3.3.72 \(\int \frac {(d \csc (a+b x))^{7/2}}{(c \sec (a+b x))^{5/2}} \, dx\) [272]

3.3.72.1 Optimal result
3.3.72.2 Mathematica [C] (verified)
3.3.72.3 Rubi [A] (verified)
3.3.72.4 Maple [B] (verified)
3.3.72.5 Fricas [C] (verification not implemented)
3.3.72.6 Sympy [F(-1)]
3.3.72.7 Maxima [F]
3.3.72.8 Giac [F]
3.3.72.9 Mupad [F(-1)]

3.3.72.1 Optimal result

Integrand size = 25, antiderivative size = 135 \[ \int \frac {(d \csc (a+b x))^{7/2}}{(c \sec (a+b x))^{5/2}} \, dx=\frac {6 d^3 \sqrt {d \csc (a+b x)}}{5 b c (c \sec (a+b x))^{3/2}}-\frac {2 d (d \csc (a+b x))^{5/2}}{5 b c (c \sec (a+b x))^{3/2}}+\frac {6 d^4 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{5 b c^2 \sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)} \sqrt {\sin (2 a+2 b x)}} \]

output
-2/5*d*(d*csc(b*x+a))^(5/2)/b/c/(c*sec(b*x+a))^(3/2)+6/5*d^3*(d*csc(b*x+a) 
)^(1/2)/b/c/(c*sec(b*x+a))^(3/2)-6/5*d^4*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a 
+1/4*Pi+b*x)*EllipticE(cos(a+1/4*Pi+b*x),2^(1/2))/b/c^2/(d*csc(b*x+a))^(1/ 
2)/(c*sec(b*x+a))^(1/2)/sin(2*b*x+2*a)^(1/2)
 
3.3.72.2 Mathematica [C] (verified)

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

Time = 2.89 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.75 \[ \int \frac {(d \csc (a+b x))^{7/2}}{(c \sec (a+b x))^{5/2}} \, dx=\frac {d^5 \left ((1-3 \cos (2 (a+b x))) \cot ^2(a+b x) \csc ^2(a+b x)+6 \sqrt [4]{-\cot ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {1}{2},\csc ^2(a+b x)\right )\right ) \sqrt {c \sec (a+b x)}}{5 b c^3 (d \csc (a+b x))^{3/2}} \]

input
Integrate[(d*Csc[a + b*x])^(7/2)/(c*Sec[a + b*x])^(5/2),x]
 
output
(d^5*((1 - 3*Cos[2*(a + b*x)])*Cot[a + b*x]^2*Csc[a + b*x]^2 + 6*(-Cot[a + 
 b*x]^2)^(1/4)*Hypergeometric2F1[-1/2, 1/4, 1/2, Csc[a + b*x]^2])*Sqrt[c*S 
ec[a + b*x]])/(5*b*c^3*(d*Csc[a + b*x])^(3/2))
 
3.3.72.3 Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3103, 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.

\(\displaystyle \int \frac {(d \csc (a+b x))^{7/2}}{(c \sec (a+b x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(d \csc (a+b x))^{7/2}}{(c \sec (a+b x))^{5/2}}dx\)

\(\Big \downarrow \) 3103

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3105

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3110

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3052

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

input
Int[(d*Csc[a + b*x])^(7/2)/(c*Sec[a + b*x])^(5/2),x]
 
output
(-2*d*(d*Csc[a + b*x])^(5/2))/(5*b*c*(c*Sec[a + b*x])^(3/2)) - (3*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*c^2)
 

3.3.72.3.1 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 3103
Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n 
_), x_Symbol] :> Simp[(-a)*(a*Csc[e + f*x])^(m - 1)*((b*Sec[e + f*x])^(n + 
1)/(f*b*(m - 1))), x] + Simp[a^2*((n + 1)/(b^2*(m - 1)))   Int[(a*Csc[e + f 
*x])^(m - 2)*(b*Sec[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && 
GtQ[m, 1] && LtQ[n, -1] && IntegersQ[2*m, 2*n]
 

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]
 
3.3.72.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(384\) vs. \(2(140)=280\).

Time = 7.82 (sec) , antiderivative size = 385, normalized size of antiderivative = 2.85

method result size
default \(-\frac {\sqrt {2}\, d^{3} \sqrt {d \csc \left (b x +a \right )}\, \left (6 \sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \operatorname {EllipticE}\left (\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )+6 \sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \operatorname {EllipticE}\left (\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right ) \sec \left (b x +a \right )-3 \sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right ) \sec \left (b x +a \right )-3 \sqrt {2}+\sqrt {2}\, \cot \left (b x +a \right ) \csc \left (b x +a \right )\right )}{5 b \sqrt {c \sec \left (b x +a \right )}\, c^{2}}\) \(385\)

input
int((d*csc(b*x+a))^(7/2)/(c*sec(b*x+a))^(5/2),x,method=_RETURNVERBOSE)
 
