\(\int \frac {\cos ^{\frac {5}{2}}(a+b x)}{\sin ^{\frac {5}{2}}(a+b x)} \, dx\) [302]

Optimal result

Integrand size = 21, antiderivative size = 153 \[ \int \frac {\cos ^{\frac {5}{2}}(a+b x)}{\sin ^{\frac {5}{2}}(a+b x)} \, dx=-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{\sqrt {2} b}+\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{\sqrt {2} b}-\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{(1+\cot (a+b x)) \sqrt {\sin (a+b x)}}\right )}{\sqrt {2} b}-\frac {2 \cos ^{\frac {3}{2}}(a+b x)}{3 b \sin ^{\frac {3}{2}}(a+b x)} \] Output:

1/2*arctan(-1+2^(1/2)*cos(b*x+a)^(1/2)/sin(b*x+a)^(1/2))*2^(1/2)/b+1/2*arc 
tan(1+2^(1/2)*cos(b*x+a)^(1/2)/sin(b*x+a)^(1/2))*2^(1/2)/b-1/2*arctanh(2^( 
1/2)*cos(b*x+a)^(1/2)/(1+cot(b*x+a))/sin(b*x+a)^(1/2))*2^(1/2)/b-2/3*cos(b 
*x+a)^(3/2)/b/sin(b*x+a)^(3/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.04 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.37 \[ \int \frac {\cos ^{\frac {5}{2}}(a+b x)}{\sin ^{\frac {5}{2}}(a+b x)} \, dx=-\frac {2 \sqrt [4]{\cos ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-\frac {3}{4},\frac {1}{4},\sin ^2(a+b x)\right )}{3 b \sqrt {\cos (a+b x)} \sin ^{\frac {3}{2}}(a+b x)} \] Input:

Integrate[Cos[a + b*x]^(5/2)/Sin[a + b*x]^(5/2),x]
 

Output:

(-2*(Cos[a + b*x]^2)^(1/4)*Hypergeometric2F1[-3/4, -3/4, 1/4, Sin[a + b*x] 
^2])/(3*b*Sqrt[Cos[a + b*x]]*Sin[a + b*x]^(3/2))
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.34, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 3047, 3042, 3055, 826, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

\(\displaystyle \int \frac {\cos ^{\frac {5}{2}}(a+b x)}{\sin ^{\frac {5}{2}}(a+b x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (a+b x)^{5/2}}{\sin (a+b x)^{5/2}}dx\)

\(\Big \downarrow \) 3047

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3055

\(\displaystyle \frac {2 \int \frac {\cot (a+b x)}{\cot ^2(a+b x)+1}d\frac {\sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}}{b}-\frac {2 \cos ^{\frac {3}{2}}(a+b x)}{3 b \sin ^{\frac {3}{2}}(a+b x)}\)

