3.630 \(\int \frac{1}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx\)

Optimal. Leaf size=138 \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{c \sec (2 a+2 b x)-c}}\right )}{b c^{3/2}}+\frac{5 \tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{2} \sqrt{c \sec (2 a+2 b x)-c}}\right )}{4 \sqrt{2} b c^{3/2}}-\frac{\tan (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}} \]

[Out]

-(ArcTanh[(Sqrt[c]*Tan[2*a + 2*b*x])/Sqrt[-c + c*Sec[2*a + 2*b*x]]]/(b*c^(3/2))) + (5*ArcTanh[(Sqrt[c]*Tan[2*a
+ 2*b*x])/(Sqrt[2]*Sqrt[-c + c*Sec[2*a + 2*b*x]])])/(4*Sqrt[2]*b*c^(3/2)) - Tan[2*a + 2*b*x]/(4*b*(-c + c*Sec
[2*a + 2*b*x])^(3/2))

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Rubi [A]  time = 0.142754, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {4397, 3777, 3920, 3774, 207, 3795} \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{c \sec (2 a+2 b x)-c}}\right )}{b c^{3/2}}+\frac{5 \tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{2} \sqrt{c \sec (2 a+2 b x)-c}}\right )}{4 \sqrt{2} b c^{3/2}}-\frac{\tan (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[(Sqrt[c]*Tan[2*a + 2*b*x])/Sqrt[-c + c*Sec[2*a + 2*b*x]]]/(b*c^(3/2))) + (5*ArcTanh[(Sqrt[c]*Tan[2*a
+ 2*b*x])/(Sqrt[2]*Sqrt[-c + c*Sec[2*a + 2*b*x]])])/(4*Sqrt[2]*b*c^(3/2)) - Tan[2*a + 2*b*x]/(4*b*(-c + c*Sec
[2*a + 2*b*x])^(3/2))

Rule 4397

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

Rule 3777

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

Rule 3920

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c/a,
Int[Sqrt[a + b*Csc[e + f*x]], x], x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 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 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 3795

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

Rubi steps

\begin{align*} \int \frac{1}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx &=\int \frac{1}{(-c+c \sec (2 a+2 b x))^{3/2}} \, dx\\ &=-\frac{\tan (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}-\frac{\int \frac{2 c+\frac{1}{2} c \sec (2 a+2 b x)}{\sqrt{-c+c \sec (2 a+2 b x)}} \, dx}{2 c^2}\\ &=-\frac{\tan (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}+\frac{\int \sqrt{-c+c \sec (2 a+2 b x)} \, dx}{c^2}-\frac{5 \int \frac{\sec (2 a+2 b x)}{\sqrt{-c+c \sec (2 a+2 b x)}} \, dx}{4 c}\\ &=-\frac{\tan (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}-\frac{\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 c}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{-2 c+x^2} \, dx,x,-\frac{c \tan (2 a+2 b x)}{\sqrt{-c+c \sec (2 a+2 b x)}}\right )}{4 b c}\\ &=-\frac{\tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{-c+c \sec (2 a+2 b x)}}\right )}{b c^{3/2}}+\frac{5 \tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{2} \sqrt{-c+c \sec (2 a+2 b x)}}\right )}{4 \sqrt{2} b c^{3/2}}-\frac{\tan (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 3.85748, size = 196, normalized size = 1.42 \[ -\frac{\cot (a+b x) \sqrt{c \tan (a+b x) \tan (2 (a+b x))} \left (\tan ^{-1}\left (\sqrt{\tan ^2(a+b x)-1}\right ) \sqrt{-\left (\tan ^2(a+b x)-1\right )^2}+\cot ^2(a+b x) \left (\cos (2 (a+b x)) \sec ^2(a+b x)\right )^{3/2}+4 \sqrt{2} \cos (2 (a+b x)) \sec ^2(a+b x) \tanh ^{-1}\left (\frac{1}{2} \sqrt{2-2 \tan ^2(a+b x)}\right )-4 \cos (2 (a+b x)) \sec ^2(a+b x) \tanh ^{-1}\left (\sqrt{1-\tan ^2(a+b x)}\right )\right )}{8 b c^2 \sqrt{1-\tan ^2(a+b x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(Cot[a + b*x]*(4*Sqrt[2]*ArcTanh[Sqrt[2 - 2*Tan[a + b*x]^2]/2]*Cos[2*(a + b*x)]*Sec[a + b*x]^2 - 4*ArcTanh[Sq
rt[1 - Tan[a + b*x]^2]]*Cos[2*(a + b*x)]*Sec[a + b*x]^2 + Cot[a + b*x]^2*(Cos[2*(a + b*x)]*Sec[a + b*x]^2)^(3/
2) + ArcTan[Sqrt[-1 + Tan[a + b*x]^2]]*Sqrt[-(-1 + Tan[a + b*x]^2)^2])*Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]])/
(8*b*c^2*Sqrt[1 - Tan[a + b*x]^2])

