Integrand size = 31, antiderivative size = 180 \[ \int \frac {\sec ^4(2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=-\frac {11 \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {-c+c \sec (2 a+2 b x)}}\right )}{4 \sqrt {2} b c^{3/2}}-\frac {\sec ^2(2 a+2 b x) \tan (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}+\frac {13 \tan (2 a+2 b x)}{6 b c \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {7 \sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{12 b c^2} \]
-11/8*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*sec(2*b*x+2*a)^2*tan(2*b*x+2*a)/b/(-c+c*sec(2*b* x+2*a))^(3/2)+13/6*tan(2*b*x+2*a)/b/c/(-c+c*sec(2*b*x+2*a))^(1/2)+7/12*(-c +c*sec(2*b*x+2*a))^(1/2)*tan(2*b*x+2*a)/b/c^2
Time = 2.97 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.56 \[ \int \frac {\sec ^4(2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=-\frac {\cot (a+b x) \left (\csc ^2(a+b x) (-24+(11+19 \cos (4 (a+b x))) \sec (2 (a+b x)))-66 \arctan \left (\sqrt {-1+\tan ^2(a+b x)}\right ) \sqrt {-1+\tan ^2(a+b x)}\right ) \sqrt {c \tan (a+b x) \tan (2 (a+b x))}}{48 b c^2} \]
-1/48*(Cot[a + b*x]*(Csc[a + b*x]^2*(-24 + (11 + 19*Cos[4*(a + b*x)])*Sec[ 2*(a + b*x)]) - 66*ArcTan[Sqrt[-1 + Tan[a + b*x]^2]]*Sqrt[-1 + Tan[a + b*x ]^2])*Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]])/(b*c^2)
Time = 1.03 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.419, Rules used = {3042, 4897, 3042, 4303, 27, 3042, 4498, 27, 3042, 4489, 3042, 4282, 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.
| \(\displaystyle \int \frac {\sec ^4(2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sec (2 a+2 b x)^4}{(c \tan (a+b x) \tan (2 a+2 b x))^{3/2}}dx\) |
| \(\Big \downarrow \) 4897 |
| \(\displaystyle \int \frac {\sec ^4(2 a+2 b x)}{(c \sec (2 a+2 b x)-c)^{3/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (2 a+2 b x+\frac {\pi }{2}\right )^4}{\left (c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4303 |
| \(\displaystyle \frac {\int \frac {\sec ^2(2 a+2 b x) (7 \sec (2 a+2 b x) c+4 c)}{2 \sqrt {c \sec (2 a+2 b x)-c}}dx}{2 c^2}-\frac {\tan (2 a+2 b x) \sec ^2(2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sec ^2(2 a+2 b x) (7 \sec (2 a+2 b x) c+4 c)}{\sqrt {c \sec (2 a+2 b x)-c}}dx}{4 c^2}-\frac {\tan (2 a+2 b x) \sec ^2(2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (2 a+2 b x+\frac {\pi }{2}\right )^2 \left (7 \csc \left (2 a+2 b x+\frac {\pi }{2}\right ) c+4 c\right )}{\sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}}dx}{4 c^2}-\frac {\tan (2 a+2 b x) \sec ^2(2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\) |
| \(\Big \downarrow \) 4498 |
| \(\displaystyle \frac {\frac {2 \int \frac {\sec (2 a+2 b x) \left (26 \sec (2 a+2 b x) c^2+7 c^2\right )}{2 \sqrt {c \sec (2 a+2 b x)-c}}dx}{3 c}+\frac {7 \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}}{4 c^2}-\frac {\tan (2 a+2 b x) \sec ^2(2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {\sec (2 a+2 b x) \left (26 \sec (2 a+2 b x) c^2+7 c^2\right )}{\sqrt {c \sec (2 a+2 b x)-c}}dx}{3 c}+\frac {7 \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}}{4 c^2}-\frac {\tan (2 a+2 b x) \sec ^2(2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\csc \left (2 a+2 b x+\frac {\pi }{2}\right ) \left (26 \csc \left (2 a+2 b x+\frac {\pi }{2}\right ) c^2+7 c^2\right )}{\sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}}dx}{3 c}+\frac {7 \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}}{4 c^2}-\frac {\tan (2 a+2 b x) \sec ^2(2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\) |