\(\Big \downarrow \) 826

\(\displaystyle \frac {2 \left (\frac {1}{2} \int \frac {\cot (a+b x)+1}{\cot ^2(a+b x)+1}d\frac {\sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}-\frac {1}{2} \int \frac {1-\cot (a+b x)}{\cot ^2(a+b x)+1}d\frac {\sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{b}-\frac {2 \cos ^{\frac {3}{2}}(a+b x)}{3 b \sin ^{\frac {3}{2}}(a+b x)}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {2 \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{\cot (a+b x)-\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}+1}d\frac {\sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}+\frac {1}{2} \int \frac {1}{\cot (a+b x)+\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}+1}d\frac {\sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )-\frac {1}{2} \int \frac {1-\cot (a+b x)}{\cot ^2(a+b x)+1}d\frac {\sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{b}-\frac {2 \cos ^{\frac {3}{2}}(a+b x)}{3 b \sin ^{\frac {3}{2}}(a+b x)}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {2 \left (\frac {1}{2} \left (\frac {\int \frac {1}{-\cot (a+b x)-1}d\left (1-\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\cot (a+b x)-1}d\left (\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}+1\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-\cot (a+b x)}{\cot ^2(a+b x)+1}d\frac {\sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{b}-\frac {2 \cos ^{\frac {3}{2}}(a+b x)}{3 b \sin ^{\frac {3}{2}}(a+b x)}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {2 \left (\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-\cot (a+b x)}{\cot ^2(a+b x)+1}d\frac {\sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{b}-\frac {2 \cos ^{\frac {3}{2}}(a+b x)}{3 b \sin ^{\frac {3}{2}}(a+b x)}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {2 \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {2}-\frac {2 \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}}{\cot (a+b x)-\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}+1}d\frac {\sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}}{2 \sqrt {2}}+\frac {\int -\frac {\sqrt {2} \left (\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}+1\right )}{\cot (a+b x)+\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}+1}d\frac {\sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{\sqrt {2}}\right )\right )}{b}-\frac {2 \cos ^{\frac {3}{2}}(a+b x)}{3 b \sin ^{\frac {3}{2}}(a+b x)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-\frac {2 \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}}{\cot (a+b x)-\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}+1}d\frac {\sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}}{2 \sqrt {2}}-\frac {\int \frac {\sqrt {2} \left (\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}+1\right )}{\cot (a+b x)+\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}+1}d\frac {\sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{\sqrt {2}}\right )\right )}{b}-\frac {2 \cos ^{\frac {3}{2}}(a+b x)}{3 b \sin ^{\frac {3}{2}}(a+b x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-\frac {2 \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}}{\cot (a+b x)-\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}+1}d\frac {\sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}}{2 \sqrt {2}}-\frac {1}{2} \int \frac {\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}+1}{\cot (a+b x)+\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}+1}d\frac {\sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{\sqrt {2}}\right )\right )}{b}-\frac {2 \cos ^{\frac {3}{2}}(a+b x)}{3 b \sin ^{\frac {3}{2}}(a+b x)}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {2 \left (\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (\cot (a+b x)-\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\cot (a+b x)+\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}+1\right )}{2 \sqrt {2}}\right )\right )}{b}-\frac {2 \cos ^{\frac {3}{2}}(a+b x)}{3 b \sin ^{\frac {3}{2}}(a+b x)}\)

Input:

Int[Cos[a + b*x]^(5/2)/Sin[a + b*x]^(5/2),x]
 

Output:

(2*((-(ArcTan[1 - (Sqrt[2]*Sqrt[Cos[a + b*x]])/Sqrt[Sin[a + b*x]]]/Sqrt[2] 
) + ArcTan[1 + (Sqrt[2]*Sqrt[Cos[a + b*x]])/Sqrt[Sin[a + b*x]]]/Sqrt[2])/2 
 + (Log[1 + Cot[a + b*x] - (Sqrt[2]*Sqrt[Cos[a + b*x]])/Sqrt[Sin[a + b*x]] 
]/(2*Sqrt[2]) - Log[1 + Cot[a + b*x] + (Sqrt[2]*Sqrt[Cos[a + b*x]])/Sqrt[S 
in[a + b*x]]]/(2*Sqrt[2]))/2))/b - (2*Cos[a + b*x]^(3/2))/(3*b*Sin[a + b*x 
]^(3/2))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 826

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 
2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s)   Int[(r + s*x^2)/(a + b*x^ 
4), x], x] - Simp[1/(2*s)   Int[(r - s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{ 
a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] 
 && AtomQ[SplitProduct[SumBaseQ, b]]))
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

rule 3042

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

rule 3047

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_), x_Symbol] :> Simp[a*(a*Cos[e + f*x])^(m - 1)*((b*Sin[e + f*x])^(n + 1)/ 
(b*f*(n + 1))), x] + Simp[a^2*((m - 1)/(b^2*(n + 1)))   Int[(a*Cos[e + f*x] 
)^(m - 2)*(b*Sin[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ 
[m, 1] && LtQ[n, -1] && (IntegersQ[2*m, 2*n] || EqQ[m + n, 0])
 

rule 3055

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_), x_Symbol] :> With[{k = Denominator[m]}, Simp[(-k)*a*(b/f)   Subst[Int[x 
^(k*(m + 1) - 1)/(a^2 + b^2*x^(2*k)), x], x, (a*Cos[e + f*x])^(1/k)/(b*Sin[ 
e + f*x])^(1/k)], x]] /; FreeQ[{a, b, e, f}, x] && EqQ[m + n, 0] && GtQ[m, 
0] && LtQ[m, 1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(430\) vs. \(2(123)=246\).

