\(\int \frac {\cos ^{\frac {3}{2}}(a+b x)}{\sin ^{\frac {3}{2}}(a+b x)} \, dx\) [301]

Optimal result

Integrand size = 21, antiderivative size = 150 \[ \int \frac {\cos ^{\frac {3}{2}}(a+b x)}{\sin ^{\frac {3}{2}}(a+b x)} \, dx=\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {\sin (a+b x)}}{\sqrt {\cos (a+b x)}}\right )}{\sqrt {2} b}-\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {\sin (a+b x)}}{\sqrt {\cos (a+b x)}}\right )}{\sqrt {2} b}+\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt {\sin (a+b x)}}{\sqrt {\cos (a+b x)} (1+\tan (a+b x))}\right )}{\sqrt {2} b}-\frac {2 \sqrt {\cos (a+b x)}}{b \sqrt {\sin (a+b x)}} \] Output:

1/2*arctan(1-2^(1/2)*sin(b*x+a)^(1/2)/cos(b*x+a)^(1/2))*2^(1/2)/b-1/2*arct 
an(1+2^(1/2)*sin(b*x+a)^(1/2)/cos(b*x+a)^(1/2))*2^(1/2)/b+1/2*arctanh(2^(1 
/2)*sin(b*x+a)^(1/2)/cos(b*x+a)^(1/2)/(1+tan(b*x+a)))*2^(1/2)/b-2*cos(b*x+ 
a)^(1/2)/b/sin(b*x+a)^(1/2)
 

Mathematica [C] (verified)

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

Time = 0.04 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.37 \[ \int \frac {\cos ^{\frac {3}{2}}(a+b x)}{\sin ^{\frac {3}{2}}(a+b x)} \, dx=-\frac {2 \cos ^2(a+b x)^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},-\frac {1}{4},\frac {3}{4},\sin ^2(a+b x)\right )}{b \cos ^{\frac {3}{2}}(a+b x) \sqrt {\sin (a+b x)}} \] Input:

Integrate[Cos[a + b*x]^(3/2)/Sin[a + b*x]^(3/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.35, 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, 3054, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3047

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3054

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

Int[Cos[a + b*x]^(3/2)/Sin[a + b*x]^(3/2),x]
 

Output:

(-2*((-(ArcTan[1 - (Sqrt[2]*Sqrt[Sin[a + b*x]])/Sqrt[Cos[a + b*x]]]/Sqrt[2 
]) + ArcTan[1 + (Sqrt[2]*Sqrt[Sin[a + b*x]])/Sqrt[Cos[a + b*x]]]/Sqrt[2])/ 
2 + (Log[1 - (Sqrt[2]*Sqrt[Sin[a + b*x]])/Sqrt[Cos[a + b*x]] + Tan[a + b*x 
]]/(2*Sqrt[2]) - Log[1 + (Sqrt[2]*Sqrt[Sin[a + b*x]])/Sqrt[Cos[a + b*x]] + 
 Tan[a + b*x]]/(2*Sqrt[2]))/2))/b - (2*Sqrt[Cos[a + b*x]])/(b*Sqrt[Sin[a + 
 b*x]])
 

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 3054

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_), 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*Sin[e + f*x])^(1/k)/(b*Cos[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. \(428\) vs. \(2(124)=248\).

Time = 1.57 (sec) , antiderivative size = 429, normalized size of antiderivative = 2.86

method	result	size
default	\(-\frac {\sqrt {2}\, \left (\left (\cos \left (b x +a \right )-1\right ) \ln \left (-\frac {2 \sqrt {2}\, \sqrt {\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \sin \left (b x +a \right )-\cos \left (b x +a \right ) \cot \left (b x +a \right )+2 \cot \left (b x +a \right )-\csc \left (b x +a \right )-2 \cos \left (b x +a \right )+\sin \left (b x +a \right )+2}{\cos \left (b x +a \right )-1}\right )+\left (1-\cos \left (b x +a \right )\right ) \ln \left (\frac {2 \sqrt {2}\, \sqrt {\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \sin \left (b x +a \right )+\cos \left (b x +a \right ) \cot \left (b x +a \right )-2 \cot \left (b x +a \right )+\csc \left (b x +a \right )+2 \cos \left (b x +a \right )-\sin \left (b x +a \right )-2}{\cos \left (b x +a \right )-1}\right )+\left (-2 \cos \left (b x +a \right )+2\right ) \arctan \left (\frac {\sqrt {2}\, \sqrt {\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \sin \left (b x +a \right )-\cos \left (b x +a \right )+1}{\cos \left (b x +a \right )-1}\right )+\left (-2 \cos \left (b x +a \right )+2\right ) \arctan \left (\frac {\sqrt {2}\, \sqrt {\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \sin \left (b x +a \right )+\cos \left (b x +a \right )-1}{\cos \left (b x +a \right )-1}\right )+4 \sqrt {2}\, \sqrt {\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \sin \left (b x +a \right )\right ) \sqrt {\cos \left (b x +a \right )}}{4 b \sin \left (b x +a \right )^{\frac {3}{2}} \sqrt {\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}}\)	\(429\)

Input:

int(cos(b*x+a)^(3/2)/sin(b*x+a)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/4/b*2^(1/2)*((cos(b*x+a)-1)*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))+(1-cos(b*x+a))*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))+( 
-2*cos(b*x+a)+2)*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))+(-2*cos(b*x+a)+2)*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))+4*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)^(1/2)/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. 422 vs. \(2 (124) = 248\).

