Integrand size = 23, antiderivative size = 77 \[ \int \frac {\sec ^6(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {(a-b)^2 \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} b^{5/2} d}-\frac {(a-2 b) \tan (c+d x)}{b^2 d}+\frac {\tan ^3(c+d x)}{3 b d} \]
(a-b)^2*arctan(b^(1/2)*tan(d*x+c)/a^(1/2))/b^(5/2)/d/a^(1/2)-(a-2*b)*tan(d *x+c)/b^2/d+1/3*tan(d*x+c)^3/b/d
Time = 0.79 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.96 \[ \int \frac {\sec ^6(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {\frac {3 (a-b)^2 \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a}}+\sqrt {b} \left (-3 a+5 b+b \sec ^2(c+d x)\right ) \tan (c+d x)}{3 b^{5/2} d} \]
((3*(a - b)^2*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/Sqrt[a] + Sqrt[b]*(- 3*a + 5*b + b*Sec[c + d*x]^2)*Tan[c + d*x])/(3*b^(5/2)*d)
Time = 0.28 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4158, 300, 2009}
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.
\(\displaystyle \int \frac {\sec ^6(c+d x)}{a+b \tan ^2(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (c+d x)^6}{a+b \tan (c+d x)^2}dx\) |
\(\Big \downarrow \) 4158 |
\(\displaystyle \frac {\int \frac {\left (\tan ^2(c+d x)+1\right )^2}{b \tan ^2(c+d x)+a}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 300 |
\(\displaystyle \frac {\int \left (\frac {\tan ^2(c+d x)}{b}+\frac {a^2-2 b a+b^2}{b^2 \left (b \tan ^2(c+d x)+a\right )}-\frac {a-2 b}{b^2}\right )d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {(a-b)^2 \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} b^{5/2}}-\frac {(a-2 b) \tan (c+d x)}{b^2}+\frac {\tan ^3(c+d x)}{3 b}}{d}\) |
(((a - b)^2*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/(Sqrt[a]*b^(5/2)) - (( a - 2*b)*Tan[c + d*x])/b^2 + Tan[c + d*x]^3/(3*b))/d
3.5.57.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int [PolynomialDivide[(a + b*x^2)^p, (c + d*x^2)^(-q), x], x] /; FreeQ[{a, b, c , d}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_ )])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[ff/(c^(m - 1)*f) Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)^n)^ p, x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && I ntegerQ[m/2] && (IntegersQ[n, p] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])
Time = 12.51 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.96
method | result | size |
derivativedivides | \(\frac {-\frac {-\frac {b \tan \left (d x +c \right )^{3}}{3}+a \tan \left (d x +c \right )-2 b \tan \left (d x +c \right )}{b^{2}}+\frac {\left (a^{2}-2 a b +b^{2}\right ) \arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{b^{2} \sqrt {a b}}}{d}\) | \(74\) |
default | \(\frac {-\frac {-\frac {b \tan \left (d x +c \right )^{3}}{3}+a \tan \left (d x +c \right )-2 b \tan \left (d x +c \right )}{b^{2}}+\frac {\left (a^{2}-2 a b +b^{2}\right ) \arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{b^{2} \sqrt {a b}}}{d}\) | \(74\) |
risch | \(-\frac {2 i \left (3 a \,{\mathrm e}^{4 i \left (d x +c \right )}-3 b \,{\mathrm e}^{4 i \left (d x +c \right )}+6 a \,{\mathrm e}^{2 i \left (d x +c \right )}-12 b \,{\mathrm e}^{2 i \left (d x +c \right )}+3 a -5 b \right )}{3 d \,b^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right ) a^{2}}{2 \sqrt {-a b}\, d \,b^{2}}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right ) a}{\sqrt {-a b}\, d b}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right )}{2 \sqrt {-a b}\, d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {-2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right ) a^{2}}{2 \sqrt {-a b}\, d \,b^{2}}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {-2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right ) a}{\sqrt {-a b}\, d b}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {-2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right )}{2 \sqrt {-a b}\, d}\) | \(446\) |