Time = 1.00 (sec) , antiderivative size = 431, normalized size of antiderivative = 2.82

method	result	size
default	\(-\frac {\sqrt {2}\, \left (\left (-3 \cos \left (b x +a \right )+3\right ) \ln \left (-\frac {2 \sqrt {2}\, \sqrt {\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \sin \left (b x +a \right )-\cos \left (b x +a \right ) \cot \left (b x +a \right )+2 \cot \left (b x +a \right )-\csc \left (b x +a \right )-2 \cos \left (b x +a \right )+\sin \left (b x +a \right )+2}{\cos \left (b x +a \right )-1}\right )+\left (3 \cos \left (b x +a \right )-3\right ) \ln \left (\frac {2 \sqrt {2}\, \sqrt {\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \sin \left (b x +a \right )+\cos \left (b x +a \right ) \cot \left (b x +a \right )-2 \cot \left (b x +a \right )+\csc \left (b x +a \right )+2 \cos \left (b x +a \right )-\sin \left (b x +a \right )-2}{\cos \left (b x +a \right )-1}\right )+\left (-6 \cos \left (b x +a \right )+6\right ) \arctan \left (\frac {\sqrt {2}\, \sqrt {\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \sin \left (b x +a \right )-\cos \left (b x +a \right )+1}{\cos \left (b x +a \right )-1}\right )+\left (-6 \cos \left (b x +a \right )+6\right ) \arctan \left (\frac {\sqrt {2}\, \sqrt {\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \sin \left (b x +a \right )+\cos \left (b x +a \right )-1}{\cos \left (b x +a \right )-1}\right )+4 \cos \left (b x +a \right ) \sqrt {\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \sqrt {2}\right ) \sqrt {\cos \left (b x +a \right )}}{12 b \sin \left (b x +a \right )^{\frac {3}{2}} \sqrt {\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}}\)	\(431\)

Input:

int(cos(b*x+a)^(5/2)/sin(b*x+a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/12/b*2^(1/2)*((-3*cos(b*x+a)+3)*ln(-(2*2^(1/2)*(sin(b*x+a)*cos(b*x+a)/( 
cos(b*x+a)+1)^2)^(1/2)*sin(b*x+a)-cos(b*x+a)*cot(b*x+a)+2*cot(b*x+a)-csc(b 
*x+a)-2*cos(b*x+a)+sin(b*x+a)+2)/(cos(b*x+a)-1))+(3*cos(b*x+a)-3)*ln((2*2^ 
(1/2)*(sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*sin(b*x+a)+cos(b*x+a) 
*cot(b*x+a)-2*cot(b*x+a)+csc(b*x+a)+2*cos(b*x+a)-sin(b*x+a)-2)/(cos(b*x+a) 
-1))+(-6*cos(b*x+a)+6)*arctan((2^(1/2)*(sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+ 
1)^2)^(1/2)*sin(b*x+a)-cos(b*x+a)+1)/(cos(b*x+a)-1))+(-6*cos(b*x+a)+6)*arc 
tan((2^(1/2)*(sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*sin(b*x+a)+cos 
(b*x+a)-1)/(cos(b*x+a)-1))+4*cos(b*x+a)*(sin(b*x+a)*cos(b*x+a)/(cos(b*x+a) 
+1)^2)^(1/2)*2^(1/2))*cos(b*x+a)^(1/2)/sin(b*x+a)^(3/2)/(sin(b*x+a)*cos(b* 
x+a)/(cos(b*x+a)+1)^2)^(1/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 475 vs. \(2 (123) = 246\).

