Integrand size = 25, antiderivative size = 114 \[ \int \frac {\csc ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {(3 a-4 b) \cot (e+f x)}{3 a^2 f \sqrt {a+b \tan ^2(e+f x)}}-\frac {\cot ^3(e+f x)}{3 a f \sqrt {a+b \tan ^2(e+f x)}}-\frac {2 (3 a-4 b) b \tan (e+f x)}{3 a^3 f \sqrt {a+b \tan ^2(e+f x)}} \]
-1/3*(3*a-4*b)*cot(f*x+e)/a^2/f/(a+b*tan(f*x+e)^2)^(1/2)-1/3*cot(f*x+e)^3/ a/f/(a+b*tan(f*x+e)^2)^(1/2)-2/3*(3*a-4*b)*b*tan(f*x+e)/a^3/f/(a+b*tan(f*x +e)^2)^(1/2)
Time = 1.54 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.04 \[ \int \frac {\csc ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\frac {\left (-3 a^2-7 a b+12 b^2-2 \left (a^2-6 a b+8 b^2\right ) \cos (2 (e+f x))+\left (a^2-5 a b+4 b^2\right ) \cos (4 (e+f x))\right ) \csc ^3(e+f x) \sec (e+f x)}{6 \sqrt {2} a^3 f \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}} \]
((-3*a^2 - 7*a*b + 12*b^2 - 2*(a^2 - 6*a*b + 8*b^2)*Cos[2*(e + f*x)] + (a^ 2 - 5*a*b + 4*b^2)*Cos[4*(e + f*x)])*Csc[e + f*x]^3*Sec[e + f*x])/(6*Sqrt[ 2]*a^3*f*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])
Time = 0.29 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4146, 359, 245, 208}
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.
\(\displaystyle \int \frac {\csc ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (e+f x)^4 \left (a+b \tan (e+f x)^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4146 |
\(\displaystyle \frac {\int \frac {\cot ^4(e+f x) \left (\tan ^2(e+f x)+1\right )}{\left (b \tan ^2(e+f x)+a\right )^{3/2}}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 359 |
\(\displaystyle \frac {\frac {(3 a-4 b) \int \frac {\cot ^2(e+f x)}{\left (b \tan ^2(e+f x)+a\right )^{3/2}}d\tan (e+f x)}{3 a}-\frac {\cot ^3(e+f x)}{3 a \sqrt {a+b \tan ^2(e+f x)}}}{f}\) |
\(\Big \downarrow \) 245 |
\(\displaystyle \frac {\frac {(3 a-4 b) \left (-\frac {2 b \int \frac {1}{\left (b \tan ^2(e+f x)+a\right )^{3/2}}d\tan (e+f x)}{a}-\frac {\cot (e+f x)}{a \sqrt {a+b \tan ^2(e+f x)}}\right )}{3 a}-\frac {\cot ^3(e+f x)}{3 a \sqrt {a+b \tan ^2(e+f x)}}}{f}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {\frac {(3 a-4 b) \left (-\frac {2 b \tan (e+f x)}{a^2 \sqrt {a+b \tan ^2(e+f x)}}-\frac {\cot (e+f x)}{a \sqrt {a+b \tan ^2(e+f x)}}\right )}{3 a}-\frac {\cot ^3(e+f x)}{3 a \sqrt {a+b \tan ^2(e+f x)}}}{f}\) |
(-1/3*Cot[e + f*x]^3/(a*Sqrt[a + b*Tan[e + f*x]^2]) + ((3*a - 4*b)*(-(Cot[ e + f*x]/(a*Sqrt[a + b*Tan[e + f*x]^2])) - (2*b*Tan[e + f*x])/(a^2*Sqrt[a + b*Tan[e + f*x]^2])))/(3*a))/f
3.2.38.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + 2*(p + 1) + 1)/(a*(m + 1))) Int[x^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, m, p}, x] && ILtQ[Si mplify[(m + 1)/2 + p + 1], 0] && NeQ[m, -1]
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 = 2.88 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.16
method | result | size |
default | \(\frac {\left (a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}\right ) \left (8 \sin \left (f x +e \right )^{4} b^{2}+10 \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{2} a b +2 a^{2} \cos \left (f x +e \right )^{4}-6 a b \sin \left (f x +e \right )^{2}-3 \cos \left (f x +e \right )^{2} a^{2}\right ) \sec \left (f x +e \right )^{3} \csc \left (f x +e \right )^{3}}{3 f \,a^{3} \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}\) | \(132\) |
1/3/f/a^3*(a*cos(f*x+e)^2+b*sin(f*x+e)^2)*(8*sin(f*x+e)^4*b^2+10*cos(f*x+e )^2*sin(f*x+e)^2*a*b+2*a^2*cos(f*x+e)^4-6*a*b*sin(f*x+e)^2-3*cos(f*x+e)^2* a^2)/(a+b*tan(f*x+e)^2)^(3/2)*sec(f*x+e)^3*csc(f*x+e)^3
