Integrand size = 31, antiderivative size = 148 \[ \int \sec ^3(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=-\frac {76 c^2 \tan (2 a+2 b x)}{105 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {19 c \sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{105 b}+\frac {2 (-c+c \sec (2 a+2 b x))^{3/2} \tan (2 a+2 b x)}{35 b}+\frac {(-c+c \sec (2 a+2 b x))^{5/2} \tan (2 a+2 b x)}{7 b c} \] Output:
-76/105*c^2*tan(2*b*x+2*a)/b/(-c+c*sec(2*b*x+2*a))^(1/2)+19/105*c*(-c+c*se c(2*b*x+2*a))^(1/2)*tan(2*b*x+2*a)/b+2/35*(-c+c*sec(2*b*x+2*a))^(3/2)*tan( 2*b*x+2*a)/b+1/7*(-c+c*sec(2*b*x+2*a))^(5/2)*tan(2*b*x+2*a)/b/c
Time = 0.70 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.49 \[ \int \sec ^3(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=-\frac {\cot (a+b x) \left (-28+76 \cot (a+b x) \cot (2 (a+b x))+24 \sec (2 (a+b x))-15 \sec ^2(2 (a+b x))\right ) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2}}{105 b} \] Input:
Integrate[Sec[2*(a + b*x)]^3*(c*Tan[a + b*x]*Tan[2*(a + b*x)])^(3/2),x]
Output:
-1/105*(Cot[a + b*x]*(-28 + 76*Cot[a + b*x]*Cot[2*(a + b*x)] + 24*Sec[2*(a + b*x)] - 15*Sec[2*(a + b*x)]^2)*(c*Tan[a + b*x]*Tan[2*(a + b*x)])^(3/2)) /b
Time = 0.89 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.10, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {3042, 4897, 3042, 4287, 27, 3042, 4489, 3042, 4280, 3042, 4279}
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.
\(\displaystyle \int \sec ^3(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (2 a+2 b x)^3 (c \tan (a+b x) \tan (2 a+2 b x))^{3/2}dx\) |
\(\Big \downarrow \) 4897 |
\(\displaystyle \int \sec ^3(2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (2 a+2 b x+\frac {\pi }{2}\right )^3 \left (c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c\right )^{3/2}dx\) |
\(\Big \downarrow \) 4287 |
\(\displaystyle \frac {2 \int \frac {1}{2} \sec (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2} (2 \sec (2 a+2 b x) c+5 c)dx}{7 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{5/2}}{7 b c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \sec (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2} (2 \sec (2 a+2 b x) c+5 c)dx}{7 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{5/2}}{7 b c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \csc \left (2 a+2 b x+\frac {\pi }{2}\right ) \left (c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c\right )^{3/2} \left (2 \csc \left (2 a+2 b x+\frac {\pi }{2}\right ) c+5 c\right )dx}{7 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{5/2}}{7 b c}\) |
\(\Big \downarrow \) 4489 |
\(\displaystyle \frac {\frac {19}{5} c \int \sec (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}dx+\frac {2 c \tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b}}{7 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{5/2}}{7 b c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {19}{5} c \int \csc \left (2 a+2 b x+\frac {\pi }{2}\right ) \left (c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c\right )^{3/2}dx+\frac {2 c \tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b}}{7 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{5/2}}{7 b c}\) |
