Integrand size = 21, antiderivative size = 136 \[ \int \cos ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\frac {d^2 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sec (a+b x) \sqrt {\sin (2 a+2 b x)}}{24 b \sqrt {d \tan (a+b x)}}+\frac {d \cos (a+b x) \sqrt {d \tan (a+b x)}}{12 b}+\frac {d \cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{30 b}-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b} \]
-1/24*d^2*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticF(cos(a+1/ 4*Pi+b*x),2^(1/2))*sec(b*x+a)*sin(2*b*x+2*a)^(1/2)/b/(d*tan(b*x+a))^(1/2)+ 1/12*d*cos(b*x+a)*(d*tan(b*x+a))^(1/2)/b+1/30*d*cos(b*x+a)^3*(d*tan(b*x+a) )^(1/2)/b-1/5*d*cos(b*x+a)^5*(d*tan(b*x+a))^(1/2)/b
Result contains complex when optimal does not.
Time = 2.08 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.96 \[ \int \cos ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\frac {\cos (2 (a+b x)) \csc (a+b x) \left (10 \sqrt [4]{-1} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (a+b x)}\right ),-1\right ) \sqrt {\sec ^2(a+b x)}+(-3+10 \cos (2 (a+b x))+3 \cos (4 (a+b x))) \sqrt {\tan (a+b x)}\right ) (d \tan (a+b x))^{3/2}}{120 b \sqrt {\tan (a+b x)} \left (-1+\tan ^2(a+b x)\right )} \]
(Cos[2*(a + b*x)]*Csc[a + b*x]*(10*(-1)^(1/4)*EllipticF[I*ArcSinh[(-1)^(1/ 4)*Sqrt[Tan[a + b*x]]], -1]*Sqrt[Sec[a + b*x]^2] + (-3 + 10*Cos[2*(a + b*x )] + 3*Cos[4*(a + b*x)])*Sqrt[Tan[a + b*x]])*(d*Tan[a + b*x])^(3/2))/(120* b*Sqrt[Tan[a + b*x]]*(-1 + Tan[a + b*x]^2))
Time = 0.74 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.08, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 3090, 3042, 3092, 3042, 3092, 3042, 3094, 3042, 3053, 3042, 3120}
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.
\(\displaystyle \int \cos ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(d \tan (a+b x))^{3/2}}{\sec (a+b x)^5}dx\) |
\(\Big \downarrow \) 3090 |
\(\displaystyle \frac {1}{10} d^2 \int \frac {\cos ^3(a+b x)}{\sqrt {d \tan (a+b x)}}dx-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{10} d^2 \int \frac {1}{\sec (a+b x)^3 \sqrt {d \tan (a+b x)}}dx-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}\) |
\(\Big \downarrow \) 3092 |
\(\displaystyle \frac {1}{10} d^2 \left (\frac {5}{6} \int \frac {\cos (a+b x)}{\sqrt {d \tan (a+b x)}}dx+\frac {\cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{3 b d}\right )-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{10} d^2 \left (\frac {5}{6} \int \frac {1}{\sec (a+b x) \sqrt {d \tan (a+b x)}}dx+\frac {\cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{3 b d}\right )-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}\) |
\(\Big \downarrow \) 3092 |
\(\displaystyle \frac {1}{10} d^2 \left (\frac {5}{6} \left (\frac {1}{2} \int \frac {\sec (a+b x)}{\sqrt {d \tan (a+b x)}}dx+\frac {\cos (a+b x) \sqrt {d \tan (a+b x)}}{b d}\right )+\frac {\cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{3 b d}\right )-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{10} d^2 \left (\frac {5}{6} \left (\frac {1}{2} \int \frac {\sec (a+b x)}{\sqrt {d \tan (a+b x)}}dx+\frac {\cos (a+b x) \sqrt {d \tan (a+b x)}}{b d}\right )+\frac {\cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{3 b d}\right )-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}\) |
\(\Big \downarrow \) 3094 |
\(\displaystyle \frac {1}{10} d^2 \left (\frac {5}{6} \left (\frac {\sqrt {\sin (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}}dx}{2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}+\frac {\cos (a+b x) \sqrt {d \tan (a+b x)}}{b d}\right )+\frac {\cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{3 b d}\right )-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{10} d^2 \left (\frac {5}{6} \left (\frac {\sqrt {\sin (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}}dx}{2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}+\frac {\cos (a+b x) \sqrt {d \tan (a+b x)}}{b d}\right )+\frac {\cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{3 b d}\right )-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}\) |
\(\Big \downarrow \) 3053 |
