\(\int \frac {(d \csc (a+b x))^{7/2}}{\sqrt {c \sec (a+b x)}} \, dx\) [255]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [C] (verification not implemented)
Sympy [F(-1)]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 25, antiderivative size = 128 \[ \int \frac {(d \csc (a+b x))^{7/2}}{\sqrt {c \sec (a+b x)}} \, dx=-\frac {4 c d^3 \sqrt {d \csc (a+b x)}}{5 b (c \sec (a+b x))^{3/2}}-\frac {2 c d (d \csc (a+b x))^{5/2}}{5 b (c \sec (a+b x))^{3/2}}-\frac {4 d^4 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{5 b \sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)} \sqrt {\sin (2 a+2 b x)}} \] Output:

-4/5*c*d^3*(d*csc(b*x+a))^(1/2)/b/(c*sec(b*x+a))^(3/2)-2/5*c*d*(d*csc(b*x+ 
a))^(5/2)/b/(c*sec(b*x+a))^(3/2)+4/5*d^4*EllipticE(cos(a+1/4*Pi+b*x),2^(1/ 
2))/b/(d*csc(b*x+a))^(1/2)/(c*sec(b*x+a))^(1/2)/sin(2*b*x+2*a)^(1/2)
 

Mathematica [C] (verified)

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

Time = 1.11 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.81 \[ \int \frac {(d \csc (a+b x))^{7/2}}{\sqrt {c \sec (a+b x)}} \, dx=-\frac {2 d^2 (d \csc (a+b x))^{3/2} \left (-\left ((-2+\cos (2 (a+b x))) \cot ^3(a+b x)\right )+\sqrt [4]{-\cot ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {1}{2},\csc ^2(a+b x)\right ) \sin (2 (a+b x))\right ) \tan ^2(a+b x)}{5 b \sqrt {c \sec (a+b x)}} \] Input:

Integrate[(d*Csc[a + b*x])^(7/2)/Sqrt[c*Sec[a + b*x]],x]
 

Output:

(-2*d^2*(d*Csc[a + b*x])^(3/2)*(-((-2 + Cos[2*(a + b*x)])*Cot[a + b*x]^3) 
+ (-Cot[a + b*x]^2)^(1/4)*Hypergeometric2F1[-1/2, 1/4, 1/2, Csc[a + b*x]^2 
]*Sin[2*(a + b*x)])*Tan[a + b*x]^2)/(5*b*Sqrt[c*Sec[a + b*x]])
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3105, 3042, 3105, 3042, 3110, 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.

\(\displaystyle \int \frac {(d \csc (a+b x))^{7/2}}{\sqrt {c \sec (a+b x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(d \csc (a+b x))^{7/2}}{\sqrt {c \sec (a+b x)}}dx\)

\(\Big \downarrow \) 3105

\(\displaystyle \frac {2}{5} d^2 \int \frac {(d \csc (a+b x))^{3/2}}{\sqrt {c \sec (a+b x)}}dx-\frac {2 c d (d \csc (a+b x))^{5/2}}{5 b (c \sec (a+b x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2}{5} d^2 \int \frac {(d \csc (a+b x))^{3/2}}{\sqrt {c \sec (a+b x)}}dx-\frac {2 c d (d \csc (a+b x))^{5/2}}{5 b (c \sec (a+b x))^{3/2}}\)

\(\Big \downarrow \) 3105

\(\displaystyle \frac {2}{5} d^2 \left (-2 d^2 \int \frac {1}{\sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)}}dx-\frac {2 c d \sqrt {d \csc (a+b x)}}{b (c \sec (a+b x))^{3/2}}\right )-\frac {2 c d (d \csc (a+b x))^{5/2}}{5 b (c \sec (a+b x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2}{5} d^2 \left (-2 d^2 \int \frac {1}{\sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)}}dx-\frac {2 c d \sqrt {d \csc (a+b x)}}{b (c \sec (a+b x))^{3/2}}\right )-\frac {2 c d (d \csc (a+b x))^{5/2}}{5 b (c \sec (a+b x))^{3/2}}\)

\(\Big \downarrow \) 3110

\(\displaystyle \frac {2}{5} d^2 \left (-\frac {2 d^2 \int \sqrt {c \cos (a+b x)} \sqrt {d \sin (a+b x)}dx}{\sqrt {c \cos (a+b x)} \sqrt {c \sec (a+b x)} \sqrt {d \sin (a+b x)} \sqrt {d \csc (a+b x)}}-\frac {2 c d \sqrt {d \csc (a+b x)}}{b (c \sec (a+b x))^{3/2}}\right )-\frac {2 c d (d \csc (a+b x))^{5/2}}{5 b (c \sec (a+b x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2}{5} d^2 \left (-\frac {2 d^2 \int \sqrt {c \cos (a+b x)} \sqrt {d \sin (a+b x)}dx}{\sqrt {c \cos (a+b x)} \sqrt {c \sec (a+b x)} \sqrt {d \sin (a+b x)} \sqrt {d \csc (a+b x)}}-\frac {2 c d \sqrt {d \csc (a+b x)}}{b (c \sec (a+b x))^{3/2}}\right )-\frac {2 c d (d \csc (a+b x))^{5/2}}{5 b (c \sec (a+b x))^{3/2}}\)