\(\Big \downarrow \) 4280 |
\(\displaystyle \frac {\frac {19}{5} c \left (\frac {c \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}-\frac {4}{3} c \int \sec (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}dx\right )+\frac {2 c \tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b}}{7 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{5/2}}{7 b c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {19}{5} c \left (\frac {c \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}-\frac {4}{3} c \int \csc \left (2 a+2 b x+\frac {\pi }{2}\right ) \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}dx\right )+\frac {2 c \tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b}}{7 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{5/2}}{7 b c}\) |
\(\Big \downarrow \) 4279 |
\(\displaystyle \frac {\frac {19}{5} c \left (\frac {c \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}-\frac {4 c^2 \tan (2 a+2 b x)}{3 b \sqrt {c \sec (2 a+2 b x)-c}}\right )+\frac {2 c \tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b}}{7 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{5/2}}{7 b c}\) |
Input:
Int[Sec[2*(a + b*x)]^3*(c*Tan[a + b*x]*Tan[2*(a + b*x)])^(3/2),x]
Output:
((-c + c*Sec[2*a + 2*b*x])^(5/2)*Tan[2*a + 2*b*x])/(7*b*c) + ((2*c*(-c + c *Sec[2*a + 2*b*x])^(3/2)*Tan[2*a + 2*b*x])/(5*b) + (19*c*((-4*c^2*Tan[2*a + 2*b*x])/(3*b*Sqrt[-c + c*Sec[2*a + 2*b*x]]) + (c*Sqrt[-c + c*Sec[2*a + 2 *b*x]]*Tan[2*a + 2*b*x])/(3*b)))/5)/(7*c)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2*b*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]])), x] /; Free Q[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_ Symbol] :> Simp[(-b)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m - 1)/(f*m)), x] + Simp[a*((2*m - 1)/m) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] && Intege rQ[2*m]
Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-Cot[e + f*x])*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2 ))), x] + Simp[1/(b*(m + 2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(b*( m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[(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)]
Time = 1.17 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.61
method | result | size |
default | \(-\frac {\sqrt {2}\, \sqrt {4}\, \left (416 \cos \left (b x +a \right )^{6}-728 \cos \left (b x +a \right )^{4}+455 \cos \left (b x +a \right )^{2}-105\right ) \sqrt {\frac {c \sin \left (b x +a \right )^{2}}{2 \cos \left (b x +a \right )^{2}-1}}\, c \cot \left (b x +a \right )}{105 b \left (2 \cos \left (b x +a \right )^{2}-1\right )^{3}}\) | \(91\) |
Input:
int(sec(2*b*x+2*a)^3*(c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x,method=_RETURNV ERBOSE)
Output:
-1/105*2^(1/2)/b*4^(1/2)*(416*cos(b*x+a)^6-728*cos(b*x+a)^4+455*cos(b*x+a) ^2-105)*(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(1/2)*c/(2*cos(b*x+a)^2-1)^3*c ot(b*x+a)
