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

Optimal. Leaf size=133 \[ -\frac{7 c^2 \sin (2 a+2 b x)}{8 b \sqrt{c \sec (2 a+2 b x)-c}}+\frac{c^2 \sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt{c \sec (2 a+2 b x)-c}}+\frac{7 c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{c \sec (2 a+2 b x)-c}}\right )}{8 b} \]

[Out]

(7*c^(3/2)*ArcTanh[(Sqrt[c]*Tan[2*a + 2*b*x])/Sqrt[-c + c*Sec[2*a + 2*b*x]]])/(8*b) - (7*c^2*Sin[2*a + 2*b*x])
/(8*b*Sqrt[-c + c*Sec[2*a + 2*b*x]]) + (c^2*Cos[2*a + 2*b*x]*Sin[2*a + 2*b*x])/(4*b*Sqrt[-c + c*Sec[2*a + 2*b*
x]])

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Rubi [A]  time = 0.256441, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {4397, 3813, 21, 3805, 3774, 207} \[ -\frac{7 c^2 \sin (2 a+2 b x)}{8 b \sqrt{c \sec (2 a+2 b x)-c}}+\frac{c^2 \sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt{c \sec (2 a+2 b x)-c}}+\frac{7 c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{c \sec (2 a+2 b x)-c}}\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*c^(3/2)*ArcTanh[(Sqrt[c]*Tan[2*a + 2*b*x])/Sqrt[-c + c*Sec[2*a + 2*b*x]]])/(8*b) - (7*c^2*Sin[2*a + 2*b*x])
/(8*b*Sqrt[-c + c*Sec[2*a + 2*b*x]]) + (c^2*Cos[2*a + 2*b*x]*Sin[2*a + 2*b*x])/(4*b*Sqrt[-c + c*Sec[2*a + 2*b*
x]])

Rule 4397

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

Rule 3813

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(b^2*C
ot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[a/(d*n), Int[(a + b*Csc[e + f*x]
)^(m - 2)*(d*Csc[e + f*x])^(n + 1)*(b*(m - 2*n - 2) - a*(m + 2*n - 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d,
e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1] && (LtQ[n, -1] || (EqQ[m, 3/2] && EqQ[n, -2^(-1)])) && IntegerQ[2
*m]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx &=\int \cos ^2(2 a+2 b x) (-c+c \sec (2 a+2 b x))^{3/2} \, dx\\ &=\frac{c^2 \cos (2 a+2 b x) \sin (2 a+2 b x)}{4 b \sqrt{-c+c \sec (2 a+2 b x)}}-\frac{1}{2} c \int \frac{\cos (2 a+2 b x) \left (\frac{7 c}{2}-\frac{7}{2} c \sec (2 a+2 b x)\right )}{\sqrt{-c+c \sec (2 a+2 b x)}} \, dx\\ &=\frac{c^2 \cos (2 a+2 b x) \sin (2 a+2 b x)}{4 b \sqrt{-c+c \sec (2 a+2 b x)}}+\frac{1}{4} (7 c) \int \cos (2 a+2 b x) \sqrt{-c+c \sec (2 a+2 b x)} \, dx\\ &=-\frac{7 c^2 \sin (2 a+2 b x)}{8 b \sqrt{-c+c \sec (2 a+2 b x)}}+\frac{c^2 \cos (2 a+2 b x) \sin (2 a+2 b x)}{4 b \sqrt{-c+c \sec (2 a+2 b x)}}-\frac{1}{8} (7 c) \int \sqrt{-c+c \sec (2 a+2 b x)} \, dx\\ &=-\frac{7 c^2 \sin (2 a+2 b x)}{8 b \sqrt{-c+c \sec (2 a+2 b x)}}+\frac{c^2 \cos (2 a+2 b x) \sin (2 a+2 b x)}{4 b \sqrt{-c+c \sec (2 a+2 b x)}}+\frac{\left (7 c^2\right ) \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{7 c^{3/2} \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{7 c^2 \sin (2 a+2 b x)}{8 b \sqrt{-c+c \sec (2 a+2 b x)}}+\frac{c^2 \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*}

Mathematica [A]  time = 0.287793, size = 105, normalized size = 0.79 \[ \frac{c \csc (a+b x) \sqrt{c \tan (a+b x) \tan (2 (a+b x))} \left (-5 \cos (a+b x)-6 \cos (3 (a+b x))+\cos (5 (a+b x))+7 \sqrt{2} \sqrt{\cos (2 (a+b x))} \tanh ^{-1}\left (\frac{\sqrt{2} \cos (a+b x)}{\sqrt{\cos (2 (a+b x))}}\right )\right )}{16 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(-5*Cos[a + b*x] + 7*Sqrt[2]*ArcTanh[(Sqrt[2]*Cos[a + b*x])/Sqrt[Cos[2*(a + b*x)]]]*Sqrt[Cos[2*(a + b*x)]]
- 6*Cos[3*(a + b*x)] + Cos[5*(a + b*x)])*Csc[a + b*x]*Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]])/(16*b)

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Maple [B]  time = 0.455, size = 792, normalized size = 6. \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)^2*(c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x)

[Out]

2^(1/2)/b/(2^(1/2)-2)/(2+2^(1/2))*(2*cos(b*x+a)^2-1)*(2^(1/2)*cos(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))+2^(1/2)*((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))-2*cos(b*x+a))*(c*sin(b*x+a)^2/(2*cos(b*x+a)^2
-1))^(3/2)/sin(b*x+a)^3+4*2^(1/2)/b/(2^(1/2)-2)^3/(2+2^(1/2))^3*(2*cos(b*x+a)^2-1)*(2^(1/2)*cos(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*c
os(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)*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))+4*cos(b*x+a)^3+
2*cos(b*x+a))*(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(3/2)/sin(b*x+a)^3-2*2^(1/2)/b/(2+2^(1/2))^5/(2^(1/2)-2)^5*(
2*cos(b*x+a)^2-1)*(16*cos(b*x+a)^5+9*2^(1/2)*cos(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))+9*2^(1/
2)*((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))-12*cos(b*x+a)^3+18*cos(b*x+a))*(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(-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)^2*(c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 2.37623, size = 1126, normalized size = 8.47 \begin{align*} \left [\frac{7 \,{\left (c \tan \left (b x + a\right )^{5} + 2 \, c \tan \left (b x + a\right )^{3} + c \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 (9 \, c \tan \left (b x + a\right )^{4} - 4 \, c \tan \left (b x + a\right )^{2} - 5 \, c\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{7 \,{\left (c \tan \left (b x + a\right )^{5} + 2 \, c \tan \left (b x + a\right )^{3} + c \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 (9 \, c \tan \left (b x + a\right )^{4} - 4 \, c \tan \left (b x + a\right )^{2} - 5 \, c\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 ] \end{align*}

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))^(3/2),x, algorithm="fricas")

[Out]

[1/32*(7*(c*tan(b*x + a)^5 + 2*c*tan(b*x + a)^3 + c*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)*(9*c*tan(b*x + a)^4 -
4*c*tan(b*x + a)^2 - 5*c)*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*(7*(c*tan(b*x + a)^5 + 2*c*tan(b*x + a)^3 + c*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)*(9*c*tan(b*x + a)^4 - 4*c*tan(b*x + a)^2 - 5*c)*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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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)**2*(c*tan(b*x+a)*tan(2*b*x+2*a))**(3/2),x)

[Out]

Timed out

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

[Out]

Timed out