Integrand size = 25, antiderivative size = 167 \[ \int \csc ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=-\frac {\sqrt {a} (a+3 b) \text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{2 f}+\frac {\sqrt {b} (3 a+b) \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{2 f}+\frac {b \sec (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{f}-\frac {\cot (e+f x) \csc (e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{2 f} \]
-1/2*cot(f*x+e)*csc(f*x+e)*(a-b+b*sec(f*x+e)^2)^(3/2)/f-1/2*(a+3*b)*arctan h(sec(f*x+e)*a^(1/2)/(a-b+b*sec(f*x+e)^2)^(1/2))*a^(1/2)/f+1/2*(3*a+b)*arc tanh(sec(f*x+e)*b^(1/2)/(a-b+b*sec(f*x+e)^2)^(1/2))*b^(1/2)/f+b*sec(f*x+e) *(a-b+b*sec(f*x+e)^2)^(1/2)/f
Leaf count is larger than twice the leaf count of optimal. \(1012\) vs. \(2(167)=334\).
Time = 7.58 (sec) , antiderivative size = 1012, normalized size of antiderivative = 6.06 \[ \int \csc ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\frac {\sqrt {\frac {a+b+a \cos (2 (e+f x))-b \cos (2 (e+f x))}{1+\cos (2 (e+f x))}} \left (-\frac {1}{2} a \cot (e+f x) \csc (e+f x)+\frac {1}{2} b \sec (e+f x)\right )}{f}+\frac {\frac {\left (a^2-b^2\right ) (1+\cos (e+f x)) \sqrt {\frac {1+\cos (2 (e+f x))}{(1+\cos (e+f x))^2}} \sqrt {\frac {a+b+(a-b) \cos (2 (e+f x))}{1+\cos (2 (e+f x))}} \left (4 \sqrt {a} \text {arctanh}\left (\frac {-\sqrt {a} \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+\sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}}{2 \sqrt {b}}\right )-\sqrt {b} \left (2 \text {arctanh}\left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-\frac {\sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}}{\sqrt {a}}\right )+\log \left (a-2 b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}\right )\right )\right ) \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \left (1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {\frac {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}{\left (1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}}}{4 \sqrt {a} \sqrt {b} \sqrt {a+b+(a-b) \cos (2 (e+f x))} \sqrt {\left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2} \sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}}-\frac {\left (a^2+6 a b+b^2\right ) (1+\cos (e+f x)) \sqrt {\frac {1+\cos (2 (e+f x))}{(1+\cos (e+f x))^2}} \sqrt {\frac {a+b+(a-b) \cos (2 (e+f x))}{1+\cos (2 (e+f x))}} \left (4 \sqrt {a} \text {arctanh}\left (\frac {-\sqrt {a} \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+\sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}}{2 \sqrt {b}}\right )+\sqrt {b} \left (2 \text {arctanh}\left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-\frac {\sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}}{\sqrt {a}}\right )+\log \left (a-2 b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}\right )\right )\right ) \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \left (1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {\frac {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}{\left (1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}}}{4 \sqrt {a} \sqrt {b} \sqrt {a+b+(a-b) \cos (2 (e+f x))} \sqrt {\left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2} \sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}}}{2 f} \]
(Sqrt[(a + b + a*Cos[2*(e + f*x)] - b*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f* x)])]*(-1/2*(a*Cot[e + f*x]*Csc[e + f*x]) + (b*Sec[e + f*x])/2))/f + (((a^ 2 - b^2)*(1 + Cos[e + f*x])*Sqrt[(1 + Cos[2*(e + f*x)])/(1 + Cos[e + f*x]) ^2]*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*(4*Sqr t[a]*ArcTanh[(-(Sqrt[a]*(-1 + Tan[(e + f*x)/2]^2)) + Sqrt[4*b*Tan[(e + f*x )/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2])/(2*Sqrt[b])] - Sqrt[b]*(2*ArcTanh [Tan[(e + f*x)/2]^2 - Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/ 2]^2)^2]/Sqrt[a]] + Log[a - 