\(\int \frac {\cos (2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx\) [631]

Optimal result

Integrand size = 29, antiderivative size = 178 \[ \int \frac {\cos (2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=-\frac {3 \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}}\right )}{2 b c^{3/2}}+\frac {9 \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {-c+c \sec (2 a+2 b x)}}\right )}{4 \sqrt {2} b c^{3/2}}-\frac {\sin (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}-\frac {3 \sin (2 a+2 b x)}{4 b c \sqrt {-c+c \sec (2 a+2 b x)}} \]

[Out]

-3/2*arctanh(c^(1/2)*tan(2*b*x+2*a)/(-c+c*sec(2*b*x+2*a))^(1/2))/b/c^(3/2)-1/4*sin(2*b*x+2*a)/b/(-c+c*sec(2*b*
x+2*a))^(3/2)+9/8*arctanh(1/2*c^(1/2)*tan(2*b*x+2*a)*2^(1/2)/(-c+c*sec(2*b*x+2*a))^(1/2))/b/c^(3/2)*2^(1/2)-3/
4*sin(2*b*x+2*a)/b/c/(-c+c*sec(2*b*x+2*a))^(1/2)

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {4482, 3902, 4107, 4005, 3859, 213, 3880} \[ \int \frac {\cos (2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=-\frac {3 \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{2 b c^{3/2}}+\frac {9 \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {c \sec (2 a+2 b x)-c}}\right )}{4 \sqrt {2} b c^{3/2}}-\frac {3 \sin (2 a+2 b x)}{4 b c \sqrt {c \sec (2 a+2 b x)-c}}-\frac {\sin (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}} \]

[In]

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

[Out]

(-3*ArcTanh[(Sqrt[c]*Tan[2*a + 2*b*x])/Sqrt[-c + c*Sec[2*a + 2*b*x]]])/(2*b*c^(3/2)) + (9*ArcTanh[(Sqrt[c]*Tan
[2*a + 2*b*x])/(Sqrt[2]*Sqrt[-c + c*Sec[2*a + 2*b*x]])])/(4*Sqrt[2]*b*c^(3/2)) - Sin[2*a + 2*b*x]/(4*b*(-c + c
*Sec[2*a + 2*b*x])^(3/2)) - (3*Sin[2*a + 2*b*x])/(4*b*c*Sqrt[-c + c*Sec[2*a + 2*b*x]])

Rule 213

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

Rule 3859

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2*(b/d), Subst[Int[1/(a + x^2), x], x, b*(C
ot[c + d*x]/Sqrt[a + b*Csc[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 3880

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2/f, Subst[Int[1/(2
*a + x^2), x], x, b*(Cot[e + f*x]/Sqrt[a + b*Csc[e + f*x]])], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0
]

Rule 3902

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-Cot[
e + f*x])*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*(2*m + 1))), x] + Dist[1/(a^2*(2*m + 1)), Int[(a + b*C
sc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*(a*(2*m + n + 1) - b*(m + n + 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b,
d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m])

Rule 4005

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c/a,
Int[Sqrt[a + b*Csc[e + f*x]], x], x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]

Rule 4107

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Dist[1
/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*B*n - A*b*(m + n + 1)*Csc[e + f*x
], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]

