\(\int \frac {\sec ^2(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx\) [621]

Optimal result

Integrand size = 31, antiderivative size = 88 \[ \int \frac {\sec ^2(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {-c+c \sec (2 a+2 b x)}}\right )}{\sqrt {2} b \sqrt {c}}+\frac {\tan (2 a+2 b x)}{b \sqrt {-c+c \sec (2 a+2 b x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {4482, 3883, 3880, 213} \[ \int \frac {\sec ^2(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\frac {\tan (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {\text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {c \sec (2 a+2 b x)-c}}\right )}{\sqrt {2} b \sqrt {c}} \]

[In]

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

[Out]

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

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 3880

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 3883

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

Rule 4482

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {\sec ^2(2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}} \, dx \\ & = \frac {\tan (2 a+2 b x)}{b \sqrt {-c+c \sec (2 a+2 b x)}}+\int \frac {\sec (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}} \, dx \\ & = \frac {\tan (2 a+2 b x)}{b \sqrt {-c+c \sec (2 a+2 b x)}}-\frac {\text {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 {\text {arctanh}\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}}+\frac {\tan (2 a+2 b x)}{b \sqrt {-c+c \sec (2 a+2 b x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.74 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.76 \[ \int \frac {\sec ^2(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\frac {\left (2+\arctan \left (\sqrt {-1+\tan ^2(a+b x)}\right ) \sqrt {-1+\tan ^2(a+b x)}\right ) \tan (2 (a+b x))}{2 b \sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(369\) vs. \(2(77)=154\).

Time = 4.09 (sec) , antiderivative size = 370, normalized size of antiderivative = 4.20

method	result	size
default	\(\frac {\sqrt {2}\, \left (\cos \left (x b +a \right )-1\right ) \left (\operatorname {arctanh}\left (\frac {2 \cos \left (x b +a \right )-1}{\left (1+\cos \left (x b +a \right )\right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}}\right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}\, \sin \left (x b +a \right )^{2}-\ln \left (\frac {2 \cos \left (x b +a \right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}+2 \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}-4 \cos \left (x b +a \right )-2}{1+\cos \left (x b +a \right )}\right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}\, \sin \left (x b +a \right )^{2}-2 \sin \left (x b +a \right )^{2}+2 \cos \left (x b +a \right )^{2}-4 \cos \left (x b +a \right )+2\right ) \sin \left (x b +a \right ) \sqrt {4}}{2 b \sqrt {\frac {c \sin \left (x b +a \right )^{2}}{2 \cos \left (x b +a \right )^{2}-1}}\, \left (\sin \left (x b +a \right )^{4}-6 \cos \left (x b +a \right )^{2} \sin \left (x b +a \right )^{2}+\cos \left (x b +a \right )^{4}+12 \cos \left (x b +a \right ) \sin \left (x b +a \right )^{2}-4 \cos \left (x b +a \right )^{3}-6 \sin \left (x b +a \right )^{2}+6 \cos \left (x b +a \right )^{2}-4 \cos \left (x b +a \right )+1\right )}\)	\(370\)

[In]

int(sec(2*b*x+2*a)^2/(c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/2*2^(1/2)/b*(cos(b*x+a)-1)*(arctanh((2*cos(b*x+a)-1)/(1+cos(b*x+a))/((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1
/2))*((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)*sin(b*x+a)^2-ln(2*(cos(b*x+a)*((2*cos(b*x+a)^2-1)/(1+cos(b*x+
a))^2)^(1/2)+((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)-2*cos(b*x+a)-1)/(1+cos(b*x+a)))*((2*cos(b*x+a)^2-1)/(
1+cos(b*x+a))^2)^(1/2)*sin(b*x+a)^2-2*sin(b*x+a)^2+2*cos(b*x+a)^2-4*cos(b*x+a)+2)*sin(b*x+a)/(c*sin(b*x+a)^2/(
2*cos(b*x+a)^2-1))^(1/2)/(sin(b*x+a)^4-6*cos(b*x+a)^2*sin(b*x+a)^2+cos(b*x+a)^4+12*cos(b*x+a)*sin(b*x+a)^2-4*c
os(b*x+a)^3-6*sin(b*x+a)^2+6*cos(b*x+a)^2-4*cos(b*x+a)+1)*4^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 245, normalized size of antiderivative = 2.78 \[ \int \frac {\sec ^2(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\left [\frac {\sqrt {2} \sqrt {c} \log \left (\frac {\tan \left (b x + a\right )^{3} - \frac {2 \, \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )}}{\sqrt {c}} - 2 \, \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{3}}\right ) \tan \left (b x + a\right ) + 4 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{4 \, b c \tan \left (b x + a\right )}, -\frac {\sqrt {2} c \sqrt {-\frac {1}{c}} \arctan \left (\frac {\sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-\frac {1}{c}}}{\tan \left (b x + a\right )}\right ) \tan \left (b x + a\right ) - 2 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{2 \, b c \tan \left (b x + a\right )}\right ] \]

[In]

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

[Out]

[1/4*(sqrt(2)*sqrt(c)*log((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*tan(b*x + a))/tan(b*x + a)^3)*tan(b*x + a) + 4*sqrt(2)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 -
1)))/(b*c*tan(b*x + a)), -1/2*(sqrt(2)*c*sqrt(-1/c)*arctan(sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*(tan(b
*x + a)^2 - 1)*sqrt(-1/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)))
/(b*c*tan(b*x + a))]

Sympy [F(-1)]

Timed out. \[ \int \frac {\sec ^2(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {\sec ^2(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\int { \frac {\sec \left (2 \, b x + 2 \, a\right )^{2}}{\sqrt {c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )}} \,d x } \]

[In]

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

[Out]

integrate(sec(2*b*x + 2*a)^2/sqrt(c*tan(2*b*x + 2*a)*tan(b*x + a)), x)

Giac [F(-1)]

Timed out. \[ \int \frac {\sec ^2(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^2(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\int \frac {1}{{\cos \left (2\,a+2\,b\,x\right )}^2\,\sqrt {c\,\mathrm {tan}\left (a+b\,x\right )\,\mathrm {tan}\left (2\,a+2\,b\,x\right )}} \,d x \]

[In]

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

[Out]

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