Integrand size = 25, antiderivative size = 166 \[ \int \frac {\cot ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{\sqrt {a} f}-\frac {\left (8 a^2+20 a b+15 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a+b}}\right )}{8 (a+b)^{5/2} f}+\frac {(4 a+7 b) \cot ^2(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{8 (a+b)^2 f}-\frac {\cot ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{4 (a+b) f} \]
-1/8*(8*a^2+20*a*b+15*b^2)*arctanh((a+b*sec(f*x+e)^2)^(1/2)/(a+b)^(1/2))/( a+b)^(5/2)/f+arctanh((a+b*sec(f*x+e)^2)^(1/2)/a^(1/2))/f/a^(1/2)+1/8*(4*a+ 7*b)*cot(f*x+e)^2*(a+b*sec(f*x+e)^2)^(1/2)/(a+b)^2/f-1/4*cot(f*x+e)^4*(a+b *sec(f*x+e)^2)^(1/2)/(a+b)/f
\[ \int \frac {\cot ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {\cot ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx \]
Time = 0.38 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.17, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 4627, 25, 354, 114, 27, 168, 27, 174, 73, 221}
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.
\(\displaystyle \int \frac {\cot ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan (e+f x)^5 \sqrt {a+b \sec (e+f x)^2}}dx\) |
\(\Big \downarrow \) 4627 |
\(\displaystyle \frac {\int -\frac {\cos (e+f x)}{\left (1-\sec ^2(e+f x)\right )^3 \sqrt {b \sec ^2(e+f x)+a}}d\sec (e+f x)}{f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {\cos (e+f x)}{\left (1-\sec ^2(e+f x)\right )^3 \sqrt {b \sec ^2(e+f x)+a}}d\sec (e+f x)}{f}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle -\frac {\int \frac {\cos (e+f x)}{\left (1-\sec ^2(e+f x)\right )^3 \sqrt {b \sec ^2(e+f x)+a}}d\sec ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 114 |
\(\displaystyle -\frac {\frac {\sqrt {a+b \sec ^2(e+f x)}}{2 (a+b) \left (1-\sec ^2(e+f x)\right )^2}-\frac {\int -\frac {\cos (e+f x) \left (3 b \sec ^2(e+f x)+4 (a+b)\right )}{2 \left (1-\sec ^2(e+f x)\right )^2 \sqrt {b \sec ^2(e+f x)+a}}d\sec ^2(e+f x)}{2 (a+b)}}{2 f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\frac {\int \frac {\cos (e+f x) \left (3 b \sec ^2(e+f x)+4 (a+b)\right )}{\left (1-\sec ^2(e+f x)\right )^2 \sqrt {b \sec ^2(e+f x)+a}}d\sec ^2(e+f x)}{4 (a+b)}+\frac {\sqrt {a+b \sec ^2(e+f x)}}{2 (a+b) \left (1-\sec ^2(e+f x)\right )^2}}{2 f}\) |
\(\Big \downarrow \) 168 |
\(\displaystyle -\frac {\frac {\frac {(4 a+7 b) \sqrt {a+b \sec ^2(e+f x)}}{(a+b) \left (1-\sec ^2(e+f x)\right )}-\frac {\int -\frac {\cos (e+f x) \left (8 (a+b)^2+b (4 a+7 b) \sec ^2(e+f x)\right )}{2 \left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec ^2(e+f x)}{a+b}}{4 (a+b)}+\frac {\sqrt {a+b \sec ^2(e+f x)}}{2 (a+b) \left (1-\sec ^2(e+f x)\right )^2}}{2 f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\frac {\frac {\int \frac {\cos (e+f x) \left (8 (a+b)^2+b (4 a+7 b) \sec ^2(e+f x)\right )}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec ^2(e+f x)}{2 (a+b)}+\frac {(4 a+7 b) \sqrt {a+b \sec ^2(e+f x)}}{(a+b) \left (1-\sec ^2(e+f x)\right )}}{4 (a+b)}+\frac {\sqrt {a+b \sec ^2(e+f x)}}{2 (a+b) \left (1-\sec ^2(e+f x)\right )^2}}{2 f}\) |
\(\Big \downarrow \) 174 |
\(\displaystyle -\frac {\frac {\frac {\left (8 a^2+20 a b+15 b^2\right ) \int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec ^2(e+f x)+8 (a+b)^2 \int \frac {\cos (e+f x)}{\sqrt {b \sec ^2(e+f x)+a}}d\sec ^2(e+f x)}{2 (a+b)}+\frac {(4 a+7 b) \sqrt {a+b \sec ^2(e+f x)}}{(a+b) \left (1-\sec ^2(e+f x)\right )}}{4 (a+b)}+\frac {\sqrt {a+b \sec ^2(e+f x)}}{2 (a+b) \left (1-\sec ^2(e+f x)\right )^2}}{2 f}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {\frac {\frac {\frac {2 \left (8 a^2+20 a b+15 b^2\right ) \int \frac {1}{\frac {a+b}{b}-\frac {\sec ^4(e+f x)}{b}}d\sqrt {b \sec ^2(e+f x)+a}}{b}+\frac {16 (a+b)^2 \int \frac {1}{\frac {\sec ^4(e+f x)}{b}-\frac {a}{b}}d\sqrt {b \sec ^2(e+f x)+a}}{b}}{2 (a+b)}+\frac {(4 a+7 b) \sqrt {a+b \sec ^2(e+f x)}}{(a+b) \left (1-\sec ^2(e+f x)\right )}}{4 (a+b)}+\frac {\sqrt {a+b \sec ^2(e+f x)}}{2 (a+b) \left (1-\sec ^2(e+f x)\right )^2}}{2 f}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {\frac {\frac {\frac {2 \left (8 a^2+20 a b+15 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}}-\frac {16 (a+b)^2 \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{\sqrt {a}}}{2 (a+b)}+\frac {(4 a+7 b) \sqrt {a+b \sec ^2(e+f x)}}{(a+b) \left (1-\sec ^2(e+f x)\right )}}{4 (a+b)}+\frac {\sqrt {a+b \sec ^2(e+f x)}}{2 (a+b) \left (1-\sec ^2(e+f x)\right )^2}}{2 f}\) |
-1/2*(Sqrt[a + b*Sec[e + f*x]^2]/(2*(a + b)*(1 - Sec[e + f*x]^2)^2) + (((- 16*(a + b)^2*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]])/Sqrt[a] + (2*(8* a^2 + 20*a*b + 15*b^2)*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a + b]])/Sq rt[a + b])/(2*(a + b)) + ((4*a + 7*b)*Sqrt[a + b*Sec[e + f*x]^2])/((a + b) *(1 - Sec[e + f*x]^2)))/(4*(a + b)))/f
3.5.7.3.1 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[1/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] && (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n *(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]
Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* ((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d) Int[(e + f*x)^ p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d) Int[(e + f*x)^p/(c + d *x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + ( f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Si mp[1/f Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p/x), x] , x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[( m - 1)/2] && (GtQ[m, 0] || EqQ[n, 2] || EqQ[n, 4] || IGtQ[p, 0] || Integers Q[2*n, p])
Leaf count of result is larger than twice the leaf count of optimal. \(3595\) vs. \(2(144)=288\).
Time = 1.55 (sec) , antiderivative size = 3596, normalized size of antiderivative = 21.66
-1/64/f/(a+b)^(9/2)/a^(1/2)*(a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+ e))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc(f*x+e)^2+2*b*(1-cos(f*x+e))^2* csc(f*x+e)^2+a+b)^(1/2)/((a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e)) ^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc(f*x+e)^2+2*b*(1-cos(f*x+e))^2*csc (f*x+e)^2+a+b)/((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^2)^(1/2)/((1-cos(f*x+e))^ 2*csc(f*x+e)^2-1)/(1-cos(f*x+e))^4*(252*a^(5/2)*ln((a*(1-cos(f*x+e))^2*csc (f*x+e)^2+b*(1-cos(f*x+e))^2*csc(f*x+e)^2+(a*(1-cos(f*x+e))^4*csc(f*x+e)^4 +b*(1-cos(f*x+e))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc(f*x+e)^2+2*b*(1- cos(f*x+e))^2*csc(f*x+e)^2+a+b)^(1/2)*(a+b)^(1/2)-a+b)/(a+b)^(1/2))*b^2*(1 -cos(f*x+e))^4-252*a^(5/2)*ln(2/(1-cos(f*x+e))^2*(-a*(1-cos(f*x+e))^2+b*(1 -cos(f*x+e))^2+(a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e))^4*csc(f*x +e)^4-2*a*(1-cos(f*x+e))^2*csc(f*x+e)^2+2*b*(1-cos(f*x+e))^2*csc(f*x+e)^2+ a+b)^(1/2)*(a+b)^(1/2)*sin(f*x+e)^2+a*sin(f*x+e)^2+b*sin(f*x+e)^2))*b^2*(1 -cos(f*x+e))^4+(1-cos(f*x+e))^6*(a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos( f*x+e))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc(f*x+e)^2+2*b*(1-cos(f*x+e) )^2*csc(f*x+e)^2+a+b)^(1/2)*(a+b)^(5/2)*a^(3/2)*csc(f*x+e)^2+(1-cos(f*x+e) )^6*(a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e))^4*csc(f*x+e)^4-2*a*( 1-cos(f*x+e))^2*csc(f*x+e)^2+2*b*(1-cos(f*x+e))^2*csc(f*x+e)^2+a+b)^(1/2)* (a+b)^(5/2)*a^(1/2)*b*csc(f*x+e)^2-11*a^(3/2)*(a*(1-cos(f*x+e))^4*csc(f*x+ e)^4+b*(1-cos(f*x+e))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc(f*x+e)^2+...
