Integrand size = 21, antiderivative size = 112 \[ \int \frac {\csc ^3(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\frac {2 \csc (a+b x)}{21 b d \sqrt {d \tan (a+b x)}}-\frac {2 \csc ^3(a+b x)}{7 b d \sqrt {d \tan (a+b x)}}-\frac {2 \csc (a+b x) \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}{21 b d^2} \] Output:
2/21*csc(b*x+a)/b/d/(d*tan(b*x+a))^(1/2)-2/7*csc(b*x+a)^3/b/d/(d*tan(b*x+a ))^(1/2)-2/21*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/d^2
Result contains complex when optimal does not.
Time = 2.05 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.21 \[ \int \frac {\csc ^3(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\frac {\csc ^3(a+b x) \left ((1+10 \cos (2 (a+b x))+\cos (4 (a+b x))) \sec ^2(a+b x)^{3/2}-8 \sqrt [4]{-1} \cos (2 (a+b x)) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (a+b x)}\right ),-1\right ) \tan ^{\frac {7}{2}}(a+b x)\right )}{42 b d \sqrt {\sec ^2(a+b x)} \sqrt {d \tan (a+b x)} \left (-1+\tan ^2(a+b x)\right )} \] Input:
Integrate[Csc[a + b*x]^3/(d*Tan[a + b*x])^(3/2),x]
Output:
(Csc[a + b*x]^3*((1 + 10*Cos[2*(a + b*x)] + Cos[4*(a + b*x)])*(Sec[a + b*x ]^2)^(3/2) - 8*(-1)^(1/4)*Cos[2*(a + b*x)]*EllipticF[I*ArcSinh[(-1)^(1/4)* Sqrt[Tan[a + b*x]]], -1]*Tan[a + b*x]^(7/2)))/(42*b*d*Sqrt[Sec[a + b*x]^2] *Sqrt[d*Tan[a + b*x]]*(-1 + Tan[a + b*x]^2))
Time = 0.56 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3077, 3042, 3079, 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.
\(\displaystyle \int \frac {\csc ^3(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (a+b x)^3 (d \tan (a+b x))^{3/2}}dx\) |
\(\Big \downarrow \) 3077 |
\(\displaystyle -\frac {\int \csc ^3(a+b x) \sqrt {d \tan (a+b x)}dx}{7 d^2}-\frac {2 \csc ^3(a+b x)}{7 b d \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \frac {\sqrt {d \tan (a+b x)}}{\sin (a+b x)^3}dx}{7 d^2}-\frac {2 \csc ^3(a+b x)}{7 b d \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3079 |
\(\displaystyle -\frac {\frac {2}{3} \int \csc (a+b x) \sqrt {d \tan (a+b x)}dx-\frac {2 d \csc (a+b x)}{3 b \sqrt {d \tan (a+b x)}}}{7 d^2}-\frac {2 \csc ^3(a+b x)}{7 b d \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {2}{3} \int \frac {\sqrt {d \tan (a+b x)}}{\sin (a+b x)}dx-\frac {2 d \csc (a+b x)}{3 b \sqrt {d \tan (a+b x)}}}{7 d^2}-\frac {2 \csc ^3(a+b x)}{7 b d \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3081 |
\(\displaystyle -\frac {\frac {2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}}dx}{3 \sqrt {\sin (a+b x)}}-\frac {2 d \csc (a+b x)}{3 b \sqrt {d \tan (a+b x)}}}{7 d^2}-\frac {2 \csc ^3(a+b x)}{7 b d \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}}dx}{3 \sqrt {\sin (a+b x)}}-\frac {2 d \csc (a+b x)}{3 b \sqrt {d \tan (a+b x)}}}{7 d^2}-\frac {2 \csc ^3(a+b x)}{7 b d \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3053 |
\(\displaystyle -\frac {\frac {2}{3} \sqrt {\sin (2 a+2 b x)} \csc (a+b x) \sqrt {d \tan (a+b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {2 d \csc (a+b x)}{3 b \sqrt {d \tan (a+b x)}}}{7 d^2}-\frac {2 \csc ^3(a+b x)}{7 b d \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {2}{3} \sqrt {\sin (2 a+2 b x)} \csc (a+b x) \sqrt {d \tan (a+b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {2 d \csc (a+b x)}{3 b \sqrt {d \tan (a+b x)}}}{7 d^2}-\frac {2 \csc ^3(a+b x)}{7 b d \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle -\frac {\frac {2 \sqrt {\sin (2 a+2 b x)} \csc (a+b x) \operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right ) \sqrt {d \tan (a+b x)}}{3 b}-\frac {2 d \csc (a+b x)}{3 b \sqrt {d \tan (a+b x)}}}{7 d^2}-\frac {2 \csc ^3(a+b x)}{7 b d \sqrt {d \tan (a+b x)}}\) |
Input:
Int[Csc[a + b*x]^3/(d*Tan[a + b*x])^(3/2),x]
Output:
(-2*Csc[a + b*x]^3)/(7*b*d*Sqrt[d*Tan[a + b*x]]) - ((-2*d*Csc[a + b*x])/(3 *b*Sqrt[d*Tan[a + b*x]]) + (2*Csc[a + b*x]*EllipticF[a - Pi/4 + b*x, 2]*Sq rt[Sin[2*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]])/(3*b))/(7*d^2)
