Integrand size = 21, antiderivative size = 142 \[ \int \frac {\cos ^5(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=-\frac {2 \cos ^5(a+b x)}{b d \sqrt {d \tan (a+b x)}}-\frac {77 \cos (a+b x) E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sqrt {d \tan (a+b x)}}{20 b d^2 \sqrt {\sin (2 a+2 b x)}}-\frac {77 \cos ^3(a+b x) (d \tan (a+b x))^{3/2}}{30 b d^3}-\frac {11 \cos ^5(a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3} \]
-2*cos(b*x+a)^5/b/d/(d*tan(b*x+a))^(1/2)+77/20*cos(b*x+a)*(sin(a+1/4*Pi+b* x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticE(cos(a+1/4*Pi+b*x),2^(1/2))*(d*tan( b*x+a))^(1/2)/b/d^2/sin(2*b*x+2*a)^(1/2)-77/30*cos(b*x+a)^3*(d*tan(b*x+a)) ^(3/2)/b/d^3-11/5*cos(b*x+a)^5*(d*tan(b*x+a))^(3/2)/b/d^3
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.91 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.63 \[ \int \frac {\cos ^5(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\frac {\sin (a+b x) \left (-277+34 \cos (2 (a+b x))+3 \cos (4 (a+b x))-308 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\tan ^2(a+b x)\right ) \sqrt {\sec ^2(a+b x)} \tan ^2(a+b x)\right )}{120 b (d \tan (a+b x))^{3/2}} \]
(Sin[a + b*x]*(-277 + 34*Cos[2*(a + b*x)] + 3*Cos[4*(a + b*x)] - 308*Hyper geometric2F1[3/4, 3/2, 7/4, -Tan[a + b*x]^2]*Sqrt[Sec[a + b*x]^2]*Tan[a + b*x]^2))/(120*b*(d*Tan[a + b*x])^(3/2))
Time = 0.76 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.06, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 3089, 3042, 3092, 3042, 3092, 3042, 3095, 3042, 3052, 3042, 3119}
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.
\(\displaystyle \int \frac {\cos ^5(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sec (a+b x)^5 (d \tan (a+b x))^{3/2}}dx\) |
\(\Big \downarrow \) 3089 |
\(\displaystyle -\frac {11 \int \cos ^5(a+b x) \sqrt {d \tan (a+b x)}dx}{d^2}-\frac {2 \cos ^5(a+b x)}{b d \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {11 \int \frac {\sqrt {d \tan (a+b x)}}{\sec (a+b x)^5}dx}{d^2}-\frac {2 \cos ^5(a+b x)}{b d \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3092 |
\(\displaystyle -\frac {11 \left (\frac {7}{10} \int \cos ^3(a+b x) \sqrt {d \tan (a+b x)}dx+\frac {\cos ^5(a+b x) (d \tan (a+b x))^{3/2}}{5 b d}\right )}{d^2}-\frac {2 \cos ^5(a+b x)}{b d \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {11 \left (\frac {7}{10} \int \frac {\sqrt {d \tan (a+b x)}}{\sec (a+b x)^3}dx+\frac {\cos ^5(a+b x) (d \tan (a+b x))^{3/2}}{5 b d}\right )}{d^2}-\frac {2 \cos ^5(a+b x)}{b d \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3092 |
\(\displaystyle -\frac {11 \left (\frac {7}{10} \left (\frac {1}{2} \int \cos (a+b x) \sqrt {d \tan (a+b x)}dx+\frac {\cos ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b d}\right )+\frac {\cos ^5(a+b x) (d \tan (a+b x))^{3/2}}{5 b d}\right )}{d^2}-\frac {2 \cos ^5(a+b x)}{b d \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {11 \left (\frac {7}{10} \left (\frac {1}{2} \int \frac {\sqrt {d \tan (a+b x)}}{\sec (a+b x)}dx+\frac {\cos ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b d}\right )+\frac {\cos ^5(a+b x) (d \tan (a+b x))^{3/2}}{5 b d}\right )}{d^2}-\frac {2 \cos ^5(a+b x)}{b d \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3095 |
\(\displaystyle -\frac {11 \left (\frac {7}{10} \left (\frac {\sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)} \int \sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}dx}{2 \sqrt {\sin (a+b x)}}+\frac {\cos ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b d}\right )+\frac {\cos ^5(a+b x) (d \tan (a+b x))^{3/2}}{5 b d}\right )}{d^2}-\frac {2 \cos ^5(a+b x)}{b d \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {11 \left (\frac {7}{10} \left (\frac {\sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)} \int \sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}dx}{2 \sqrt {\sin (a+b x)}}+\frac {\cos ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b d}\right )+\frac {\cos ^5(a+b x) (d \tan (a+b x))^{3/2}}{5 b d}\right )}{d^2}-\frac {2 \cos ^5(a+b x)}{b d \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3052 |
\(\displaystyle -\frac {11 \left (\frac {7}{10} \left (\frac {\cos (a+b x) \sqrt {d \tan (a+b x)} \int \sqrt {\sin (2 a+2 b x)}dx}{2 \sqrt {\sin (2 a+2 b x)}}+\frac {\cos ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b d}\right )+\frac {\cos ^5(a+b x) (d \tan (a+b x))^{3/2}}{5 b d}\right )}{d^2}-\frac {2 \cos ^5(a+b x)}{b d \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {11 \left (\frac {7}{10} \left (\frac {\cos (a+b x) \sqrt {d \tan (a+b x)} \int \sqrt {\sin (2 a+2 b x)}dx}{2 \sqrt {\sin (2 a+2 b x)}}+\frac {\cos ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b d}\right )+\frac {\cos ^5(a+b x) (d \tan (a+b x))^{3/2}}{5 b d}\right )}{d^2}-\frac {2 \cos ^5(a+b x)}{b d \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle -\frac {11 \left (\frac {\cos ^5(a+b x) (d \tan (a+b x))^{3/2}}{5 b d}+\frac {7}{10} \left (\frac {\cos ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b d}+\frac {\cos (a+b x) E\left (\left .a+b x-\frac {\pi }{4}\right |2\right ) \sqrt {d \tan (a+b x)}}{2 b \sqrt {\sin (2 a+2 b x)}}\right )\right )}{d^2}-\frac {2 \cos ^5(a+b x)}{b d \sqrt {d \tan (a+b x)}}\) |
