Integrand size = 23, antiderivative size = 136 \[ \int \frac {\csc (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{a^{5/2} f}-\frac {b \sec (e+f x)}{3 a (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac {(5 a-3 b) b \sec (e+f x)}{3 a^2 (a-b)^2 f \sqrt {a-b+b \sec ^2(e+f x)}} \]
-arctanh(sec(f*x+e)*a^(1/2)/(a-b+b*sec(f*x+e)^2)^(1/2))/a^(5/2)/f-1/3*b*se c(f*x+e)/a/(a-b)/f/(a-b+b*sec(f*x+e)^2)^(3/2)-1/3*(5*a-3*b)*b*sec(f*x+e)/a ^2/(a-b)^2/f/(a-b+b*sec(f*x+e)^2)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(300\) vs. \(2(136)=272\).
Time = 10.68 (sec) , antiderivative size = 300, normalized size of antiderivative = 2.21 \[ \int \frac {\csc (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\frac {\cos (e+f x) \left (-\frac {2 \sqrt {2} \sqrt {a} b \left (6 a^2+a b-3 b^2+3 \left (2 a^2-3 a b+b^2\right ) \cos (2 (e+f x))\right )}{(a-b)^2 (a+b+(a-b) \cos (2 (e+f x)))^2}+\frac {3 \left (2 \text {arctanh}\left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-\frac {\sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}}{\sqrt {a}}\right )+\log \left (a-2 b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}\right )\right ) \sec ^2\left (\frac {1}{2} (e+f x)\right )}{\sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^4\left (\frac {1}{2} (e+f x)\right )}}\right ) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}{6 a^{5/2} f} \]
(Cos[e + f*x]*((-2*Sqrt[2]*Sqrt[a]*b*(6*a^2 + a*b - 3*b^2 + 3*(2*a^2 - 3*a *b + b^2)*Cos[2*(e + f*x)]))/((a - b)^2*(a + b + (a - b)*Cos[2*(e + f*x)]) ^2) + (3*(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]])*S ec[(e + f*x)/2]^2)/Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[(e + f*x)/2 ]^4])*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])/(6*a^(5/2)* f)
Time = 0.33 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.09, 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, 316, 25, 402, 27, 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.
| \(\displaystyle \int \frac {\csc (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (e+f x) \left (a+b \tan (e+f x)^2\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4147 |
| \(\displaystyle \frac {\int -\frac {1}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )^{5/2}}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )^{5/2}}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {\frac {\int -\frac {-2 b \sec ^2(e+f x)+3 a-b}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )^{3/2}}d\sec (e+f x)}{3 a (a-b)}-\frac {b \sec (e+f x)}{3 a (a-b) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {\int \frac {-2 b \sec ^2(e+f x)+3 a-b}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )^{3/2}}d\sec (e+f x)}{3 a (a-b)}-\frac {b \sec (e+f x)}{3 a (a-b) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {-\frac {\frac {b (5 a-3 b) \sec (e+f x)}{a (a-b) \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {\int -\frac {3 (a-b)^2}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)}{a (a-b)}}{3 a (a-b)}-\frac {b \sec (e+f x)}{3 a (a-b) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {\frac {3 (a-b) \int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)}{a}+\frac {b (5 a-3 b) \sec (e+f x)}{a (a-b) \sqrt {a+b \sec ^2(e+f x)-b}}}{3 a (a-b)}-\frac {b \sec (e+f x)}{3 a (a-b) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {-\frac {\frac {3 (a-b) \int \frac {1}{1-\frac {a \sec ^2(e+f x)}{b \sec ^2(e+f x)+a-b}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a-b}}}{a}+\frac {b (5 a-3 b) \sec (e+f x)}{a (a-b) \sqrt {a+b \sec ^2(e+f x)-b}}}{3 a (a-b)}-\frac {b \sec (e+f x)}{3 a (a-b) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {-\frac {\frac {3 (a-b) \text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{a^{3/2}}+\frac {b (5 a-3 b) \sec (e+f x)}{a (a-b) \sqrt {a+b \sec ^2(e+f x)-b}}}{3 a (a-b)}-\frac {b \sec (e+f x)}{3 a (a-b) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\) |
(-1/3*(b*Sec[e + f*x])/(a*(a - b)*(a - b + b*Sec[e + f*x]^2)^(3/2)) - ((3* (a - b)*ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/a^ (3/2) + ((5*a - 3*b)*b*Sec[e + f*x])/(a*(a - b)*Sqrt[a - b + b*Sec[e + f*x ]^2]))/(3*a*(a - b)))/f
3.2.43.3.1 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]*((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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d)) Int[(a + b*x^2)^(p + 1)*(c + d*x ^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x ], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ! ( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, p, q, 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]
Leaf count of result is larger than twice the leaf count of optimal. \(17399\) vs. \(2(122)=244\).
