\(\int \frac {\sec ^3(2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx\) [627]

Optimal result

Integrand size = 31, antiderivative size = 128 \[ \int \frac {\sec ^3(2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=-\frac {7 \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {-c+c \sec (2 a+2 b x)}}\right )}{4 \sqrt {2} b c^{3/2}}-\frac {\tan (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}+\frac {\tan (2 a+2 b x)}{b c \sqrt {-c+c \sec (2 a+2 b x)}} \]

[Out]

-7/8*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/c^(3/2)*2^(1/2)-1/4*tan(2*b*x+2
*a)/b/(-c+c*sec(2*b*x+2*a))^(3/2)+tan(2*b*x+2*a)/b/c/(-c+c*sec(2*b*x+2*a))^(1/2)

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {4482, 3884, 4086, 3880, 213} \[ \int \frac {\sec ^3(2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=-\frac {7 \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {c \sec (2 a+2 b x)-c}}\right )}{4 \sqrt {2} b c^{3/2}}+\frac {\tan (2 a+2 b x)}{b c \sqrt {c \sec (2 a+2 b x)-c}}-\frac {\tan (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}} \]

[In]

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

[Out]

(-7*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)) + Tan[2*a + 2*b*x]/(b*c*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 3884

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[b*Cot[e + f*x]*((
a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x] - Dist[1/(a^2*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m
+ 1)*(a*m - b*(2*m + 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)
]

Rule 4086

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> Simp[(-B)*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Dist[(a*B*m + A*b*(m + 1))/(b
*(m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B, e, f, m}, x] && NeQ[A*b - a*B
, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 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 ^3(2 a+2 b x)}{(-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 {\sec (2 a+2 b x) \left (\frac {3 c}{2}+2 c \sec (2 a+2 b x)\right )}{\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 {\tan (2 a+2 b x)}{b c \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {7 \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 {\tan (2 a+2 b x)}{b c \sqrt {-c+c \sec (2 a+2 b x)}}-\frac {7 \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 )}{4 b c} \\ & = -\frac {7 \text {arctanh}\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}}+\frac {\tan (2 a+2 b x)}{b c \sqrt {-c+c \sec (2 a+2 b x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.66 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.73 \[ \int \frac {\sec ^3(2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\frac {\left (-5+4 \sec (2 (a+b x))+7 \arctan \left (\sqrt {-1+\tan ^2(a+b x)}\right ) \sec (2 (a+b x)) \sin ^2(a+b x) \sqrt {-1+\tan ^2(a+b x)}\right ) \tan (2 (a+b x))}{4 b (c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \]

[In]

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

[Out]

((-5 + 4*Sec[2*(a + b*x)] + 7*ArcTan[Sqrt[-1 + Tan[a + b*x]^2]]*Sec[2*(a + b*x)]*Sin[a + b*x]^2*Sqrt[-1 + Tan[
a + b*x]^2])*Tan[2*(a + b*x)])/(4*b*(c*Tan[a + b*x]*Tan[2*(a + b*x)])^(3/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(494\) vs. \(2(111)=222\).

Time = 3.91 (sec) , antiderivative size = 495, normalized size of antiderivative = 3.87

method	result	size
default	\(\frac {\sqrt {2}\, \csc \left (x b +a \right ) \left (1-\cos \left (x b +a \right )\right ) \left (\csc \left (x b +a \right )^{6} \left (1-\cos \left (x b +a \right )\right )^{6}+14 \csc \left (x b +a \right )^{2} \ln \left (\csc \left (x b +a \right )^{2} \left (1-\cos \left (x b +a \right )\right )^{2}-3+\sqrt {\csc \left (x b +a \right )^{4} \left (1-\cos \left (x b +a \right )\right )^{4}-6 \csc \left (x b +a \right )^{2} \left (1-\cos \left (x b +a \right )\right )^{2}+1}\right ) \left (1-\cos \left (x b +a \right )\right )^{2} \sqrt {\csc \left (x b +a \right )^{4} \left (1-\cos \left (x b +a \right )\right )^{4}-6 \csc \left (x b +a \right )^{2} \left (1-\cos \left (x b +a \right )\right )^{2}+1}+14 \csc \left (x b +a \right )^{2} \operatorname {arctanh}\left (\frac {3 \csc \left (x b +a \right )^{2} \left (1-\cos \left (x b +a \right )\right )^{2}-1}{\sqrt {\csc \left (x b +a \right )^{4} \left (1-\cos \left (x b +a \right )\right )^{4}-6 \csc \left (x b +a \right )^{2} \left (1-\cos \left (x b +a \right )\right )^{2}+1}}\right ) \left (1-\cos \left (x b +a \right )\right )^{2} \sqrt {\csc \left (x b +a \right )^{4} \left (1-\cos \left (x b +a \right )\right )^{4}-6 \csc \left (x b +a \right )^{2} \left (1-\cos \left (x b +a \right )\right )^{2}+1}-39 \csc \left (x b +a \right )^{4} \left (1-\cos \left (x b +a \right )\right )^{4}+39 \csc \left (x b +a \right )^{2} \left (1-\cos \left (x b +a \right )\right )^{2}-1\right ) \sqrt {4}}{64 b \left (\frac {\csc \left (x b +a \right )^{2} \left (1-\cos \left (x b +a \right )\right )^{2} c}{\csc \left (x b +a \right )^{4} \left (1-\cos \left (x b +a \right )\right )^{4}-6 \csc \left (x b +a \right )^{2} \left (1-\cos \left (x b +a \right )\right )^{2}+1}\right )^{\frac {3}{2}} \left (\csc \left (x b +a \right )^{4} \left (1-\cos \left (x b +a \right )\right )^{4}-6 \csc \left (x b +a \right )^{2} \left (1-\cos \left (x b +a \right )\right )^{2}+1\right )^{2}}\)	\(495\)