2*b - a*Tan[(e + f*x)/2]^2 + Sqrt[a]*Sqrt[4*b* Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2]]))*(-1 + Tan[(e + f*x) /2]^2)*(1 + Tan[(e + f*x)/2]^2)*Sqrt[(4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan [(e + f*x)/2]^2)^2)/(1 + Tan[(e + f*x)/2]^2)^2])/(4*Sqrt[a]*Sqrt[b]*Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]]*Sqrt[(-1 + Tan[(e + f*x)/2]^2)^2]*Sqrt[4* b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2]) - ((a^2 + 6*a*b + b ^2)*(1 + Cos[e + f*x])*Sqrt[(1 + Cos[2*(e + f*x)])/(1 + Cos[e + f*x])^2]*S qrt[(a + b + (a - b)*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*(4*Sqrt[a]* ArcTanh[(-(Sqrt[a]*(-1 + Tan[(e + f*x)/2]^2)) + Sqrt[4*b*Tan[(e + f*x)/2]^ 2 + a*(-1 + Tan[(e + f*x)/2]^2)^2])/(2*Sqrt[b])] + Sqrt[b]*(2*ArcTanh[Tan[ (e + f*x)/2]^2 - Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2) ^2]/Sqrt[a]] + Log[a - 2*b - a*Tan[(e + f*x)/2]^2 + Sqrt[a]*Sqrt[4*b*Tan[( e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2]]))*(-1 + Tan[(e + f*x)/2...
Time = 0.39 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.01, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4147, 369, 403, 27, 398, 224, 219, 291, 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.
| \(\displaystyle \int \csc ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \tan (e+f x)^2\right )^{3/2}}{\sin (e+f x)^3}dx\) |
| \(\Big \downarrow \) 4147 |
| \(\displaystyle \frac {\int \frac {\sec ^2(e+f x) \left (b \sec ^2(e+f x)+a-b\right )^{3/2}}{\left (1-\sec ^2(e+f x)\right )^2}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 369 |
| \(\displaystyle \frac {\frac {\sec (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{2 \left (1-\sec ^2(e+f x)\right )}-\frac {1}{2} \int \frac {\sqrt {b \sec ^2(e+f x)+a-b} \left (4 b \sec ^2(e+f x)+a-b\right )}{1-\sec ^2(e+f x)}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 403 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{2} \int -\frac {2 \left (a^2-b^2+b (3 a+b) \sec ^2(e+f x)\right )}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)+2 b \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}\right )+\frac {\sec (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{2 \left (1-\sec ^2(e+f x)\right )}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{2} \left (2 b \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}-\int \frac {a^2-b^2+b (3 a+b) \sec ^2(e+f x)}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)\right )+\frac {\sec (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{2 \left (1-\sec ^2(e+f x)\right )}}{f}\) |
| \(\Big \downarrow \) 398 |
| \(\displaystyle \frac {\frac {1}{2} \left (b (3 a+b) \int \frac {1}{\sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)-a (a+3 b) \int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)+2 b \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}\right )+\frac {\sec (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{2 \left (1-\sec ^2(e+f x)\right )}}{f}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {1}{2} \left (-a (a+3 b) \int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)+b (3 a+b) \int \frac {1}{1-\frac {b \sec ^2(e+f x)}{b \sec ^2(e+f x)+a-b}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a-b}}+2 b \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}\right )+\frac {\sec (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{2 \left (1-\sec ^2(e+f x)\right )}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {1}{2} \left (-a (a+3 b) \int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)+\sqrt {b} (3 a+b) \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )+2 b \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}\right )+\frac {\sec (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{2 \left (1-\sec ^2(e+f x)\right )}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {1}{2} \left (-a (a+3 b) \int \frac {1}{1-\frac {a \sec ^2(e+f x)}{b \sec ^2(e+f x)+a-b}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a-b}}+\sqrt {b} (3 a+b) \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )+2 b \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}\right )+\frac {\sec (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{2 \left (1-\sec ^2(e+f x)\right )}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {1}{2} \left (-\sqrt {a} (a+3 b) \text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )+\sqrt {b} (3 a+b) \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )+2 b \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}\right )+\frac {\sec (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{2 \left (1-\sec ^2(e+f x)\right )}}{f}\) |
((Sec[e + f*x]*(a - b + b*Sec[e + f*x]^2)^(3/2))/(2*(1 - Sec[e + f*x]^2)) + (-(Sqrt[a]*(a + 3*b)*ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]]) + Sqrt[b]*(3*a + b)*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]] + 2*b*Sec[e + f*x]*Sqrt[a - b + b*Sec[e + f*x]^2])/2 )/f
3.2.8.3.1 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[(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[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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/(2* b*(p + 1))), x] - Simp[e^2/(2*b*(p + 1)) Int[(e*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 1)*Simp[c*(m - 1) + d*(m + 2*q - 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 0 ] && GtQ[m, 1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]) , x_Symbol] :> Simp[f/b Int[1/Sqrt[c + d*x^2], x], x] + Simp[(b*e - a*f)/ b Int[1/((a + b*x^2)*Sqrt[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[f*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(b*(2*(p + q + 1) + 1))), x] + Simp[1/(b*(2*(p + q + 1) + 1)) Int[(a + b*x^2)^p*(c + d*x^2)^(q - 1)*Simp[c*(b*e - a*f + b*e*2*(p + q + 1)) + (d*(b*e - a*f) + f*2*q*(b*c - a*d) + b*d*e*2*(p + q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && GtQ[q, 0] && NeQ[2*(p + q + 1) + 1, 0]
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]
Leaf count of result is larger than twice the leaf count of optimal. \(1727\) vs. \(2(145)=290\).
Time = 2.57 (sec) , antiderivative size = 1728, normalized size of antiderivative = 10.35
1/4/f/a/b*(-2*cos(f*x+e)^3*ln(-4*b^(1/2)*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b )/(cos(f*x+e)+1)^2)^(1/2)-4*b^(1/2)*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(co s(f*x+e)+1)^2)^(1/2)*sec(f*x+e)-4*b*sec(f*x+e))*b^(5/2)*a+cos(f*x+e)^3*ln( 2/a^(1/2)*(cos(f*x+e)*a^(1/2)*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+ e)+1)^2)^(1/2)+((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)* a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+b)/(cos(f*x+e)+1))*a^(5/2)*b+cos(f*x+e)^ 3*ln(-4*(cos(f*x+e)*a^(1/2)*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e) +1)^2)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/ (cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/(cos(f*x+e)-1))*a^(5/2)*b-6*cos(f*x+e)^ 3*ln(-4*b^(1/2)*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2) -4*b^(1/2)*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*sec( f*x+e)-4*b*sec(f*x+e))*b^(3/2)*a^2+3*cos(f*x+e)^3*ln(2/a^(1/2)*(cos(f*x+e) *a^(1/2)*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)+((a*co s(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+ b*cos(f*x+e)+b)/(cos(f*x+e)+1))*a^(3/2)*b^2+3*cos(f*x+e)^3*ln(-4*(cos(f*x+ e)*a^(1/2)*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)+cos( f*x+e)*a-b*cos(f*x+e)+((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2) ^(1/2)*a^(1/2)+b)/(cos(f*x+e)-1))*a^(3/2)*b^2+2*cos(f*x+e)^2*ln(-4*b^(1/2) *((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)-4*b^(1/2)*((a* cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*sec(f*x+e)-4*b*s...
