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

Optimal. Leaf size=178 \[ -\frac{3 \tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{c \sec (2 a+2 b x)-c}}\right )}{2 b c^{3/2}}+\frac{9 \tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{2} \sqrt{c \sec (2 a+2 b x)-c}}\right )}{4 \sqrt{2} b c^{3/2}}-\frac{3 \sin (2 a+2 b x)}{4 b c \sqrt{c \sec (2 a+2 b x)-c}}-\frac{\sin (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}} \]

[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]])

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Rubi [A]  time = 0.319622, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {4397, 3817, 4022, 3920, 3774, 207, 3795} \[ -\frac{3 \tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{c \sec (2 a+2 b x)-c}}\right )}{2 b c^{3/2}}+\frac{9 \tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{2} \sqrt{c \sec (2 a+2 b x)-c}}\right )}{4 \sqrt{2} b c^{3/2}}-\frac{3 \sin (2 a+2 b x)}{4 b c \sqrt{c \sec (2 a+2 b x)-c}}-\frac{\sin (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}} \]

Antiderivative was successfully verified.

[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 4397

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

Rule 3817

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*Csc
[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 4022

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 3920

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 3774

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 207

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

Rule 3795

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
]

Rubi steps

\begin{align*} \int \frac{\cos (2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx &=\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 \operatorname{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 \operatorname{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 \tanh ^{-1}\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 \tanh ^{-1}\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]  time = 6.19842, size = 342, normalized size = 1.92 \[ \frac{\tan ^2(a+b x) \tan ^2(2 (a+b x)) \left (\frac{1}{2} \sin (2 (a+b x))-\frac{1}{4} \cot (a+b x)-\frac{1}{8} \cot (a+b x) \csc ^2(a+b x)\right )}{b (c \tan (a+b x) \tan (2 (a+b x)))^{3/2}}-\frac{3 \tan ^{\frac{3}{2}}(a+b x) \tan ^{\frac{3}{2}}(2 (a+b x)) \left (\frac{\tan ^{-1}\left (\sqrt{\tan ^2(a+b x)-1}\right ) \tan ^{\frac{3}{2}}(a+b x) \sqrt{\tan ^2(a+b x)-1} \sqrt{\tan (2 (a+b x))} \csc ^2(a+b x) \sec ^2(a+b x)}{\left (\tan ^2(a+b x)+1\right )^2}+\frac{\sqrt{2} \cos (2 (a+b x)) \tan ^{\frac{3}{2}}(a+b x) \sqrt{\tan (2 (a+b x))} \csc ^2(a+b x) \sec ^2(a+b x) \left (2 \tanh ^{-1}\left (\frac{1}{2} \sqrt{2-2 \tan ^2(a+b x)}\right )-\sqrt{2} \tanh ^{-1}\left (\sqrt{1-\tan ^2(a+b x)}\right )\right )}{\sqrt{1-\tan ^2(a+b x)} \left (\tan ^2(a+b x)+1\right )}\right )}{8 b (c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.43, size = 1157, normalized size = 6.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*2^(1/2)/b*4^(1/2)*(-1+cos(b*x+a))^2*(8*arctanh(1/2*2^(1/2)*cos(b*x+a)*4^(1/2)*(-1+cos(b*x+a))/sin(b*x+a)^
2/((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2))*cos(b*x+a)*2^(1/2)-8*2^(1/2)*arctanh(1/2*2^(1/2)*cos(b*x+a)*4^(
1/2)*(-1+cos(b*x+a))/sin(b*x+a)^2/((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2))-5*ln(-2*(cos(b*x+a)^2*((2*cos(b
*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)-2*cos(b*x+a)^2+cos(b*x+a)-((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)+1)/si
n(b*x+a)^2)*cos(b*x+a)-5*arctanh(1/2*4^(1/2)*(2*cos(b*x+a)^2-3*cos(b*x+a)+1)/sin(b*x+a)^2/((2*cos(b*x+a)^2-1)/
(cos(b*x+a)+1)^2)^(1/2))*cos(b*x+a)+2*cos(b*x+a)*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)+5*ln(-2*(cos(b*x+
a)^2*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)-2*cos(b*x+a)^2+cos(b*x+a)-((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^
2)^(1/2)+1)/sin(b*x+a)^2)+5*arctanh(1/2*4^(1/2)*(2*cos(b*x+a)^2-3*cos(b*x+a)+1)/sin(b*x+a)^2/((2*cos(b*x+a)^2-
1)/(cos(b*x+a)+1)^2)^(1/2)))/(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(3/2)/sin(b*x+a)^3/((2*cos(b*x+a)^2-1)/(cos(b
*x+a)+1)^2)^(3/2)-1/16*2^(1/2)/b*4^(1/2)*(-1+cos(b*x+a))^2*(-4*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*cos
(b*x+a)^3+10*arctanh(1/2*2^(1/2)*cos(b*x+a)*4^(1/2)*(-1+cos(b*x+a))/sin(b*x+a)^2/((2*cos(b*x+a)^2-1)/(cos(b*x+
a)+1)^2)^(1/2))*cos(b*x+a)*2^(1/2)-10*2^(1/2)*arctanh(1/2*2^(1/2)*cos(b*x+a)*4^(1/2)*(-1+cos(b*x+a))/sin(b*x+a
)^2/((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2))-7*ln(-2*(cos(b*x+a)^2*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(
1/2)-2*cos(b*x+a)^2+cos(b*x+a)-((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)+1)/sin(b*x+a)^2)*cos(b*x+a)+6*cos(b
*x+a)*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)-7*arctanh(1/2*4^(1/2)*(2*cos(b*x+a)^2-3*cos(b*x+a)+1)/sin(b*
x+a)^2/((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2))*cos(b*x+a)+7*ln(-2*(cos(b*x+a)^2*((2*cos(b*x+a)^2-1)/(cos(
b*x+a)+1)^2)^(1/2)-2*cos(b*x+a)^2+cos(b*x+a)-((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)+1)/sin(b*x+a)^2)+7*ar
ctanh(1/2*4^(1/2)*(2*cos(b*x+a)^2-3*cos(b*x+a)+1)/sin(b*x+a)^2/((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)))/(
(2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(3/2)/(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(3/2)/sin(b*x+a)^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 2.91226, size = 1372, normalized size = 7.71 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

Exception raised: TypeError