\(\int \csc ^5(a+b x) \sqrt {d \tan (a+b x)} \, dx\) [63]

Optimal result

Integrand size = 21, antiderivative size = 105 \[ \int \csc ^5(a+b x) \sqrt {d \tan (a+b x)} \, dx=-\frac {4 d \csc (a+b x)}{7 b \sqrt {d \tan (a+b x)}}-\frac {2 d \csc ^3(a+b x)}{7 b \sqrt {d \tan (a+b x)}}+\frac {4 \csc (a+b x) \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}{7 b} \] Output:

-4/7*d*csc(b*x+a)/b/(d*tan(b*x+a))^(1/2)-2/7*d*csc(b*x+a)^3/b/(d*tan(b*x+a 
))^(1/2)+4/7*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
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 1.16 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.18 \[ \int \csc ^5(a+b x) \sqrt {d \tan (a+b x)} \, dx=-\frac {2 d \cos (2 (a+b x)) \csc ^3(a+b x) \left ((-2+\cos (2 (a+b x))) \sec ^2(a+b x)^{3/2}-4 \sqrt [4]{-1} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (a+b x)}\right ),-1\right ) \tan ^{\frac {7}{2}}(a+b x)\right )}{7 b \sqrt {\sec ^2(a+b x)} \sqrt {d \tan (a+b x)} \left (-1+\tan ^2(a+b x)\right )} \] Input:

Integrate[Csc[a + b*x]^5*Sqrt[d*Tan[a + b*x]],x]
 

Output:

(-2*d*Cos[2*(a + b*x)]*Csc[a + b*x]^3*((-2 + Cos[2*(a + b*x)])*(Sec[a + b* 
x]^2)^(3/2) - 4*(-1)^(1/4)*EllipticF[I*ArcSinh[(-1)^(1/4)*Sqrt[Tan[a + b*x 
]]], -1]*Tan[a + b*x]^(7/2)))/(7*b*Sqrt[Sec[a + b*x]^2]*Sqrt[d*Tan[a + b*x 
]]*(-1 + Tan[a + b*x]^2))
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3079, 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.

\(\displaystyle \int \csc ^5(a+b x) \sqrt {d \tan (a+b x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {d \tan (a+b x)}}{\sin (a+b x)^5}dx\)

\(\Big \downarrow \) 3079

\(\displaystyle \frac {6}{7} \int \csc ^3(a+b x) \sqrt {d \tan (a+b x)}dx-\frac {2 d \csc ^3(a+b x)}{7 b \sqrt {d \tan (a+b x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {6}{7} \int \frac {\sqrt {d \tan (a+b x)}}{\sin (a+b x)^3}dx-\frac {2 d \csc ^3(a+b x)}{7 b \sqrt {d \tan (a+b x)}}\)

\(\Big \downarrow \) 3079

\(\displaystyle \frac {6}{7} \left (\frac {2}{3} \int \csc (a+b x) \sqrt {d \tan (a+b x)}dx-\frac {2 d \csc (a+b x)}{3 b \sqrt {d \tan (a+b x)}}\right )-\frac {2 d \csc ^3(a+b x)}{7 b \sqrt {d \tan (a+b x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {6}{7} \left (\frac {2}{3} \int \frac {\sqrt {d \tan (a+b x)}}{\sin (a+b x)}dx-\frac {2 d \csc (a+b x)}{3 b \sqrt {d \tan (a+b x)}}\right )-\frac {2 d \csc ^3(a+b x)}{7 b \sqrt {d \tan (a+b x)}}\)

\(\Big \downarrow \) 3081

\(\displaystyle \frac {6}{7} \left (\frac {2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}}dx}{3 \sqrt {\sin (a+b x)}}-\frac {2 d \csc (a+b x)}{3 b \sqrt {d \tan (a+b x)}}\right )-\frac {2 d \csc ^3(a+b x)}{7 b \sqrt {d \tan (a+b x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {6}{7} \left (\frac {2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}}dx}{3 \sqrt {\sin (a+b x)}}-\frac {2 d \csc (a+b x)}{3 b \sqrt {d \tan (a+b x)}}\right )-\frac {2 d \csc ^3(a+b x)}{7 b \sqrt {d \tan (a+b x)}}\)

\(\Big \downarrow \) 3053

\(\displaystyle \frac {6}{7} \left (\frac {2}{3} \sqrt {\sin (2 a+2 b x)} \csc (a+b x) \sqrt {d \tan (a+b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {2 d \csc (a+b x)}{3 b \sqrt {d \tan (a+b x)}}\right )-\frac {2 d \csc ^3(a+b x)}{7 b \sqrt {d \tan (a+b x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {6}{7} \left (\frac {2}{3} \sqrt {\sin (2 a+2 b x)} \csc (a+b x) \sqrt {d \tan (a+b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {2 d \csc (a+b x)}{3 b \sqrt {d \tan (a+b x)}}\right )-\frac {2 d \csc ^3(a+b x)}{7 b \sqrt {d \tan (a+b x)}}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {6}{7} \left (\frac {2 \sqrt {\sin (2 a+2 b x)} \csc (a+b x) \operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right ) \sqrt {d \tan (a+b x)}}{3 b}-\frac {2 d \csc (a+b x)}{3 b \sqrt {d \tan (a+b x)}}\right )-\frac {2 d \csc ^3(a+b x)}{7 b \sqrt {d \tan (a+b x)}}\)