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[(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[b*(a*Sin[e + f*x])^(m + 2)*((b*Tan[e + f*x])^(n - 1) /(a^2*f*(m + n + 1))), x] + Simp[(m + 2)/(a^2*(m + n + 1)) Int[(a*Sin[e + f*x])^(m + 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && L tQ[m, -1] && NeQ[m + n + 1, 0] && IntegersQ[2*m, 2*n]
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 = 0.89 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.18
method | result | size |
default | \(\frac {\frac {2 \sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}\, \sqrt {-2 \csc \left (b x +a \right )+2 \cot \left (b x +a \right )+2}\, \sqrt {-\csc \left (b x +a \right )+\cot \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (-1-\sec \left (b x +a \right )\right )}{21}+\frac {2 \csc \left (b x +a \right )^{3} \left (-\cos \left (b x +a \right )^{2}-2\right )}{21}}{b \sqrt {d \tan \left (b x +a \right )}\, d}\) | \(132\) |
Input:
int(csc(b*x+a)^3/(d*tan(b*x+a))^(3/2),x,method=_RETURNVERBOSE)
Output:
1/b*(2/21*(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))*(-1-sec(b*x+a))+2/21*csc(b*x+a)^3*(-cos(b*x+a)^2-2))/(d*t an(b*x+a))^(1/2)/d
Result contains complex when optimal does not.
Time = 0.13 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.46 \[ \int \frac {\csc ^3(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\frac {2 \, {\left ({\left (\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1\right )} \sqrt {i \, d} F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) + {\left (\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1\right )} \sqrt {-i \, d} F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) - {\left (\cos \left (b x + a\right )^{3} + 2 \, \cos \left (b x + a\right )\right )} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}\right )}}{21 \, {\left (b d^{2} \cos \left (b x + a\right )^{4} - 2 \, b d^{2} \cos \left (b x + a\right )^{2} + b d^{2}\right )}} \] Input:
integrate(csc(b*x+a)^3/(d*tan(b*x+a))^(3/2),x, algorithm="fricas")
Output:
2/21*((cos(b*x + a)^4 - 2*cos(b*x + a)^2 + 1)*sqrt(I*d)*elliptic_f(arcsin( cos(b*x + a) + I*sin(b*x + a)), -1) + (cos(b*x + a)^4 - 2*cos(b*x + a)^2 + 1)*sqrt(-I*d)*elliptic_f(arcsin(cos(b*x + a) - I*sin(b*x + a)), -1) - (co s(b*x + a)^3 + 2*cos(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)))/(b*d^2*c os(b*x + a)^4 - 2*b*d^2*cos(b*x + a)^2 + b*d^2)
\[ \int \frac {\csc ^3(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\int \frac {\csc ^{3}{\left (a + b x \right )}}{\left (d \tan {\left (a + b x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(csc(b*x+a)**3/(d*tan(b*x+a))**(3/2),x)
Output:
Integral(csc(a + b*x)**3/(d*tan(a + b*x))**(3/2), x)
\[ \int \frac {\csc ^3(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\int { \frac {\csc \left (b x + a\right )^{3}}{\left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(csc(b*x+a)^3/(d*tan(b*x+a))^(3/2),x, algorithm="maxima")
Output:
integrate(csc(b*x + a)^3/(d*tan(b*x + a))^(3/2), x)
\[ \int \frac {\csc ^3(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\int { \frac {\csc \left (b x + a\right )^{3}}{\left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(csc(b*x+a)^3/(d*tan(b*x+a))^(3/2),x, algorithm="giac")
Output:
integrate(csc(b*x + a)^3/(d*tan(b*x + a))^(3/2), x)
Timed out. \[ \int \frac {\csc ^3(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\int \frac {1}{{\sin \left (a+b\,x\right )}^3\,{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2}} \,d x \] Input:
int(1/(sin(a + b*x)^3*(d*tan(a + b*x))^(3/2)),x)
Output:
int(1/(sin(a + b*x)^3*(d*tan(a + b*x))^(3/2)), x)
\[ \int \frac {\csc ^3(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\frac {\sqrt {d}\, \left (\int \frac {\sqrt {\tan \left (b x +a \right )}\, \csc \left (b x +a \right )^{3}}{\tan \left (b x +a \right )^{2}}d x \right )}{d^{2}} \] Input:
int(csc(b*x+a)^3/(d*tan(b*x+a))^(3/2),x)
Output:
(sqrt(d)*int((sqrt(tan(a + b*x))*csc(a + b*x)**3)/tan(a + b*x)**2,x))/d**2