Time = 0.08 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.75 \[ \int \sec ^3(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=-\frac {2 \, \sqrt {2} {\left (105 \, c \tan \left (b x + a\right )^{6} - 140 \, c \tan \left (b x + a\right )^{4} + 133 \, c \tan \left (b x + a\right )^{2} - 38 \, c\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{105 \, {\left (b \tan \left (b x + a\right )^{7} - 3 \, b \tan \left (b x + a\right )^{5} + 3 \, b \tan \left (b x + a\right )^{3} - b \tan \left (b x + a\right )\right )}} \] Input:
integrate(sec(2*b*x+2*a)^3*(c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x, algorith m="fricas")
Output:
-2/105*sqrt(2)*(105*c*tan(b*x + a)^6 - 140*c*tan(b*x + a)^4 + 133*c*tan(b* x + a)^2 - 38*c)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))/(b*tan(b*x + a)^7 - 3*b*tan(b*x + a)^5 + 3*b*tan(b*x + a)^3 - b*tan(b*x + a))
Timed out. \[ \int \sec ^3(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\text {Timed out} \] Input:
integrate(sec(2*b*x+2*a)**3*(c*tan(b*x+a)*tan(2*b*x+2*a))**(3/2),x)
Output:
Timed out
\[ \int \sec ^3(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\int { \left (c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )\right )^{\frac {3}{2}} \sec \left (2 \, b x + 2 \, a\right )^{3} \,d x } \] Input:
integrate(sec(2*b*x+2*a)^3*(c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x, algorith m="maxima")
Output:
4/105*(210*(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)^(3/4)*(3*(b*c*cos(4*b*x + 4*a)^2 + b*c*sin(4*b*x + 4*a)^2 + 2*b*c*cos( 4*b*x + 4*a) + b*c)*integrate(-(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(16*b*x + 16*a)*cos(4*b*x + 4*a) + 3*c os(12*b*x + 12*a)*cos(4*b*x + 4*a) + 3*cos(8*b*x + 8*a)*cos(4*b*x + 4*a) + cos(4*b*x + 4*a)^2 + sin(16*b*x + 16*a)*sin(4*b*x + 4*a) + 3*sin(12*b*x + 12*a)*sin(4*b*x + 4*a) + 3*sin(8*b*x + 8*a)*sin(4*b*x + 4*a) + sin(4*b*x + 4*a)^2)*cos(3/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)) + (cos (4*b*x + 4*a)*sin(16*b*x + 16*a) + 3*cos(4*b*x + 4*a)*sin(12*b*x + 12*a) + 3*cos(4*b*x + 4*a)*sin(8*b*x + 8*a) - cos(16*b*x + 16*a)*sin(4*b*x + 4*a) - 3*cos(12*b*x + 12*a)*sin(4*b*x + 4*a) - 3*cos(8*b*x + 8*a)*sin(4*b*x + 4*a))*sin(3/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)))*cos(5/2*a rctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a))) + ((cos(4*b*x + 4*a)*sin(16*b* x + 16*a) + 3*cos(4*b*x + 4*a)*sin(12*b*x + 12*a) + 3*cos(4*b*x + 4*a)*sin (8*b*x + 8*a) - cos(16*b*x + 16*a)*sin(4*b*x + 4*a) - 3*cos(12*b*x + 12*a) *sin(4*b*x + 4*a) - 3*cos(8*b*x + 8*a)*sin(4*b*x + 4*a))*cos(3/2*arctan2(s in(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)) - (cos(16*b*x + 16*a)*cos(4*b*x + 4*a) + 3*cos(12*b*x + 12*a)*cos(4*b*x + 4*a) + 3*cos(8*b*x + 8*a)*cos(4*b *x + 4*a) + cos(4*b*x + 4*a)^2 + sin(16*b*x + 16*a)*sin(4*b*x + 4*a) + 3*s in(12*b*x + 12*a)*sin(4*b*x + 4*a) + 3*sin(8*b*x + 8*a)*sin(4*b*x + 4*a...
Leaf count of result is larger than twice the leaf count of optimal. 5862 vs. \(2 (132) = 264\).
Time = 8.67 (sec) , antiderivative size = 5862, normalized size of antiderivative = 39.61 \[ \int \sec ^3(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\text {Too large to display} \] Input:
integrate(sec(2*b*x+2*a)^3*(c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x, algorith m="giac")