| \(\Big \downarrow \) 4489 |
| \(\displaystyle \frac {\frac {33 c^2 \int \frac {\sec (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}dx+\frac {26 c^2 \tan (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}}{3 c}+\frac {7 \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}}{4 c^2}-\frac {\tan (2 a+2 b x) \sec ^2(2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {33 c^2 \int \frac {\csc \left (2 a+2 b x+\frac {\pi }{2}\right )}{\sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}}dx+\frac {26 c^2 \tan (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}}{3 c}+\frac {7 \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}}{4 c^2}-\frac {\tan (2 a+2 b x) \sec ^2(2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\) |
| \(\Big \downarrow \) 4282 |
| \(\displaystyle \frac {\frac {\frac {26 c^2 \tan (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {33 c^2 \int \frac {1}{\frac {c^2 \tan ^2(2 a+2 b x)}{c \sec (2 a+2 b x)-c}-2 c}d\left (-\frac {c \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{b}}{3 c}+\frac {7 \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}}{4 c^2}-\frac {\tan (2 a+2 b x) \sec ^2(2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle \frac {\frac {\frac {26 c^2 \tan (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {33 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {c \sec (2 a+2 b x)-c}}\right )}{\sqrt {2} b}}{3 c}+\frac {7 \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}}{4 c^2}-\frac {\tan (2 a+2 b x) \sec ^2(2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\) |
-1/4*(Sec[2*a + 2*b*x]^2*Tan[2*a + 2*b*x])/(b*(-c + c*Sec[2*a + 2*b*x])^(3 /2)) + ((7*Sqrt[-c + c*Sec[2*a + 2*b*x]]*Tan[2*a + 2*b*x])/(3*b) + ((-33*c ^(3/2)*ArcTanh[(Sqrt[c]*Tan[2*a + 2*b*x])/(Sqrt[2]*Sqrt[-c + c*Sec[2*a + 2 *b*x]])])/(Sqrt[2]*b) + (26*c^2*Tan[2*a + 2*b*x])/(b*Sqrt[-c + c*Sec[2*a + 2*b*x]]))/(3*c))/(4*c^2)
3.7.26.3.1 Defintions of rubi rules used
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[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-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]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[(-d^2)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d *Csc[e + f*x])^(n - 2)/(f*(2*m + 1))), x] + Simp[d^2/(a*b*(2*m + 1)) Int[ (a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 2)*(b*(n - 2) + a*(m - n + 2)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[n, 2] && (IntegersQ[2*m, 2*n] || IntegerQ[m])
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)]
Int[csc[(e_.) + (f_.)*(x_)]^2*(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 + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int [Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B) *Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a *B, 0] && !LtQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(1069\) vs. \(2(157)=314\).
