Integrand size = 20, antiderivative size = 80 \[ \int (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\frac {c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}}\right )}{b}+\frac {c^2 \tan (2 a+2 b x)}{b \sqrt {-c+c \sec (2 a+2 b x)}} \] Output:
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)
Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.08 \[ \int (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\frac {c \left (2 \cot (a+b x)+\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \cos (a+b x)}{\sqrt {\cos (2 (a+b x))}}\right ) \sqrt {\cos (2 (a+b x))} \csc (a+b x)\right ) \sqrt {c \tan (a+b x) \tan (2 (a+b x))}}{2 b} \] Input:
Integrate[(c*Tan[a + b*x]*Tan[2*(a + b*x)])^(3/2),x]
Output:
(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)
Time = 0.38 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4897, 3042, 4262, 27, 3042, 4261, 220}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (c \tan (a+b x) \tan (2 a+2 b x))^{3/2}dx\) |
\(\Big \downarrow \) 4897 |
\(\displaystyle \int (c \sec (2 a+2 b x)-c)^{3/2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c\right )^{3/2}dx\) |
\(\Big \downarrow \) 4262 |
\(\displaystyle \frac {c^2 \tan (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}-2 c \int \frac {1}{2} \sqrt {c \sec (2 a+2 b x)-c}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {c^2 \tan (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}-c \int \sqrt {c \sec (2 a+2 b x)-c}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {c^2 \tan (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}-c \int \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}dx\) |
\(\Big \downarrow \) 4261 |
\(\displaystyle \frac {c^2 \int \frac {1}{\frac {c^2 \tan ^2(2 a+2 b x)}{c \sec (2 a+2 b x)-c}-c}d\left (-\frac {c \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{b}+\frac {c^2 \tan (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{b}+\frac {c^2 \tan (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}\) |
Input:
Int[(c*Tan[a + b*x]*Tan[2*(a + b*x)])^(3/2),x]
Output:
(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]])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(b/d) Subst[Int[1/(a + x^2), x], x, b*(Cot[c + d*x]/Sqrt[a + b*Csc[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[(-b^2)*C ot[c + d*x]*((a + b*Csc[c + d*x])^(n - 2)/(d*(n - 1))), x] + Simp[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] && Inte gerQ[2*n]
Leaf count of result is larger than twice the leaf count of optimal. \(223\) vs. \(2(72)=144\).
Time = 1.58 (sec) , antiderivative size = 224, normalized size of antiderivative = 2.80
method | result | size |
default | \(-\frac {\sqrt {2}\, \sqrt {4}\, c \sqrt {\frac {c \sin \left (b x +a \right )^{2}}{2 \cos \left (b x +a \right )^{2}-1}}\, \left (\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right ) \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}}\right ) \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \cot \left (b x +a \right )+\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right ) \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}}\right ) \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \csc \left (b x +a \right )+2 \cot \left (b x +a \right )\right )}{2 b \left (2+\sqrt {2}\right ) \left (-2+\sqrt {2}\right )}\) | \(224\) |
Input:
int((c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/2*2^(1/2)/b*4^(1/2)/(2+2^(1/2))/(-2+2^(1/2))*c*(c*sin(b*x+a)^2/(2*cos(b *x+a)^2-1))^(1/2)*(2^(1/2)*arctanh(2^(1/2)*cos(b*x+a)/(cos(b*x+a)+1)/((2*c os(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2))*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1) ^2)^(1/2)*cot(b*x+a)+2^(1/2)*arctanh(2^(1/2)*cos(b*x+a)/(cos(b*x+a)+1)/((2 *cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2))*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+ 1)^2)^(1/2)*csc(b*x+a)+2*cot(b*x+a))
Time = 0.11 (sec) , antiderivative size = 296, normalized size of antiderivative = 3.70 \[ \int (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\left [\frac {c^{\frac {3}{2}} \log \left (-\frac {c \tan \left (b x + a\right )^{5} - 14 \, c \tan \left (b x + a\right )^{3} + 4 \, \sqrt {2} {\left (\tan \left (b x + a\right )^{4} - 4 \, \tan \left (b x + a\right )^{2} + 3\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} \sqrt {c} + 17 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{5} + 2 \, \tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )}\right ) \tan \left (b x + a\right ) + 4 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} c}{4 \, b \tan \left (b x + a\right )}, -\frac {\sqrt {-c} c \arctan \left (\frac {2 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-c}}{c \tan \left (b x + a\right )^{3} - 3 \, c \tan \left (b x + a\right )}\right ) \tan \left (b x + a\right ) - 2 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} c}{2 \, b \tan \left (b x + a\right )}\right ] \] Input:
integrate((c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x, algorithm="fricas")
Output:
[1/4*(c^(3/2)*log(-(c*tan(b*x + a)^5 - 14*c*tan(b*x + a)^3 + 4*sqrt(2)*(ta n(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))]
Timed out. \[ \int (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\text {Timed out} \] Input:
integrate((c*tan(b*x+a)*tan(2*b*x+2*a))**(3/2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 1317 vs. \(2 (72) = 144\).
Time = 0.25 (sec) , antiderivative size = 1317, normalized size of antiderivative = 16.46 \[ \int (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\text {Too large to display} \] Input:
integrate((c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x, algorithm="maxima")
Output:
-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 + sqr t(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(s in(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)*c os(1/2*arctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a)))^2 + (cos(1/2*arctan2(s in(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...
Exception generated. \[ \int (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\int {\left (c\,\mathrm {tan}\left (a+b\,x\right )\,\mathrm {tan}\left (2\,a+2\,b\,x\right )\right )}^{3/2} \,d x \] Input:
int((c*tan(a + b*x)*tan(2*a + 2*b*x))^(3/2),x)
Output:
int((c*tan(a + b*x)*tan(2*a + 2*b*x))^(3/2), x)
\[ \int (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\sqrt {c}\, \left (\int \sqrt {\tan \left (b x +a \right )}\, \sqrt {\tan \left (2 b x +2 a \right )}\, \tan \left (2 b x +2 a \right ) \tan \left (b x +a \right )d x \right ) c \] Input:
int((c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x)
Output:
sqrt(c)*int(sqrt(tan(a + b*x))*sqrt(tan(2*a + 2*b*x))*tan(2*a + 2*b*x)*tan (a + b*x),x)*c