Integrand size = 23, antiderivative size = 104 \[ \int \frac {\sec ^6(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=-\frac {\left (3 a^2-2 a b-b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} b^{5/2} d}+\frac {\tan (c+d x)}{b^2 d}+\frac {(a-b)^2 \tan (c+d x)}{2 a b^2 d \left (a+b \tan ^2(c+d x)\right )} \]
-1/2*(3*a^2-2*a*b-b^2)*arctan(b^(1/2)*tan(d*x+c)/a^(1/2))/a^(3/2)/b^(5/2)/ d+tan(d*x+c)/b^2/d+1/2*(a-b)^2*tan(d*x+c)/a/b^2/d/(a+b*tan(d*x+c)^2)
Time = 1.31 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00 \[ \int \frac {\sec ^6(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\frac {-\frac {(a-b) (3 a+b) \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{3/2}}+\frac {(a-b)^2 \sqrt {b} \sin (2 (c+d x))}{a (a+b+(a-b) \cos (2 (c+d x)))}+2 \sqrt {b} \tan (c+d x)}{2 b^{5/2} d} \]
(-(((a - b)*(3*a + b)*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/a^(3/2)) + ( (a - b)^2*Sqrt[b]*Sin[2*(c + d*x)])/(a*(a + b + (a - b)*Cos[2*(c + d*x)])) + 2*Sqrt[b]*Tan[c + d*x])/(2*b^(5/2)*d)
Time = 0.31 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.95, 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.
| \(\displaystyle \int \frac {\sec ^6(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sec (c+d x)^6}{\left (a+b \tan (c+d x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 4158 |
| \(\displaystyle \frac {\int \frac {\left (\tan ^2(c+d x)+1\right )^2}{\left (b \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 300 |
| \(\displaystyle \frac {\int \left (\frac {1}{b^2}-\frac {a^2-b^2+2 (a-b) b \tan ^2(c+d x)}{b^2 \left (b \tan ^2(c+d x)+a\right )^2}\right )d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {\left (3 a^2-2 a b-b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} b^{5/2}}+\frac {(a-b)^2 \tan (c+d x)}{2 a b^2 \left (a+b \tan ^2(c+d x)\right )}+\frac {\tan (c+d x)}{b^2}}{d}\) |
(-1/2*((3*a^2 - 2*a*b - b^2)*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/(a^(3 /2)*b^(5/2)) + Tan[c + d*x]/b^2 + ((a - b)^2*Tan[c + d*x])/(2*a*b^2*(a + b *Tan[c + d*x]^2)))/d
3.5.69.3.1 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 = 31.44 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.93
| method | result | size |
| derivativedivides | \(\frac {\frac {\tan \left (d x +c \right )}{b^{2}}-\frac {-\frac {\left (a^{2}-2 a b +b^{2}\right ) \tan \left (d x +c \right )}{2 a \left (a +b \tan \left (d x +c \right )^{2}\right )}+\frac {\left (3 a^{2}-2 a b -b^{2}\right ) \arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{2 a \sqrt {a b}}}{b^{2}}}{d}\) | \(97\) |
| default | \(\frac {\frac {\tan \left (d x +c \right )}{b^{2}}-\frac {-\frac {\left (a^{2}-2 a b +b^{2}\right ) \tan \left (d x +c \right )}{2 a \left (a +b \tan \left (d x +c \right )^{2}\right )}+\frac {\left (3 a^{2}-2 a b -b^{2}\right ) \arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{2 a \sqrt {a b}}}{b^{2}}}{d}\) | \(97\) |
| risch | \(\frac {i \left (-3 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+2 a b \,{\mathrm e}^{4 i \left (d x +c \right )}+b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-6 \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2}-2 a b \,{\mathrm e}^{2 i \left (d x +c \right )}-3 a^{2}+4 a b -b^{2}\right )}{a \,b^{2} d \left (-a \,{\mathrm e}^{4 i \left (d x +c \right )}+b \,{\mathrm e}^{4 i \left (d x +c \right )}-2 a \,{\mathrm e}^{2 i \left (d x +c \right )}-2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-a +b \right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {3 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {-2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right )}{4 \sqrt {-a b}\, d \,b^{2}}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {-2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right )}{2 \sqrt {-a b}\, d b}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {-2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right )}{4 \sqrt {-a b}\, d a}+\frac {3 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right )}{4 \sqrt {-a b}\, d \,b^{2}}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right )}{2 \sqrt {-a b}\, d b}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right )}{4 \sqrt {-a b}\, d a}\) | \(531\) |
1/d*(1/b^2*tan(d*x+c)-1/b^2*(-1/2*(a^2-2*a*b+b^2)/a*tan(d*x+c)/(a+b*tan(d* x+c)^2)+1/2*(3*a^2-2*a*b-b^2)/a/(a*b)^(1/2)*arctan(b*tan(d*x+c)/(a*b)^(1/2 ))))
Leaf count of result is larger than twice the leaf count of optimal. 199 vs. \(2 (92) = 184\).
