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

Optimal. Leaf size=45 \[ -\frac {\sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4397, 3774, 207} \[ -\frac {\sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 4397

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

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 73, normalized size = 1.62 \[ -\frac {\sqrt {\cos (2 (a+b x))} \csc (a+b x) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \tanh ^{-1}\left (\frac {\sqrt {2} \cos (a+b x)}{\sqrt {\cos (2 (a+b x))}}\right )}{\sqrt {2} b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((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)]])/(Sqrt[2]*b))

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fricas [A]  time = 1.46, size = 201, normalized size = 4.47 \[ \left [\frac {\sqrt {c} \log \left (-\frac {c \tan \left (b x + a\right )^{5} - 14 \, c \tan \left (b x + a\right )^{3} - 4 \, \sqrt {2} {\left (\tan \left (b x + a\right )^{4} - 4 \, \tan \left (b x + a\right )^{2} + 3\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} \sqrt {c} + 17 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{5} + 2 \, \tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )}\right )}{4 \, b}, \frac {\sqrt {-c} \arctan \left (\frac {2 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-c}}{c \tan \left (b x + a\right )^{3} - 3 \, c \tan \left (b x + a\right )}\right )}{2 \, b}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*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)))/b, 1/2*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)))/b]

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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((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 = 0.94, size = 136, normalized size = 3.02 \[ -\frac {\sqrt {\frac {c \left (1-\left (\cos ^{2}\left (b x +a \right )\right )\right )}{2 \left (\cos ^{2}\left (b x +a \right )\right )-1}}\, \sin \left (b x +a \right ) \sqrt {\frac {2 \left (\cos ^{2}\left (b x +a \right )\right )-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \arctanh \left (\frac {\cos \left (b x +a \right ) \sqrt {4}\, \left (-1+\cos \left (b x +a \right )\right ) \sqrt {2}}{2 \sin \left (b x +a \right )^{2} \sqrt {\frac {2 \left (\cos ^{2}\left (b x +a \right )\right )-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}}\right ) \sqrt {4}}{2 b \left (-1+\cos \left (b x +a \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.50, size = 430, normalized size = 9.56 \[ \frac {\sqrt {c} {\left (\log \left (4 \, \sqrt {\cos \left (4 \, b x + 4 \, a\right )^{2} + \sin \left (4 \, b x + 4 \, a\right )^{2} + 2 \, \cos \left (4 \, b x + 4 \, a\right ) + 1} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, b x + 4 \, a\right ), \cos \left (4 \, b x + 4 \, a\right ) + 1\right )\right )^{2} + 4 \, \sqrt {\cos \left (4 \, b x + 4 \, a\right )^{2} + \sin \left (4 \, b x + 4 \, a\right )^{2} + 2 \, \cos \left (4 \, b x + 4 \, a\right ) + 1} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, b x + 4 \, a\right ), \cos \left (4 \, b x + 4 \, a\right ) + 1\right )\right )^{2} + 8 \, {\left (\cos \left (4 \, b x + 4 \, a\right )^{2} + \sin \left (4 \, b x + 4 \, a\right )^{2} + 2 \, \cos \left (4 \, b x + 4 \, a\right ) + 1\right )}^{\frac {1}{4}} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, b x + 4 \, a\right ), \cos \left (4 \, b x + 4 \, a\right ) + 1\right )\right ) + 4\right ) - \log \left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + \sqrt {\cos \left (4 \, b x + 4 \, a\right )^{2} + \sin \left (4 \, b x + 4 \, a\right )^{2} + 2 \, \cos \left (4 \, b x + 4 \, a\right ) + 1} {\left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, b x + 4 \, a\right ), \cos \left (4 \, b x + 4 \, a\right ) + 1\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, b x + 4 \, a\right ), \cos \left (4 \, b x + 4 \, a\right ) + 1\right )\right )^{2}\right )} + 2 \, {\left (\cos \left (4 \, b x + 4 \, a\right )^{2} + \sin \left (4 \, b x + 4 \, a\right )^{2} + 2 \, \cos \left (4 \, b x + 4 \, a\right ) + 1\right )}^{\frac {1}{4}} {\left (\cos \left (2 \, b x + 2 \, a\right ) \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, b x + 4 \, a\right ), \cos \left (4 \, b x + 4 \, a\right ) + 1\right )\right ) + \sin \left (2 \, b x + 2 \, a\right ) \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, b x + 4 \, a\right ), \cos \left (4 \, b x + 4 \, a\right ) + 1\right )\right )\right )}\right )\right )}}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(c)*(log(4*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 + 4*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 + 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)*cos(1/2*arctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a) + 1)) + 4) - log(cos(2*b*x +
2*a)^2 + sin(2*b*x + 2*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)*(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) + 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(2*b*x + 2*a)*cos(1/2*arct
an2(sin(4*b*x + 4*a), cos(4*b*x + 4*a) + 1)) + sin(2*b*x + 2*a)*sin(1/2*arctan2(sin(4*b*x + 4*a), cos(4*b*x +
4*a) + 1)))))/b

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \sqrt {c\,\mathrm {tan}\left (a+b\,x\right )\,\mathrm {tan}\left (2\,a+2\,b\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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