Input:

Int[Csc[a + b*x]^5*Sqrt[d*Tan[a + b*x]],x]
 

Output:

(-2*d*Csc[a + b*x]^3)/(7*b*Sqrt[d*Tan[a + b*x]]) + (6*((-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] 
*Sqrt[Sin[2*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]])/(3*b)))/7
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3053

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]
 

rule 3079

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]
 

rule 3081

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])
 

rule 3120

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]
 

Maple [A] (verified)

Time = 0.90 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.34

method	result	size
default	\(-\frac {2 \sqrt {d \tan \left (b x +a \right )}\, \left (\sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}\, \sqrt {-2 \csc \left (b x +a \right )+2 \cot \left (b x +a \right )+2}\, \sqrt {-\csc \left (b x +a \right )+\cot \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (-2 \cot \left (b x +a \right )-2 \csc \left (b x +a \right )\right )+\cot \left (b x +a \right ) \csc \left (b x +a \right )^{3} \left (-2 \cos \left (b x +a \right )^{2}+3\right )\right )}{7 b}\)	\(141\)

Input:

int(csc(b*x+a)^5*(d*tan(b*x+a))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-2/7/b*(d*tan(b*x+a))^(1/2)*((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))*(-2*cot(b*x+a)-2*csc(b*x+a))+cot(b*x+a 
)*csc(b*x+a)^3*(-2*cos(b*x+a)^2+3))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.50 \[ \int \csc ^5(a+b x) \sqrt {d \tan (a+b x)} \, dx=-\frac {2 \, {\left (2 \, {\left (\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1\right )} \sqrt {i \, d} F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) + 2 \, {\left (\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1\right )} \sqrt {-i \, d} F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) - {\left (2 \, \cos \left (b x + a\right )^{3} - 3 \, \cos \left (b x + a\right )\right )} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}\right )}}{7 \, {\left (b \cos \left (b x + a\right )^{4} - 2 \, b \cos \left (b x + a\right )^{2} + b\right )}} \] Input:

integrate(csc(b*x+a)^5*(d*tan(b*x+a))^(1/2),x, algorithm="fricas")
 

Output:

-2/7*(2*(cos(b*x + a)^4 - 2*cos(b*x + a)^2 + 1)*sqrt(I*d)*elliptic_f(arcsi 
n(cos(b*x + a) + I*sin(b*x + a)), -1) + 2*(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) - 
 (2*cos(b*x + a)^3 - 3*cos(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)))/(b 
*cos(b*x + a)^4 - 2*b*cos(b*x + a)^2 + b)
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \csc ^5(a+b x) \sqrt {d \tan (a+b x)} \, dx=\int \sqrt {d \tan {\left (a + b x \right )}} \csc ^{5}{\left (a + b x \right )}\, dx \] Input:

integrate(csc(b*x+a)**5*(d*tan(b*x+a))**(1/2),x)
 

Output:

Integral(sqrt(d*tan(a + b*x))*csc(a + b*x)**5, x)
 

Maxima [F]

\[ \int \csc ^5(a+b x) \sqrt {d \tan (a+b x)} \, dx=\int { \sqrt {d \tan \left (b x + a\right )} \csc \left (b x + a\right )^{5} \,d x } \] Input:

integrate(csc(b*x+a)^5*(d*tan(b*x+a))^(1/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(d*tan(b*x + a))*csc(b*x + a)^5, x)
 

Giac [F]

\[ \int \csc ^5(a+b x) \sqrt {d \tan (a+b x)} \, dx=\int { \sqrt {d \tan \left (b x + a\right )} \csc \left (b x + a\right )^{5} \,d x } \] Input:

integrate(csc(b*x+a)^5*(d*tan(b*x+a))^(1/2),x, algorithm="giac")
 

Output:

integrate(sqrt(d*tan(b*x + a))*csc(b*x + a)^5, x)
 

Mupad [F(-1)]

Timed out. \[ \int \csc ^5(a+b x) \sqrt {d \tan (a+b x)} \, dx=\int \frac {\sqrt {d\,\mathrm {tan}\left (a+b\,x\right )}}{{\sin \left (a+b\,x\right )}^5} \,d x \] Input:

int((d*tan(a + b*x))^(1/2)/sin(a + b*x)^5,x)
 

Output:

int((d*tan(a + b*x))^(1/2)/sin(a + b*x)^5, x)
 

Reduce [F]

\[ \int \csc ^5(a+b x) \sqrt {d \tan (a+b x)} \, dx=\sqrt {d}\, \left (\int \sqrt {\tan \left (b x +a \right )}\, \csc \left (b x +a \right )^{5}d x \right ) \] Input:

int(csc(b*x+a)^5*(d*tan(b*x+a))^(1/2),x)
 

Output:

sqrt(d)*int(sqrt(tan(a + b*x))*csc(a + b*x)**5,x)