(-2*Cos[a + b*x]^5)/(b*d*Sqrt[d*Tan[a + b*x]]) - (11*((Cos[a + b*x]^5*(d*T an[a + b*x])^(3/2))/(5*b*d) + (7*((Cos[a + b*x]*EllipticE[a - Pi/4 + b*x, 2]*Sqrt[d*Tan[a + b*x]])/(2*b*Sqrt[Sin[2*a + 2*b*x]]) + (Cos[a + b*x]^3*(d *Tan[a + b*x])^(3/2))/(3*b*d)))/10))/d^2
3.3.68.3.1 Defintions of rubi rules used
Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]] , x_Symbol] :> Simp[Sqrt[a*Sin[e + f*x]]*(Sqrt[b*Cos[e + f*x]]/Sqrt[Sin[2*e + 2*f*x]]) Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f}, x]
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[(a*Sec[e + f*x])^m*((b*Tan[e + f*x])^(n + 1)/(b*f*(n + 1))), x] - Simp[(m + n + 1)/(b^2*(n + 1)) Int[(a*Sec[e + f*x])^m*(b*Ta n[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, -1] && I ntegersQ[2*m, 2*n]
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[(-(a*Sec[e + f*x])^m)*((b*Tan[e + f*x])^(n + 1)/(b*f* m)), x] + Simp[(m + n + 1)/(a^2*m) Int[(a*Sec[e + f*x])^(m + 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (LtQ[m, -1] || (EqQ[m, -1 ] && EqQ[n, -2^(-1)])) && IntegersQ[2*m, 2*n]
Int[Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]]/sec[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[Sqrt[Cos[e + f*x]]*(Sqrt[b*Tan[e + f*x]]/Sqrt[Sin[e + f*x]]) Int[ Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]], x], x] /; FreeQ[{b, e, f}, x]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Leaf count of result is larger than twice the leaf count of optimal. \(392\) vs. \(2(151)=302\).
Time = 1.39 (sec) , antiderivative size = 393, normalized size of antiderivative = 2.77
method | result | size |
default | \(\frac {\left (12 \left (\cos ^{5}\left (b x +a \right )\right ) \sqrt {2}+22 \left (\cos ^{3}\left (b x +a \right )\right ) \sqrt {2}+462 \sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}\, \sqrt {-\csc \left (b x +a \right )+1+\cot \left (b x +a \right )}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, E\left (\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )-231 \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \sqrt {-\csc \left (b x +a \right )+1+\cot \left (b x +a \right )}\, \sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}\, F\left (\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )+462 \sec \left (b x +a \right ) \sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}\, \sqrt {-\csc \left (b x +a \right )+1+\cot \left (b x +a \right )}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, E\left (\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )-231 \sec \left (b x +a \right ) \sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}\, \sqrt {-\csc \left (b x +a \right )+1+\cot \left (b x +a \right )}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, F\left (\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )+77 \sqrt {2}\, \cos \left (b x +a \right )-231 \sqrt {2}\right ) \sqrt {2}}{120 b \sqrt {d \tan \left (b x +a \right )}\, d}\) | \(393\) |
1/120/b/(d*tan(b*x+a))^(1/2)/d*(12*cos(b*x+a)^5*2^(1/2)+22*cos(b*x+a)^3*2^ (1/2)+462*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*(-csc(b*x+a)+1+cot(b*x+a))^(1/2) *(cot(b*x+a)-csc(b*x+a))^(1/2)*EllipticE((1+csc(b*x+a)-cot(b*x+a))^(1/2),1 /2*2^(1/2))-231*(cot(b*x+a)-csc(b*x+a))^(1/2)*(-csc(b*x+a)+1+cot(b*x+a))^( 1/2)*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*EllipticF((1+csc(b*x+a)-cot(b*x+a))^( 1/2),1/2*2^(1/2))+462*sec(b*x+a)*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*(-csc(b*x +a)+1+cot(b*x+a))^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*EllipticE((1+csc(b*x +a)-cot(b*x+a))^(1/2),1/2*2^(1/2))-231*sec(b*x+a)*(1+csc(b*x+a)-cot(b*x+a) )^(1/2)*(-csc(b*x+a)+1+cot(b*x+a))^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*Ell ipticF((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2*2^(1/2))+77*2^(1/2)*cos(b*x+a)- 231*2^(1/2))*2^(1/2)
\[ \int \frac {\cos ^5(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\int { \frac {\cos \left (b x + a\right )^{5}}{\left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\cos ^5(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^5(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\int { \frac {\cos \left (b x + a\right )^{5}}{\left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {\cos ^5(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\int { \frac {\cos \left (b x + a\right )^{5}}{\left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\cos ^5(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\int \frac {{\cos \left (a+b\,x\right )}^5}{{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2}} \,d x \]