Integrand size = 23, antiderivative size = 90 \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {x}{a^2}+\frac {(a-2 b) \sqrt {a+b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 a^2 b^{3/2} f}-\frac {(a+b) \tan (e+f x)}{2 a b f \left (a+b+b \tan ^2(e+f x)\right )} \]
x/a^2+1/2*(a-2*b)*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1/2))*(a+b)^(1/2)/a^2/b ^(3/2)/f-1/2*(a+b)*tan(f*x+e)/a/b/f/(a+b+b*tan(f*x+e)^2)
Result contains complex when optimal does not.
Time = 3.73 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.77 \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^4(e+f x) \left (2 x (a+2 b+a \cos (2 (e+f x)))+\frac {\left (-a^2+a b+2 b^2\right ) \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (a+2 b+a \cos (2 (e+f x))) (\cos (2 e)-i \sin (2 e))}{b \sqrt {a+b} f \sqrt {b (\cos (e)-i \sin (e))^4}}+\frac {(a+b) ((a+2 b) \sin (2 e)-a \sin (2 f x))}{b f (\cos (e)-\sin (e)) (\cos (e)+\sin (e))}\right )}{8 a^2 \left (a+b \sec ^2(e+f x)\right )^2} \]
((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^4*(2*x*(a + 2*b + a*Cos[2*(e + f*x)]) + ((-a^2 + a*b + 2*b^2)*ArcTan[(Sec[f*x]*(Cos[2*e] - I*Sin[2*e])* (-((a + 2*b)*Sin[f*x]) + a*Sin[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])]*(a + 2*b + a*Cos[2*(e + f*x)])*(Cos[2*e] - I*Sin[2*e]))/( b*Sqrt[a + b]*f*Sqrt[b*(Cos[e] - I*Sin[e])^4]) + ((a + b)*((a + 2*b)*Sin[2 *e] - a*Sin[2*f*x]))/(b*f*(Cos[e] - Sin[e])*(Cos[e] + Sin[e]))))/(8*a^2*(a + b*Sec[e + f*x]^2)^2)
Time = 0.33 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.16, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4629, 2075, 372, 397, 216, 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.
| \(\displaystyle \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (e+f x)^4}{\left (a+b \sec (e+f x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 4629 |
| \(\displaystyle \frac {\int \frac {\tan ^4(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (a+b \left (\tan ^2(e+f x)+1\right )\right )^2}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 2075 |
| \(\displaystyle \frac {\int \frac {\tan ^4(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 372 |
| \(\displaystyle \frac {\frac {\int \frac {(a-b) \tan ^2(e+f x)+a+b}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{2 a b}-\frac {(a+b) \tan (e+f x)}{2 a b \left (a+b \tan ^2(e+f x)+b\right )}}{f}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\frac {2 b \int \frac {1}{\tan ^2(e+f x)+1}d\tan (e+f x)}{a}+\frac {(a-2 b) (a+b) \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}}{2 a b}-\frac {(a+b) \tan (e+f x)}{2 a b \left (a+b \tan ^2(e+f x)+b\right )}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {(a-2 b) (a+b) \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}+\frac {2 b \arctan (\tan (e+f x))}{a}}{2 a b}-\frac {(a+b) \tan (e+f x)}{2 a b \left (a+b \tan ^2(e+f x)+b\right )}}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {2 b \arctan (\tan (e+f x))}{a}+\frac {(a-2 b) \sqrt {a+b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a \sqrt {b}}}{2 a b}-\frac {(a+b) \tan (e+f x)}{2 a b \left (a+b \tan ^2(e+f x)+b\right )}}{f}\) |
(((2*b*ArcTan[Tan[e + f*x]])/a + ((a - 2*b)*Sqrt[a + b]*ArcTan[(Sqrt[b]*Ta n[e + f*x])/Sqrt[a + b]])/(a*Sqrt[b]))/(2*a*b) - ((a + b)*Tan[e + f*x])/(2 *a*b*(a + b + b*Tan[e + f*x]^2)))/f
3.4.57.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 )^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 )) Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 3] && IntBinomialQ[a , b, c, d, e, m, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*Expa ndToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{e, m, p, q}, x] && Binomi alQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] && ! BinomialMatchQ[{u, v}, x]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f _.)*(x_)])^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[ff/f Subst[Int[(d*ff*x)^m*((a + b*(1 + ff^2*x^2)^(n/2))^p/(1 + ff^2*x^2 )), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && Inte gerQ[n/2] && (IntegerQ[m/2] || EqQ[n, 2])
