3.609 \(\int \cos ^2(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx\)

Optimal. Leaf size=129 \[ \frac {3 c \sin (2 a+2 b x)}{8 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {c \sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {3 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{8 b} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {4397, 3805, 3774, 207} \[ \frac {3 c \sin (2 a+2 b x)}{8 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {c \sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {3 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Sqrt[c]*ArcTanh[(Sqrt[c]*Tan[2*a + 2*b*x])/Sqrt[-c + c*Sec[2*a + 2*b*x]]])/(8*b) + (3*c*Sin[2*a + 2*b*x])/
(8*b*Sqrt[-c + c*Sec[2*a + 2*b*x]]) - (c*Cos[2*a + 2*b*x]*Sin[2*a + 2*b*x])/(4*b*Sqrt[-c + c*Sec[2*a + 2*b*x]]
)

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 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 3805

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

Rule 4397

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

Rubi steps

\begin {align*} \int \cos ^2(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx &=\int \cos ^2(2 a+2 b x) \sqrt {-c+c \sec (2 a+2 b x)} \, dx\\ &=-\frac {c \cos (2 a+2 b x) \sin (2 a+2 b x)}{4 b \sqrt {-c+c \sec (2 a+2 b x)}}-\frac {3}{4} \int \cos (2 a+2 b x) \sqrt {-c+c \sec (2 a+2 b x)} \, dx\\ &=\frac {3 c \sin (2 a+2 b x)}{8 b \sqrt {-c+c \sec (2 a+2 b x)}}-\frac {c \cos (2 a+2 b x) \sin (2 a+2 b x)}{4 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {3}{8} \int \sqrt {-c+c \sec (2 a+2 b x)} \, dx\\ &=\frac {3 c \sin (2 a+2 b x)}{8 b \sqrt {-c+c \sec (2 a+2 b x)}}-\frac {c \cos (2 a+2 b x) \sin (2 a+2 b x)}{4 b \sqrt {-c+c \sec (2 a+2 b x)}}-\frac {(3 c) \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 )}{8 b}\\ &=-\frac {3 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}}\right )}{8 b}+\frac {3 c \sin (2 a+2 b x)}{8 b \sqrt {-c+c \sec (2 a+2 b x)}}-\frac {c \cos (2 a+2 b x) \sin (2 a+2 b x)}{4 b \sqrt {-c+c \sec (2 a+2 b x)}}\\ \end {align*}

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Mathematica [A]  time = 0.25, size = 105, normalized size = 0.81 \[ \frac {\sqrt {c \tan (a+b x) \tan (2 (a+b x))} \left (2 (-\sin (2 (a+b x))+\sin (4 (a+b x))+\cot (a+b x))-3 \sqrt {2} \sqrt {\cos (2 (a+b x))} \csc (a+b x) \tanh ^{-1}\left (\frac {\sqrt {2} \cos (a+b x)}{\sqrt {\cos (2 (a+b x))}}\right )\right )}{16 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*Sqrt[2]*ArcTanh[(Sqrt[2]*Cos[a + b*x])/Sqrt[Cos[2*(a + b*x)]]]*Sqrt[Cos[2*(a + b*x)]]*Csc[a + b*x] + 2*(C
ot[a + b*x] - Sin[2*(a + b*x)] + Sin[4*(a + b*x)]))*Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]])/(16*b)

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fricas [A]  time = 0.95, size = 419, normalized size = 3.25 \[ \left [\frac {3 \, {\left (\tan \left (b x + a\right )^{5} + 2 \, \tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )\right )} \sqrt {c} \log \left (-\frac {c \tan \left (b x + a\right )^{5} - 14 \, c \tan \left (b x + a\right )^{3} - 4 \, \sqrt {2} {\left (\tan \left (b x + a\right )^{4} - 4 \, \tan \left (b x + a\right )^{2} + 3\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} \sqrt {c} + 17 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{5} + 2 \, \tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )}\right ) - 4 \, \sqrt {2} {\left (5 \, \tan \left (b x + a\right )^{4} - 4 \, \tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{32 \, {\left (b \tan \left (b x + a\right )^{5} + 2 \, b \tan \left (b x + a\right )^{3} + b \tan \left (b x + a\right )\right )}}, \frac {3 \, {\left (\tan \left (b x + a\right )^{5} + 2 \, \tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )\right )} \sqrt {-c} \arctan \left (\frac {2 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-c}}{c \tan \left (b x + a\right )^{3} - 3 \, c \tan \left (b x + a\right )}\right ) - 2 \, \sqrt {2} {\left (5 \, \tan \left (b x + a\right )^{4} - 4 \, \tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{16 \, {\left (b \tan \left (b x + a\right )^{5} + 2 \, b \tan \left (b x + a\right )^{3} + b \tan \left (b x + a\right )\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(2*b*x+2*a)^2*(c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x, algorithm="fricas")

[Out]