Time = 2.13 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.36 \[ \int \frac {\csc ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {{\left (2 \, {\left (a^{2} - 5 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - {\left (3 \, a^{2} - 16 \, a b + 16 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - 2 \, {\left (3 \, a b - 4 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{3 \, {\left ({\left (a^{4} - a^{3} b\right )} f \cos \left (f x + e\right )^{4} - a^{3} b f - {\left (a^{4} - 2 \, a^{3} b\right )} f \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )} \]
-1/3*(2*(a^2 - 5*a*b + 4*b^2)*cos(f*x + e)^5 - (3*a^2 - 16*a*b + 16*b^2)*c os(f*x + e)^3 - 2*(3*a*b - 4*b^2)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e) ^2 + b)/cos(f*x + e)^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)*sin(f*x + e))
\[ \int \frac {\csc ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\csc ^{4}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
Time = 0.24 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.24 \[ \int \frac {\csc ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\frac {6 \, b \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a^{2}} - \frac {8 \, b^{2} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a^{3}} + \frac {3}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a \tan \left (f x + e\right )} - \frac {4 \, b}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a^{2} \tan \left (f x + e\right )} + \frac {1}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a \tan \left (f x + e\right )^{3}}}{3 \, f} \]
-1/3*(6*b*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a)*a^2) - 8*b^2*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a)*a^3) + 3/(sqrt(b*tan(f*x + e)^2 + a)*a*tan( f*x + e)) - 4*b/(sqrt(b*tan(f*x + e)^2 + a)*a^2*tan(f*x + e)) + 1/(sqrt(b* tan(f*x + e)^2 + a)*a*tan(f*x + e)^3))/f
\[ \int \frac {\csc ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\csc \left (f x + e\right )^{4}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]
Time = 34.89 (sec) , antiderivative size = 269040, normalized size of antiderivative = 2360.00 \[ \int \frac {\csc ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display} \]
((a + (b*(exp(e*2i + f*x*2i)*1i - 1i)^2)/(exp(e*2i + f*x*2i) + 1)^2)^(1/2) *(2*exp(e*2i + f*x*2i) + exp(e*4i + f*x*4i) + 1)*(((a + 3*b)*(((a + 3*b)*( ((a + 3*b)*(((a + 3*b)*(((a + 3*b)*((((((a - b)*(a - 2*b) - (a + 2*b)^2)*( a - b)^2*(a + 2*b))/(a*b - a^2) + (((a - b)*(a + 2*b) - (a - b)*(a + 3*b)) *(a - b)*(a + 2*b)^2)/(a*b - a^2))*(a - b)^5)/(3072*a^4*f*(a*b^2 - a^2*b)* (a + 2*b)*(a*1i - b*1i)) + ((a - b)^7*(a + 2*b)*(a + 3*b))/(3072*a^4*f*(a* b^2 - a^2*b)*(a*b - a^2)*(a*1i - b*1i)) - ((a - b)^7*(a + 2*b)*(3*a + b))/ (1024*a^4*f*(a*b^2 - a^2*b)*(a*b - a^2)*(a*1i - b*1i))))/(a - b) + ((a + 3 *b)*(((a + 3*b)*((((((a - b)*(a - 2*b) - (a + 2*b)^2)*(a - b)^2*(a + 2*b)) /(a*b - a^2) + (((a - b)*(a + 2*b) - (a - b)*(a + 3*b))*(a - b)*(a + 2*b)^ 2)/(a*b - a^2))*(a - b)^5)/(3072*a^4*f*(a*b^2 - a^2*b)*(a + 2*b)*(a*1i - b *1i)) + ((a - b)^7*(a + 2*b)*(a + 3*b))/(3072*a^4*f*(a*b^2 - a^2*b)*(a*b - a^2)*(a*1i - b*1i)) - ((a - b)^7*(a + 2*b)*(3*a + b))/(1024*a^4*f*(a*b^2 - a^2*b)*(a*b - a^2)*(a*1i - b*1i))))/(a - b) + (((a + 2*b)^3 + (((a - b)* (a - 2*b) - (a + 2*b)^2)*((a - b)*(a + 2*b) - (a - b)*(a + 3*b))*(a + 2*b) )/(a*b - a^2))*(a - b)^5)/(3072*a^4*f*(a*b^2 - a^2*b)*(a + 2*b)*(a*1i - b* 1i)) + ((a - b)^6*(a + 2*b)*(9*a + 4*b))/(768*a^3*f*(a*b^2 - a^2*b)*(a*b - a^2)*(a*1i - b*1i)) - (((((a - b)*(a - 2*b) - (a + 2*b)^2)*(a - b)^2*(a + 2*b))/(a*b - a^2) + (((a - b)*(a + 2*b) - (a - b)*(a + 3*b))*(a - b)*(a + 2*b)^2)/(a*b - a^2))*(a - b)^4*(3*a + b))/(1024*a^4*f*(a*b^2 - a^2*b)*...