Integrand size = 25, antiderivative size = 153 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{a^{3/2} f}+\frac {(2 a+5 b) \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a+b}}\right )}{2 (a+b)^{5/2} f}-\frac {(a-2 b) b}{2 a (a+b)^2 f \sqrt {a+b \sec ^2(e+f x)}}-\frac {\cot ^2(e+f x)}{2 (a+b) f \sqrt {a+b \sec ^2(e+f x)}} \]
-arctanh((a+b*sec(f*x+e)^2)^(1/2)/a^(1/2))/a^(3/2)/f+1/2*(2*a+5*b)*arctanh ((a+b*sec(f*x+e)^2)^(1/2)/(a+b)^(1/2))/(a+b)^(5/2)/f-1/2*(a-2*b)*b/a/(a+b) ^2/f/(a+b*sec(f*x+e)^2)^(1/2)-1/2*cot(f*x+e)^2/(a+b)/f/(a+b*sec(f*x+e)^2)^ (1/2)
\[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx \]
Time = 0.39 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.13, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4627, 354, 114, 27, 169, 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.
| \(\displaystyle \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\tan (e+f x)^3 \left (a+b \sec (e+f x)^2\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4627 |
| \(\displaystyle \frac {\int \frac {\cos (e+f x)}{\left (1-\sec ^2(e+f x)\right )^2 \left (b \sec ^2(e+f x)+a\right )^{3/2}}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {\int \frac {\cos (e+f x)}{\left (1-\sec ^2(e+f x)\right )^2 \left (b \sec ^2(e+f x)+a\right )^{3/2}}d\sec ^2(e+f x)}{2 f}\) |
| \(\Big \downarrow \) 114 |
| \(\displaystyle \frac {\frac {1}{(a+b) \left (1-\sec ^2(e+f x)\right ) \sqrt {a+b \sec ^2(e+f x)}}-\frac {\int -\frac {\cos (e+f x) \left (3 b \sec ^2(e+f x)+2 a+2 b\right )}{2 \left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a\right )^{3/2}}d\sec ^2(e+f x)}{a+b}}{2 f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {\cos (e+f x) \left (3 b \sec ^2(e+f x)+2 (a+b)\right )}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a\right )^{3/2}}d\sec ^2(e+f x)}{2 (a+b)}+\frac {1}{(a+b) \left (1-\sec ^2(e+f x)\right ) \sqrt {a+b \sec ^2(e+f x)}}}{2 f}\) |
| \(\Big \downarrow \) 169 |
| \(\displaystyle \frac {\frac {\frac {2 \int \frac {\cos (e+f x) \left (2 (a+b)^2+(a-2 b) b \sec ^2(e+f x)\right )}{2 \left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec ^2(e+f x)}{a (a+b)}-\frac {2 b (a-2 b)}{a (a+b) \sqrt {a+b \sec ^2(e+f x)}}}{2 (a+b)}+\frac {1}{(a+b) \left (1-\sec ^2(e+f x)\right ) \sqrt {a+b \sec ^2(e+f x)}}}{2 f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {\cos (e+f x) \left (2 (a+b)^2+(a-2 b) b \sec ^2(e+f x)\right )}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec ^2(e+f x)}{a (a+b)}-\frac {2 b (a-2 b)}{a (a+b) \sqrt {a+b \sec ^2(e+f x)}}}{2 (a+b)}+\frac {1}{(a+b) \left (1-\sec ^2(e+f x)\right ) \sqrt {a+b \sec ^2(e+f x)}}}{2 f}\) |
| \(\Big \downarrow \) 174 |