Time = 0.34 (sec) , antiderivative size = 479, normalized size of antiderivative = 4.61 \[ \int \frac {\sec ^6(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\left [\frac {{\left ({\left (3 \, a^{3} - 5 \, a^{2} b + a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, a^{2} b - 2 \, a b^{2} - b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a b} \log \left (\frac {{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left ({\left (a + b\right )} \cos \left (d x + c\right )^{3} - b \cos \left (d x + c\right )\right )} \sqrt {-a b} \sin \left (d x + c\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) + 4 \, {\left (2 \, a^{2} b^{2} + {\left (3 \, a^{3} b - 4 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{8 \, {\left (a^{2} b^{4} d \cos \left (d x + c\right ) + {\left (a^{3} b^{3} - a^{2} b^{4}\right )} d \cos \left (d x + c\right )^{3}\right )}}, \frac {{\left ({\left (3 \, a^{3} - 5 \, a^{2} b + a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, a^{2} b - 2 \, a b^{2} - b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {a b} \arctan \left (\frac {{\left ({\left (a + b\right )} \cos \left (d x + c\right )^{2} - b\right )} \sqrt {a b}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) + 2 \, {\left (2 \, a^{2} b^{2} + {\left (3 \, a^{3} b - 4 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{2} b^{4} d \cos \left (d x + c\right ) + {\left (a^{3} b^{3} - a^{2} b^{4}\right )} d \cos \left (d x + c\right )^{3}\right )}}\right ] \]
[1/8*(((3*a^3 - 5*a^2*b + a*b^2 + b^3)*cos(d*x + c)^3 + (3*a^2*b - 2*a*b^2 - b^3)*cos(d*x + c))*sqrt(-a*b)*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*(2*a^2*b^2 + (3*a^3*b - 4*a^2*b^2 + a*b^3)*cos(d*x + c)^2)*sin(d*x + c))/(a^2*b^4*d*cos(d*x + c) + (a^3*b^3 - a^2*b^4)*d*cos(d*x + c)^3), 1/4*(((3*a^3 - 5*a^2*b + a*b^2 + b^3)*cos(d* x + c)^3 + (3*a^2*b - 2*a*b^2 - b^3)*cos(d*x + c))*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))) + 2* (2*a^2*b^2 + (3*a^3*b - 4*a^2*b^2 + a*b^3)*cos(d*x + c)^2)*sin(d*x + c))/( a^2*b^4*d*cos(d*x + c) + (a^3*b^3 - a^2*b^4)*d*cos(d*x + c)^3)]
\[ \int \frac {\sec ^6(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\int \frac {\sec ^{6}{\left (c + d x \right )}}{\left (a + b \tan ^{2}{\left (c + d x \right )}\right )^{2}}\, dx \]
Time = 0.31 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.96 \[ \int \frac {\sec ^6(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\frac {\frac {{\left (a^{2} - 2 \, a b + b^{2}\right )} \tan \left (d x + c\right )}{a b^{3} \tan \left (d x + c\right )^{2} + a^{2} b^{2}} + \frac {2 \, \tan \left (d x + c\right )}{b^{2}} - \frac {{\left (3 \, a^{2} - 2 \, a b - b^{2}\right )} \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )}{\sqrt {a b} a b^{2}}}{2 \, d} \]
1/2*((a^2 - 2*a*b + b^2)*tan(d*x + c)/(a*b^3*tan(d*x + c)^2 + a^2*b^2) + 2 *tan(d*x + c)/b^2 - (3*a^2 - 2*a*b - b^2)*arctan(b*tan(d*x + c)/sqrt(a*b)) /(sqrt(a*b)*a*b^2))/d
Time = 0.61 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.23 \[ \int \frac {\sec ^6(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\frac {\frac {2 \, \tan \left (d x + c\right )}{b^{2}} - \frac {{\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )\right )} {\left (3 \, a^{2} - 2 \, a b - b^{2}\right )}}{\sqrt {a b} a b^{2}} + \frac {a^{2} \tan \left (d x + c\right ) - 2 \, a b \tan \left (d x + c\right ) + b^{2} \tan \left (d x + c\right )}{{\left (b \tan \left (d x + c\right )^{2} + a\right )} a b^{2}}}{2 \, d} \]
1/2*(2*tan(d*x + c)/b^2 - (pi*floor((d*x + c)/pi + 1/2)*sgn(b) + arctan(b* tan(d*x + c)/sqrt(a*b)))*(3*a^2 - 2*a*b - b^2)/(sqrt(a*b)*a*b^2) + (a^2*ta n(d*x + c) - 2*a*b*tan(d*x + c) + b^2*tan(d*x + c))/((b*tan(d*x + c)^2 + a )*a*b^2))/d
Time = 11.68 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.14 \[ \int \frac {\sec ^6(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\frac {\mathrm {tan}\left (c+d\,x\right )}{b^2\,d}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (a^2-2\,a\,b+b^2\right )}{2\,a\,d\,\left (b^3\,{\mathrm {tan}\left (c+d\,x\right )}^2+a\,b^2\right )}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (c+d\,x\right )\,\left (a-b\right )\,\left (3\,a+b\right )}{\sqrt {a}\,\left (-3\,a^2+2\,a\,b+b^2\right )}\right )\,\left (a-b\right )\,\left (3\,a+b\right )}{2\,a^{3/2}\,b^{5/2}\,d} \]