Leaf count of result is larger than twice the leaf count of optimal. 518 vs. \(2 (144) = 288\).
Time = 1.38 (sec) , antiderivative size = 2257, normalized size of antiderivative = 13.60 \[ \int \frac {\cot ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\text {Too large to display} \]
[1/32*(4*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(f*x + e)^4 + a^3 + 3*a^2*b + 3*a*b^2 + b^3 - 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(f*x + e)^2)*sqrt(a) *log(128*a^4*cos(f*x + e)^8 + 256*a^3*b*cos(f*x + e)^6 + 160*a^2*b^2*cos(f *x + e)^4 + 32*a*b^3*cos(f*x + e)^2 + b^4 + 8*(16*a^3*cos(f*x + e)^8 + 24* a^2*b*cos(f*x + e)^6 + 10*a*b^2*cos(f*x + e)^4 + b^3*cos(f*x + e)^2)*sqrt( a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)) + ((8*a^3 + 20*a^2*b + 15* a*b^2)*cos(f*x + e)^4 + 8*a^3 + 20*a^2*b + 15*a*b^2 - 2*(8*a^3 + 20*a^2*b + 15*a*b^2)*cos(f*x + e)^2)*sqrt(a + b)*log(2*((8*a^2 + 8*a*b + b^2)*cos(f *x + e)^4 + 2*(4*a*b + 3*b^2)*cos(f*x + e)^2 + b^2 - 4*((2*a + b)*cos(f*x + e)^4 + b*cos(f*x + e)^2)*sqrt(a + b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)) - 4*(3*(2*a^3 + 5*a^2*b + 3*a*b^2)*cos(f*x + e)^4 - (4*a^3 + 11*a^2*b + 7*a*b^2)*cos(f*x + e)^2)* sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*f*cos(f*x + e)^4 - 2*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*f*cos(f*x + e)^2 + (a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*f), 1/16*(((8*a^3 + 20*a^2*b + 15*a*b^2)*cos(f*x + e)^4 + 8*a^3 + 20*a^2*b + 15*a*b^2 - 2*(8*a^3 + 20* a^2*b + 15*a*b^2)*cos(f*x + e)^2)*sqrt(-a - b)*arctan(1/2*((2*a + b)*cos(f *x + e)^2 + b)*sqrt(-a - b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(( a^2 + a*b)*cos(f*x + e)^2 + a*b + b^2)) + 2*((a^3 + 3*a^2*b + 3*a*b^2 + b^ 3)*cos(f*x + e)^4 + a^3 + 3*a^2*b + 3*a*b^2 + b^3 - 2*(a^3 + 3*a^2*b + ...
\[ \int \frac {\cot ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {\cot ^{5}{\left (e + f x \right )}}{\sqrt {a + b \sec ^{2}{\left (e + f x \right )}}}\, dx \]
\[ \int \frac {\cot ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int { \frac {\cot \left (f x + e\right )^{5}}{\sqrt {b \sec \left (f x + e\right )^{2} + a}} \,d x } \]
\[ \int \frac {\cot ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int { \frac {\cot \left (f x + e\right )^{5}}{\sqrt {b \sec \left (f x + e\right )^{2} + a}} \,d x } \]
Timed out. \[ \int \frac {\cot ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {{\mathrm {cot}\left (e+f\,x\right )}^5}{\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}} \,d x \]