| \(\displaystyle \frac {\frac {\frac {a (2 a+5 b) \int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec ^2(e+f x)+2 (a+b)^2 \int \frac {\cos (e+f x)}{\sqrt {b \sec ^2(e+f x)+a}}d\sec ^2(e+f x)}{a (a+b)}-\frac {2 b (a-2 b)}{a (a+b) \sqrt {a+b \sec ^2(e+f x)}}}{2 (a+b)}+\frac {1}{(a+b) \left (1-\sec ^2(e+f x)\right ) \sqrt {a+b \sec ^2(e+f x)}}}{2 f}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {\frac {\frac {\frac {4 (a+b)^2 \int \frac {1}{\frac {\sec ^4(e+f x)}{b}-\frac {a}{b}}d\sqrt {b \sec ^2(e+f x)+a}}{b}+\frac {2 a (2 a+5 b) \int \frac {1}{\frac {a+b}{b}-\frac {\sec ^4(e+f x)}{b}}d\sqrt {b \sec ^2(e+f x)+a}}{b}}{a (a+b)}-\frac {2 b (a-2 b)}{a (a+b) \sqrt {a+b \sec ^2(e+f x)}}}{2 (a+b)}+\frac {1}{(a+b) \left (1-\sec ^2(e+f x)\right ) \sqrt {a+b \sec ^2(e+f x)}}}{2 f}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {\frac {\frac {2 a (2 a+5 b) \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}}-\frac {4 (a+b)^2 \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{\sqrt {a}}}{a (a+b)}-\frac {2 b (a-2 b)}{a (a+b) \sqrt {a+b \sec ^2(e+f x)}}}{2 (a+b)}+\frac {1}{(a+b) \left (1-\sec ^2(e+f x)\right ) \sqrt {a+b \sec ^2(e+f x)}}}{2 f}\) |
(1/((a + b)*(1 - Sec[e + f*x]^2)*Sqrt[a + b*Sec[e + f*x]^2]) + (((-4*(a + b)^2*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]])/Sqrt[a] + (2*a*(2*a + 5* b)*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a + b]])/Sqrt[a + b])/(a*(a + b )) - (2*(a - 2*b)*b)/(a*(a + b)*Sqrt[a + b*Sec[e + f*x]^2]))/(2*(a + b)))/ (2*f)
3.5.19.3.1 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] && LtQ[m, -1] && IntegersQ[ 2*m, 2*n, 2*p]
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. \(13609\) vs. \(2(131)=262\).
Time = 1.16 (sec) , antiderivative size = 13610, normalized size of antiderivative = 88.95
Leaf count of result is larger than twice the leaf count of optimal. 540 vs. \(2 (131) = 262\).
Time = 1.49 (sec) , antiderivative size = 2347, normalized size of antiderivative = 15.34 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display} \]
[1/8*(((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cos(f*x + e)^4 - a^3*b - 3*a^2* b^2 - 3*a*b^3 - b^4 - (a^4 + 2*a^3*b - 2*a*b^3 - b^4)*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*co s(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)*sq rt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)) + ((2*a^4 + 5*a^3*b)*co s(f*x + e)^4 - 2*a^3*b - 5*a^2*b^2 - (2*a^4 + 3*a^3*b - 5*a^2*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*((a^4 + a^3*b + 2*a^2*b^2 + 2*a*b^3)*c os(f*x + e)^4 + (a^3*b - a^2*b^2 - 2*a*b^3)*cos(f*x + e)^2)*sqrt((a*cos(f* x + e)^2 + b)/cos(f*x + e)^2))/((a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*f*co s(f*x + e)^4 - (a^6 + 2*a^5*b - 2*a^3*b^3 - a^2*b^4)*f*cos(f*x + e)^2 - (a ^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*f), -1/8*(2*((2*a^4 + 5*a^3*b)*cos (f*x + e)^4 - 2*a^3*b - 5*a^2*b^2 - (2*a^4 + 3*a^3*b - 5*a^2*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)) - ((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cos(f*x + e)^4 - a^3* b - 3*a^2*b^2 - 3*a*b^3 - b^4 - (a^4 + 2*a^3*b - 2*a*b^3 - b^4)*cos(f*x...
\[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\cot ^{3}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
Timed out. \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\cot \left (f x + e\right )^{3}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {{\mathrm {cot}\left (e+f\,x\right )}^3}{{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}} \,d x \]