Integrand size = 23, antiderivative size = 236 \[ \int \frac {\cot ^3(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {(4 a-7 b) \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a-b}}\right )}{4 (a-b)^{5/2} d}+\frac {(4 a+7 b) \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )}{4 (a+b)^{5/2} d}+\frac {2 b^4}{a \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {\sqrt {a+b \sec (c+d x)}}{4 (a+b)^2 d (1-\sec (c+d x))}+\frac {\sqrt {a+b \sec (c+d x)}}{4 (a-b)^2 d (1+\sec (c+d x))} \]
-2*arctanh((a+b*sec(d*x+c))^(1/2)/a^(1/2))/a^(3/2)/d+1/4*(4*a-7*b)*arctanh ((a+b*sec(d*x+c))^(1/2)/(a-b)^(1/2))/(a-b)^(5/2)/d+1/4*(4*a+7*b)*arctanh(( a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2))/(a+b)^(5/2)/d+2*b^4/a/(a^2-b^2)^2/d/(a+ b*sec(d*x+c))^(1/2)+1/4*(a+b*sec(d*x+c))^(1/2)/(a+b)^2/d/(1-sec(d*x+c))+1/ 4*(a+b*sec(d*x+c))^(1/2)/(a-b)^2/d/(1+sec(d*x+c))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 1.09 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.33 \[ \int \frac {\cot ^3(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}}-\frac {2 a \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \sec (c+d x)}{a-b}\right )}{(a-b) \sqrt {a+b \sec (c+d x)}}+\frac {2 a \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \sec (c+d x)}{a+b}\right )}{(a+b) \sqrt {a+b \sec (c+d x)}}+\frac {4 b \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},1+\frac {b \sec (c+d x)}{a}\right )}{a \sqrt {a+b \sec (c+d x)}}+\frac {b^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},2,\frac {1}{2},\frac {a+b \sec (c+d x)}{a-b}\right )}{(a-b)^2 \sqrt {a+b \sec (c+d x)}}-\frac {b^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},2,\frac {1}{2},\frac {a+b \sec (c+d x)}{a+b}\right )}{(a+b)^2 \sqrt {a+b \sec (c+d x)}}}{2 b d} \]
((-2*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a - b]])/Sqrt[a - b] + (2*ArcTa nh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]])/Sqrt[a + b] - (2*a*Hypergeometri c2F1[-1/2, 1, 1/2, (a + b*Sec[c + d*x])/(a - b)])/((a - b)*Sqrt[a + b*Sec[ c + d*x]]) + (2*a*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Sec[c + d*x])/(a + b)])/((a + b)*Sqrt[a + b*Sec[c + d*x]]) + (4*b*Hypergeometric2F1[-1/2, 1 , 1/2, 1 + (b*Sec[c + d*x])/a])/(a*Sqrt[a + b*Sec[c + d*x]]) + (b^2*Hyperg eometric2F1[-1/2, 2, 1/2, (a + b*Sec[c + d*x])/(a - b)])/((a - b)^2*Sqrt[a + b*Sec[c + d*x]]) - (b^2*Hypergeometric2F1[-1/2, 2, 1/2, (a + b*Sec[c + d*x])/(a + b)])/((a + b)^2*Sqrt[a + b*Sec[c + d*x]]))/(2*b*d)
Time = 0.59 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.31, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 25, 4373, 561, 25, 1674, 2009}
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.
\(\displaystyle \int \frac {\cot ^3(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {1}{\cot \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {1}{\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^3 \left (a+b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4373 |
\(\displaystyle \frac {b^4 \int \frac {\cos (c+d x)}{b (a+b \sec (c+d x))^{3/2} \left (b^2-b^2 \sec ^2(c+d x)\right )^2}d(b \sec (c+d x))}{d}\) |
\(\Big \downarrow \) 561 |
\(\displaystyle \frac {2 b^4 \int -\frac {\cos ^2(c+d x)}{b^2 \left (a-b^2 \sec ^2(c+d x)\right ) \left (b^4 \sec ^4(c+d x)-2 a b^2 \sec ^2(c+d x)+a^2-b^2\right )^2}d\sqrt {a+b \sec (c+d x)}}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {2 b^4 \int \frac {\cos ^2(c+d x)}{b^2 \left (a-b^2 \sec ^2(c+d x)\right ) \left (b^4 \sec ^4(c+d x)-2 a b^2 \sec ^2(c+d x)+a^2-b^2\right )^2}d\sqrt {a+b \sec (c+d x)}}{d}\) |
\(\Big \downarrow \) 1674 |
\(\displaystyle -\frac {2 b^4 \int \left (\frac {\cos ^2(c+d x)}{a (a-b)^2 b^2 (a+b)^2}+\frac {1}{a b^4 \left (a-b^2 \sec ^2(c+d x)\right )}+\frac {3 b-2 a}{4 (a-b)^2 b^4 \left (-b^2 \sec ^2(c+d x)+a-b\right )}+\frac {-2 a-3 b}{4 b^4 (a+b)^2 \left (-b^2 \sec ^2(c+d x)+a+b\right )}+\frac {1}{4 (a-b) b^3 \left (-b^2 \sec ^2(c+d x)+a-b\right )^2}-\frac {1}{4 b^3 (a+b) \left (-b^2 \sec ^2(c+d x)+a+b\right )^2}\right )d\sqrt {a+b \sec (c+d x)}}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 b^4 \left (-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} b^4}+\frac {\cos (c+d x)}{a b \left (a^2-b^2\right )^2}+\frac {(2 a-3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a-b}}\right )}{4 b^4 (a-b)^{5/2}}+\frac {(2 a+3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )}{4 b^4 (a+b)^{5/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a-b}}\right )}{8 b^3 (a-b)^{5/2}}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )}{8 b^3 (a+b)^{5/2}}-\frac {\sec (c+d x)}{8 b^2 (a-b)^2 \left (a-b^2 \sec ^2(c+d x)-b\right )}+\frac {\sec (c+d x)}{8 b^2 (a+b)^2 \left (a-b^2 \sec ^2(c+d x)+b\right )}\right )}{d}\) |
(2*b^4*(-(ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a]]/(a^(3/2)*b^4)) + ((2*a - 3*b)*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a - b]])/(4*(a - b)^(5/2)*b^ 4) - ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a - b]]/(8*(a - b)^(5/2)*b^3) + ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]]/(8*b^3*(a + b)^(5/2)) + ((2 *a + 3*b)*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]])/(4*b^4*(a + b)^(5 /2)) + Cos[c + d*x]/(a*b*(a^2 - b^2)^2) - Sec[c + d*x]/(8*(a - b)^2*b^2*(a - b - b^2*Sec[c + d*x]^2)) + Sec[c + d*x]/(8*b^2*(a + b)^2*(a + b - b^2*S ec[c + d*x]^2))))/d
3.4.39.3.1 Defintions of rubi rules used
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> With[{k = Denominator[n]}, Simp[k/d Subst[Int[x^(k*(n + 1) - 1)*(-c /d + x^k/d)^m*Simp[(b*c^2 + a*d^2)/d^2 - 2*b*c*(x^k/d^2) + b*(x^(2*k)/d^2), x]^p, x], x, (c + d*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, m, p}, x] && Frac tionQ[n] && IntegerQ[p] && IntegerQ[m]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* (a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p, q}, x] && N eQ[b^2 - 4*a*c, 0] && (IGtQ[p, 0] || IGtQ[q, 0] || IntegersQ[m, q])
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _), x_Symbol] :> Simp[-(-1)^((m - 1)/2)/(d*b^(m - 1)) Subst[Int[(b^2 - x^ 2)^((m - 1)/2)*((a + x)^n/x), x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(6366\) vs. \(2(204)=408\).
Time = 2.34 (sec) , antiderivative size = 6367, normalized size of antiderivative = 26.98
Leaf count of result is larger than twice the leaf count of optimal. 936 vs. \(2 (202) = 404\).
Time = 48.26 (sec) , antiderivative size = 8098, normalized size of antiderivative = 34.31 \[ \int \frac {\cot ^3(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\text {Too large to display} \]
\[ \int \frac {\cot ^3(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {\cot ^{3}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]
Timed out. \[ \int \frac {\cot ^3(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\cot ^3(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{%%%{32,[2,6]%%%}+%%%{-32,[1,7]%%%},[6,1]%%%}+%%%{%%{[%%%{6 4,[2,6]%%
Timed out. \[ \int \frac {\cot ^3(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {{\mathrm {cot}\left (c+d\,x\right )}^3}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]