Rule 4482

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\cos (2 a+2 b x)}{(-c+c \sec (2 a+2 b x))^{3/2}} \, dx \\ & = -\frac {\sin (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}-\frac {\int \frac {\cos (2 a+2 b x) \left (3 c+\frac {3}{2} c \sec (2 a+2 b x)\right )}{\sqrt {-c+c \sec (2 a+2 b x)}} \, dx}{2 c^2} \\ & = -\frac {\sin (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}-\frac {3 \sin (2 a+2 b x)}{4 b c \sqrt {-c+c \sec (2 a+2 b x)}}-\frac {\int \frac {3 c^2+\frac {3}{2} c^2 \sec (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}} \, dx}{2 c^3} \\ & = -\frac {\sin (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}-\frac {3 \sin (2 a+2 b x)}{4 b c \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {3 \int \sqrt {-c+c \sec (2 a+2 b x)} \, dx}{2 c^2}-\frac {9 \int \frac {\sec (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}} \, dx}{4 c} \\ & = -\frac {\sin (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}-\frac {3 \sin (2 a+2 b x)}{4 b c \sqrt {-c+c \sec (2 a+2 b x)}}-\frac {3 \text {Subst}\left (\int \frac {1}{-c+x^2} \, dx,x,-\frac {c \tan (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}}\right )}{2 b c}+\frac {9 \text {Subst}\left (\int \frac {1}{-2 c+x^2} \, dx,x,-\frac {c \tan (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}}\right )}{4 b c} \\ & = -\frac {3 \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}}\right )}{2 b c^{3/2}}+\frac {9 \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {-c+c \sec (2 a+2 b x)}}\right )}{4 \sqrt {2} b c^{3/2}}-\frac {\sin (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}-\frac {3 \sin (2 a+2 b x)}{4 b c \sqrt {-c+c \sec (2 a+2 b x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.86 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.22 \[ \int \frac {\cos (2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\frac {\csc (a+b x) \sec (a+b x) \left (6 \text {arctanh}\left (\sqrt {1-\tan ^2(a+b x)}\right ) \cos (2 (a+b x))-6 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1-\tan ^2(a+b x)}}{\sqrt {2}}\right ) \cos (2 (a+b x))+(1-3 \cos (2 (a+b x))+\cos (4 (a+b x))) \cot ^2(a+b x) \sqrt {\cos (2 (a+b x)) \sec ^2(a+b x)}-3 \arctan \left (\sqrt {-1+\tan ^2(a+b x)}\right ) \cos ^2(a+b x) \sqrt {-\left (-1+\tan ^2(a+b x)\right )^2}\right ) \sqrt {c \tan (a+b x) \tan (2 (a+b x))}}{8 b c^2 \sqrt {1-\tan ^2(a+b x)}} \]

[In]

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

[Out]

(Csc[a + b*x]*Sec[a + b*x]*(6*ArcTanh[Sqrt[1 - Tan[a + b*x]^2]]*Cos[2*(a + b*x)] - 6*Sqrt[2]*ArcTanh[Sqrt[1 -
Tan[a + b*x]^2]/Sqrt[2]]*Cos[2*(a + b*x)] + (1 - 3*Cos[2*(a + b*x)] + Cos[4*(a + b*x)])*Cot[a + b*x]^2*Sqrt[Co
s[2*(a + b*x)]*Sec[a + b*x]^2] - 3*ArcTan[Sqrt[-1 + Tan[a + b*x]^2]]*Cos[a + b*x]^2*Sqrt[-(-1 + Tan[a + b*x]^2
)^2])*Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]])/(8*b*c^2*Sqrt[1 - Tan[a + b*x]^2])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1074\) vs. \(2(151)=302\).

Time = 2.96 (sec) , antiderivative size = 1075, normalized size of antiderivative = 6.04

method	result	size
default	\(\text {Expression too large to display}\)	\(1075\)

[In]

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

[Out]