[1/32*(3*(tan(b*x + a)^5 + 2*tan(b*x + a)^3 + tan(b*x + a))*sqrt(c)*log(-(c*tan(b*x + a)^5 - 14*c*tan(b*x + a)
^3 - 4*sqrt(2)*(tan(b*x + a)^4 - 4*tan(b*x + a)^2 + 3)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*sqrt(c) +
17*c*tan(b*x + a))/(tan(b*x + a)^5 + 2*tan(b*x + a)^3 + tan(b*x + a))) - 4*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*tan(b*x + a)^5 + 2*b*tan(b*x + a)^3 + b*tan(b*
x + a)), 1/16*(3*(tan(b*x + a)^5 + 2*tan(b*x + a)^3 + tan(b*x + a))*sqrt(-c)*arctan(2*sqrt(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)^3 - 3*c*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*tan(b*x + a)^5 + 2*
b*tan(b*x + a)^3 + b*tan(b*x + a))]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(2*b*x+2*a)^2*(c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 1.22, size = 657, normalized size = 5.09 \[ -\frac {\sqrt {\frac {c \left (1-\left (\cos ^{2}\left (b x +a \right )\right )\right )}{2 \left (\cos ^{2}\left (b x +a \right )\right )-1}}\, \sin \left (b x +a \right ) \sqrt {\frac {2 \left (\cos ^{2}\left (b x +a \right )\right )-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \arctanh \left (\frac {\cos \left (b x +a \right ) \sqrt {4}\, \left (-1+\cos \left (b x +a \right )\right ) \sqrt {2}}{2 \sin \left (b x +a \right )^{2} \sqrt {\frac {2 \left (\cos ^{2}\left (b x +a \right )\right )-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}}\right ) \sqrt {4}}{2 b \left (-1+\cos \left (b x +a \right )\right )}+\frac {\sqrt {2}\, \sin \left (b x +a \right ) \sqrt {\frac {c \left (1-\left (\cos ^{2}\left (b x +a \right )\right )\right )}{2 \left (\cos ^{2}\left (b x +a \right )\right )-1}}\, \left (\sqrt {2}\, \sqrt {\frac {2 \left (\cos ^{2}\left (b x +a \right )\right )-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \arctanh \left (\frac {\cos \left (b x +a \right ) \sqrt {4}\, \left (-1+\cos \left (b x +a \right )\right ) \sqrt {2}}{2 \sin \left (b x +a \right )^{2} \sqrt {\frac {2 \left (\cos ^{2}\left (b x +a \right )\right )-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}}\right ) \cos \left (b x +a \right )+\sqrt {2}\, \sqrt {\frac {2 \left (\cos ^{2}\left (b x +a \right )\right )-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \arctanh \left (\frac {\cos \left (b x +a \right ) \sqrt {4}\, \left (-1+\cos \left (b x +a \right )\right ) \sqrt {2}}{2 \sin \left (b x +a \right )^{2} \sqrt {\frac {2 \left (\cos ^{2}\left (b x +a \right )\right )-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}}\right )-4 \left (\cos ^{3}\left (b x +a \right )\right )+2 \cos \left (b x +a \right )\right ) \sqrt {4}}{4 b \left (-1+\cos ^{2}\left (b x +a \right )\right )}-\frac {\sqrt {2}\, \sin \left (b x +a \right ) \sqrt {\frac {c \left (1-\left (\cos ^{2}\left (b x +a \right )\right )\right )}{2 \left (\cos ^{2}\left (b x +a \right )\right )-1}}\, \left (-16 \left (\cos ^{5}\left (b x +a \right )\right )+3 \sqrt {2}\, \sqrt {\frac {2 \left (\cos ^{2}\left (b x +a \right )\right )-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \arctanh \left (\frac {\cos \left (b x +a \right ) \sqrt {4}\, \left (-1+\cos \left (b x +a \right )\right ) \sqrt {2}}{2 \sin \left (b x +a \right )^{2} \sqrt {\frac {2 \left (\cos ^{2}\left (b x +a \right )\right )-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}}\right ) \cos \left (b x +a \right )+3 \sqrt {2}\, \sqrt {\frac {2 \left (\cos ^{2}\left (b x +a \right )\right )-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \arctanh \left (\frac {\cos \left (b x +a \right ) \sqrt {4}\, \left (-1+\cos \left (b x +a \right )\right ) \sqrt {2}}{2 \sin \left (b x +a \right )^{2} \sqrt {\frac {2 \left (\cos ^{2}\left (b x +a \right )\right )-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}}\right )-4 \left (\cos ^{3}\left (b x +a \right )\right )+6 \cos \left (b x +a \right )\right ) \sqrt {4}}{32 b \left (-1+\cos ^{2}\left (b x +a \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(2*b*x+2*a)^2*(c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x)

[Out]