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Maple [B]  time = 0.388, size = 561, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x)

[Out]

-1/32*2^(1/2)/b*4^(1/2)*(-1+cos(b*x+a))^2*(8*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))*cos(b*x+a)*2^(1/2)-8*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))-5*ln(-2*(cos(b*x+a)^2*((2*cos(
b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)-2*cos(b*x+a)^2+cos(b*x+a)-((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)+1)/s
in(b*x+a)^2)*cos(b*x+a)-5*arctanh(1/2*4^(1/2)*(2*cos(b*x+a)^2-3*cos(b*x+a)+1)/sin(b*x+a)^2/((2*cos(b*x+a)^2-1)
/(cos(b*x+a)+1)^2)^(1/2))*cos(b*x+a)+2*cos(b*x+a)*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)+5*ln(-2*(cos(b*x
+a)^2*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)-2*cos(b*x+a)^2+cos(b*x+a)-((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)
^2)^(1/2)+1)/sin(b*x+a)^2)+5*arctanh(1/2*4^(1/2)*(2*cos(b*x+a)^2-3*cos(b*x+a)+1)/sin(b*x+a)^2/((2*cos(b*x+a)^2
-1)/(cos(b*x+a)+1)^2)^(1/2)))/(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(3/2)/sin(b*x+a)^3/((2*cos(b*x+a)^2-1)/(cos(
b*x+a)+1)^2)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.97344, size = 1148, normalized size = 8.32 \begin{align*} \left [\frac{5 \, \sqrt{2} \sqrt{c} \log \left (\frac{c \tan \left (b x + a\right )^{3} + 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} - 2 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{3}}\right ) \tan \left (b x + a\right )^{3} + 8 \, \sqrt{c} \log \left (\frac{c \tan \left (b x + a\right )^{3} - 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} - 3 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )}\right ) \tan \left (b x + a\right )^{3} + 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 )}}{16 \, b c^{2} \tan \left (b x + a\right )^{3}}, \frac{5 \, \sqrt{2} \sqrt{-c} \arctan \left (\frac{\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 )}\right ) \tan \left (b x + a\right )^{3} - 8 \, \sqrt{-c} \arctan \left (\frac{\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}}{2 \, c \tan \left (b x + a\right )}\right ) \tan \left (b x + a\right )^{3} + \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 \, b c^{2} \tan \left (b x + a\right )^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x, algorithm="fricas")

[Out]

[1/16*(5*sqrt(2)*sqrt(c)*log((c*tan(b*x + a)^3 + 2*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*(tan(b*x + a)^
2 - 1)*sqrt(c) - 2*c*tan(b*x + a))/tan(b*x + a)^3)*tan(b*x + a)^3 + 8*sqrt(c)*log((c*tan(b*x + a)^3 - 2*sqrt(2
)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*(tan(b*x + a)^2 - 1)*sqrt(c) - 3*c*tan(b*x + a))/(tan(b*x + a)^
3 + tan(b*x + a)))*tan(b*x + a)^3 + 2*sqrt(2)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*(tan(b*x + a)^2 - 1
))/(b*c^2*tan(b*x + a)^3), 1/8*(5*sqrt(2)*sqrt(-c)*arctan(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)))*tan(b*x + a)^3 - 8*sqrt(-c)*arctan(1/2*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)))*tan(b*x + a)^3 + sqrt(2)*sqrt(-c*tan(b*
x + a)^2/(tan(b*x + a)^2 - 1))*(tan(b*x + a)^2 - 1))/(b*c^2*tan(b*x + a)^3)]

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

[Out]

Timed out