Integrand size = 23, antiderivative size = 77 \[ \int \frac {\sec ^4(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\frac {(a+b) \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} b^{3/2} d}-\frac {(a-b) \tan (c+d x)}{2 a b d \left (a+b \tan ^2(c+d x)\right )} \]
1/2*(a+b)*arctan(b^(1/2)*tan(d*x+c)/a^(1/2))/a^(3/2)/b^(3/2)/d-1/2*(a-b)*t an(d*x+c)/a/b/d/(a+b*tan(d*x+c)^2)
Time = 0.87 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.08 \[ \int \frac {\sec ^4(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\frac {(a+b) \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )+\frac {\sqrt {a} \sqrt {b} (-a+b) \sin (2 (c+d x))}{a+b+(a-b) \cos (2 (c+d x))}}{2 a^{3/2} b^{3/2} d} \]
((a + b)*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]] + (Sqrt[a]*Sqrt[b]*(-a + b )*Sin[2*(c + d*x)])/(a + b + (a - b)*Cos[2*(c + d*x)]))/(2*a^(3/2)*b^(3/2) *d)
Time = 0.26 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4158, 298, 218}
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.
\(\displaystyle \int \frac {\sec ^4(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (c+d x)^4}{\left (a+b \tan (c+d x)^2\right )^2}dx\) |
\(\Big \downarrow \) 4158 |
\(\displaystyle \frac {\int \frac {\tan ^2(c+d x)+1}{\left (b \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 298 |
\(\displaystyle \frac {\frac {(a+b) \int \frac {1}{b \tan ^2(c+d x)+a}d\tan (c+d x)}{2 a b}-\frac {(a-b) \tan (c+d x)}{2 a b \left (a+b \tan ^2(c+d x)\right )}}{d}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {(a+b) \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} b^{3/2}}-\frac {(a-b) \tan (c+d x)}{2 a b \left (a+b \tan ^2(c+d x)\right )}}{d}\) |
(((a + b)*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/(2*a^(3/2)*b^(3/2)) - (( a - b)*Tan[c + d*x])/(2*a*b*(a + b*Tan[c + d*x]^2)))/d
3.5.70.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
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 = 9.35 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.90
method | result | size |
derivativedivides | \(\frac {-\frac {\left (a -b \right ) \tan \left (d x +c \right )}{2 a b \left (a +b \tan \left (d x +c \right )^{2}\right )}+\frac {\left (a +b \right ) \arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{2 a b \sqrt {a b}}}{d}\) | \(69\) |
default | \(\frac {-\frac {\left (a -b \right ) \tan \left (d x +c \right )}{2 a b \left (a +b \tan \left (d x +c \right )^{2}\right )}+\frac {\left (a +b \right ) \arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{2 a b \sqrt {a b}}}{d}\) | \(69\) |
risch | \(-\frac {i \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+b \,{\mathrm e}^{2 i \left (d x +c \right )}+a -b \right )}{a b d \left (a \,{\mathrm e}^{4 i \left (d x +c \right )}-b \,{\mathrm e}^{4 i \left (d x +c \right )}+2 a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{2 i \left (d x +c \right )}+a -b \right )}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right )}{4 \sqrt {-a b}\, d b}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right )}{4 \sqrt {-a b}\, d a}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i a b -\sqrt {-a b}\, a -\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right )}{4 \sqrt {-a b}\, d b}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i a b -\sqrt {-a b}\, a -\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right )}{4 \sqrt {-a b}\, d a}\) | \(345\) |
1/d*(-1/2*(a-b)/a/b*tan(d*x+c)/(a+b*tan(d*x+c)^2)+1/2*(a+b)/a/b/(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. 143 vs. \(2 (65) = 130\).
Time = 0.31 (sec) , antiderivative size = 367, normalized size of antiderivative = 4.77 \[ \int \frac {\sec ^4(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\left [-\frac {4 \, {\left (a^{2} b - a b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left ({\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + a b + b^{2}\right )} \sqrt {-a b} \log \left (\frac {{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left ({\left (a + b\right )} \cos \left (d x + c\right )^{3} - b \cos \left (d x + c\right )\right )} \sqrt {-a b} \sin \left (d x + c\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right )}{8 \, {\left (a^{2} b^{3} d + {\left (a^{3} b^{2} - a^{2} b^{3}\right )} d \cos \left (d x + c\right )^{2}\right )}}, -\frac {2 \, {\left (a^{2} b - a b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left ({\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + a b + b^{2}\right )} \sqrt {a b} \arctan \left (\frac {{\left ({\left (a + b\right )} \cos \left (d x + c\right )^{2} - b\right )} \sqrt {a b}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right )}{4 \, {\left (a^{2} b^{3} d + {\left (a^{3} b^{2} - a^{2} b^{3}\right )} d \cos \left (d x + c\right )^{2}\right )}}\right ] \]
[-1/8*(4*(a^2*b - a*b^2)*cos(d*x + c)*sin(d*x + c) + ((a^2 - b^2)*cos(d*x + c)^2 + a*b + b^2)*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)))/(a^2*b^3*d + (a^3*b^2 - a^2*b^3)*d*cos( d*x + c)^2), -1/4*(2*(a^2*b - a*b^2)*cos(d*x + c)*sin(d*x + c) + ((a^2 - b ^2)*cos(d*x + c)^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))))/(a^2*b^3*d + (a^3*b^2 - a^2*b^3)*d*cos(d*x + c)^2)]
\[ \int \frac {\sec ^4(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\int \frac {\sec ^{4}{\left (c + d x \right )}}{\left (a + b \tan ^{2}{\left (c + d x \right )}\right )^{2}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.90 \[ \int \frac {\sec ^4(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=-\frac {\frac {{\left (a - b\right )} \tan \left (d x + c\right )}{a b^{2} \tan \left (d x + c\right )^{2} + a^{2} b} - \frac {{\left (a + b\right )} \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )}{\sqrt {a b} a b}}{2 \, d} \]
-1/2*((a - b)*tan(d*x + c)/(a*b^2*tan(d*x + c)^2 + a^2*b) - (a + b)*arctan (b*tan(d*x + c)/sqrt(a*b))/(sqrt(a*b)*a*b))/d
Time = 0.59 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.19 \[ \int \frac {\sec ^4(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\frac {\frac {{\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )\right )} {\left (a + b\right )}}{\sqrt {a b} a b} - \frac {a \tan \left (d x + c\right ) - b \tan \left (d x + c\right )}{{\left (b \tan \left (d x + c\right )^{2} + a\right )} a b}}{2 \, d} \]
1/2*((pi*floor((d*x + c)/pi + 1/2)*sgn(b) + arctan(b*tan(d*x + c)/sqrt(a*b )))*(a + b)/(sqrt(a*b)*a*b) - (a*tan(d*x + c) - b*tan(d*x + c))/((b*tan(d* x + c)^2 + a)*a*b))/d
Time = 11.76 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.84 \[ \int \frac {\sec ^4(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (c+d\,x\right )}{\sqrt {a}}\right )\,\left (a+b\right )}{2\,a^{3/2}\,b^{3/2}\,d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (a-b\right )}{2\,a\,b\,d\,\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^2+a\right )} \]