Time = 9.49 (sec) , antiderivative size = 1070, normalized size of antiderivative = 5.94
1/96*2^(1/2)/b*csc(b*x+a)*(152*((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2) *cos(b*x+a)^5-132*arctanh((2*cos(b*x+a)-1)/(1+cos(b*x+a))/((2*cos(b*x+a)^2 -1)/(1+cos(b*x+a))^2)^(1/2))*cos(b*x+a)^5+132*ln(2*(cos(b*x+a)*((2*cos(b*x +a)^2-1)/(1+cos(b*x+a))^2)^(1/2)+((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/ 2)-2*cos(b*x+a)-1)/(1+cos(b*x+a)))*cos(b*x+a)^5+132*cos(b*x+a)^4*arctanh(( 2*cos(b*x+a)-1)/(1+cos(b*x+a))/((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2) )-132*cos(b*x+a)^4*ln(2*(cos(b*x+a)*((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^ (1/2)+((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)-2*cos(b*x+a)-1)/(1+cos(b *x+a)))-200*((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)*cos(b*x+a)^3+132*a rctanh((2*cos(b*x+a)-1)/(1+cos(b*x+a))/((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^ 2)^(1/2))*cos(b*x+a)^3-132*ln(2*(cos(b*x+a)*((2*cos(b*x+a)^2-1)/(1+cos(b*x +a))^2)^(1/2)+((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)-2*cos(b*x+a)-1)/ (1+cos(b*x+a)))*cos(b*x+a)^3-132*cos(b*x+a)^2*arctanh((2*cos(b*x+a)-1)/(1+ cos(b*x+a))/((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2))+132*cos(b*x+a)^2* ln(2*(cos(b*x+a)*((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)+((2*cos(b*x+a )^2-1)/(1+cos(b*x+a))^2)^(1/2)-2*cos(b*x+a)-1)/(1+cos(b*x+a)))+54*cos(b*x+ a)*((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)-33*arctanh((2*cos(b*x+a)-1) /(1+cos(b*x+a))/((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2))*cos(b*x+a)+33 *ln(2*(cos(b*x+a)*((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)+((2*cos(b*x+ a)^2-1)/(1+cos(b*x+a))^2)^(1/2)-2*cos(b*x+a)-1)/(1+cos(b*x+a)))*cos(b*x...
Time = 0.29 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.94 \[ \int \frac {\sec ^4(2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\left [\frac {33 \, \sqrt {2} {\left (\tan \left (b x + a\right )^{5} - \tan \left (b x + a\right )^{3}\right )} \sqrt {c} \log \left (\frac {c \tan \left (b x + a\right )^{3} - 2 \, \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {c} - 2 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{3}}\right ) + 2 \, \sqrt {2} {\left (27 \, \tan \left (b x + a\right )^{4} - 46 \, \tan \left (b x + a\right )^{2} + 3\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{48 \, {\left (b c^{2} \tan \left (b x + a\right )^{5} - b c^{2} \tan \left (b x + a\right )^{3}\right )}}, -\frac {33 \, \sqrt {2} {\left (\tan \left (b x + a\right )^{5} - \tan \left (b x + a\right )^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-c}}{c \tan \left (b x + a\right )}\right ) - \sqrt {2} {\left (27 \, \tan \left (b x + a\right )^{4} - 46 \, \tan \left (b x + a\right )^{2} + 3\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{24 \, {\left (b c^{2} \tan \left (b x + a\right )^{5} - b c^{2} \tan \left (b x + a\right )^{3}\right )}}\right ] \]
[1/48*(33*sqrt(2)*(tan(b*x + a)^5 - tan(b*x + a)^3)*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) + 2*sqrt(2)*(27*tan(b*x + a)^4 - 46*tan(b*x + a)^2 + 3)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1)) )/(b*c^2*tan(b*x + a)^5 - b*c^2*tan(b*x + a)^3), -1/24*(33*sqrt(2)*(tan(b* x + a)^5 - tan(b*x + a)^3)*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))) - sqrt(2)*(2 7*tan(b*x + a)^4 - 46*tan(b*x + a)^2 + 3)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1)))/(b*c^2*tan(b*x + a)^5 - b*c^2*tan(b*x + a)^3)]
Timed out. \[ \int \frac {\sec ^4(2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\sec ^4(2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\int { \frac {\sec \left (2 \, b x + 2 \, a\right )^{4}}{\left (c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sec ^4(2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\sec ^4(2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\int \frac {1}{{\cos \left (2\,a+2\,b\,x\right )}^4\,{\left (c\,\mathrm {tan}\left (a+b\,x\right )\,\mathrm {tan}\left (2\,a+2\,b\,x\right )\right )}^{3/2}} \,d x \]