-1/2/b*(c*(1-cos(b*x+a)^2)/(2*cos(b*x+a)^2-1))^(1/2)*sin(b*x+a)*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*ar
ctanh(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)*2^(1/2))
/(-1+cos(b*x+a))*4^(1/2)+1/4*2^(1/2)/b*sin(b*x+a)*(c*(1-cos(b*x+a)^2)/(2*cos(b*x+a)^2-1))^(1/2)*(2^(1/2)*((2*c
os(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*arctanh(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)*2^(1/2))*cos(b*x+a)+2^(1/2)*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*arcta
nh(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)*2^(1/2))-4*
cos(b*x+a)^3+2*cos(b*x+a))/(-1+cos(b*x+a)^2)*4^(1/2)-1/32*2^(1/2)/b*sin(b*x+a)*(c*(1-cos(b*x+a)^2)/(2*cos(b*x+
a)^2-1))^(1/2)*(-16*cos(b*x+a)^5+3*2^(1/2)*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*arctanh(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)*2^(1/2))*cos(b*x+a)+3*2^(1/2)
*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*arctanh(1/2*cos(b*x+a)*4^(1/2)*(-1+cos(b*x+a))/sin(b*x+a)^2/((2*c
os(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*2^(1/2))-4*cos(b*x+a)^3+6*cos(b*x+a))/(-1+cos(b*x+a)^2)*4^(1/2)

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maxima [B]  time = 0.72, size = 1421, normalized size = 11.02 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(2*b*x+2*a)^2*(c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x, algorithm="maxima")

[Out]

-1/64*(4*(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)^(1/4)*((cos(1/2*arctan2(sin(4*b*x
+ 4*a), -cos(4*b*x + 4*a) - 1))*sin(4*b*x + 4*a) - (cos(4*b*x + 4*a) - 2)*sin(1/2*arctan2(sin(4*b*x + 4*a), -c
os(4*b*x + 4*a) - 1)))*cos(1/2*arctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a))) - cos(1/2*arctan2(sin(4*b*x + 4*a)
, -cos(4*b*x + 4*a) - 1))*sin(4*b*x + 4*a) - (cos(4*b*x + 4*a) - 2)*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b
*x + 4*a) - 1)) - ((cos(4*b*x + 4*a) - 2)*cos(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)) + sin(4*b*
x + 4*a)*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)))*sin(1/2*arctan2(sin(4*b*x + 4*a), cos(4*b*
x + 4*a))))*sqrt(c) - 3*sqrt(c)*(log(sqrt(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)*co
s(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))^2 + sqrt(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*c
os(4*b*x + 4*a) + 1)*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))^2 + 2*(cos(4*b*x + 4*a)^2 + sin
(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)^(1/4)*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)) + 1)
- log(sqrt(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)*cos(1/2*arctan2(sin(4*b*x + 4*a)
, -cos(4*b*x + 4*a) - 1))^2 + sqrt(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)*sin(1/2*a
rctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))^2 - 2*(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x
+ 4*a) + 1)^(1/4)*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)) + 1) + log(((cos(1/2*arctan2(sin(4
*b*x + 4*a), -cos(4*b*x + 4*a) - 1))^2 + sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))^2)*cos(1/2*
arctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a)))^2 + (cos(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))^2
+ sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))^2)*sin(1/2*arctan2(sin(4*b*x + 4*a), cos(4*b*x + 4
*a)))^2)*sqrt(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1) + 2*(cos(4*b*x + 4*a)^2 + sin(
4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)^(1/4)*(cos(1/2*arctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a)))*sin(1/2*a
rctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)) + cos(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))*s
in(1/2*arctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a)))) + 1) - log(((cos(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x
+ 4*a) - 1))^2 + sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))^2)*cos(1/2*arctan2(sin(4*b*x + 4*a
), cos(4*b*x + 4*a)))^2 + (cos(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))^2 + sin(1/2*arctan2(sin(4
*b*x + 4*a), -cos(4*b*x + 4*a) - 1))^2)*sin(1/2*arctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a)))^2)*sqrt(cos(4*b*x
+ 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1) - 2*(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(
4*b*x + 4*a) + 1)^(1/4)*(cos(1/2*arctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a)))*sin(1/2*arctan2(sin(4*b*x + 4*a)
, -cos(4*b*x + 4*a) - 1)) + cos(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))*sin(1/2*arctan2(sin(4*b*
x + 4*a), cos(4*b*x + 4*a)))) + 1)))/b

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (2\,a+2\,b\,x\right )}^2\,\sqrt {c\,\mathrm {tan}\left (a+b\,x\right )\,\mathrm {tan}\left (2\,a+2\,b\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(2*a + 2*b*x)^2*(c*tan(a + b*x)*tan(2*a + 2*b*x))^(1/2),x)

[Out]

int(cos(2*a + 2*b*x)^2*(c*tan(a + b*x)*tan(2*a + 2*b*x))^(1/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(2*b*x+2*a)**2*(c*tan(b*x+a)*tan(2*b*x+2*a))**(1/2),x)

[Out]

Timed out

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