Integrand size = 19, antiderivative size = 80 \[ \int \csc (a+b x) (d \tan (a+b x))^{5/2} \, dx=-\frac {d^2 \csc (a+b x) \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}{3 b}+\frac {2 d \csc (a+b x) (d \tan (a+b x))^{3/2}}{3 b} \] Output:
-1/3*d^2*csc(b*x+a)*InverseJacobiAM(a-1/4*Pi+b*x,2^(1/2))*sin(2*b*x+2*a)^( 1/2)*(d*tan(b*x+a))^(1/2)/b+2/3*d*csc(b*x+a)*(d*tan(b*x+a))^(3/2)/b
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.41 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.89 \[ \int \csc (a+b x) (d \tan (a+b x))^{5/2} \, dx=\frac {2 d^2 \cos (a+b x) \left (\sec ^2(a+b x)-\operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\tan ^2(a+b x)\right ) \sqrt {\sec ^2(a+b x)}\right ) \sqrt {d \tan (a+b x)}}{3 b} \] Input:
Integrate[Csc[a + b*x]*(d*Tan[a + b*x])^(5/2),x]
Output:
(2*d^2*Cos[a + b*x]*(Sec[a + b*x]^2 - Hypergeometric2F1[1/4, 1/2, 5/4, -Ta n[a + b*x]^2]*Sqrt[Sec[a + b*x]^2])*Sqrt[d*Tan[a + b*x]])/(3*b)
Time = 0.43 (sec) , antiderivative size = 80, 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, 3074, 3042, 3081, 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.
\(\displaystyle \int \csc (a+b x) (d \tan (a+b x))^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(d \tan (a+b x))^{5/2}}{\sin (a+b x)}dx\) |
\(\Big \downarrow \) 3074 |
\(\displaystyle \frac {2 d \csc (a+b x) (d \tan (a+b x))^{3/2}}{3 b}-\frac {1}{3} d^2 \int \csc (a+b x) \sqrt {d \tan (a+b x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 d \csc (a+b x) (d \tan (a+b x))^{3/2}}{3 b}-\frac {1}{3} d^2 \int \frac {\sqrt {d \tan (a+b x)}}{\sin (a+b x)}dx\) |
\(\Big \downarrow \) 3081 |
\(\displaystyle \frac {2 d \csc (a+b x) (d \tan (a+b x))^{3/2}}{3 b}-\frac {d^2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}}dx}{3 \sqrt {\sin (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 d \csc (a+b x) (d \tan (a+b x))^{3/2}}{3 b}-\frac {d^2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}}dx}{3 \sqrt {\sin (a+b x)}}\) |
\(\Big \downarrow \) 3053 |
\(\displaystyle \frac {2 d \csc (a+b x) (d \tan (a+b x))^{3/2}}{3 b}-\frac {1}{3} d^2 \sqrt {\sin (2 a+2 b x)} \csc (a+b x) \sqrt {d \tan (a+b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 d \csc (a+b x) (d \tan (a+b x))^{3/2}}{3 b}-\frac {1}{3} d^2 \sqrt {\sin (2 a+2 b x)} \csc (a+b x) \sqrt {d \tan (a+b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {2 d \csc (a+b x) (d \tan (a+b x))^{3/2}}{3 b}-\frac {d^2 \sqrt {\sin (2 a+2 b x)} \csc (a+b x) \operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right ) \sqrt {d \tan (a+b x)}}{3 b}\) |
Input:
Int[Csc[a + b*x]*(d*Tan[a + b*x])^(5/2),x]
Output:
-1/3*(d^2*Csc[a + b*x]*EllipticF[a - Pi/4 + b*x, 2]*Sqrt[Sin[2*a + 2*b*x]] *Sqrt[d*Tan[a + b*x]])/b + (2*d*Csc[a + b*x]*(d*Tan[a + b*x])^(3/2))/(3*b)
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_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[b*(a*Sin[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(n - 1))), x] - Simp[b^2*((m + n - 1)/(n - 1)) Int[(a*Sin[e + f*x])^m*(b*Ta n[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && In tegersQ[2*m, 2*n] && !(GtQ[m, 1] && !IntegerQ[(m - 1)/2])
