3.623 \(\int \frac{1}{\sqrt{c \tan (a+b x) \tan (2 (a+b x))}} \, dx\)

Optimal. Leaf size=100 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{c \sec (2 a+2 b x)-c}}\right )}{b \sqrt{c}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{2} \sqrt{c \sec (2 a+2 b x)-c}}\right )}{\sqrt{2} b \sqrt{c}} \]

[Out]

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

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Rubi [A]  time = 0.0899419, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4397, 3776, 3774, 207, 3795} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{c \sec (2 a+2 b x)-c}}\right )}{b \sqrt{c}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{2} \sqrt{c \sec (2 a+2 b x)-c}}\right )}{\sqrt{2} b \sqrt{c}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]],x]

[Out]

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

Rule 4397

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

Rule 3776

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[1/a, Int[Sqrt[a + b*Csc[c + d*x]], x], x]
- Dist[b/a, Int[Csc[c + d*x]/Sqrt[a + b*Csc[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && 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}{\sqrt{c \tan (a+b x) \tan (2 (a+b x))}} \, dx &=\int \frac{1}{\sqrt{-c+c \sec (2 a+2 b x)}} \, dx\\ &=-\frac{\int \sqrt{-c+c \sec (2 a+2 b x)} \, dx}{c}+\int \frac{\sec (2 a+2 b x)}{\sqrt{-c+c \sec (2 a+2 b x)}} \, dx\\ &=-\frac{\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 )}{b}+\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}\\ &=\frac{\tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{-c+c \sec (2 a+2 b x)}}\right )}{b \sqrt{c}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{2} \sqrt{-c+c \sec (2 a+2 b x)}}\right )}{\sqrt{2} b \sqrt{c}}\\ \end{align*}

Mathematica [A]  time = 0.332494, size = 94, normalized size = 0.94 \[ \frac{\tan (a+b x) \left (2 \tanh ^{-1}\left (\frac{1}{2} \sqrt{2-2 \tan ^2(a+b x)}\right )-\sqrt{2} \tanh ^{-1}\left (\sqrt{1-\tan ^2(a+b x)}\right )\right )}{b \sqrt{2-2 \tan ^2(a+b x)} \sqrt{c \tan (a+b x) \tan (2 (a+b x))}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]],x]

[Out]

((2*ArcTanh[Sqrt[2 - 2*Tan[a + b*x]^2]/2] - Sqrt[2]*ArcTanh[Sqrt[1 - Tan[a + b*x]^2]])*Tan[a + b*x])/(b*Sqrt[2
- 2*Tan[a + b*x]^2]*Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]])

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Maple [B]  time = 0.363, size = 301, normalized size = 3. \begin{align*} -{\frac{\sqrt{2}\sqrt{4} \left ( \cos \left ( bx+a \right ) +1 \right ) }{8\,b\sin \left ( bx+a \right ) c}\sqrt{{\frac{2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-1}{ \left ( \cos \left ( bx+a \right ) +1 \right ) ^{2}}}}\sqrt{{\frac{c \left ( 1- \left ( \cos \left ( bx+a \right ) \right ) ^{2} \right ) }{2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-1}}} \left ( 2\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\cos \left ( bx+a \right ) \sqrt{4} \left ( -1+\cos \left ( bx+a \right ) \right ) }{ \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 ) -\ln \left ( -2\,{\frac{1}{ \left ( \sin \left ( bx+a \right ) \right ) ^{2}} \left ( \left ( \cos \left ( bx+a \right ) \right ) ^{2}\sqrt{{\frac{2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-1}{ \left ( \cos \left ( bx+a \right ) +1 \right ) ^{2}}}}-2\, \left ( \cos \left ( bx+a \right ) \right ) ^{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}}}}+1 \right ) } \right ) -{\it Artanh} \left ({\frac{\sqrt{4} \left ( 2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-3\,\cos \left ( bx+a \right ) +1 \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 ) \right ) } \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))^(1/2),x)

[Out]

-1/8*2^(1/2)/b*4^(1/2)*(cos(b*x+a)+1)*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*(c*(1-cos(b*x+a)^2)/(2*cos(b
*x+a)^2-1))^(1/2)*(2*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))-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)-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)))/sin(b*x+a)/c

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.14032, size = 807, normalized size = 8.07 \begin{align*} \left [\frac{\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 ) + 2 \, \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 )}{4 \, b c}, -\frac{\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 ) - 2 \, \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 )}{2 \, b c}\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))^(1/2),x, algorithm="fricas")

[Out]

[1/4*(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) + 2*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)))
)/(b*c), -1/2*(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))) - 2*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))))/(b*c)]

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

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \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))^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError