Integrand size = 25, antiderivative size = 114 \[ \int \cot ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=-\frac {a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{f}+\frac {(2 a-b) \sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a+b}}\right )}{2 f}-\frac {(a+b) \cot ^2(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 f} \]
-a^(3/2)*arctanh((a+b*sec(f*x+e)^2)^(1/2)/a^(1/2))/f+1/2*(2*a-b)*arctanh(( a+b*sec(f*x+e)^2)^(1/2)/(a+b)^(1/2))*(a+b)^(1/2)/f-1/2*(a+b)*cot(f*x+e)^2* (a+b*sec(f*x+e)^2)^(1/2)/f
Result contains complex when optimal does not.
Time = 6.31 (sec) , antiderivative size = 622, normalized size of antiderivative = 5.46 \[ \int \cot ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {\sqrt {2} e^{i (e+f x)} \sqrt {4 b+a e^{-2 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^2} \cos ^3(e+f x) \left (\frac {(a+b) \left (1+e^{2 i (e+f x)}\right )}{\left (-1+e^{2 i (e+f x)}\right )^2}-\frac {-2 i a^{3/2} \sqrt {a+b} f x+\left (2 a^2+a b-b^2\right ) \log \left (1-e^{2 i (e+f x)}\right )+a^{3/2} \sqrt {a+b} \log \left (a+2 b+a e^{2 i (e+f x)}+\sqrt {a} \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}\right )+a^{3/2} \sqrt {a+b} \log \left (a+a e^{2 i (e+f x)}+2 b e^{2 i (e+f x)}+\sqrt {a} \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}\right )-2 a^2 \log \left (a+b+a e^{2 i (e+f x)}+b e^{2 i (e+f x)}+\sqrt {a+b} \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}\right )-a b \log \left (a+b+a e^{2 i (e+f x)}+b e^{2 i (e+f x)}+\sqrt {a+b} \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}\right )+b^2 \log \left (a+b+a e^{2 i (e+f x)}+b e^{2 i (e+f x)}+\sqrt {a+b} \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}\right )}{\sqrt {a+b} \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}}\right ) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{f (a+2 b+a \cos (2 e+2 f x))^{3/2}} \]
(Sqrt[2]*E^(I*(e + f*x))*Sqrt[4*b + (a*(1 + E^((2*I)*(e + f*x)))^2)/E^((2* I)*(e + f*x))]*Cos[e + f*x]^3*(((a + b)*(1 + E^((2*I)*(e + f*x))))/(-1 + E ^((2*I)*(e + f*x)))^2 - ((-2*I)*a^(3/2)*Sqrt[a + b]*f*x + (2*a^2 + a*b - b ^2)*Log[1 - E^((2*I)*(e + f*x))] + a^(3/2)*Sqrt[a + b]*Log[a + 2*b + a*E^( (2*I)*(e + f*x)) + Sqrt[a]*Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)* (e + f*x)))^2]] + a^(3/2)*Sqrt[a + b]*Log[a + a*E^((2*I)*(e + f*x)) + 2*b* E^((2*I)*(e + f*x)) + Sqrt[a]*Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2* I)*(e + f*x)))^2]] - 2*a^2*Log[a + b + a*E^((2*I)*(e + f*x)) + b*E^((2*I)* (e + f*x)) + Sqrt[a + b]*Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2]] - a*b*Log[a + b + a*E^((2*I)*(e + f*x)) + b*E^((2*I)*(e + f* x)) + Sqrt[a + b]*Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x) ))^2]] + b^2*Log[a + b + a*E^((2*I)*(e + f*x)) + b*E^((2*I)*(e + f*x)) + S qrt[a + b]*Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2]]) /(Sqrt[a + b]*Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2 ]))*(a + b*Sec[e + f*x]^2)^(3/2))/(f*(a + 2*b + a*Cos[2*e + 2*f*x])^(3/2))
Time = 0.34 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4627, 354, 109, 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.