Time = 0.99 (sec) , antiderivative size = 994, normalized size of antiderivative = 5.95 \[ \int \csc ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\text {Too large to display} \]
[1/4*(((a + 3*b)*cos(f*x + e)^3 - (a + 3*b)*cos(f*x + e))*sqrt(a)*log(-2*( (a - b)*cos(f*x + e)^2 - 2*sqrt(a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f *x + e)^2)*cos(f*x + e) + a + b)/(cos(f*x + e)^2 - 1)) + ((3*a + b)*cos(f* x + e)^3 - (3*a + b)*cos(f*x + e))*sqrt(b)*log(-((a - b)*cos(f*x + e)^2 + 2*sqrt(b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + 2*b)/cos(f*x + e)^2) + 2*((a + b)*cos(f*x + e)^2 - b)*sqrt(((a - b)*cos(f *x + e)^2 + b)/cos(f*x + e)^2))/(f*cos(f*x + e)^3 - f*cos(f*x + e)), -1/4* (2*((3*a + b)*cos(f*x + e)^3 - (3*a + b)*cos(f*x + e))*sqrt(-b)*arctan(sqr t(-b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b) - ((a + 3*b)*cos(f*x + e)^3 - (a + 3*b)*cos(f*x + e))*sqrt(a)*log(-2*((a - b )*cos(f*x + e)^2 - 2*sqrt(a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e )^2)*cos(f*x + e) + a + b)/(cos(f*x + e)^2 - 1)) - 2*((a + b)*cos(f*x + e) ^2 - b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(f*cos(f*x + e) ^3 - f*cos(f*x + e)), 1/4*(2*((a + 3*b)*cos(f*x + e)^3 - (a + 3*b)*cos(f*x + e))*sqrt(-a)*arctan(sqrt(-a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/a) + ((3*a + b)*cos(f*x + e)^3 - (3*a + b)*cos(f*x + e))*sqrt(b)*log(-((a - b)*cos(f*x + e)^2 + 2*sqrt(b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + 2*b)/cos(f*x + e)^2) + 2*((a + b)*cos(f*x + e)^2 - b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)) /(f*cos(f*x + e)^3 - f*cos(f*x + e)), 1/2*(((a + 3*b)*cos(f*x + e)^3 - ...
Timed out. \[ \int \csc ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \]
\[ \int \csc ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \csc \left (f x + e\right )^{3} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 1467 vs. \(2 (145) = 290\).
Time = 2.44 (sec) , antiderivative size = 1467, normalized size of antiderivative = 8.78 \[ \int \csc ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\text {Too large to display} \]
1/8*(sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan( 1/2*f*x + 1/2*e)^2 + a)*a*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) - 4*(a^2*sgn(tan (1/2*f*x + 1/2*e)^2 - 1) + 3*a*b*sgn(tan(1/2*f*x + 1/2*e)^2 - 1))*arctan(- (sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan( 1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))/sqrt(-a))/sqrt(-a) + 8*(3*a*b*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) + b^2*sgn(tan(1/2*f*x + 1/2*e)^2 - 1))*arctan(-1/2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a) - sqrt(a))/sqrt(-b))/sqrt(-b) - 2*(a^2*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) + 3*a *b*sgn(tan(1/2*f*x + 1/2*e)^2 - 1))*log(abs((sqrt(a)*tan(1/2*f*x + 1/2*e)^ 2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1 /2*f*x + 1/2*e)^2 + a))*sqrt(a) - a + 2*b))/sqrt(a) + 2*((sqrt(a)*tan(1/2* f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^ 2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*a^2*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) - 2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*t an(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*a*b*sgn(tan(1/2*f *x + 1/2*e)^2 - 1) - a^(5/2)*sgn(tan(1/2*f*x + 1/2*e)^2 - 1))/((sqrt(a)*ta n(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1 /2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^2 - a) + 16*((sqrt(a)*tan(1/2*f *x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e...
Timed out. \[ \int \csc ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\int \frac {{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2}}{{\sin \left (e+f\,x\right )}^3} \,d x \]