\(\displaystyle \frac {1}{10} d^2 \left (\frac {5}{6} \left (\frac {\sqrt {\sin (2 a+2 b x)} \sec (a+b x) \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx}{2 \sqrt {d \tan (a+b x)}}+\frac {\cos (a+b x) \sqrt {d \tan (a+b x)}}{b d}\right )+\frac {\cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{3 b d}\right )-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{10} d^2 \left (\frac {5}{6} \left (\frac {\sqrt {\sin (2 a+2 b x)} \sec (a+b x) \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx}{2 \sqrt {d \tan (a+b x)}}+\frac {\cos (a+b x) \sqrt {d \tan (a+b x)}}{b d}\right )+\frac {\cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{3 b d}\right )-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {1}{10} d^2 \left (\frac {\cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{3 b d}+\frac {5}{6} \left (\frac {\cos (a+b x) \sqrt {d \tan (a+b x)}}{b d}+\frac {\sqrt {\sin (2 a+2 b x)} \sec (a+b x) \operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right )}{2 b \sqrt {d \tan (a+b x)}}\right )\right )-\frac {d \cos ^5(a+b x) \sqrt {d \tan (a+b x)}}{5 b}\) |
-1/5*(d*Cos[a + b*x]^5*Sqrt[d*Tan[a + b*x]])/b + (d^2*((Cos[a + b*x]^3*Sqr t[d*Tan[a + b*x]])/(3*b*d) + (5*((EllipticF[a - Pi/4 + b*x, 2]*Sec[a + b*x ]*Sqrt[Sin[2*a + 2*b*x]])/(2*b*Sqrt[d*Tan[a + b*x]]) + (Cos[a + b*x]*Sqrt[ d*Tan[a + b*x]])/(b*d)))/6))/10
3.3.46.3.1 Defintions of rubi rules used
Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_ )]]), x_Symbol] :> Simp[Sqrt[Sin[2*e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b *Cos[e + f*x]]) Int[1/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[b*(a*Sec[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*m)) , x] - Simp[b^2*((n - 1)/(a^2*m)) Int[(a*Sec[e + f*x])^(m + 2)*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[n, 1] && (LtQ[m, -1 ] || (EqQ[m, -1] && EqQ[n, 3/2])) && IntegersQ[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[sec[(e_.) + (f_.)*(x_)]/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[Sqrt[Sin[e + f*x]]/(Sqrt[Cos[e + f*x]]*Sqrt[b*Tan[e + f*x]]) Int[ 1/(Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]]), x], x] /; FreeQ[{b, e, f}, x]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Result contains complex when optimal does not.
Time = 6.39 (sec) , antiderivative size = 1958, normalized size of antiderivative = 14.40
1/240/b*sin(b*x+a)*(24*2^(1/2)*cos(b*x+a)^5*sin(b*x+a)+30*I*(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)*EllipticPi((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))-3 0*I*(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)*EllipticPi((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2-1/ 2*I,1/2*2^(1/2))*cos(b*x+a)-4*2^(1/2)*cos(b*x+a)^3*sin(b*x+a)+30*I*(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)*EllipticPi((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2+1/2*I,1/2*2^( 1/2))*cos(b*x+a)-30*I*(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)*EllipticPi((1+csc(b*x+a)-cot(b* x+a))^(1/2),1/2-1/2*I,1/2*2^(1/2))+50*(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+cs c(b*x+a)-cot(b*x+a))^(1/2),1/2*2^(1/2))*cos(b*x+a)-30*(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) *EllipticPi((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(b*x +a)-30*(cot(b*x+a)-csc(b*x+a))^(1/2)*(-csc(b*x+a)+1+cot(b*x+a))^(1/2)*(1+c sc(b*x+a)-cot(b*x+a))^(1/2)*EllipticPi((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2 +1/2*I,1/2*2^(1/2))*cos(b*x+a)+50*(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))-30*(cot(b*x+a)-csc(b*x+a))^(1/2)*(-...
\[ \int \cos ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} \cos \left (b x + a\right )^{5} \,d x } \]
Timed out. \[ \int \cos ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\text {Timed out} \]
\[ \int \cos ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} \cos \left (b x + a\right )^{5} \,d x } \]
Exception generated. \[ \int \cos ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:The choice was done assuming 0=[0,0 ]ext_reduce Error: Bad Argument TypeThe choice was done assuming 0=[0,0]ex t_reduce
Timed out. \[ \int \cos ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\int {\cos \left (a+b\,x\right )}^5\,{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2} \,d x \]