| \(\displaystyle \int \cot ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sec (e+f x)^2\right )^{3/2}}{\tan (e+f x)^3}dx\) |
| \(\Big \downarrow \) 4627 |
| \(\displaystyle \frac {\int \frac {\cos (e+f x) \left (b \sec ^2(e+f x)+a\right )^{3/2}}{\left (1-\sec ^2(e+f x)\right )^2}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {\int \frac {\cos (e+f x) \left (b \sec ^2(e+f x)+a\right )^{3/2}}{\left (1-\sec ^2(e+f x)\right )^2}d\sec ^2(e+f x)}{2 f}\) |
| \(\Big \downarrow \) 109 |
| \(\displaystyle \frac {\frac {(a+b) \sqrt {a+b \sec ^2(e+f x)}}{1-\sec ^2(e+f x)}-\int -\frac {\cos (e+f x) \left (2 a^2+(a-b) b \sec ^2(e+f x)\right )}{2 \left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec ^2(e+f x)}{2 f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{2} \int \frac {\cos (e+f x) \left (2 a^2+(a-b) b \sec ^2(e+f x)\right )}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec ^2(e+f x)+\frac {(a+b) \sqrt {a+b \sec ^2(e+f x)}}{1-\sec ^2(e+f x)}}{2 f}\) |
| \(\Big \downarrow \) 174 |
| \(\displaystyle \frac {\frac {1}{2} \left (2 a^2 \int \frac {\cos (e+f x)}{\sqrt {b \sec ^2(e+f x)+a}}d\sec ^2(e+f x)+(2 a-b) (a+b) \int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec ^2(e+f x)\right )+\frac {(a+b) \sqrt {a+b \sec ^2(e+f x)}}{1-\sec ^2(e+f x)}}{2 f}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {4 a^2 \int \frac {1}{\frac {\sec ^4(e+f x)}{b}-\frac {a}{b}}d\sqrt {b \sec ^2(e+f x)+a}}{b}+\frac {2 (2 a-b) (a+b) \int \frac {1}{\frac {a+b}{b}-\frac {\sec ^4(e+f x)}{b}}d\sqrt {b \sec ^2(e+f x)+a}}{b}\right )+\frac {(a+b) \sqrt {a+b \sec ^2(e+f x)}}{1-\sec ^2(e+f x)}}{2 f}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {1}{2} \left (2 (2 a-b) \sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a+b}}\right )-4 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )\right )+\frac {(a+b) \sqrt {a+b \sec ^2(e+f x)}}{1-\sec ^2(e+f x)}}{2 f}\) |
((-4*a^(3/2)*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]] + 2*(2*a - b)*Sqr t[a + b]*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a + b]])/2 + ((a + b)*Sqr t[a + b*Sec[e + f*x]^2])/(1 - Sec[e + f*x]^2))/(2*f)
3.4.93.3.1 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*c - a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((e + f *x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] + Simp[1/(b*(b*e - a*f)*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])
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. \(1568\) vs. \(2(96)=192\).
Time = 1.26 (sec) , antiderivative size = 1569, normalized size of antiderivative = 13.76
-1/4/f/(a+b)^(3/2)*(4*cos(f*x+e)*ln(4*cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+co s(f*x+e))^2)^(1/2)*a^(1/2)+4*cos(f*x+e)*a+4*a^(1/2)*((b+a*cos(f*x+e)^2)/(1 +cos(f*x+e))^2)^(1/2))*(a+b)^(3/2)*a^(3/2)-4*a^(3/2)*ln(4*cos(f*x+e)*((b+a *cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)+4*cos(f*x+e)*a+4*a^(1/2)*(( b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2))*(a+b)^(3/2)-2*cos(f*x+e)*((b+a* cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(3/2)*a-2*cos(f*x+e)*((b+a*cos (f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(3/2)*b-2*cos(f*x+e)*ln(-4*(((b+a *cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)*cos(f*x+e)+((b+a*cos(f* x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+cos(f*x+e)*a+b)/(-1+cos(f*x+e) ))*a^3-3*cos(f*x+e)*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+ b)^(1/2)*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2 )+cos(f*x+e)*a+b)/(-1+cos(f*x+e)))*a^2*b+cos(f*x+e)*ln(-4*(((b+a*cos(f*x+e )^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1 +cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+cos(f*x+e)*a+b)/(-1+cos(f*x+e)))*b^3+2*c os(f*x+e)*ln(2/(a+b)^(1/2)*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a +b)^(1/2)*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/ 2)-cos(f*x+e)*a+b)/(1+cos(f*x+e)))*a^3+3*cos(f*x+e)*ln(2/(a+b)^(1/2)*(((b+ a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)*cos(f*x+e)+((b+a*cos(f *x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-cos(f*x+e)*a+b)/(1+cos(f*x+e) ))*a^2*b-cos(f*x+e)*ln(2/(a+b)^(1/2)*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e)...
Leaf count of result is larger than twice the leaf count of optimal. 278 vs. \(2 (96) = 192\).
Time = 0.60 (sec) , antiderivative size = 1300, normalized size of antiderivative = 11.40 \[ \int \cot ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\text {Too large to display} \]
[1/8*(4*(a + b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)^2 + (a*cos(f*x + e)^2 - a)*sqrt(a)*log(128*a^4*cos(f*x + e)^8 + 256*a^3*b*c os(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)) - ((2*a - b)*cos(f*x + e)^2 - 2*a + b)*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)))/(f*c os(f*x + e)^2 - f), 1/8*(4*(a + b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e )^2)*cos(f*x + e)^2 - 2*((2*a - b)*cos(f*x + e)^2 - 2*a + b)*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*co s(f*x + e)^2 - a)*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)))/ (f*cos(f*x + e)^2 - f), 1/8*(4*(a + b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)^2 + 2*(a*cos(f*x + e)^2 - a)*sqrt(-a)*arctan(1/4*(8* a^2*cos(f*x + e)^4 + 8*a*b*cos(f*x + e)^2 + b^2)*sqrt(-a)*sqrt((a*cos(f...
\[ \int \cot ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}} \cot ^{3}{\left (e + f x \right )}\, dx \]
\[ \int \cot ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \cot \left (f x + e\right )^{3} \,d x } \]
Exception generated. \[ \int \cot ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
Timed out. \[ \int \cot ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int {\mathrm {cot}\left (e+f\,x\right )}^3\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2} \,d x \]