Integrand size = 25, antiderivative size = 143 \[ \int \frac {\csc ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {3 (a-b)^2 \text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{8 a^{5/2} f}-\frac {(5 a-3 b) \cot (e+f x) \csc (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{8 a^2 f}-\frac {\cot ^3(e+f x) \csc (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{4 a f} \]
-3/8*(a-b)^2*arctanh(sec(f*x+e)*a^(1/2)/(a-b+b*sec(f*x+e)^2)^(1/2))/a^(5/2 )/f-1/8*(5*a-3*b)*cot(f*x+e)*csc(f*x+e)*(a-b+b*sec(f*x+e)^2)^(1/2)/a^2/f-1 /4*cot(f*x+e)^3*csc(f*x+e)*(a-b+b*sec(f*x+e)^2)^(1/2)/a/f
Time = 4.69 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.91 \[ \int \frac {\csc ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\frac {\left (-\sqrt {2} \sqrt {a} \cot (e+f x) \csc (e+f x) \left (3 a-3 b+2 a \csc ^2(e+f x)\right )+\frac {3 (a-b)^2 \cos (e+f x) \left (2 \text {arctanh}\left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-\frac {\sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}}{\sqrt {a}}\right )+\log \left (a-2 b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}\right )\right ) \sec ^2\left (\frac {1}{2} (e+f x)\right )}{\sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^4\left (\frac {1}{2} (e+f x)\right )}}\right ) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}{16 a^{5/2} f} \]
((-(Sqrt[2]*Sqrt[a]*Cot[e + f*x]*Csc[e + f*x]*(3*a - 3*b + 2*a*Csc[e + f*x ]^2)) + (3*(a - b)^2*Cos[e + f*x]*(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 + T an[(e + f*x)/2]^2)^2]])*Sec[(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])/(16*a^(5/2)*f)
Time = 0.36 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4147, 25, 372, 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.
| \(\displaystyle \int \frac {\csc ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (e+f x)^5 \sqrt {a+b \tan (e+f x)^2}}dx\) |
| \(\Big \downarrow \) 4147 |
| \(\displaystyle \frac {\int -\frac {\sec ^4(e+f x)}{\left (1-\sec ^2(e+f x)\right )^3 \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \frac {\sec ^4(e+f x)}{\left (1-\sec ^2(e+f x)\right )^3 \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 372 |
| \(\displaystyle \frac {\frac {\int \frac {2 (2 a-b) \sec ^2(e+f x)+a-b}{\left (1-\sec ^2(e+f x)\right )^2 \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)}{4 a}-\frac {\sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{4 a \left (1-\sec ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\int -\frac {3 (a-b)^2}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)}{2 a}+\frac {(5 a-3 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{2 a \left (1-\sec ^2(e+f x)\right )}}{4 a}-\frac {\sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{4 a \left (1-\sec ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {(5 a-3 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{2 a \left (1-\sec ^2(e+f x)\right )}-\frac {3 (a-b)^2 \int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)}{2 a}}{4 a}-\frac {\sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{4 a \left (1-\sec ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {\frac {(5 a-3 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{2 a \left (1-\sec ^2(e+f x)\right )}-\frac {3 (a-b)^2 \int \frac {1}{1-\frac {a \sec ^2(e+f x)}{b \sec ^2(e+f x)+a-b}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a-b}}}{2 a}}{4 a}-\frac {\sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{4 a \left (1-\sec ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\frac {(5 a-3 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{2 a \left (1-\sec ^2(e+f x)\right )}-\frac {3 (a-b)^2 \text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{2 a^{3/2}}}{4 a}-\frac {\sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{4 a \left (1-\sec ^2(e+f x)\right )^2}}{f}\) |
(-1/4*(Sec[e + f*x]*Sqrt[a - b + b*Sec[e + f*x]^2])/(a*(1 - Sec[e + f*x]^2 )^2) + ((-3*(a - b)^2*ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/(2*a^(3/2)) + ((5*a - 3*b)*Sec[e + f*x]*Sqrt[a - b + b*Sec[e + f*x]^2])/(2*a*(1 - Sec[e + f*x]^2)))/(4*a))/f
3.2.21.3.1 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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 )^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 )) Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 3] && IntBinomialQ[a , b, c, d, e, m, 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. \(1626\) vs. \(2(127)=254\).