Output:
2/105*((((((((38*sqrt(2)*c^4*sgn(tan(b*x)^2*tan(a)^2 - tan(b*x)^2 - 4*tan( b*x)*tan(a) - tan(a)^2 + 1)*sgn(tan(b*x) + tan(a))*tan(a)^23 + 171*sqrt(2) *c^4*sgn(tan(b*x)^2*tan(a)^2 - tan(b*x)^2 - 4*tan(b*x)*tan(a) - tan(a)^2 + 1)*sgn(tan(b*x) + tan(a))*tan(a)^21 + 140*sqrt(2)*c^4*sgn(tan(b*x)^2*tan( a)^2 - tan(b*x)^2 - 4*tan(b*x)*tan(a) - tan(a)^2 + 1)*sgn(tan(b*x) + tan(a ))*tan(a)^19 - 581*sqrt(2)*c^4*sgn(tan(b*x)^2*tan(a)^2 - tan(b*x)^2 - 4*ta n(b*x)*tan(a) - tan(a)^2 + 1)*sgn(tan(b*x) + tan(a))*tan(a)^17 - 1708*sqrt (2)*c^4*sgn(tan(b*x)^2*tan(a)^2 - tan(b*x)^2 - 4*tan(b*x)*tan(a) - tan(a)^ 2 + 1)*sgn(tan(b*x) + tan(a))*tan(a)^15 - 2282*sqrt(2)*c^4*sgn(tan(b*x)^2* tan(a)^2 - tan(b*x)^2 - 4*tan(b*x)*tan(a) - tan(a)^2 + 1)*sgn(tan(b*x) + t an(a))*tan(a)^13 - 2464*sqrt(2)*c^4*sgn(tan(b*x)^2*tan(a)^2 - tan(b*x)^2 - 4*tan(b*x)*tan(a) - tan(a)^2 + 1)*sgn(tan(b*x) + tan(a))*tan(a)^11 - 2930 *sqrt(2)*c^4*sgn(tan(b*x)^2*tan(a)^2 - tan(b*x)^2 - 4*tan(b*x)*tan(a) - ta n(a)^2 + 1)*sgn(tan(b*x) + tan(a))*tan(a)^9 - 2986*sqrt(2)*c^4*sgn(tan(b*x )^2*tan(a)^2 - tan(b*x)^2 - 4*tan(b*x)*tan(a) - tan(a)^2 + 1)*sgn(tan(b*x) + tan(a))*tan(a)^7 - 1953*sqrt(2)*c^4*sgn(tan(b*x)^2*tan(a)^2 - tan(b*x)^ 2 - 4*tan(b*x)*tan(a) - tan(a)^2 + 1)*sgn(tan(b*x) + tan(a))*tan(a)^5 - 70 0*sqrt(2)*c^4*sgn(tan(b*x)^2*tan(a)^2 - tan(b*x)^2 - 4*tan(b*x)*tan(a) - t an(a)^2 + 1)*sgn(tan(b*x) + tan(a))*tan(a)^3 - 105*sqrt(2)*c^4*sgn(tan(b*x )^2*tan(a)^2 - tan(b*x)^2 - 4*tan(b*x)*tan(a) - tan(a)^2 + 1)*sgn(tan(b...
Time = 23.55 (sec) , antiderivative size = 479, normalized size of antiderivative = 3.24 \[ \int \sec ^3(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx =\text {Too large to display} \] Input:
int((c*tan(a + b*x)*tan(2*a + 2*b*x))^(3/2)/cos(2*a + 2*b*x)^3,x)
Output:
(((c*8i)/(7*b) - (c*exp(a*2i + b*x*2i)*8i)/(7*b))*((c*(exp(a*2i + b*x*2i)* 1i - 1i)*(exp(a*4i + b*x*4i)*1i - 1i))/((exp(a*2i + b*x*2i) + 1)*(exp(a*4i + b*x*4i) + 1)))^(1/2))/((exp(a*2i + b*x*2i) - 1)*(exp(a*4i + b*x*4i) + 1 )^3) - (((c*4i)/(5*b) - (c*exp(a*2i + b*x*2i)*92i)/(35*b))*((c*(exp(a*2i + b*x*2i)*1i - 1i)*(exp(a*4i + b*x*4i)*1i - 1i))/((exp(a*2i + b*x*2i) + 1)* (exp(a*4i + b*x*4i) + 1)))^(1/2))/((exp(a*2i + b*x*2i) - 1)*(exp(a*4i + b* x*4i) + 1)^2) - (((c*4i)/(3*b) + (c*exp(a*2i + b*x*2i)*52i)/(105*b))*((c*( exp(a*2i + b*x*2i)*1i - 1i)*(exp(a*4i + b*x*4i)*1i - 1i))/((exp(a*2i + b*x *2i) + 1)*(exp(a*4i + b*x*4i) + 1)))^(1/2))/((exp(a*2i + b*x*2i) - 1)*(exp (a*4i + b*x*4i) + 1)) - (c*exp(a*2i + b*x*2i)*((c*(exp(a*2i + b*x*2i)*1i - 1i)*(exp(a*4i + b*x*4i)*1i - 1i))/((exp(a*2i + b*x*2i) + 1)*(exp(a*4i + b *x*4i) + 1)))^(1/2)*104i)/(105*b*(exp(a*2i + b*x*2i) - 1))
\[ \int \sec ^3(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\sqrt {c}\, \left (\int \sqrt {\tan \left (b x +a \right )}\, \sqrt {\tan \left (2 b x +2 a \right )}\, \sec \left (2 b x +2 a \right )^{3} \tan \left (2 b x +2 a \right ) \tan \left (b x +a \right )d x \right ) c \] Input:
int(sec(2*b*x+2*a)^3*(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))*sec(2*a + 2*b*x)**3* tan(2*a + 2*b*x)*tan(a + b*x),x)*c