output
-1/5/b*2^(1/2)*d^3*(d*csc(b*x+a))^(1/2)/(c*sec(b*x+a))^(1/2)/c^2*(6*(1+csc 
(b*x+a)-cot(b*x+a))^(1/2)*(cot(b*x+a)-csc(b*x+a)+1)^(1/2)*(cot(b*x+a)-csc( 
b*x+a))^(1/2)*EllipticE((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2*2^(1/2))-3*(1+ 
csc(b*x+a)-cot(b*x+a))^(1/2)*(cot(b*x+a)-csc(b*x+a)+1)^(1/2)*(cot(b*x+a)-c 
sc(b*x+a))^(1/2)*EllipticF((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2*2^(1/2))+6* 
(1+csc(b*x+a)-cot(b*x+a))^(1/2)*(cot(b*x+a)-csc(b*x+a)+1)^(1/2)*(cot(b*x+a 
)-csc(b*x+a))^(1/2)*EllipticE((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2*2^(1/2)) 
*sec(b*x+a)-3*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*(cot(b*x+a)-csc(b*x+a)+1)^(1 
/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*EllipticF((1+csc(b*x+a)-cot(b*x+a))^(1/2 
),1/2*2^(1/2))*sec(b*x+a)-3*2^(1/2)+2^(1/2)*cot(b*x+a)*csc(b*x+a))
 
3.3.72.5 Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.88 \[ \int \frac {(d \csc (a+b x))^{7/2}}{(c \sec (a+b x))^{5/2}} \, dx=-\frac {3 \, {\left (d^{3} \cos \left (b x + a\right )^{2} - d^{3}\right )} \sqrt {-4 i \, c d} E(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) + 3 \, {\left (d^{3} \cos \left (b x + a\right )^{2} - d^{3}\right )} \sqrt {4 i \, c d} E(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) - 3 \, {\left (d^{3} \cos \left (b x + a\right )^{2} - d^{3}\right )} \sqrt {-4 i \, c d} F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) - 3 \, {\left (d^{3} \cos \left (b x + a\right )^{2} - d^{3}\right )} \sqrt {4 i \, c d} F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) - 4 \, {\left (3 \, d^{3} \cos \left (b x + a\right )^{4} - 2 \, d^{3} \cos \left (b x + a\right )^{2}\right )} \sqrt {\frac {c}{\cos \left (b x + a\right )}} \sqrt {\frac {d}{\sin \left (b x + a\right )}}}{10 \, {\left (b c^{3} \cos \left (b x + a\right )^{2} - b c^{3}\right )}} \]

input
integrate((d*csc(b*x+a))^(7/2)/(c*sec(b*x+a))^(5/2),x, algorithm="fricas")
 
output
-1/10*(3*(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) + 3*(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) - 3*(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) - 3*(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) - 4*(3*d^3*cos(b*x + a)^4 - 2*d^3*cos(b*x 
 + a)^2)*sqrt(c/cos(b*x + a))*sqrt(d/sin(b*x + a)))/(b*c^3*cos(b*x + a)^2 
- b*c^3)
 
3.3.72.6 Sympy [F(-1)]

Timed out. \[ \int \frac {(d \csc (a+b x))^{7/2}}{(c \sec (a+b x))^{5/2}} \, dx=\text {Timed out} \]

input
integrate((d*csc(b*x+a))**(7/2)/(c*sec(b*x+a))**(5/2),x)
 
output
Timed out
 
3.3.72.7 Maxima [F]

\[ \int \frac {(d \csc (a+b x))^{7/2}}{(c \sec (a+b x))^{5/2}} \, dx=\int { \frac {\left (d \csc \left (b x + a\right )\right )^{\frac {7}{2}}}{\left (c \sec \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \]

input
integrate((d*csc(b*x+a))^(7/2)/(c*sec(b*x+a))^(5/2),x, algorithm="maxima")
 
output
integrate((d*csc(b*x + a))^(7/2)/(c*sec(b*x + a))^(5/2), x)
 
3.3.72.8 Giac [F]

\[ \int \frac {(d \csc (a+b x))^{7/2}}{(c \sec (a+b x))^{5/2}} \, dx=\int { \frac {\left (d \csc \left (b x + a\right )\right )^{\frac {7}{2}}}{\left (c \sec \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \]

input
integrate((d*csc(b*x+a))^(7/2)/(c*sec(b*x+a))^(5/2),x, algorithm="giac")
 
output
integrate((d*csc(b*x + a))^(7/2)/(c*sec(b*x + a))^(5/2), x)
 
3.3.72.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(d \csc (a+b x))^{7/2}}{(c \sec (a+b x))^{5/2}} \, dx=\int \frac {{\left (\frac {d}{\sin \left (a+b\,x\right )}\right )}^{7/2}}{{\left (\frac {c}{\cos \left (a+b\,x\right )}\right )}^{5/2}} \,d x \]

input
int((d/sin(a + b*x))^(7/2)/(c/cos(a + b*x))^(5/2),x)
 
output
int((d/sin(a + b*x))^(7/2)/(c/cos(a + b*x))^(5/2), x)