Time = 3.23 (sec) , antiderivative size = 17400, normalized size of antiderivative = 127.94
Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (122) = 244\).
Time = 0.40 (sec) , antiderivative size = 696, normalized size of antiderivative = 5.12 \[ \int \frac {\csc (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\left [\frac {3 \, {\left ({\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{4} + a^{2} b^{2} - 2 \, a b^{3} + b^{4} + 2 \, {\left (a^{3} b - 3 \, a^{2} b^{2} + 3 \, a b^{3} - b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {a} \log \left (-\frac {2 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt {a} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) - 2 \, {\left (3 \, {\left (2 \, a^{3} b - 3 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (5 \, a^{2} b^{2} - 3 \, a b^{3}\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}}}}{6 \, {\left ({\left (a^{7} - 4 \, a^{6} b + 6 \, a^{5} b^{2} - 4 \, a^{4} b^{3} + a^{3} b^{4}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{6} b - 3 \, a^{5} b^{2} + 3 \, a^{4} b^{3} - a^{3} b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} b^{2} - 2 \, a^{4} b^{3} + a^{3} b^{4}\right )} f\right )}}, \frac {3 \, {\left ({\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{4} + a^{2} b^{2} - 2 \, a b^{3} + b^{4} + 2 \, {\left (a^{3} b - 3 \, a^{2} b^{2} + 3 \, a b^{3} - b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a}\right ) - {\left (3 \, {\left (2 \, a^{3} b - 3 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (5 \, a^{2} b^{2} - 3 \, a b^{3}\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^{7} - 4 \, a^{6} b + 6 \, a^{5} b^{2} - 4 \, a^{4} b^{3} + a^{3} b^{4}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{6} b - 3 \, a^{5} b^{2} + 3 \, a^{4} b^{3} - a^{3} b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} b^{2} - 2 \, a^{4} b^{3} + a^{3} b^{4}\right )} f\right )}}\right ] \]
[1/6*(3*((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x + e)^4 + a^2* b^2 - 2*a*b^3 + b^4 + 2*(a^3*b - 3*a^2*b^2 + 3*a*b^3 - b^4)*cos(f*x + e)^2 )*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*(3*(2*a^3*b - 3*a^2*b^2 + a*b^3)*cos(f*x + e)^3 + (5*a^2*b^2 - 3*a*b^3) *cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^7 - 4*a^6*b + 6*a^5*b^2 - 4*a^4*b^3 + a^3*b^4)*f*cos(f*x + e)^4 + 2*(a^6*b - 3 *a^5*b^2 + 3*a^4*b^3 - a^3*b^4)*f*cos(f*x + e)^2 + (a^5*b^2 - 2*a^4*b^3 + a^3*b^4)*f), 1/3*(3*((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x + e)^4 + a^2*b^2 - 2*a*b^3 + b^4 + 2*(a^3*b - 3*a^2*b^2 + 3*a*b^3 - b^4)*co s(f*x + e)^2)*sqrt(-a)*arctan(sqrt(-a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/c os(f*x + e)^2)*cos(f*x + e)/a) - (3*(2*a^3*b - 3*a^2*b^2 + a*b^3)*cos(f*x + e)^3 + (5*a^2*b^2 - 3*a*b^3)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^7 - 4*a^6*b + 6*a^5*b^2 - 4*a^4*b^3 + a^3*b^4)*f *cos(f*x + e)^4 + 2*(a^6*b - 3*a^5*b^2 + 3*a^4*b^3 - a^3*b^4)*f*cos(f*x + e)^2 + (a^5*b^2 - 2*a^4*b^3 + a^3*b^4)*f)]
\[ \int \frac {\csc (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\csc {\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {\csc (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\csc \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {\csc (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\csc \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\csc (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {1}{\sin \left (e+f\,x\right )\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \]