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

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Rubi [A]  time = 0.09, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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 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 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
]

Rule 4397

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

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*}

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Mathematica [A]  time = 0.31, 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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fricas [A]  time = 1.90, size = 309, normalized size = 3.09 \[ \left [\frac {\sqrt {2} \sqrt {c} \log \left (\frac {c \tan \left (b x + a\right )^{3} - 2 \, \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {c} - 2 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{3}}\right ) + 2 \, \sqrt {c} \log \left (\frac {c \tan \left (b x + a\right )^{3} + 2 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {c} - 3 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )}\right )}{4 \, b c}, -\frac {\sqrt {2} \sqrt {-c} \arctan \left (\frac {\sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-c}}{c \tan \left (b x + a\right )}\right ) - 2 \, \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-c}}{2 \, c \tan \left (b x + a\right )}\right )}{2 \, b c}\right ] \]

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

[Out]

Timed out

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maple [B]  time = 1.01, size = 301, normalized size = 3.01 \[ -\frac {\sqrt {2}\, \left (\cos \left (b x +a \right )+1\right ) \sqrt {\frac {2 \left (\cos ^{2}\left (b x +a \right )\right )-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \sqrt {\frac {c \left (1-\left (\cos ^{2}\left (b x +a \right )\right )\right )}{2 \left (\cos ^{2}\left (b x +a \right )\right )-1}}\, \left (2 \sqrt {2}\, \arctanh \left (\frac {\cos \left (b x +a \right ) \sqrt {4}\, \left (-1+\cos \left (b x +a \right )\right ) \sqrt {2}}{2 \sin \left (b x +a \right )^{2} \sqrt {\frac {2 \left (\cos ^{2}\left (b x +a \right )\right )-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}}\right )-\ln \left (-\frac {2 \left (\left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {\frac {2 \left (\cos ^{2}\left (b x +a \right )\right )-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}-2 \left (\cos ^{2}\left (b x +a \right )\right )+\cos \left (b x +a \right )-\sqrt {\frac {2 \left (\cos ^{2}\left (b x +a \right )\right )-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}+1\right )}{\sin \left (b x +a \right )^{2}}\right )-\arctanh \left (\frac {\sqrt {4}\, \left (2 \left (\cos ^{2}\left (b x +a \right )\right )-3 \cos \left (b x +a \right )+1\right )}{2 \sqrt {\frac {2 \left (\cos ^{2}\left (b x +a \right )\right )-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \sin \left (b x +a \right )^{2}}\right )\right ) \sqrt {4}}{8 b \sin \left (b x +a \right ) c} \]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

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: RuntimeError >> ECL says: Error executing code in Maxima: sign: argument cannot be imaginary
; found %i

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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