-1/64*2^(1/2)/b*csc(b*x+a)/(csc(b*x+a)^2*(1-cos(b*x+a))^2*c/(csc(b*x+a)^4*(1-cos(b*x+a))^4-6*csc(b*x+a)^2*(1-c
os(b*x+a))^2+1))^(3/2)*(1-cos(b*x+a))*(16*csc(b*x+a)^2*2^(1/2)*arctanh((csc(b*x+a)^2*(1-cos(b*x+a))^2-1)*2^(1/
2)/(csc(b*x+a)^4*(1-cos(b*x+a))^4-6*csc(b*x+a)^2*(1-cos(b*x+a))^2+1)^(1/2))*(1-cos(b*x+a))^2-10*csc(b*x+a)^2*l
n(csc(b*x+a)^2*(1-cos(b*x+a))^2-3+(csc(b*x+a)^4*(1-cos(b*x+a))^4-6*csc(b*x+a)^2*(1-cos(b*x+a))^2+1)^(1/2))*(1-
cos(b*x+a))^2-10*csc(b*x+a)^2*arctanh((3*csc(b*x+a)^2*(1-cos(b*x+a))^2-1)/(csc(b*x+a)^4*(1-cos(b*x+a))^4-6*csc
(b*x+a)^2*(1-cos(b*x+a))^2+1)^(1/2))*(1-cos(b*x+a))^2+csc(b*x+a)^2*(1-cos(b*x+a))^2*(csc(b*x+a)^4*(1-cos(b*x+a
))^4-6*csc(b*x+a)^2*(1-cos(b*x+a))^2+1)^(1/2)-(csc(b*x+a)^4*(1-cos(b*x+a))^4-6*csc(b*x+a)^2*(1-cos(b*x+a))^2+1
)^(1/2))/(csc(b*x+a)^4*(1-cos(b*x+a))^4-6*csc(b*x+a)^2*(1-cos(b*x+a))^2+1)^(3/2)*4^(1/2)+1/16*2^(1/2)/b*csc(b*
x+a)*(4*((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)*cos(b*x+a)^3+10*2^(1/2)*arctanh(cos(b*x+a)/(1+cos(b*x+a))/
((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)*2^(1/2))*cos(b*x+a)-10*2^(1/2)*arctanh(cos(b*x+a)/(1+cos(b*x+a))/(
(2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)*2^(1/2))-6*cos(b*x+a)*((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)-7
*arctanh((2*cos(b*x+a)-1)/(1+cos(b*x+a))/((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2))*cos(b*x+a)+7*ln(2*(cos(b
*x+a)*((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)+((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)-2*cos(b*x+a)-1)/
(1+cos(b*x+a)))*cos(b*x+a)+7*arctanh((2*cos(b*x+a)-1)/(1+cos(b*x+a))/((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/
2))-7*ln(2*(cos(b*x+a)*((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)+((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)
-2*cos(b*x+a)-1)/(1+cos(b*x+a))))/((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)/(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-
1))^(1/2)/c*4^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 528, normalized size of antiderivative = 2.97 \[ \int \frac {\cos (2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\left [\frac {9 \, \sqrt {2} {\left (\tan \left (b x + a\right )^{5} + \tan \left (b x + a\right )^{3}\right )} \sqrt {c} \log \left (\frac {c \tan \left (b x + a\right )^{3} + 2 \, \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {c} - 2 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{3}}\right ) + 12 \, {\left (\tan \left (b x + a\right )^{5} + \tan \left (b x + a\right )^{3}\right )} \sqrt {c} \log \left (\frac {c \tan \left (b x + a\right )^{3} - 2 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {c} - 3 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )}\right ) + 2 \, \sqrt {2} {\left (5 \, \tan \left (b x + a\right )^{4} - 4 \, \tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{16 \, {\left (b c^{2} \tan \left (b x + a\right )^{5} + b c^{2} \tan \left (b x + a\right )^{3}\right )}}, \frac {9 \, \sqrt {2} {\left (\tan \left (b x + a\right )^{5} + \tan \left (b x + a\right )^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-c}}{c \tan \left (b x + a\right )}\right ) - 12 \, {\left (\tan \left (b x + a\right )^{5} + \tan \left (b x + a\right )^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-c}}{2 \, c \tan \left (b x + a\right )}\right ) + \sqrt {2} {\left (5 \, \tan \left (b x + a\right )^{4} - 4 \, \tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{8 \, {\left (b c^{2} \tan \left (b x + a\right )^{5} + b c^{2} \tan \left (b x + a\right )^{3}\right )}}\right ] \]

[In]

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

[Out]

[1/16*(9*sqrt(2)*(tan(b*x + a)^5 + tan(b*x + a)^3)*sqrt(c)*log((c*tan(b*x + a)^3 + 2*sqrt(-c*tan(b*x + a)^2/(t
an(b*x + a)^2 - 1))*(tan(b*x + a)^2 - 1)*sqrt(c) - 2*c*tan(b*x + a))/tan(b*x + a)^3) + 12*(tan(b*x + a)^5 + ta
n(b*x + a)^3)*sqrt(c)*log((c*tan(b*x + a)^3 - 2*sqrt(2)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*(tan(b*x
+ a)^2 - 1)*sqrt(c) - 3*c*tan(b*x + a))/(tan(b*x + a)^3 + tan(b*x + a))) + 2*sqrt(2)*(5*tan(b*x + a)^4 - 4*tan
(b*x + a)^2 - 1)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1)))/(b*c^2*tan(b*x + a)^5 + b*c^2*tan(b*x + a)^3),
1/8*(9*sqrt(2)*(tan(b*x + a)^5 + tan(b*x + a)^3)*sqrt(-c)*arctan(sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*
(tan(b*x + a)^2 - 1)*sqrt(-c)/(c*tan(b*x + a))) - 12*(tan(b*x + a)^5 + tan(b*x + a)^3)*sqrt(-c)*arctan(1/2*sqr
t(2)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*(tan(b*x + a)^2 - 1)*sqrt(-c)/(c*tan(b*x + a))) + sqrt(2)*(5
*tan(b*x + a)^4 - 4*tan(b*x + a)^2 - 1)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1)))/(b*c^2*tan(b*x + a)^5 +
b*c^2*tan(b*x + a)^3)]

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos (2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-1)]

Timed out. \[ \int \frac {\cos (2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [F(-1)]

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

[In]

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

[Out]

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