\(\int \csc ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx\) [72]

Optimal result

Integrand size = 21, antiderivative size = 102 \[ \int \csc ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx=-\frac {4 d^2 \cos (a+b x)}{b \sqrt {d \tan (a+b x)}}-\frac {4 d^2 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sin (a+b x)}{b \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}+\frac {2 d \csc (a+b x) \sqrt {d \tan (a+b x)}}{b} \] Output:

-4*d^2*cos(b*x+a)/b/(d*tan(b*x+a))^(1/2)+4*d^2*EllipticE(cos(a+1/4*Pi+b*x) 
,2^(1/2))*sin(b*x+a)/b/sin(2*b*x+2*a)^(1/2)/(d*tan(b*x+a))^(1/2)+2*d*csc(b 
*x+a)*(d*tan(b*x+a))^(1/2)/b
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.54 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.70 \[ \int \csc ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx=-\frac {2 \cos (a+b x) \left (-6+3 \csc ^2(a+b x)+4 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\tan ^2(a+b x)\right ) \sqrt {\sec ^2(a+b x)}\right ) (d \tan (a+b x))^{3/2}}{3 b} \] Input:

Integrate[Csc[a + b*x]^3*(d*Tan[a + b*x])^(3/2),x]
 

Output:

(-2*Cos[a + b*x]*(-6 + 3*Csc[a + b*x]^2 + 4*Hypergeometric2F1[3/4, 3/2, 7/ 
4, -Tan[a + b*x]^2]*Sqrt[Sec[a + b*x]^2])*(d*Tan[a + b*x])^(3/2))/(3*b)
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.35, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3073, 3042, 3081, 3042, 3050, 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.

\(\displaystyle \int \csc ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(d \tan (a+b x))^{3/2}}{\sin (a+b x)^3}dx\)

\(\Big \downarrow \) 3073

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

\(\Big \downarrow \) 3042

\(\displaystyle 2 d^2 \int \frac {1}{\sin (a+b x) \sqrt {d \tan (a+b x)}}dx+\frac {2 d \csc (a+b x) \sqrt {d \tan (a+b x)}}{b}\)

\(\Big \downarrow \) 3081

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3050

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3052

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle \frac {2 d^2 \sqrt {\sin (a+b x)} \left (-\frac {2 \cos ^{\frac {3}{2}}(a+b x)}{b \sqrt {\sin (a+b x)}}-\frac {2 \sqrt {\sin (a+b x)} \sqrt {\cos (a+b x)} E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{b \sqrt {\sin (2 a+2 b x)}}\right )}{\sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}+\frac {2 d \csc (a+b x) \sqrt {d \tan (a+b x)}}{b}\)

Input:

Int[Csc[a + b*x]^3*(d*Tan[a + b*x])^(3/2),x]
 

Output:

(2*d^2*Sqrt[Sin[a + b*x]]*((-2*Cos[a + b*x]^(3/2))/(b*Sqrt[Sin[a + b*x]]) 
- (2*Sqrt[Cos[a + b*x]]*EllipticE[a - Pi/4 + b*x, 2]*Sqrt[Sin[a + b*x]])/( 
b*Sqrt[Sin[2*a + 2*b*x]])))/(Sqrt[Cos[a + b*x]]*Sqrt[d*Tan[a + b*x]]) + (2 
*d*Csc[a + b*x]*Sqrt[d*Tan[a + b*x]])/b
 

Defintions of rubi rules used

rule 3042

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

rule 3050

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_), x_Symbol] :> Simp[(b*Cos[e + f*x])^(n + 1)*((a*Sin[e + f*x])^(m + 1)/(a 
*b*f*(m + 1))), x] + Simp[(m + n + 2)/(a^2*(m + 1))   Int[(b*Cos[e + f*x])^ 
n*(a*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, - 
1] && IntegersQ[2*m, 2*n]
 

rule 3052

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]
 

rule 3073

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*(n - 1))), x] - Simp[b^2*((m + 2)/(a^2*(n - 1)))   Int[(a*Sin[e + f* 
x])^(m + 2)*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f}, x] && G 
tQ[n, 1] && (LtQ[m, -1] || (EqQ[m, -1] && EqQ[n, 3/2])) && 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 3119

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(211\) vs. \(2(95)=190\).

