\right )}^3}-\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )}{\sqrt {a}}\right )\,\left (a-b\right )}{a^{5/2}\,f} \]