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

Optimal. Leaf size=86 \[ \frac {c^2 \sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {3 c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {4397, 3814, 21, 3805, 3774, 207} \[ \frac {c^2 \sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {3 c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{2 b} \]

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

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 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 3814

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(b^2*
Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n)/(f*(m + n - 1)), x] + Dist[b/(m + n - 1), Int[(a
+ b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n*(b*(m + 2*n - 1) + a*(3*m + 2*n - 4)*Csc[e + f*x]), x], x] /; Fr
eeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1] && NeQ[m + n - 1, 0] && IntegerQ[2*m]

Rule 4397

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

Rubi steps

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

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Mathematica [A]  time = 0.24, size = 93, normalized size = 1.08 \[ \frac {c \csc (a+b x) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \left (\cos (a+b x)+\cos (3 (a+b x))-3 \sqrt {2} \sqrt {\cos (2 (a+b x))} \tanh ^{-1}\left (\frac {\sqrt {2} \cos (a+b x)}{\sqrt {\cos (2 (a+b x))}}\right )\right )}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(Cos[a + b*x] - 3*Sqrt[2]*ArcTanh[(Sqrt[2]*Cos[a + b*x])/Sqrt[Cos[2*(a + b*x)]]]*Sqrt[Cos[2*(a + b*x)]] + C
os[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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fricas [B]  time = 2.09, size = 369, normalized size = 4.29 \[ \left [\frac {3 \, {\left (c \tan \left (b x + a\right )^{3} + c \tan \left (b x + a\right )\right )} \sqrt {c} \log \left (-\frac {c \tan \left (b x + a\right )^{5} - 14 \, c \tan \left (b x + a\right )^{3} - 4 \, \sqrt {2} {\left (\tan \left (b x + a\right )^{4} - 4 \, \tan \left (b x + a\right )^{2} + 3\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} \sqrt {c} + 17 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{5} + 2 \, \tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )}\right ) - 4 \, \sqrt {2} {\left (c \tan \left (b x + a\right )^{2} - c\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{8 \, {\left (b \tan \left (b x + a\right )^{3} + b \tan \left (b x + a\right )\right )}}, \frac {3 \, {\left (c \tan \left (b x + a\right )^{3} + c \tan \left (b x + a\right )\right )} \sqrt {-c} \arctan \left (\frac {2 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-c}}{c \tan \left (b x + a\right )^{3} - 3 \, c \tan \left (b x + a\right )}\right ) - 2 \, \sqrt {2} {\left (c \tan \left (b x + a\right )^{2} - c\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{4 \, {\left (b \tan \left (b x + a\right )^{3} + b \tan \left (b x + a\right )\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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maple [B]  time = 1.09, size = 518, normalized size = 6.02 \[ -\frac {\sqrt {2}\, \left (2 \left (\cos ^{2}\left (b x +a \right )\right )-1\right ) \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 )-2 \cos \left (b x +a \right )\right ) \left (\frac {c \left (\sin ^{2}\left (b x +a \right )\right )}{2 \left (\cos ^{2}\left (b x +a \right )\right )-1}\right )^{\frac {3}{2}}}{b \sin \left (b x +a \right )^{3} \left (2+\sqrt {2}\right ) \left (\sqrt {2}-2\right )}-\frac {2 \sqrt {2}\, \left (2 \left (\cos ^{2}\left (b x +a \right )\right )-1\right ) \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 ) \left (\frac {c \left (\sin ^{2}\left (b x +a \right )\right )}{2 \left (\cos ^{2}\left (b x +a \right )\right )-1}\right )^{\frac {3}{2}}}{b \sin \left (b x +a \right )^{3} \left (2+\sqrt {2}\right )^{3} \left (\sqrt {2}-2\right )^{3}} \]

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]

-2^(1/2)/b*(2*cos(b*x+a)^2-1)*(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)+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))-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))/(2^(1/2)-2)-2*2^(1/2)/b*(2*cos(b*x+a)^2-1)*(2^(1/2)*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*a
rctanh(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)*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))+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))^3/(2^(1/2)-2)^3

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maxima [B]  time = 1.15, size = 1058, normalized size = 12.30 \[ \text {result too large to display} \]

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]

-1/16*(4*(c*cos(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))*sin(2*b*x + 2*a) + (c*cos(2*b*x + 2*a) +
c)*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)))*(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*co
s(4*b*x + 4*a) + 1)^(1/4)*sqrt(c) - 3*(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) - 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) + c*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
+ 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) - 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) - c*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)*s
in(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) - 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*arc
tan2(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))*sqrt(c))/b

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \cos \left (2\,a+2\,b\,x\right )\,{\left (c\,\mathrm {tan}\left (a+b\,x\right )\,\mathrm {tan}\left (2\,a+2\,b\,x\right )\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

[Out]

Timed out

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