1/d*(-1/b^2*(-1/3*b*tan(d*x+c)^3+a*tan(d*x+c)-2*b*tan(d*x+c))+(a^2-2*a*b+b ^2)/b^2/(a*b)^(1/2)*arctan(b*tan(d*x+c)/(a*b)^(1/2)))
Time = 0.29 (sec) , antiderivative size = 339, normalized size of antiderivative = 4.40 \[ \int \frac {\sec ^6(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\left [-\frac {3 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \sqrt {-a b} \cos \left (d x + c\right )^{3} \log \left (\frac {{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left ({\left (a + b\right )} \cos \left (d x + c\right )^{3} - b \cos \left (d x + c\right )\right )} \sqrt {-a b} \sin \left (d x + c\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) - 4 \, {\left (a b^{2} - {\left (3 \, a^{2} b - 5 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, a b^{3} d \cos \left (d x + c\right )^{3}}, -\frac {3 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \sqrt {a b} \arctan \left (\frac {{\left ({\left (a + b\right )} \cos \left (d x + c\right )^{2} - b\right )} \sqrt {a b}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{3} - 2 \, {\left (a b^{2} - {\left (3 \, a^{2} b - 5 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, a b^{3} d \cos \left (d x + c\right )^{3}}\right ] \]
[-1/12*(3*(a^2 - 2*a*b + b^2)*sqrt(-a*b)*cos(d*x + c)^3*log(((a^2 + 6*a*b + b^2)*cos(d*x + c)^4 - 2*(3*a*b + b^2)*cos(d*x + c)^2 + 4*((a + b)*cos(d* x + c)^3 - b*cos(d*x + c))*sqrt(-a*b)*sin(d*x + c) + b^2)/((a^2 - 2*a*b + b^2)*cos(d*x + c)^4 + 2*(a*b - b^2)*cos(d*x + c)^2 + b^2)) - 4*(a*b^2 - (3 *a^2*b - 5*a*b^2)*cos(d*x + c)^2)*sin(d*x + c))/(a*b^3*d*cos(d*x + c)^3), -1/6*(3*(a^2 - 2*a*b + b^2)*sqrt(a*b)*arctan(1/2*((a + b)*cos(d*x + c)^2 - b)*sqrt(a*b)/(a*b*cos(d*x + c)*sin(d*x + c)))*cos(d*x + c)^3 - 2*(a*b^2 - (3*a^2*b - 5*a*b^2)*cos(d*x + c)^2)*sin(d*x + c))/(a*b^3*d*cos(d*x + c)^3 )]
\[ \int \frac {\sec ^6(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\int \frac {\sec ^{6}{\left (c + d x \right )}}{a + b \tan ^{2}{\left (c + d x \right )}}\, dx \]
Time = 0.31 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.90 \[ \int \frac {\sec ^6(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {\frac {3 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} + \frac {b \tan \left (d x + c\right )^{3} - 3 \, {\left (a - 2 \, b\right )} \tan \left (d x + c\right )}{b^{2}}}{3 \, d} \]
1/3*(3*(a^2 - 2*a*b + b^2)*arctan(b*tan(d*x + c)/sqrt(a*b))/(sqrt(a*b)*b^2 ) + (b*tan(d*x + c)^3 - 3*(a - 2*b)*tan(d*x + c))/b^2)/d
Time = 0.49 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.25 \[ \int \frac {\sec ^6(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {\frac {3 \, {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )\right )} {\left (a^{2} - 2 \, a b + b^{2}\right )}}{\sqrt {a b} b^{2}} + \frac {b^{2} \tan \left (d x + c\right )^{3} - 3 \, a b \tan \left (d x + c\right ) + 6 \, b^{2} \tan \left (d x + c\right )}{b^{3}}}{3 \, d} \]
1/3*(3*(pi*floor((d*x + c)/pi + 1/2)*sgn(b) + arctan(b*tan(d*x + c)/sqrt(a *b)))*(a^2 - 2*a*b + b^2)/(sqrt(a*b)*b^2) + (b^2*tan(d*x + c)^3 - 3*a*b*ta n(d*x + c) + 6*b^2*tan(d*x + c))/b^3)/d
Time = 11.98 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.17 \[ \int \frac {\sec ^6(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,b\,d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {a}{b^2}-\frac {2}{b}\right )}{d}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (c+d\,x\right )\,{\left (a-b\right )}^2}{\sqrt {a}\,\left (a^2-2\,a\,b+b^2\right )}\right )\,{\left (a-b\right )}^2}{\sqrt {a}\,b^{5/2}\,d} \]