\(\int (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx\) [615]
Optimal result
Integrand size = 20, antiderivative size = 80 \[
\int (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\frac {c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}}\right )}{b}+\frac {c^2 \tan (2 a+2 b x)}{b \sqrt {-c+c \sec (2 a+2 b x)}}
\]
[Out]
c^(3/2)*arctanh(c^(1/2)*tan(2*b*x+2*a)/(-c+c*sec(2*b*x+2*a))^(1/2))/b+c^2*tan(2*b*x+2*a)/b/(-c+c*sec(2*b*x+2*a
))^(1/2)
Rubi [A] (verified)
Time = 0.07 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of
steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4482, 3860, 21, 3859, 213}
\[
\int (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\frac {c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{b}+\frac {c^2 \tan (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}
\]
[In]
Int[(c*Tan[a + b*x]*Tan[2*(a + b*x)])^(3/2),x]
[Out]
(c^(3/2)*ArcTanh[(Sqrt[c]*Tan[2*a + 2*b*x])/Sqrt[-c + c*Sec[2*a + 2*b*x]]])/b + (c^2*Tan[2*a + 2*b*x])/(b*Sqrt
[-c + c*Sec[2*a + 2*b*x]])
Rule 21
Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + 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 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 3860
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[(-b^2)*Cot[c + d*x]*((a + b*Csc[c + d*x])^(
n - 2)/(d*(n - 1))), x] + Dist[a/(n - 1), Int[(a + b*Csc[c + d*x])^(n - 2)*(a*(n - 1) + b*(3*n - 4)*Csc[c + d*
x]), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]
Rule 4482
Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]
Rubi steps \begin{align*}
\text {integral}& = \int (-c+c \sec (2 a+2 b x))^{3/2} \, dx \\ & = \frac {c^2 \tan (2 a+2 b x)}{b \sqrt {-c+c \sec (2 a+2 b x)}}-(2 c) \int \frac {-\frac {c}{2}+\frac {1}{2} c \sec (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}} \, dx \\ & = \frac {c^2 \tan (2 a+2 b x)}{b \sqrt {-c+c \sec (2 a+2 b x)}}-c \int \sqrt {-c+c \sec (2 a+2 b x)} \, dx \\ & = \frac {c^2 \tan (2 a+2 b x)}{b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {c^2 \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} \\ & = \frac {c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}}\right )}{b}+\frac {c^2 \tan (2 a+2 b x)}{b \sqrt {-c+c \sec (2 a+2 b x)}} \\
\end{align*}
Mathematica [A] (verified)
Time = 0.32 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.08
\[
\int (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\frac {c \left (2 \cot (a+b x)+\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \cos (a+b x)}{\sqrt {\cos (2 (a+b x))}}\right ) \sqrt {\cos (2 (a+b x))} \csc (a+b x)\right ) \sqrt {c \tan (a+b x) \tan (2 (a+b x))}}{2 b}
\]
[In]
Integrate[(c*Tan[a + b*x]*Tan[2*(a + b*x)])^(3/2),x]
[Out]
(c*(2*Cot[a + b*x] + Sqrt[2]*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)]])/(2*b)
Maple [B] (verified)
Leaf count of result is larger than twice the leaf count of optimal. \(223\) vs. \(2(72)=144\).
Time = 17.21 (sec) , antiderivative size = 224, normalized size of antiderivative =
2.80
| | |
| method | result | size |
| | |
| default |
\(-\frac {\sqrt {2}\, c \sqrt {\frac {c \sin \left (x b +a \right )^{2}}{2 \cos \left (x b +a \right )^{2}-1}}\, \left (\cot \left (x b +a \right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}\, \operatorname {arctanh}\left (\frac {\cos \left (x b +a \right ) \sqrt {2}}{\left (1+\cos \left (x b +a \right )\right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}}\right ) \sqrt {2}+\csc \left (x b +a \right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}\, \operatorname {arctanh}\left (\frac {\cos \left (x b +a \right ) \sqrt {2}}{\left (1+\cos \left (x b +a \right )\right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}}\right ) \sqrt {2}+2 \cot \left (x b +a \right )\right ) \sqrt {4}}{2 b \left (2+\sqrt {2}\right ) \left (-2+\sqrt {2}\right )}\) |
\(224\) |
| | |
|
|
|
|
[In]
int((c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x,method=_RETURNVERBOSE)
[Out]
-1/2*2^(1/2)/b*c*(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(1/2)*(cot(b*x+a)*((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(
1/2)*arctanh(cos(b*x+a)/(1+cos(b*x+a))/((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)*2^(1/2))*2^(1/2)+csc(b*x+a)
*((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)*arctanh(cos(b*x+a)/(1+cos(b*x+a))/((2*cos(b*x+a)^2-1)/(1+cos(b*x+
a))^2)^(1/2)*2^(1/2))*2^(1/2)+2*cot(b*x+a))*4^(1/2)/(2+2^(1/2))/(-2+2^(1/2))
Fricas [A] (verification not implemented)
none
Time = 0.29 (sec) , antiderivative size = 296, normalized size of antiderivative = 3.70
\[
\int (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\left [\frac {c^{\frac {3}{2}} \log \left (-\frac {c \tan \left (b x + a\right )^{5} - 14 \, c \tan \left (b x + a\right )^{3} + 4 \, \sqrt {2} {\left (\tan \left (b x + a\right )^{4} - 4 \, \tan \left (b x + a\right )^{2} + 3\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} \sqrt {c} + 17 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{5} + 2 \, \tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )}\right ) \tan \left (b x + a\right ) + 4 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} c}{4 \, b \tan \left (b x + a\right )}, -\frac {\sqrt {-c} c \arctan \left (\frac {2 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-c}}{c \tan \left (b x + a\right )^{3} - 3 \, c \tan \left (b x + a\right )}\right ) \tan \left (b x + a\right ) - 2 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} c}{2 \, b \tan \left (b x + a\right )}\right ]
\]
[In]
integrate((c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x, algorithm="fricas")
[Out]
[1/4*(c^(3/2)*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)))*tan(b*x + a) + 4*sqrt(2)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*c)/(b*tan(b*x + a)), -1
/2*(sqrt(-c)*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)))*tan(b*x + a) - 2*sqrt(2)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*c)/(
b*tan(b*x + a))]
Sympy [F(-1)]
Timed out. \[
\int (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\text {Timed out}
\]
[In]
integrate((c*tan(b*x+a)*tan(2*b*x+2*a))**(3/2),x)
[Out]
Timed out
Maxima [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 1317 vs. \(2 (72) = 144\).
Time = 0.47 (sec) , antiderivative size = 1317, normalized size of antiderivative = 16.46
\[
\int (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\text {Too large to display}
\]
[In]
integrate((c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x, algorithm="maxima")
[Out]
-1/8*((cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)^(1/4)*(c*log(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 + s
qrt(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 + 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)*sin(1/2*arcta
n2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)) + 1) - c*log(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 + 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 - 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)*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*
b*x + 4*a) - 1)) + 1) + c*log(((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)*cos(1/2*arctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a)))^2 + (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)*sin(1/2*arctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*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) + 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(1/2*a
rctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a)))*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)) + cos(1/
2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))*sin(1/2*arctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a)))) + 1)
- c*log(((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), -co
s(4*b*x + 4*a) - 1))^2)*cos(1/2*arctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a)))^2 + (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)*sin(1/2*arctan2
(sin(4*b*x + 4*a), cos(4*b*x + 4*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
) - 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(1/2*arctan2(sin(4*b*x + 4*
a), cos(4*b*x + 4*a)))*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)) + cos(1/2*arctan2(sin(4*b*x +
4*a), -cos(4*b*x + 4*a) - 1))*sin(1/2*arctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a)))) + 1))*sqrt(c) + 8*(c*cos(
1/2*arctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a)))*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)) + c
*cos(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))*sin(1/2*arctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a))
) + c*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)))*sqrt(c))/((cos(4*b*x + 4*a)^2 + sin(4*b*x + 4
*a)^2 + 2*cos(4*b*x + 4*a) + 1)^(1/4)*b)
Giac [F(-1)]
Timed out. \[
\int (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\text {Timed out}
\]
[In]
integrate((c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x, algorithm="giac")
[Out]
Timed out
Mupad [F(-1)]
Timed out. \[
\int (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\int {\left (c\,\mathrm {tan}\left (a+b\,x\right )\,\mathrm {tan}\left (2\,a+2\,b\,x\right )\right )}^{3/2} \,d x
\]
[In]
int((c*tan(a + b*x)*tan(2*a + 2*b*x))^(3/2),x)
[Out]
int((c*tan(a + b*x)*tan(2*a + 2*b*x))^(3/2), x)