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[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]
Time = 0.94 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.56
method | result | size |
default | \(\frac {\sqrt {d \tan \left (b x +a \right )}\, d^{2} \left (\sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}\, \sqrt {-2 \csc \left (b x +a \right )+2 \cot \left (b x +a \right )+2}\, \sqrt {-\csc \left (b x +a \right )+\cot \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (-\cot \left (b x +a \right )-\csc \left (b x +a \right )\right )+2 \sec \left (b x +a \right )\right )}{3 b}\) | \(125\) |
Input:
int(csc(b*x+a)*(d*tan(b*x+a))^(5/2),x,method=_RETURNVERBOSE)
Output:
1/3/b*(d*tan(b*x+a))^(1/2)*d^2*((csc(b*x+a)-cot(b*x+a)+1)^(1/2)*(-2*csc(b* x+a)+2*cot(b*x+a)+2)^(1/2)*(-csc(b*x+a)+cot(b*x+a))^(1/2)*EllipticF((csc(b *x+a)-cot(b*x+a)+1)^(1/2),1/2*2^(1/2))*(-cot(b*x+a)-csc(b*x+a))+2*sec(b*x+ a))
Result contains complex when optimal does not.
Time = 0.11 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.29 \[ \int \csc (a+b x) (d \tan (a+b x))^{5/2} \, dx=\frac {\sqrt {i \, d} d^{2} \cos \left (b x + a\right ) F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) + \sqrt {-i \, d} d^{2} \cos \left (b x + a\right ) F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) + 2 \, d^{2} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{3 \, b \cos \left (b x + a\right )} \] Input:
integrate(csc(b*x+a)*(d*tan(b*x+a))^(5/2),x, algorithm="fricas")
Output:
1/3*(sqrt(I*d)*d^2*cos(b*x + a)*elliptic_f(arcsin(cos(b*x + a) + I*sin(b*x + a)), -1) + sqrt(-I*d)*d^2*cos(b*x + a)*elliptic_f(arcsin(cos(b*x + a) - I*sin(b*x + a)), -1) + 2*d^2*sqrt(d*sin(b*x + a)/cos(b*x + a)))/(b*cos(b* x + a))
Timed out. \[ \int \csc (a+b x) (d \tan (a+b x))^{5/2} \, dx=\text {Timed out} \] Input:
integrate(csc(b*x+a)*(d*tan(b*x+a))**(5/2),x)
Output:
Timed out
\[ \int \csc (a+b x) (d \tan (a+b x))^{5/2} \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{\frac {5}{2}} \csc \left (b x + a\right ) \,d x } \] Input:
integrate(csc(b*x+a)*(d*tan(b*x+a))^(5/2),x, algorithm="maxima")
Output:
integrate((d*tan(b*x + a))^(5/2)*csc(b*x + a), x)
\[ \int \csc (a+b x) (d \tan (a+b x))^{5/2} \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{\frac {5}{2}} \csc \left (b x + a\right ) \,d x } \] Input:
integrate(csc(b*x+a)*(d*tan(b*x+a))^(5/2),x, algorithm="giac")
Output:
integrate((d*tan(b*x + a))^(5/2)*csc(b*x + a), x)
Timed out. \[ \int \csc (a+b x) (d \tan (a+b x))^{5/2} \, dx=\int \frac {{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{5/2}}{\sin \left (a+b\,x\right )} \,d x \] Input:
int((d*tan(a + b*x))^(5/2)/sin(a + b*x),x)
Output:
int((d*tan(a + b*x))^(5/2)/sin(a + b*x), x)
\[ \int \csc (a+b x) (d \tan (a+b x))^{5/2} \, dx=\sqrt {d}\, \left (\int \sqrt {\tan \left (b x +a \right )}\, \csc \left (b x +a \right ) \tan \left (b x +a \right )^{2}d x \right ) d^{2} \] Input:
int(csc(b*x+a)*(d*tan(b*x+a))^(5/2),x)
Output:
sqrt(d)*int(sqrt(tan(a + b*x))*csc(a + b*x)*tan(a + b*x)**2,x)*d**2