Time = 0.89 (sec) , antiderivative size = 1627, normalized size of antiderivative = 11.38
1/64/f/a^(7/2)*(a*(-cos(f*x+e)+1)^4*csc(f*x+e)^4-2*a*(-cos(f*x+e)+1)^2*csc (f*x+e)^2+4*b*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+a)^(1/2)/((a*(-cos(f*x+e)+1)^ 4*csc(f*x+e)^4-2*a*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+4*b*(-cos(f*x+e)+1)^2*cs c(f*x+e)^2+a)/((-cos(f*x+e)+1)^2*csc(f*x+e)^2-1)^2)^(1/2)/((-cos(f*x+e)+1) ^2*csc(f*x+e)^2-1)/(-cos(f*x+e)+1)^4*((-cos(f*x+e)+1)^6*(a*(-cos(f*x+e)+1) ^4*csc(f*x+e)^4-2*a*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+4*b*(-cos(f*x+e)+1)^2*c sc(f*x+e)^2+a)^(1/2)*a^(5/2)*csc(f*x+e)^2+12*ln((a*(-cos(f*x+e)+1)^2*csc(f *x+e)^2+(a*(-cos(f*x+e)+1)^4*csc(f*x+e)^4-2*a*(-cos(f*x+e)+1)^2*csc(f*x+e) ^2+4*b*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+a)^(1/2)*a^(1/2)-a+2*b)/a^(1/2))*a^3 *(-cos(f*x+e)+1)^4+12*ln(2/(-cos(f*x+e)+1)^2*(-a*(-cos(f*x+e)+1)^2+2*b*(-c os(f*x+e)+1)^2+(a*(-cos(f*x+e)+1)^4*csc(f*x+e)^4-2*a*(-cos(f*x+e)+1)^2*csc (f*x+e)^2+4*b*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+a)^(1/2)*a^(1/2)*sin(f*x+e)^2 +a*sin(f*x+e)^2))*a^3*(-cos(f*x+e)+1)^4+9*(a*(-cos(f*x+e)+1)^4*csc(f*x+e)^ 4-2*a*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+4*b*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+a) ^(1/2)*a^(5/2)*(-cos(f*x+e)+1)^4-6*a^(3/2)*(a*(-cos(f*x+e)+1)^4*csc(f*x+e) ^4-2*a*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+4*b*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+a )^(1/2)*b*(-cos(f*x+e)+1)^4-24*ln((a*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+(a*(-c os(f*x+e)+1)^4*csc(f*x+e)^4-2*a*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+4*b*(-cos(f *x+e)+1)^2*csc(f*x+e)^2+a)^(1/2)*a^(1/2)-a+2*b)/a^(1/2))*a^2*(-cos(f*x+e)+ 1)^4*b+12*ln((a*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+(a*(-cos(f*x+e)+1)^4*csc...
Time = 0.37 (sec) , antiderivative size = 437, normalized size of antiderivative = 3.06 \[ \int \frac {\csc ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\left [\frac {3 \, {\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} - 2 \, a b + b^{2}\right )} \sqrt {a} \log \left (-\frac {2 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt {a} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) + 2 \, {\left (3 \, {\left (a^{2} - a b\right )} \cos \left (f x + e\right )^{3} - {\left (5 \, a^{2} - 3 \, a b\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{16 \, {\left (a^{3} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} f \cos \left (f x + e\right )^{2} + a^{3} f\right )}}, \frac {3 \, {\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} - 2 \, a b + b^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a}\right ) + {\left (3 \, {\left (a^{2} - a b\right )} \cos \left (f x + e\right )^{3} - {\left (5 \, a^{2} - 3 \, a b\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{8 \, {\left (a^{3} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} f \cos \left (f x + e\right )^{2} + a^{3} f\right )}}\right ] \]
[1/16*(3*((a^2 - 2*a*b + b^2)*cos(f*x + e)^4 - 2*(a^2 - 2*a*b + b^2)*cos(f *x + e)^2 + a^2 - 2*a*b + b^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*(a^2 - a*b)*cos(f*x + e)^3 - (5*a^2 - 3 *a*b)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(a^ 3*f*cos(f*x + e)^4 - 2*a^3*f*cos(f*x + e)^2 + a^3*f), 1/8*(3*((a^2 - 2*a*b + b^2)*cos(f*x + e)^4 - 2*(a^2 - 2*a*b + b^2)*cos(f*x + e)^2 + a^2 - 2*a* b + b^2)*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^2 - a*b)*cos(f*x + e)^3 - (5*a^2 - 3*a*b )*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(a^3*f* cos(f*x + e)^4 - 2*a^3*f*cos(f*x + e)^2 + a^3*f)]
\[ \int \frac {\csc ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int \frac {\csc ^{5}{\left (e + f x \right )}}{\sqrt {a + b \tan ^{2}{\left (e + f x \right )}}}\, dx \]
\[ \int \frac {\csc ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int { \frac {\csc \left (f x + e\right )^{5}}{\sqrt {b \tan \left (f x + e\right )^{2} + a}} \,d x } \]
\[ \int \frac {\csc ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int { \frac {\csc \left (f x + e\right )^{5}}{\sqrt {b \tan \left (f x + e\right )^{2} + a}} \,d x } \]
Timed out. \[ \int \frac {\csc ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int \frac {1}{{\sin \left (e+f\,x\right )}^5\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}} \,d x \]