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

Optimal. Leaf size=80 \[ \frac{c^2 \tan (2 a+2 b x)}{b \sqrt{c \sec (2 a+2 b x)-c}}+\frac{c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{c \sec (2 a+2 b x)-c}}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.0597835, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4397, 3775, 21, 3774, 207} \[ \frac{c^2 \tan (2 a+2 b x)}{b \sqrt{c \sec (2 a+2 b x)-c}}+\frac{c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{c \sec (2 a+2 b x)-c}}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4397

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

Rule 3775

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Simp[(b^2*Cot[c + d*x]*(a + b*Csc[c + d*x])^(n
- 2))/(d*(n - 1)), x] + Dist[a/(n - 1), Int[(a + b*Csc[c + d*x])^(n - 2)*(a*(n - 1) + b*(3*n - 4)*Csc[c + d*x]
), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]

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

Mathematica [A]  time = 0.157503, size = 86, normalized size = 1.08 \[ \frac{c \sqrt{c \tan (a+b x) \tan (2 (a+b x))} \left (2 \cot (a+b x)+\sqrt{2} \sqrt{\cos (2 (a+b x))} \csc (a+b x) \tanh ^{-1}\left (\frac{\sqrt{2} \cos (a+b x)}{\sqrt{\cos (2 (a+b x))}}\right )\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(2*Cot[a + b*x] + 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])*Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]])/(2*b)

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Maple [B]  time = 0.514, size = 253, normalized size = 3.2 \begin{align*}{\frac{\sqrt{2} \left ( 2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-1 \right ) }{b \left ( \sqrt{2}-2 \right ) \left ( 2+\sqrt{2} \right ) \left ( \sin \left ( bx+a \right ) \right ) ^{3}} \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 ) -2\,\cos \left ( bx+a \right ) \right ) \left ({\frac{c \left ( \sin \left ( bx+a \right ) \right ) ^{2}}{2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-1}} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((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

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Maxima [B]  time = 2.1117, size = 1778, normalized size = 22.22 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*((cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)^(1/4)*(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 + s
qrt(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*arcta
n2(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(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*a
rctan2(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)
- c*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), -co
s(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))*sqrt(c) + 8*(c*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)) + c
*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))
) + c*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)))*sqrt(c))/((cos(4*b*x + 4*a)^2 + sin(4*b*x + 4
*a)^2 + 2*cos(4*b*x + 4*a) + 1)^(1/4)*b)

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Fricas [A]  time = 2.42018, size = 770, normalized size = 9.62 \begin{align*} \left [\frac{c^{\frac{3}{2}} \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 ) \tan \left (b x + a\right ) + 4 \, \sqrt{2} \sqrt{-\frac{c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} c}{4 \, b \tan \left (b x + a\right )}, -\frac{\sqrt{-c} 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 ) \tan \left (b x + a\right ) - 2 \, \sqrt{2} \sqrt{-\frac{c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} c}{2 \, b \tan \left (b x + a\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(c^(3/2)*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)))*tan(b*x + a) + 4*sqrt(2)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*c)/(b*tan(b*x + a)), -1
/2*(sqrt(-c)*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)))*tan(b*x + a) - 2*sqrt(2)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*c)/(
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((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((c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x, algorithm="giac")

[Out]

Timed out