Time = 0.17 (sec) , antiderivative size = 475, normalized size of antiderivative = 3.10 \[ \int \frac {\cos ^{\frac {5}{2}}(a+b x)}{\sin ^{\frac {5}{2}}(a+b x)} \, dx=\frac {3 \, {\left (\sqrt {2} \cos \left (b x + a\right )^{2} - \sqrt {2}\right )} \arctan \left (\frac {2 \, \cos \left (b x + a\right )^{3} - 2 \, \cos \left (b x + a\right )^{2} \sin \left (b x + a\right ) + \sqrt {2} \sqrt {\cos \left (b x + a\right )} \sqrt {\sin \left (b x + a\right )} - 2 \, \cos \left (b x + a\right )}{2 \, {\left (\cos \left (b x + a\right )^{3} + \cos \left (b x + a\right )^{2} \sin \left (b x + a\right ) - \cos \left (b x + a\right )\right )}}\right ) + 3 \, {\left (\sqrt {2} \cos \left (b x + a\right )^{2} - \sqrt {2}\right )} \arctan \left (-\frac {2 \, \cos \left (b x + a\right )^{3} - 2 \, \cos \left (b x + a\right )^{2} \sin \left (b x + a\right ) - \sqrt {2} \sqrt {\cos \left (b x + a\right )} \sqrt {\sin \left (b x + a\right )} - 2 \, \cos \left (b x + a\right )}{2 \, {\left (\cos \left (b x + a\right )^{3} + \cos \left (b x + a\right )^{2} \sin \left (b x + a\right ) - \cos \left (b x + a\right )\right )}}\right ) - 6 \, {\left (\sqrt {2} \cos \left (b x + a\right )^{2} - \sqrt {2}\right )} \arctan \left (-\frac {\sqrt {2} \cos \left (b x + a\right ) - \sqrt {2} \sin \left (b x + a\right )}{2 \, \sqrt {\cos \left (b x + a\right )} \sqrt {\sin \left (b x + a\right )}}\right ) - 3 \, {\left (\sqrt {2} \cos \left (b x + a\right )^{2} - \sqrt {2}\right )} \log \left (2 \, {\left (\sqrt {2} \cos \left (b x + a\right ) + \sqrt {2} \sin \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right )} \sqrt {\sin \left (b x + a\right )} + 4 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right ) + 3 \, {\left (\sqrt {2} \cos \left (b x + a\right )^{2} - \sqrt {2}\right )} \log \left (-2 \, {\left (\sqrt {2} \cos \left (b x + a\right ) + \sqrt {2} \sin \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right )} \sqrt {\sin \left (b x + a\right )} + 4 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right ) + 16 \, \cos \left (b x + a\right )^{\frac {3}{2}} \sqrt {\sin \left (b x + a\right )}}{24 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )}} \] Input:

integrate(cos(b*x+a)^(5/2)/sin(b*x+a)^(5/2),x, algorithm="fricas")
 

Output:

1/24*(3*(sqrt(2)*cos(b*x + a)^2 - sqrt(2))*arctan(1/2*(2*cos(b*x + a)^3 - 
2*cos(b*x + a)^2*sin(b*x + a) + sqrt(2)*sqrt(cos(b*x + a))*sqrt(sin(b*x + 
a)) - 2*cos(b*x + a))/(cos(b*x + a)^3 + cos(b*x + a)^2*sin(b*x + a) - cos( 
b*x + a))) + 3*(sqrt(2)*cos(b*x + a)^2 - sqrt(2))*arctan(-1/2*(2*cos(b*x + 
 a)^3 - 2*cos(b*x + a)^2*sin(b*x + a) - sqrt(2)*sqrt(cos(b*x + a))*sqrt(si 
n(b*x + a)) - 2*cos(b*x + a))/(cos(b*x + a)^3 + cos(b*x + a)^2*sin(b*x + a 
) - cos(b*x + a))) - 6*(sqrt(2)*cos(b*x + a)^2 - sqrt(2))*arctan(-1/2*(sqr 
t(2)*cos(b*x + a) - sqrt(2)*sin(b*x + a))/(sqrt(cos(b*x + a))*sqrt(sin(b*x 
 + a)))) - 3*(sqrt(2)*cos(b*x + a)^2 - sqrt(2))*log(2*(sqrt(2)*cos(b*x + a 
) + sqrt(2)*sin(b*x + a))*sqrt(cos(b*x + a))*sqrt(sin(b*x + a)) + 4*cos(b* 
x + a)*sin(b*x + a) + 1) + 3*(sqrt(2)*cos(b*x + a)^2 - sqrt(2))*log(-2*(sq 
rt(2)*cos(b*x + a) + sqrt(2)*sin(b*x + a))*sqrt(cos(b*x + a))*sqrt(sin(b*x 
 + a)) + 4*cos(b*x + a)*sin(b*x + a) + 1) + 16*cos(b*x + a)^(3/2)*sqrt(sin 
(b*x + a)))/(b*cos(b*x + a)^2 - b)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(a+b x)}{\sin ^{\frac {5}{2}}(a+b x)} \, dx=\text {Timed out} \] Input:

integrate(cos(b*x+a)**(5/2)/sin(b*x+a)**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\cos ^{\frac {5}{2}}(a+b x)}{\sin ^{\frac {5}{2}}(a+b x)} \, dx=\int { \frac {\cos \left (b x + a\right )^{\frac {5}{2}}}{\sin \left (b x + a\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate(cos(b*x+a)^(5/2)/sin(b*x+a)^(5/2),x, algorithm="maxima")
 

Output:

integrate(cos(b*x + a)^(5/2)/sin(b*x + a)^(5/2), x)
 

Giac [F]

\[ \int \frac {\cos ^{\frac {5}{2}}(a+b x)}{\sin ^{\frac {5}{2}}(a+b x)} \, dx=\int { \frac {\cos \left (b x + a\right )^{\frac {5}{2}}}{\sin \left (b x + a\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate(cos(b*x+a)^(5/2)/sin(b*x+a)^(5/2),x, algorithm="giac")
 

Output:

integrate(cos(b*x + a)^(5/2)/sin(b*x + a)^(5/2), x)
 

Mupad [B] (verification not implemented)

Time = 26.87 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.29 \[ \int \frac {\cos ^{\frac {5}{2}}(a+b x)}{\sin ^{\frac {5}{2}}(a+b x)} \, dx=-\frac {2\,{\cos \left (a+b\,x\right )}^{7/2}\,{\left ({\sin \left (a+b\,x\right )}^2\right )}^{3/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {7}{4},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (a+b\,x\right )}^2\right )}{7\,b\,{\sin \left (a+b\,x\right )}^{3/2}} \] Input:

int(cos(a + b*x)^(5/2)/sin(a + b*x)^(5/2),x)
 

Output:

-(2*cos(a + b*x)^(7/2)*(sin(a + b*x)^2)^(3/4)*hypergeom([7/4, 7/4], 11/4, 
cos(a + b*x)^2))/(7*b*sin(a + b*x)^(3/2))
 

Reduce [F]

\[ \int \frac {\cos ^{\frac {5}{2}}(a+b x)}{\sin ^{\frac {5}{2}}(a+b x)} \, dx=\int \frac {\sqrt {\sin \left (b x +a \right )}\, \sqrt {\cos \left (b x +a \right )}\, \cos \left (b x +a \right )^{2}}{\sin \left (b x +a \right )^{3}}d x \] Input:

int(cos(b*x+a)^(5/2)/sin(b*x+a)^(5/2),x)
 

Output:

int((sqrt(sin(a + b*x))*sqrt(cos(a + b*x))*cos(a + b*x)**2)/sin(a + b*x)** 
3,x)