\(\Big \downarrow \) 3052

\(\displaystyle \frac {2}{5} d^2 \left (-\frac {2 d^2 \int \sqrt {\sin (2 a+2 b x)}dx}{\sqrt {\sin (2 a+2 b x)} \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}-\frac {2 c d \sqrt {d \csc (a+b x)}}{b (c \sec (a+b x))^{3/2}}\right )-\frac {2 c d (d \csc (a+b x))^{5/2}}{5 b (c \sec (a+b x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2}{5} d^2 \left (-\frac {2 d^2 \int \sqrt {\sin (2 a+2 b x)}dx}{\sqrt {\sin (2 a+2 b x)} \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}-\frac {2 c d \sqrt {d \csc (a+b x)}}{b (c \sec (a+b x))^{3/2}}\right )-\frac {2 c d (d \csc (a+b x))^{5/2}}{5 b (c \sec (a+b x))^{3/2}}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {2}{5} d^2 \left (-\frac {2 d^2 E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{b \sqrt {\sin (2 a+2 b x)} \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}-\frac {2 c d \sqrt {d \csc (a+b x)}}{b (c \sec (a+b x))^{3/2}}\right )-\frac {2 c d (d \csc (a+b x))^{5/2}}{5 b (c \sec (a+b x))^{3/2}}\)

Input:

Int[(d*Csc[a + b*x])^(7/2)/Sqrt[c*Sec[a + b*x]],x]
 

Output:

(-2*c*d*(d*Csc[a + b*x])^(5/2))/(5*b*(c*Sec[a + b*x])^(3/2)) + (2*d^2*((-2 
*c*d*Sqrt[d*Csc[a + b*x]])/(b*(c*Sec[a + b*x])^(3/2)) - (2*d^2*EllipticE[a 
 - Pi/4 + b*x, 2])/(b*Sqrt[d*Csc[a + b*x]]*Sqrt[c*Sec[a + b*x]]*Sqrt[Sin[2 
*a + 2*b*x]])))/5
 

Defintions of rubi rules used

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

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 3105
Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n 
_.), x_Symbol] :> Simp[(-a)*b*(a*Csc[e + f*x])^(m - 1)*((b*Sec[e + f*x])^(n 
 - 1)/(f*(m - 1))), x] + Simp[a^2*((m + n - 2)/(m - 1))   Int[(a*Csc[e + f* 
x])^(m - 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[ 
m, 1] && IntegersQ[2*m, 2*n] &&  !GtQ[n, m]
 

rule 3110
Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n 
_), x_Symbol] :> Simp[(a*Csc[e + f*x])^m*(b*Sec[e + f*x])^n*(a*Sin[e + f*x] 
)^m*(b*Cos[e + f*x])^n   Int[1/((a*Sin[e + f*x])^m*(b*Cos[e + f*x])^n), x], 
 x] /; FreeQ[{a, b, e, f, m, n}, x] && IntegerQ[m - 1/2] && IntegerQ[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. \(221\) vs. \(2(109)=218\).

Time = 1.42 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.73

method result size
default \(-\frac {2 \sqrt {d \csc \left (b x +a \right )}\, d^{3} \left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}\, \sqrt {2 \cot \left (b x +a \right )-2 \csc \left (b x +a \right )+2}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \operatorname {EllipticE}\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (-2-2 \sec \left (b x +a \right )\right )+\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}\, \sqrt {2 \cot \left (b x +a \right )-2 \csc \left (b x +a \right )+2}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (1+\sec \left (b x +a \right )\right )+2+\csc \left (b x +a \right ) \cot \left (b x +a \right )\right )}{5 b \sqrt {c \sec \left (b x +a \right )}}\) \(222\)

Input:

int((d*csc(b*x+a))^(7/2)/(c*sec(b*x+a))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-2/5/b*(d*csc(b*x+a))^(1/2)*d^3/(c*sec(b*x+a))^(1/2)*((-cot(b*x+a)+csc(b*x 
+a)+1)^(1/2)*(2*cot(b*x+a)-2*csc(b*x+a)+2)^(1/2)*(cot(b*x+a)-csc(b*x+a))^( 
1/2)*EllipticE((-cot(b*x+a)+csc(b*x+a)+1)^(1/2),1/2*2^(1/2))*(-2-2*sec(b*x 
+a))+(-cot(b*x+a)+csc(b*x+a)+1)^(1/2)*(2*cot(b*x+a)-2*csc(b*x+a)+2)^(1/2)* 
(cot(b*x+a)-csc(b*x+a))^(1/2)*EllipticF((-cot(b*x+a)+csc(b*x+a)+1)^(1/2),1 
/2*2^(1/2))*(1+sec(b*x+a))+2+csc(b*x+a)*cot(b*x+a))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.94 \[ \int \frac {(d \csc (a+b x))^{7/2}}{\sqrt {c \sec (a+b x)}} \, dx=\frac {{\left (d^{3} \cos \left (b x + a\right )^{2} - d^{3}\right )} \sqrt {-4 i \, c d} E(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) + {\left (d^{3} \cos \left (b x + a\right )^{2} - d^{3}\right )} \sqrt {4 i \, c d} E(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) - {\left (d^{3} \cos \left (b x + a\right )^{2} - d^{3}\right )} \sqrt {-4 i \, c d} F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) - {\left (d^{3} \cos \left (b x + a\right )^{2} - d^{3}\right )} \sqrt {4 i \, c d} F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) - 2 \, {\left (2 \, d^{3} \cos \left (b x + a\right )^{4} - 3 \, d^{3} \cos \left (b x + a\right )^{2}\right )} \sqrt {\frac {c}{\cos \left (b x + a\right )}} \sqrt {\frac {d}{\sin \left (b x + a\right )}}}{5 \, {\left (b c \cos \left (b x + a\right )^{2} - b c\right )}} \] Input:

integrate((d*csc(b*x+a))^(7/2)/(c*sec(b*x+a))^(1/2),x, algorithm="fricas")
 

Output:

1/5*((d^3*cos(b*x + a)^2 - d^3)*sqrt(-4*I*c*d)*elliptic_e(arcsin(cos(b*x + 
 a) + I*sin(b*x + a)), -1) + (d^3*cos(b*x + a)^2 - d^3)*sqrt(4*I*c*d)*elli 
ptic_e(arcsin(cos(b*x + a) - I*sin(b*x + a)), -1) - (d^3*cos(b*x + a)^2 - 
d^3)*sqrt(-4*I*c*d)*elliptic_f(arcsin(cos(b*x + a) + I*sin(b*x + a)), -1) 
- (d^3*cos(b*x + a)^2 - d^3)*sqrt(4*I*c*d)*elliptic_f(arcsin(cos(b*x + a) 
- I*sin(b*x + a)), -1) - 2*(2*d^3*cos(b*x + a)^4 - 3*d^3*cos(b*x + a)^2)*s 
qrt(c/cos(b*x + a))*sqrt(d/sin(b*x + a)))/(b*c*cos(b*x + a)^2 - b*c)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(d \csc (a+b x))^{7/2}}{\sqrt {c \sec (a+b x)}} \, dx=\text {Timed out} \] Input:

integrate((d*csc(b*x+a))**(7/2)/(c*sec(b*x+a))**(1/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(d \csc (a+b x))^{7/2}}{\sqrt {c \sec (a+b x)}} \, dx=\int { \frac {\left (d \csc \left (b x + a\right )\right )^{\frac {7}{2}}}{\sqrt {c \sec \left (b x + a\right )}} \,d x } \] Input:

integrate((d*csc(b*x+a))^(7/2)/(c*sec(b*x+a))^(1/2),x, algorithm="maxima")
 

Output:

integrate((d*csc(b*x + a))^(7/2)/sqrt(c*sec(b*x + a)), x)
 

Giac [F]

\[ \int \frac {(d \csc (a+b x))^{7/2}}{\sqrt {c \sec (a+b x)}} \, dx=\int { \frac {\left (d \csc \left (b x + a\right )\right )^{\frac {7}{2}}}{\sqrt {c \sec \left (b x + a\right )}} \,d x } \] Input:

integrate((d*csc(b*x+a))^(7/2)/(c*sec(b*x+a))^(1/2),x, algorithm="giac")
 

Output:

integrate((d*csc(b*x + a))^(7/2)/sqrt(c*sec(b*x + a)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(d \csc (a+b x))^{7/2}}{\sqrt {c \sec (a+b x)}} \, dx=\int \frac {{\left (\frac {d}{\sin \left (a+b\,x\right )}\right )}^{7/2}}{\sqrt {\frac {c}{\cos \left (a+b\,x\right )}}} \,d x \] Input:

int((d/sin(a + b*x))^(7/2)/(c/cos(a + b*x))^(1/2),x)
 

Output:

int((d/sin(a + b*x))^(7/2)/(c/cos(a + b*x))^(1/2), x)
 

Reduce [F]

\[ \int \frac {(d \csc (a+b x))^{7/2}}{\sqrt {c \sec (a+b x)}} \, dx=\frac {\sqrt {d}\, \sqrt {c}\, \left (\int \frac {\sqrt {\sec \left (b x +a \right )}\, \sqrt {\csc \left (b x +a \right )}\, \csc \left (b x +a \right )^{3}}{\sec \left (b x +a \right )}d x \right ) d^{3}}{c} \] Input:

int((d*csc(b*x+a))^(7/2)/(c*sec(b*x+a))^(1/2),x)
 

Output:

(sqrt(d)*sqrt(c)*int((sqrt(sec(a + b*x))*sqrt(csc(a + b*x))*csc(a + b*x)** 
3)/sec(a + b*x),x)*d**3)/c