Time = 0.97 (sec) , antiderivative size = 212, normalized size of antiderivative = 2.08

method	result	size
default	\(\frac {\left (-4 \cos \left (b x +a \right )+2+\left (4 \cos \left (b x +a \right )+4\right ) \sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}\, \sqrt {-2 \csc \left (b x +a \right )+2 \cot \left (b x +a \right )+2}\, \sqrt {-\csc \left (b x +a \right )+\cot \left (b x +a \right )}\, \operatorname {EllipticE}\left (\sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )+\left (-2 \cos \left (b x +a \right )-2\right ) \sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}\, \sqrt {-2 \csc \left (b x +a \right )+2 \cot \left (b x +a \right )+2}\, \sqrt {-\csc \left (b x +a \right )+\cot \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )\right ) \csc \left (b x +a \right ) \sqrt {d \tan \left (b x +a \right )}\, d}{b}\)	\(212\)

Input:

int(csc(b*x+a)^3*(d*tan(b*x+a))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/b*(-4*cos(b*x+a)+2+(4*cos(b*x+a)+4)*(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)*EllipticE( 
(csc(b*x+a)-cot(b*x+a)+1)^(1/2),1/2*2^(1/2))+(-2*cos(b*x+a)-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)))*csc 
(b*x+a)*(d*tan(b*x+a))^(1/2)*d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.73 \[ \int \csc ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx=-\frac {2 \, {\left (i \, \sqrt {i \, d} d E(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) \sin \left (b x + a\right ) - i \, \sqrt {-i \, d} d E(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) \sin \left (b x + a\right ) - i \, \sqrt {i \, d} d F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) \sin \left (b x + a\right ) + i \, \sqrt {-i \, d} d F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) \sin \left (b x + a\right ) + {\left (2 \, d \cos \left (b x + a\right )^{2} - d\right )} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}\right )}}{b \sin \left (b x + a\right )} \] Input:

integrate(csc(b*x+a)^3*(d*tan(b*x+a))^(3/2),x, algorithm="fricas")
 

Output:

-2*(I*sqrt(I*d)*d*elliptic_e(arcsin(cos(b*x + a) + I*sin(b*x + a)), -1)*si 
n(b*x + a) - I*sqrt(-I*d)*d*elliptic_e(arcsin(cos(b*x + a) - I*sin(b*x + a 
)), -1)*sin(b*x + a) - I*sqrt(I*d)*d*elliptic_f(arcsin(cos(b*x + a) + I*si 
n(b*x + a)), -1)*sin(b*x + a) + I*sqrt(-I*d)*d*elliptic_f(arcsin(cos(b*x + 
 a) - I*sin(b*x + a)), -1)*sin(b*x + a) + (2*d*cos(b*x + a)^2 - d)*sqrt(d* 
sin(b*x + a)/cos(b*x + a)))/(b*sin(b*x + a))
 

Sympy [F(-1)]

Timed out. \[ \int \csc ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx=\text {Timed out} \] Input:

integrate(csc(b*x+a)**3*(d*tan(b*x+a))**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \csc ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} \csc \left (b x + a\right )^{3} \,d x } \] Input:

integrate(csc(b*x+a)^3*(d*tan(b*x+a))^(3/2),x, algorithm="maxima")
 

Output:

integrate((d*tan(b*x + a))^(3/2)*csc(b*x + a)^3, x)
 

Giac [F]

\[ \int \csc ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} \csc \left (b x + a\right )^{3} \,d x } \] Input:

integrate(csc(b*x+a)^3*(d*tan(b*x+a))^(3/2),x, algorithm="giac")
 

Output:

integrate((d*tan(b*x + a))^(3/2)*csc(b*x + a)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \csc ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx=\int \frac {{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2}}{{\sin \left (a+b\,x\right )}^3} \,d x \] Input:

int((d*tan(a + b*x))^(3/2)/sin(a + b*x)^3,x)
 

Output:

int((d*tan(a + b*x))^(3/2)/sin(a + b*x)^3, x)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int \csc ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx=\sqrt {d}\, \left (\int \sqrt {\tan \left (b x +a \right )}\, \csc \left (b x +a \right )^{3} \tan \left (b x +a \right )d x \right ) d \] 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),x)*d