Integrand size = 19, antiderivative size = 78 \[ \int \frac {\sin (a+b x)}{(d \tan (a+b x))^{5/2}} \, dx=-\frac {2 \sin (a+b x)}{b d (d \tan (a+b x))^{3/2}}-\frac {3 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sin (a+b x)}{b d^2 \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}} \]
3*(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))*sin(b*x+a)/b/d^2/sin(2*b*x+2*a)^(1/2)/(d*tan(b*x+a))^(1/2)-2*si n(b*x+a)/b/d/(d*tan(b*x+a))^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.49 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.88 \[ \int \frac {\sin (a+b x)}{(d \tan (a+b x))^{5/2}} \, dx=-\frac {2 \cos (a+b x) \left (1+\operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\tan ^2(a+b x)\right ) \sqrt {\sec ^2(a+b x)} \tan ^2(a+b x)\right )}{b d^2 \sqrt {d \tan (a+b x)}} \]
(-2*Cos[a + b*x]*(1 + Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[a + b*x]^2]*Sq rt[Sec[a + b*x]^2]*Tan[a + b*x]^2))/(b*d^2*Sqrt[d*Tan[a + b*x]])
Time = 0.45 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {3042, 3077, 3042, 3081, 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.
\(\displaystyle \int \frac {\sin (a+b x)}{(d \tan (a+b x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (a+b x)}{(d \tan (a+b x))^{5/2}}dx\) |
\(\Big \downarrow \) 3077 |
\(\displaystyle -\frac {3 \int \frac {\sin (a+b x)}{\sqrt {d \tan (a+b x)}}dx}{d^2}-\frac {2 \sin (a+b x)}{b d (d \tan (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {3 \int \frac {\sin (a+b x)}{\sqrt {d \tan (a+b x)}}dx}{d^2}-\frac {2 \sin (a+b x)}{b d (d \tan (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 3081 |
\(\displaystyle -\frac {3 \sqrt {\sin (a+b x)} \int \sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}dx}{d^2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}-\frac {2 \sin (a+b x)}{b d (d \tan (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {3 \sqrt {\sin (a+b x)} \int \sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}dx}{d^2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}-\frac {2 \sin (a+b x)}{b d (d \tan (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 3052 |
\(\displaystyle -\frac {3 \sin (a+b x) \int \sqrt {\sin (2 a+2 b x)}dx}{d^2 \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}-\frac {2 \sin (a+b x)}{b d (d \tan (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {3 \sin (a+b x) \int \sqrt {\sin (2 a+2 b x)}dx}{d^2 \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}-\frac {2 \sin (a+b x)}{b d (d \tan (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle -\frac {3 \sin (a+b x) E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{b d^2 \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}-\frac {2 \sin (a+b x)}{b d (d \tan (a+b x))^{3/2}}\) |
(-2*Sin[a + b*x])/(b*d*(d*Tan[a + b*x])^(3/2)) - (3*EllipticE[a - Pi/4 + b *x, 2]*Sin[a + b*x])/(b*d^2*Sqrt[Sin[2*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]])
3.2.11.3.1 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_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[(a*Sin[e + f*x])^m*((b*Tan[e + f*x])^(n + 1)/(b*f*(m + n + 1))), x] - Simp[(n + 1)/(b^2*(m + n + 1)) Int[(a*Sin[e + f*x])^m*( b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, -1] && NeQ[m + n + 1, 0] && IntegersQ[2*m, 2*n] && !(EqQ[n, -3/2] && EqQ[m, 1] )
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[Cos[e + f*x]^n*((b*Tan[e + f*x])^n/(a*Sin[e + f*x])^ n) Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] && !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(- 1)]) || IntegersQ[m - 1/2, n - 1/2])
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. \(365\) vs. \(2(97)=194\).
Time = 0.95 (sec) , antiderivative size = 366, normalized size of antiderivative = 4.69
method | result | size |
default | \(\frac {\left (6 \sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}\, \sqrt {-\csc \left (b x +a \right )+1+\cot \left (b x +a \right )}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, E\left (\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \sqrt {-\csc \left (b x +a \right )+1+\cot \left (b x +a \right )}\, \sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}\, F\left (\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )+6 \sec \left (b x +a \right ) \sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}\, \sqrt {-\csc \left (b x +a \right )+1+\cot \left (b x +a \right )}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, E\left (\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )-3 \sec \left (b x +a \right ) \sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}\, \sqrt {-\csc \left (b x +a \right )+1+\cot \left (b x +a \right )}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, F\left (\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )+\sqrt {2}\, \cos \left (b x +a \right )-3 \sqrt {2}\right ) \sqrt {2}}{2 b \sqrt {d \tan \left (b x +a \right )}\, d^{2}}\) | \(366\) |
1/2/b/(d*tan(b*x+a))^(1/2)/d^2*(6*(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))-3*(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))+6*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))-3*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)*EllipticF((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2*2^(1/2))+2^(1 /2)*cos(b*x+a)-3*2^(1/2))*2^(1/2)
\[ \int \frac {\sin (a+b x)}{(d \tan (a+b x))^{5/2}} \, dx=\int { \frac {\sin \left (b x + a\right )}{\left (d \tan \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {\sin (a+b x)}{(d \tan (a+b x))^{5/2}} \, dx=\int \frac {\sin {\left (a + b x \right )}}{\left (d \tan {\left (a + b x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {\sin (a+b x)}{(d \tan (a+b x))^{5/2}} \, dx=\int { \frac {\sin \left (b x + a\right )}{\left (d \tan \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {\sin (a+b x)}{(d \tan (a+b x))^{5/2}} \, dx=\int { \frac {\sin \left (b x + a\right )}{\left (d \tan \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sin (a+b x)}{(d \tan (a+b x))^{5/2}} \, dx=\int \frac {\sin \left (a+b\,x\right )}{{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{5/2}} \,d x \]