\(\int \frac {\cos ^2(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx\) [262]

Optimal result

Integrand size = 21, antiderivative size = 191 \[ \int \frac {\cos ^2(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\frac {5 \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b d^{3/2}}-\frac {5 \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b d^{3/2}}+\frac {5 \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d} (1+\tan (a+b x))}\right )}{4 \sqrt {2} b d^{3/2}}-\frac {5}{2 b d \sqrt {d \tan (a+b x)}}+\frac {\cos ^2(a+b x)}{2 b d \sqrt {d \tan (a+b x)}} \] Output:

5/8*arctan(1-2^(1/2)*(d*tan(b*x+a))^(1/2)/d^(1/2))*2^(1/2)/b/d^(3/2)-5/8*a 
rctan(1+2^(1/2)*(d*tan(b*x+a))^(1/2)/d^(1/2))*2^(1/2)/b/d^(3/2)+5/8*arctan 
h(2^(1/2)*(d*tan(b*x+a))^(1/2)/d^(1/2)/(1+tan(b*x+a)))*2^(1/2)/b/d^(3/2)-5 
/2/b/d/(d*tan(b*x+a))^(1/2)+1/2*cos(b*x+a)^2/b/d/(d*tan(b*x+a))^(1/2)
 

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.60 \[ \int \frac {\cos ^2(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\frac {\csc (a+b x) \left (-17 \cos (a+b x)+\cos (3 (a+b x))+5 \arcsin (\cos (a+b x)-\sin (a+b x)) \sqrt {\sin (2 (a+b x))}+5 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right ) \sqrt {\sin (2 (a+b x))}\right ) \sqrt {d \tan (a+b x)}}{8 b d^2} \] Input:

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

Output:

(Csc[a + b*x]*(-17*Cos[a + b*x] + Cos[3*(a + b*x)] + 5*ArcSin[Cos[a + b*x] 
 - Sin[a + b*x]]*Sqrt[Sin[2*(a + b*x)]] + 5*Log[Cos[a + b*x] + Sin[a + b*x 
] + Sqrt[Sin[2*(a + b*x)]]]*Sqrt[Sin[2*(a + b*x)]])*Sqrt[d*Tan[a + b*x]])/ 
(8*b*d^2)
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.29, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 3087, 253, 264, 266, 27, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sec (a+b x)^2 (d \tan (a+b x))^{3/2}}dx\)

\(\Big \downarrow \) 3087

\(\displaystyle \frac {\int \frac {1}{(d \tan (a+b x))^{3/2} \left (\tan ^2(a+b x)+1\right )^2}d\tan (a+b x)}{b}\)

\(\Big \downarrow \) 253

\(\displaystyle \frac {\frac {5}{4} \int \frac {1}{(d \tan (a+b x))^{3/2} \left (\tan ^2(a+b x)+1\right )}d\tan (a+b x)+\frac {1}{2 d \left (\tan ^2(a+b x)+1\right ) \sqrt {d \tan (a+b x)}}}{b}\)

\(\Big \downarrow \) 264

\(\displaystyle \frac {\frac {5}{4} \left (-\frac {\int \frac {\sqrt {d \tan (a+b x)}}{\tan ^2(a+b x)+1}d\tan (a+b x)}{d^2}-\frac {2}{d \sqrt {d \tan (a+b x)}}\right )+\frac {1}{2 d \left (\tan ^2(a+b x)+1\right ) \sqrt {d \tan (a+b x)}}}{b}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {\frac {5}{4} \left (-\frac {2 \int \frac {d^3 \tan (a+b x)}{\tan ^2(a+b x) d^2+d^2}d\sqrt {d \tan (a+b x)}}{d^3}-\frac {2}{d \sqrt {d \tan (a+b x)}}\right )+\frac {1}{2 d \left (\tan ^2(a+b x)+1\right ) \sqrt {d \tan (a+b x)}}}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {5}{4} \left (-\frac {2 \int \frac {d \tan (a+b x)}{\tan ^2(a+b x) d^2+d^2}d\sqrt {d \tan (a+b x)}}{d}-\frac {2}{d \sqrt {d \tan (a+b x)}}\right )+\frac {1}{2 d \left (\tan ^2(a+b x)+1\right ) \sqrt {d \tan (a+b x)}}}{b}\)