Time = 0.16 (sec) , antiderivative size = 422, normalized size of antiderivative = 2.81 \[ \int \frac {\cos ^{\frac {3}{2}}(a+b x)}{\sin ^{\frac {3}{2}}(a+b x)} \, dx=\frac {\sqrt {2} \arctan \left (\frac {2 \, \cos \left (b x + a\right )^{3} - 2 \, \cos \left (b x + a\right )^{2} \sin \left (b x + a\right ) + \sqrt {2} \sqrt {\cos \left (b x + a\right )} \sqrt {\sin \left (b x + a\right )} - 2 \, \cos \left (b x + a\right )}{2 \, {\left (\cos \left (b x + a\right )^{3} + \cos \left (b x + a\right )^{2} \sin \left (b x + a\right ) - \cos \left (b x + a\right )\right )}}\right ) \sin \left (b x + a\right ) + \sqrt {2} \arctan \left (-\frac {2 \, \cos \left (b x + a\right )^{3} - 2 \, \cos \left (b x + a\right )^{2} \sin \left (b x + a\right ) - \sqrt {2} \sqrt {\cos \left (b x + a\right )} \sqrt {\sin \left (b x + a\right )} - 2 \, \cos \left (b x + a\right )}{2 \, {\left (\cos \left (b x + a\right )^{3} + \cos \left (b x + a\right )^{2} \sin \left (b x + a\right ) - \cos \left (b x + a\right )\right )}}\right ) \sin \left (b x + a\right ) - 2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} \cos \left (b x + a\right ) - \sqrt {2} \sin \left (b x + a\right )}{2 \, \sqrt {\cos \left (b x + a\right )} \sqrt {\sin \left (b x + a\right )}}\right ) \sin \left (b x + a\right ) + \sqrt {2} \log \left (2 \, {\left (\sqrt {2} \cos \left (b x + a\right ) + \sqrt {2} \sin \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right )} \sqrt {\sin \left (b x + a\right )} + 4 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - \sqrt {2} \log \left (-2 \, {\left (\sqrt {2} \cos \left (b x + a\right ) + \sqrt {2} \sin \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right )} \sqrt {\sin \left (b x + a\right )} + 4 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 16 \, \sqrt {\cos \left (b x + a\right )} \sqrt {\sin \left (b x + a\right )}}{8 \, b \sin \left (b x + a\right )} \] Input:

integrate(cos(b*x+a)^(3/2)/sin(b*x+a)^(3/2),x, algorithm="fricas")
 

Output:

1/8*(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)))*sin(b*x + a) + 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)))*sin(b*x + a) - 2*sqrt(2)*ar 
ctan(-1/2*(sqrt(2)*cos(b*x + a) - sqrt(2)*sin(b*x + a))/(sqrt(cos(b*x + a) 
)*sqrt(sin(b*x + a))))*sin(b*x + a) + 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)*sin(b*x + a) - 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)*sin(b*x + a) - 16*sqrt(cos(b*x + a))*sqrt(sin(b*x 
+ a)))/(b*sin(b*x + a))
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(cos(b*x+a)^(3/2)/sin(b*x+a)^(3/2),x, algorithm="maxima")
 

Output:

integrate(cos(b*x + a)^(3/2)/sin(b*x + a)^(3/2), x)
 

Giac [F]

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

integrate(cos(b*x+a)^(3/2)/sin(b*x+a)^(3/2),x, algorithm="giac")
 

Output:

integrate(cos(b*x + a)^(3/2)/sin(b*x + a)^(3/2), x)
 

Mupad [B] (verification not implemented)

Time = 26.47 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.29 \[ \int \frac {\cos ^{\frac {3}{2}}(a+b x)}{\sin ^{\frac {3}{2}}(a+b x)} \, dx=-\frac {2\,{\cos \left (a+b\,x\right )}^{5/2}\,{\left ({\sin \left (a+b\,x\right )}^2\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {5}{4},\frac {5}{4};\ \frac {9}{4};\ {\cos \left (a+b\,x\right )}^2\right )}{5\,b\,\sqrt {\sin \left (a+b\,x\right )}} \] Input:

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

Output:

-(2*cos(a + b*x)^(5/2)*(sin(a + b*x)^2)^(1/4)*hypergeom([5/4, 5/4], 9/4, c 
os(a + b*x)^2))/(5*b*sin(a + b*x)^(1/2))
 

Reduce [F]

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

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

Output:

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