[In]

int(sec(2*b*x+2*a)^3/(c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/64*2^(1/2)/b*csc(b*x+a)/(csc(b*x+a)^2*(1-cos(b*x+a))^2*c/(csc(b*x+a)^4*(1-cos(b*x+a))^4-6*csc(b*x+a)^2*(1-co
s(b*x+a))^2+1))^(3/2)*(1-cos(b*x+a))*(csc(b*x+a)^6*(1-cos(b*x+a))^6+14*csc(b*x+a)^2*ln(csc(b*x+a)^2*(1-cos(b*x
+a))^2-3+(csc(b*x+a)^4*(1-cos(b*x+a))^4-6*csc(b*x+a)^2*(1-cos(b*x+a))^2+1)^(1/2))*(1-cos(b*x+a))^2*(csc(b*x+a)
^4*(1-cos(b*x+a))^4-6*csc(b*x+a)^2*(1-cos(b*x+a))^2+1)^(1/2)+14*csc(b*x+a)^2*arctanh((3*csc(b*x+a)^2*(1-cos(b*
x+a))^2-1)/(csc(b*x+a)^4*(1-cos(b*x+a))^4-6*csc(b*x+a)^2*(1-cos(b*x+a))^2+1)^(1/2))*(1-cos(b*x+a))^2*(csc(b*x+
a)^4*(1-cos(b*x+a))^4-6*csc(b*x+a)^2*(1-cos(b*x+a))^2+1)^(1/2)-39*csc(b*x+a)^4*(1-cos(b*x+a))^4+39*csc(b*x+a)^
2*(1-cos(b*x+a))^2-1)/(csc(b*x+a)^4*(1-cos(b*x+a))^4-6*csc(b*x+a)^2*(1-cos(b*x+a))^2+1)^2*4^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 276, normalized size of antiderivative = 2.16 \[ \int \frac {\sec ^3(2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\left [\frac {7 \, \sqrt {2} \sqrt {c} \log \left (\frac {c \tan \left (b x + a\right )^{3} - 2 \, \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {c} - 2 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{3}}\right ) \tan \left (b x + a\right )^{3} + 2 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (9 \, \tan \left (b x + a\right )^{2} - 1\right )}}{16 \, b c^{2} \tan \left (b x + a\right )^{3}}, -\frac {7 \, \sqrt {2} \sqrt {-c} \arctan \left (\frac {\sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-c}}{c \tan \left (b x + a\right )}\right ) \tan \left (b x + a\right )^{3} - \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (9 \, \tan \left (b x + a\right )^{2} - 1\right )}}{8 \, b c^{2} \tan \left (b x + a\right )^{3}}\right ] \]

[In]

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

[Out]

[1/16*(7*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 + 2*sqrt(2)*sqrt(-c*tan(b*x + a)^2/(tan(b*x
+ a)^2 - 1))*(9*tan(b*x + a)^2 - 1))/(b*c^2*tan(b*x + a)^3), -1/8*(7*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 - sqrt(2)*sqrt(-c*t
an(b*x + a)^2/(tan(b*x + a)^2 - 1))*(9*tan(b*x + a)^2 - 1))/(b*c^2*tan(b*x + a)^3)]

Sympy [F(-1)]

Timed out. \[ \int \frac {\sec ^3(2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {\sec ^3(2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\int { \frac {\sec \left (2 \, b x + 2 \, a\right )^{3}}{\left (c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )\right )^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

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

Giac [F(-1)]

Timed out. \[ \int \frac {\sec ^3(2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^3(2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\int \frac {1}{{\cos \left (2\,a+2\,b\,x\right )}^3\,{\left (c\,\mathrm {tan}\left (a+b\,x\right )\,\mathrm {tan}\left (2\,a+2\,b\,x\right )\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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