\(\Big \downarrow \) 826

\(\displaystyle \frac {\frac {5}{4} \left (-\frac {2 \left (\frac {1}{2} \int \frac {\tan (a+b x) d+d}{\tan ^2(a+b x) d^2+d^2}d\sqrt {d \tan (a+b x)}-\frac {1}{2} \int \frac {d-d \tan (a+b x)}{\tan ^2(a+b x) d^2+d^2}d\sqrt {d \tan (a+b x)}\right )}{d}-\frac {2}{d \sqrt {d \tan (a+b x)}}\right )+\frac {1}{2 d \left (\tan ^2(a+b x)+1\right ) \sqrt {d \tan (a+b x)}}}{b}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {\frac {5}{4} \left (-\frac {2 \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{\tan (a+b x) d+d-\sqrt {2} \sqrt {d \tan (a+b x)} \sqrt {d}}d\sqrt {d \tan (a+b x)}+\frac {1}{2} \int \frac {1}{\tan (a+b x) d+d+\sqrt {2} \sqrt {d \tan (a+b x)} \sqrt {d}}d\sqrt {d \tan (a+b x)}\right )-\frac {1}{2} \int \frac {d-d \tan (a+b x)}{\tan ^2(a+b x) d^2+d^2}d\sqrt {d \tan (a+b x)}\right )}{d}-\frac {2}{d \sqrt {d \tan (a+b x)}}\right )+\frac {1}{2 d \left (\tan ^2(a+b x)+1\right ) \sqrt {d \tan (a+b x)}}}{b}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {\frac {5}{4} \left (-\frac {2 \left (\frac {1}{2} \left (\frac {\int \frac {1}{-d \tan (a+b x)-1}d\left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}-\frac {\int \frac {1}{-d \tan (a+b x)-1}d\left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}\right )-\frac {1}{2} \int \frac {d-d \tan (a+b x)}{\tan ^2(a+b x) d^2+d^2}d\sqrt {d \tan (a+b x)}\right )}{d}-\frac {2}{d \sqrt {d \tan (a+b x)}}\right )+\frac {1}{2 d \left (\tan ^2(a+b x)+1\right ) \sqrt {d \tan (a+b x)}}}{b}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {5}{4} \left (-\frac {2 \left (\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\frac {1}{2} \int \frac {d-d \tan (a+b x)}{\tan ^2(a+b x) d^2+d^2}d\sqrt {d \tan (a+b x)}\right )}{d}-\frac {2}{d \sqrt {d \tan (a+b x)}}\right )+\frac {1}{2 d \left (\tan ^2(a+b x)+1\right ) \sqrt {d \tan (a+b x)}}}{b}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {\frac {5}{4} \left (-\frac {2 \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (a+b x)}}{\tan (a+b x) d+d-\sqrt {2} \sqrt {d \tan (a+b x)} \sqrt {d}}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int -\frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{\tan (a+b x) d+d+\sqrt {2} \sqrt {d \tan (a+b x)} \sqrt {d}}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{d}-\frac {2}{d \sqrt {d \tan (a+b x)}}\right )+\frac {1}{2 d \left (\tan ^2(a+b x)+1\right ) \sqrt {d \tan (a+b x)}}}{b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {5}{4} \left (-\frac {2 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (a+b x)}}{\tan (a+b x) d+d-\sqrt {2} \sqrt {d \tan (a+b x)} \sqrt {d}}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int \frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{\tan (a+b x) d+d+\sqrt {2} \sqrt {d \tan (a+b x)} \sqrt {d}}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{d}-\frac {2}{d \sqrt {d \tan (a+b x)}}\right )+\frac {1}{2 d \left (\tan ^2(a+b x)+1\right ) \sqrt {d \tan (a+b x)}}}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {5}{4} \left (-\frac {2 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (a+b x)}}{\tan (a+b x) d+d-\sqrt {2} \sqrt {d \tan (a+b x)} \sqrt {d}}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int \frac {\sqrt {d}+\sqrt {2} \sqrt {d \tan (a+b x)}}{\tan (a+b x) d+d+\sqrt {2} \sqrt {d \tan (a+b x)} \sqrt {d}}d\sqrt {d \tan (a+b x)}}{2 \sqrt {d}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{d}-\frac {2}{d \sqrt {d \tan (a+b x)}}\right )+\frac {1}{2 d \left (\tan ^2(a+b x)+1\right ) \sqrt {d \tan (a+b x)}}}{b}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {\frac {5}{4} \left (-\frac {2 \left (\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )+\frac {1}{2} \left (\frac {\log \left (d \tan (a+b x)-\sqrt {2} \sqrt {d} \sqrt {d \tan (a+b x)}+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (d \tan (a+b x)+\sqrt {2} \sqrt {d} \sqrt {d \tan (a+b x)}+d\right )}{2 \sqrt {2} \sqrt {d}}\right )\right )}{d}-\frac {2}{d \sqrt {d \tan (a+b x)}}\right )+\frac {1}{2 d \left (\tan ^2(a+b x)+1\right ) \sqrt {d \tan (a+b x)}}}{b}\)

Input:

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

Output:

(1/(2*d*Sqrt[d*Tan[a + b*x]]*(1 + Tan[a + b*x]^2)) + (5*((-2*((-(ArcTan[1 
- (Sqrt[2]*Sqrt[d*Tan[a + b*x]])/Sqrt[d]]/(Sqrt[2]*Sqrt[d])) + ArcTan[1 + 
(Sqrt[2]*Sqrt[d*Tan[a + b*x]])/Sqrt[d]]/(Sqrt[2]*Sqrt[d]))/2 + (Log[d + d* 
Tan[a + b*x] - Sqrt[2]*Sqrt[d]*Sqrt[d*Tan[a + b*x]]]/(2*Sqrt[2]*Sqrt[d]) - 
 Log[d + d*Tan[a + b*x] + Sqrt[2]*Sqrt[d]*Sqrt[d*Tan[a + b*x]]]/(2*Sqrt[2] 
*Sqrt[d]))/2))/d - 2/(d*Sqrt[d*Tan[a + b*x]])))/4)/b
 

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 253

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(c*x 
)^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 
2*a*(p + 1))   Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m 
}, x] && LtQ[p, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 264

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

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 3087

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S 
ymbol] :> Simp[1/f   Subst[Int[(b*x)^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + 
f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n - 1) 
/2] && LtQ[0, n, m - 1])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(504\) vs. \(2(147)=294\).

Time = 6.29 (sec) , antiderivative size = 505, normalized size of antiderivative = 2.64

method	result	size
default	\(-\frac {\left (5 \ln \left (-\frac {\cot \left (b x +a \right ) \cos \left (b x +a \right )-2 \cot \left (b x +a \right )-2 \sin \left (b x +a \right ) \sqrt {-\frac {2 \sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}-2 \cos \left (b x +a \right )+\csc \left (b x +a \right )-\sin \left (b x +a \right )+2}{-1+\cos \left (b x +a \right )}\right ) \sin \left (b x +a \right )-5 \ln \left (-\frac {\cot \left (b x +a \right ) \cos \left (b x +a \right )-2 \cot \left (b x +a \right )+2 \sin \left (b x +a \right ) \sqrt {-\frac {2 \sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}-2 \cos \left (b x +a \right )+\csc \left (b x +a \right )-\sin \left (b x +a \right )+2}{-1+\cos \left (b x +a \right )}\right ) \sin \left (b x +a \right )+10 \arctan \left (\frac {\sin \left (b x +a \right ) \sqrt {-\frac {2 \sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}-\cos \left (b x +a \right )+1}{-1+\cos \left (b x +a \right )}\right ) \sin \left (b x +a \right )+10 \arctan \left (\frac {\sin \left (b x +a \right ) \sqrt {-\frac {2 \sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}+\cos \left (b x +a \right )-1}{-1+\cos \left (b x +a \right )}\right ) \sin \left (b x +a \right )+\left (-32 \cos \left (b x +a \right ) \left (\cos \left (b x +a \right )+1\right ) \sin \left (b x +a \right )-4 \left (\cos \left (b x +a \right )^{2}-5\right ) \left (\cos \left (b x +a \right )+1\right )\right ) \sqrt {-\frac {2 \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 ) \left (32 \cos \left (b x +a \right )+32\right ) \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}}}\right ) \sqrt {2}}{16 b \sqrt {-\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \left (\cos \left (b x +a \right )+1\right ) d \sqrt {d \tan \left (b x +a \right )}}\)	\(505\)

Input:

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

Output:

-1/16/b*(5*ln(-(cot(b*x+a)*cos(b*x+a)-2*cot(b*x+a)-2*sin(b*x+a)*(-2*sin(b* 
x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)-2*cos(b*x+a)+csc(b*x+a)-sin(b*x+a) 
+2)/(-1+cos(b*x+a)))*sin(b*x+a)-5*ln(-(cot(b*x+a)*cos(b*x+a)-2*cot(b*x+a)+ 
2*sin(b*x+a)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)-2*cos(b*x+a 
)+csc(b*x+a)-sin(b*x+a)+2)/(-1+cos(b*x+a)))*sin(b*x+a)+10*arctan((sin(b*x+ 
a)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)-cos(b*x+a)+1)/(-1+cos 
(b*x+a)))*sin(b*x+a)+10*arctan((sin(b*x+a)*(-2*sin(b*x+a)*cos(b*x+a)/(cos( 
b*x+a)+1)^2)^(1/2)+cos(b*x+a)-1)/(-1+cos(b*x+a)))*sin(b*x+a)+(-32*cos(b*x+ 
a)*(cos(b*x+a)+1)*sin(b*x+a)-4*(cos(b*x+a)^2-5)*(cos(b*x+a)+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)*(32*cos(b*x+ 
a)+32)*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)/(cos(b*x+a)+1)^2)^(1/2)/(cos(b*x+a)+1)/d/(d*tan(b*x+a))^(1/2 
)*2^(1/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 463 vs. \(2 (147) = 294\).

Time = 0.17 (sec) , antiderivative size = 463, normalized size of antiderivative = 2.42 \[ \int \frac {\cos ^2(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\frac {10 \, \sqrt {2} \sqrt {d} \arctan \left (-\frac {\sqrt {2} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}} \cos \left (b x + a\right )}{\sqrt {d} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )}}\right ) \sin \left (b x + a\right ) + 5 \, \sqrt {2} \sqrt {d} \arctan \left (\frac {2 \, \cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + \frac {\sqrt {2} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{\sqrt {d}} - 2}{2 \, {\left (\cos \left (b x + a\right )^{2} + \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1\right )}}\right ) \sin \left (b x + a\right ) + 5 \, \sqrt {2} \sqrt {d} \arctan \left (-\frac {2 \, \cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - \frac {\sqrt {2} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{\sqrt {d}} - 2}{2 \, {\left (\cos \left (b x + a\right )^{2} + \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1\right )}}\right ) \sin \left (b x + a\right ) + 5 \, \sqrt {2} \sqrt {d} \log \left (4 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + \frac {2 \, \sqrt {2} {\left (\cos \left (b x + a\right )^{2} + \cos \left (b x + a\right ) \sin \left (b x + a\right )\right )} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{\sqrt {d}} + 1\right ) \sin \left (b x + a\right ) - 5 \, \sqrt {2} \sqrt {d} \log \left (4 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - \frac {2 \, \sqrt {2} {\left (\cos \left (b x + a\right )^{2} + \cos \left (b x + a\right ) \sin \left (b x + a\right )\right )} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{\sqrt {d}} + 1\right ) \sin \left (b x + a\right ) + 16 \, {\left (\cos \left (b x + a\right )^{3} - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{32 \, b d^{2} \sin \left (b x + a\right )} \] Input:

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

Output:

1/32*(10*sqrt(2)*sqrt(d)*arctan(-sqrt(2)*sqrt(d*sin(b*x + a)/cos(b*x + a)) 
*cos(b*x + a)/(sqrt(d)*(cos(b*x + a) - sin(b*x + a))))*sin(b*x + a) + 5*sq 
rt(2)*sqrt(d)*arctan(1/2*(2*cos(b*x + a)^2 - 2*cos(b*x + a)*sin(b*x + a) + 
 sqrt(2)*sqrt(d*sin(b*x + a)/cos(b*x + a))/sqrt(d) - 2)/(cos(b*x + a)^2 + 
cos(b*x + a)*sin(b*x + a) - 1))*sin(b*x + a) + 5*sqrt(2)*sqrt(d)*arctan(-1 
/2*(2*cos(b*x + a)^2 - 2*cos(b*x + a)*sin(b*x + a) - sqrt(2)*sqrt(d*sin(b* 
x + a)/cos(b*x + a))/sqrt(d) - 2)/(cos(b*x + a)^2 + cos(b*x + a)*sin(b*x + 
 a) - 1))*sin(b*x + a) + 5*sqrt(2)*sqrt(d)*log(4*cos(b*x + a)*sin(b*x + a) 
 + 2*sqrt(2)*(cos(b*x + a)^2 + cos(b*x + a)*sin(b*x + a))*sqrt(d*sin(b*x + 
 a)/cos(b*x + a))/sqrt(d) + 1)*sin(b*x + a) - 5*sqrt(2)*sqrt(d)*log(4*cos( 
b*x + a)*sin(b*x + a) - 2*sqrt(2)*(cos(b*x + a)^2 + cos(b*x + a)*sin(b*x + 
 a))*sqrt(d*sin(b*x + a)/cos(b*x + a))/sqrt(d) + 1)*sin(b*x + a) + 16*(cos 
(b*x + a)^3 - 5*cos(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)))/(b*d^2*si 
n(b*x + a))
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.07 \[ \int \frac {\cos ^2(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=-\frac {\frac {10 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {10 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {5 \, \sqrt {2} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {5 \, \sqrt {2} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {8 \, {\left (5 \, d^{2} \tan \left (b x + a\right )^{2} + 4 \, d^{2}\right )}}{\left (d \tan \left (b x + a\right )\right )^{\frac {5}{2}} + \sqrt {d \tan \left (b x + a\right )} d^{2}}}{16 \, b d} \] Input:

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

Output:

-1/16*(10*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(b*x + 
 a)))/sqrt(d))/sqrt(d) + 10*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 
 2*sqrt(d*tan(b*x + a)))/sqrt(d))/sqrt(d) - 5*sqrt(2)*log(d*tan(b*x + a) + 
 sqrt(2)*sqrt(d*tan(b*x + a))*sqrt(d) + d)/sqrt(d) + 5*sqrt(2)*log(d*tan(b 
*x + a) - sqrt(2)*sqrt(d*tan(b*x + a))*sqrt(d) + d)/sqrt(d) + 8*(5*d^2*tan 
(b*x + a)^2 + 4*d^2)/((d*tan(b*x + a))^(5/2) + sqrt(d*tan(b*x + a))*d^2))/ 
(b*d)
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.32 \[ \int \frac {\cos ^2(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=-\frac {\frac {10 \, \sqrt {2} {\left | d \right |}^{\frac {3}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{b d^{2}} + \frac {10 \, \sqrt {2} {\left | d \right |}^{\frac {3}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{b d^{2}} - \frac {5 \, \sqrt {2} {\left | d \right |}^{\frac {3}{2}} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{b d^{2}} + \frac {5 \, \sqrt {2} {\left | d \right |}^{\frac {3}{2}} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{b d^{2}} + \frac {8 \, {\left (5 \, d^{2} \tan \left (b x + a\right )^{2} + 4 \, d^{2}\right )}}{{\left (\sqrt {d \tan \left (b x + a\right )} d^{2} \tan \left (b x + a\right )^{2} + \sqrt {d \tan \left (b x + a\right )} d^{2}\right )} b}}{16 \, d} \] Input:

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

Output:

-1/16*(10*sqrt(2)*abs(d)^(3/2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) + 
2*sqrt(d*tan(b*x + a)))/sqrt(abs(d)))/(b*d^2) + 10*sqrt(2)*abs(d)^(3/2)*ar 
ctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) - 2*sqrt(d*tan(b*x + a)))/sqrt(abs 
(d)))/(b*d^2) - 5*sqrt(2)*abs(d)^(3/2)*log(d*tan(b*x + a) + sqrt(2)*sqrt(d 
*tan(b*x + a))*sqrt(abs(d)) + abs(d))/(b*d^2) + 5*sqrt(2)*abs(d)^(3/2)*log 
(d*tan(b*x + a) - sqrt(2)*sqrt(d*tan(b*x + a))*sqrt(abs(d)) + abs(d))/(b*d 
^2) + 8*(5*d^2*tan(b*x + a)^2 + 4*d^2)/((sqrt(d*tan(b*x + a))*d^2*tan(b*x 
+ a)^2 + sqrt(d*tan(b*x + a))*d^2)*b))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(d)*int((sqrt(tan(a + b*x))*cos(a + b*x)**2)/tan(a + b*x)**2,x))/d**2