\(\int \frac {1}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx\) [630]

Optimal result

Integrand size = 20, antiderivative size = 138 \[ \int \frac {1}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}}\right )}{b c^{3/2}}+\frac {5 \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {-c+c \sec (2 a+2 b x)}}\right )}{4 \sqrt {2} b c^{3/2}}-\frac {\tan (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}} \]

[Out]

-arctanh(c^(1/2)*tan(2*b*x+2*a)/(-c+c*sec(2*b*x+2*a))^(1/2))/b/c^(3/2)+5/8*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)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {4482, 3862, 4005, 3859, 213, 3880} \[ \int \frac {1}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{b c^{3/2}}+\frac {5 \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {c \sec (2 a+2 b x)-c}}\right )}{4 \sqrt {2} b c^{3/2}}-\frac {\tan (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}} \]

[In]

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

[Out]

-(ArcTanh[(Sqrt[c]*Tan[2*a + 2*b*x])/Sqrt[-c + c*Sec[2*a + 2*b*x]]]/(b*c^(3/2))) + (5*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))

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 3859

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 3862

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[(-Cot[c + d*x])*((a + b*Csc[c + d*x])^n/(d*
(2*n + 1))), x] + Dist[1/(a^2*(2*n + 1)), Int[(a + b*Csc[c + d*x])^(n + 1)*(a*(2*n + 1) - b*(n + 1)*Csc[c + d*
x]), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && LeQ[n, -1] && IntegerQ[2*n]

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 4005

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c/a,
Int[Sqrt[a + b*Csc[e + f*x]], x], x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]

Rule 4482

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{(-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 {2 c+\frac {1}{2} c \sec (2 a+2 b x)}{\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 {\int \sqrt {-c+c \sec (2 a+2 b x)} \, dx}{c^2}-\frac {5 \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 {\text {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 c}+\frac {5 \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 {\text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}}\right )}{b c^{3/2}}+\frac {5 \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}} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.20 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.43 \[ \int \frac {1}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=-\frac {\cot (a+b x) \left (-4 \text {arctanh}\left (\sqrt {1-\tan ^2(a+b x)}\right ) \cos (2 (a+b x)) \sec ^2(a+b x)+4 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1-\tan ^2(a+b x)}}{\sqrt {2}}\right ) \cos (2 (a+b x)) \sec ^2(a+b x)+\cot ^2(a+b x) \left (\cos (2 (a+b x)) \sec ^2(a+b x)\right )^{3/2}+\arctan \left (\sqrt {-1+\tan ^2(a+b x)}\right ) \sqrt {-\left (-1+\tan ^2(a+b x)\right )^2}\right ) \sqrt {c \tan (a+b x) \tan (2 (a+b x))}}{8 b c^2 \sqrt {1-\tan ^2(a+b x)}} \]

[In]

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

[Out]

-1/8*(Cot[a + b*x]*(-4*ArcTanh[Sqrt[1 - Tan[a + b*x]^2]]*Cos[2*(a + b*x)]*Sec[a + b*x]^2 + 4*Sqrt[2]*ArcTanh[S
qrt[1 - Tan[a + b*x]^2]/Sqrt[2]]*Cos[2*(a + b*x)]*Sec[a + b*x]^2 + Cot[a + b*x]^2*(Cos[2*(a + b*x)]*Sec[a + b*
x]^2)^(3/2) + ArcTan[Sqrt[-1 + Tan[a + b*x]^2]]*Sqrt[-(-1 + Tan[a + b*x]^2)^2])*Sqrt[c*Tan[a + b*x]*Tan[2*(a +
b*x)]])/(b*c^2*Sqrt[1 - Tan[a + b*x]^2])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(551\) vs. \(2(117)=234\).

Time = 4.37 (sec) , antiderivative size = 552, normalized size of antiderivative = 4.00

method	result	size
default	\(\frac {\sqrt {2}\, \csc \left (x b +a \right ) \left (1-\cos \left (x b +a \right )\right ) \left (16 \csc \left (x b +a \right )^{2} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (\csc \left (x b +a \right )^{2} \left (1-\cos \left (x b +a \right )\right )^{2}-1\right ) \sqrt {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}}\right ) \left (1-\cos \left (x b +a \right )\right )^{2}-10 \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}-10 \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}+\csc \left (x b +a \right )^{2} \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}-\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 ) \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 )^{\frac {3}{2}}}\)	\(552\)

[In]

int(1/(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))*(16*csc(b*x+a)^2*2^(1/2)*arctanh((csc(b*x+a)^2*(1-cos(b*x+a))^2-1)*2^(1/2
)/(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))*(1-cos(b*x+a))^2-10*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-c
os(b*x+a))^2-10*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)^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)-(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))/(csc(b*x+a)^4*(1-cos(b*x+a))^4-6*csc(b*x+a)^2*(1-cos(b*x+a))^2+1)^(3/2)*4^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 438, normalized size of antiderivative = 3.17 \[ \int \frac {1}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\left [\frac {5 \, \sqrt {2} \sqrt {c} \log \left (\frac {c \tan \left (b x + a\right )^{3} + 2 \, \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {c} - 2 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{3}}\right ) \tan \left (b x + a\right )^{3} + 8 \, \sqrt {c} \log \left (\frac {c \tan \left (b x + a\right )^{3} - 2 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {c} - 3 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )}\right ) \tan \left (b x + a\right )^{3} + 2 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )}}{16 \, b c^{2} \tan \left (b x + a\right )^{3}}, \frac {5 \, \sqrt {2} \sqrt {-c} \arctan \left (\frac {\sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-c}}{c \tan \left (b x + a\right )}\right ) \tan \left (b x + a\right )^{3} - 8 \, \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-c}}{2 \, c \tan \left (b x + a\right )}\right ) \tan \left (b x + a\right )^{3} + \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )}}{8 \, b c^{2} \tan \left (b x + a\right )^{3}}\right ] \]

[In]

integrate(1/(c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x, algorithm="fricas")

[Out]

[1/16*(5*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 + 8*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)))*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
))/(b*c^2*tan(b*x + a)^3), 1/8*(5*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 - 8*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)))*tan(b*x + a)^3 + sqrt(2)*sqrt(-c*tan(b*
x + a)^2/(tan(b*x + a)^2 - 1))*(tan(b*x + a)^2 - 1))/(b*c^2*tan(b*x + a)^3)]

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(1/(c*tan(b*x+a)*tan(2*b*x+2*a))**(3/2),x)

[Out]

Timed out

Maxima [F]

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

[In]

integrate(1/(c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x, algorithm="maxima")

[Out]

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

Giac [F(-1)]

Timed out. \[ \int \frac {1}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(1/(c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

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

[In]

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

[Out]

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