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

Optimal. Leaf size=84 \[ \frac{\sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{c \sec (2 a+2 b x)-c}}\right )}{2 b}-\frac{c \sin (2 a+2 b x)}{2 b \sqrt{c \sec (2 a+2 b x)-c}} \]

[Out]

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

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Rubi [A]  time = 0.13946, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {4397, 3805, 3774, 207} \[ \frac{\sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{c \sec (2 a+2 b x)-c}}\right )}{2 b}-\frac{c \sin (2 a+2 b x)}{2 b \sqrt{c \sec (2 a+2 b x)-c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4397

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

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 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])

Rubi steps

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

Mathematica [A]  time = 0.244687, size = 92, normalized size = 1.1 \[ -\frac{\csc (a+b x) \sqrt{c \tan (a+b x) \tan (2 (a+b x))} \left (\cos (a+b x)+\cos (3 (a+b x))-\sqrt{2} \sqrt{\cos (2 (a+b x))} \tanh ^{-1}\left (\frac{\sqrt{2} \cos (a+b x)}{\sqrt{\cos (2 (a+b x))}}\right )\right )}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.424, size = 387, normalized size = 4.6 \begin{align*}{\frac{\sqrt{4}\sin \left ( bx+a \right ) }{2\,b \left ( -1+\cos \left ( bx+a \right ) \right ) }\sqrt{{\frac{c \left ( 1- \left ( \cos \left ( bx+a \right ) \right ) ^{2} \right ) }{2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-1}}}\sqrt{{\frac{2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-1}{ \left ( \cos \left ( bx+a \right ) +1 \right ) ^{2}}}}{\it Artanh} \left ({\frac{\sqrt{2}\cos \left ( bx+a \right ) \sqrt{4} \left ( -1+\cos \left ( bx+a \right ) \right ) }{2\, \left ( \sin \left ( bx+a \right ) \right ) ^{2}}{\frac{1}{\sqrt{{\frac{2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-1}{ \left ( \cos \left ( bx+a \right ) +1 \right ) ^{2}}}}}}} \right ) }-{\frac{\sqrt{2}\sqrt{4}\sin \left ( bx+a \right ) }{8\,b \left ( -1+ \left ( \cos \left ( bx+a \right ) \right ) ^{2} \right ) }\sqrt{{\frac{c \left ( \sin \left ( bx+a \right ) \right ) ^{2}}{2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-1}}} \left ( \sqrt{2}\cos \left ( bx+a \right ) \sqrt{{\frac{2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-1}{ \left ( \cos \left ( bx+a \right ) +1 \right ) ^{2}}}}{\it Artanh} \left ({\frac{\sqrt{2}\cos \left ( bx+a \right ) \sqrt{4} \left ( -1+\cos \left ( bx+a \right ) \right ) }{2\, \left ( \sin \left ( bx+a \right ) \right ) ^{2}}{\frac{1}{\sqrt{{\frac{2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-1}{ \left ( \cos \left ( bx+a \right ) +1 \right ) ^{2}}}}}}} \right ) +\sqrt{2}\sqrt{{\frac{2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-1}{ \left ( \cos \left ( bx+a \right ) +1 \right ) ^{2}}}}{\it Artanh} \left ({\frac{\sqrt{2}\cos \left ( bx+a \right ) \sqrt{4} \left ( -1+\cos \left ( bx+a \right ) \right ) }{2\, \left ( \sin \left ( bx+a \right ) \right ) ^{2}}{\frac{1}{\sqrt{{\frac{2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-1}{ \left ( \cos \left ( bx+a \right ) +1 \right ) ^{2}}}}}}} \right ) -4\, \left ( \cos \left ( bx+a \right ) \right ) ^{3}+2\,\cos \left ( bx+a \right ) \right ) } \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))^(1/2),x)

[Out]

1/2/b*4^(1/2)*(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)*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))/(-1+cos(b*x+a))-1/8*2^(1/2)/b*4^(1/2)*(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(1/2)*sin(b*x+a)*(2^(1/2)*co
s(b*x+a)*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*arctanh(1/2*2^(1/2)*cos(b*x+a)*4^(1/2)*(-1+cos(b*x+a))/si
n(b*x+a)^2/((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2))+2^(1/2)*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*ar
ctanh(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))
-4*cos(b*x+a)^3+2*cos(b*x+a))/(-1+cos(b*x+a)^2)

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Maxima [B]  time = 2.10735, size = 1416, normalized size = 16.86 \begin{align*} \text{result too large to display} \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))^(1/2),x, algorithm="maxima")

[Out]

1/16*(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(2*b*x + 2*a) + (cos(2*b*x + 2*a) + 1)*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos
(4*b*x + 4*a) - 1)))*sqrt(c) - 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)*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*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)*si
n(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(((cos(2*b*x + 2*
a)^2 + sin(2*b*x + 2*a)^2)*cos(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))^2 + (cos(2*b*x + 2*a)^2 +
sin(2*b*x + 2*a)^2)*sin(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 + si
n(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) - 1))*sin(2*b*x + 2*a) + cos(2*b*x + 2*a)*sin(1
/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))) + 1) - log(((cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2)*co
s(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))^2 + (cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2)*sin(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) - 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(si
n(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))*sin(2*b*x + 2*a) + cos(2*b*x + 2*a)*sin(1/2*arctan2(sin(4*b*x + 4*a),
-cos(4*b*x + 4*a) - 1))) + 1)))/b

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Fricas [B]  time = 2.40093, size = 921, normalized size = 10.96 \begin{align*} \left [\frac{{\left (\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} \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 )}}{8 \,{\left (b \tan \left (b x + a\right )^{3} + b \tan \left (b x + a\right )\right )}}, -\frac{{\left (\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} \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 )}}{4 \,{\left (b \tan \left (b x + a\right )^{3} + b \tan \left (b x + a\right )\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))^(1/2),x, algorithm="fricas")

[Out]

[1/8*((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))/(ta
n(b*x + a)^5 + 2*tan(b*x + a)^3 + tan(b*x + a))) + 4*sqrt(2)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*(tan
(b*x + a)^2 - 1))/(b*tan(b*x + a)^3 + b*tan(b*x + a)), -1/4*((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*ta
n(b*x + a))) - 2*sqrt(2)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*(tan(b*x + a)^2 - 1))/(b*tan(b*x + a)^3
+ b*tan(b*x + a))]

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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))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 )\,{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))^(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), x)