Time = 4.55 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.94
| method | result | size |
| derivativedivides | \(\frac {\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a^{2}}+\frac {\left (a +b \right ) \left (-\frac {a \tan \left (f x +e \right )}{2 b \left (a +b +b \tan \left (f x +e \right )^{2}\right )}+\frac {\left (a -2 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 b \sqrt {\left (a +b \right ) b}}\right )}{a^{2}}}{f}\) | \(85\) |
| default | \(\frac {\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a^{2}}+\frac {\left (a +b \right ) \left (-\frac {a \tan \left (f x +e \right )}{2 b \left (a +b +b \tan \left (f x +e \right )^{2}\right )}+\frac {\left (a -2 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 b \sqrt {\left (a +b \right ) b}}\right )}{a^{2}}}{f}\) | \(85\) |
| risch | \(\frac {x}{a^{2}}-\frac {i \left (a^{2} {\mathrm e}^{2 i \left (f x +e \right )}+3 a b \,{\mathrm e}^{2 i \left (f x +e \right )}+2 b^{2} {\mathrm e}^{2 i \left (f x +e \right )}+a^{2}+a b \right )}{a^{2} b f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )}+\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{4 b^{2} f a}-\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{2 b f \,a^{2}}-\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{4 b^{2} f a}+\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{2 b f \,a^{2}}\) | \(311\) |
1/f*(1/a^2*arctan(tan(f*x+e))+(a+b)/a^2*(-1/2*a/b*tan(f*x+e)/(a+b+b*tan(f* x+e)^2)+1/2*(a-2*b)/b/((a+b)*b)^(1/2)*arctan(b*tan(f*x+e)/((a+b)*b)^(1/2)) ))
Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (78) = 156\).
Time = 0.29 (sec) , antiderivative size = 393, normalized size of antiderivative = 4.37 \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\left [\frac {8 \, a b f x \cos \left (f x + e\right )^{2} + 8 \, b^{2} f x - 4 \, {\left (a^{2} + a b\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left ({\left (a^{2} - 2 \, a b\right )} \cos \left (f x + e\right )^{2} + a b - 2 \, b^{2}\right )} \sqrt {-\frac {a + b}{b}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - b^{2} \cos \left (f x + e\right )\right )} \sqrt {-\frac {a + b}{b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right )}{8 \, {\left (a^{3} b f \cos \left (f x + e\right )^{2} + a^{2} b^{2} f\right )}}, \frac {4 \, a b f x \cos \left (f x + e\right )^{2} + 4 \, b^{2} f x - 2 \, {\left (a^{2} + a b\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left ({\left (a^{2} - 2 \, a b\right )} \cos \left (f x + e\right )^{2} + a b - 2 \, b^{2}\right )} \sqrt {\frac {a + b}{b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {a + b}{b}}}{2 \, {\left (a + b\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right )}{4 \, {\left (a^{3} b f \cos \left (f x + e\right )^{2} + a^{2} b^{2} f\right )}}\right ] \]
[1/8*(8*a*b*f*x*cos(f*x + e)^2 + 8*b^2*f*x - 4*(a^2 + a*b)*cos(f*x + e)*si n(f*x + e) - ((a^2 - 2*a*b)*cos(f*x + e)^2 + a*b - 2*b^2)*sqrt(-(a + b)/b) *log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a*b + 4*b^2)*cos(f*x + e )^2 + 4*((a*b + 2*b^2)*cos(f*x + e)^3 - b^2*cos(f*x + e))*sqrt(-(a + b)/b) *sin(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)))/( a^3*b*f*cos(f*x + e)^2 + a^2*b^2*f), 1/4*(4*a*b*f*x*cos(f*x + e)^2 + 4*b^2 *f*x - 2*(a^2 + a*b)*cos(f*x + e)*sin(f*x + e) - ((a^2 - 2*a*b)*cos(f*x + e)^2 + a*b - 2*b^2)*sqrt((a + b)/b)*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)*sqrt((a + b)/b)/((a + b)*cos(f*x + e)*sin(f*x + e))))/(a^3*b*f*cos(f*x + e)^2 + a^2*b^2*f)]
\[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\int \frac {\tan ^{4}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.07 \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=-\frac {\frac {{\left (a + b\right )} \tan \left (f x + e\right )}{a b^{2} \tan \left (f x + e\right )^{2} + a^{2} b + a b^{2}} - \frac {2 \, {\left (f x + e\right )}}{a^{2}} - \frac {{\left (a^{2} - a b - 2 \, b^{2}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a^{2} b}}{2 \, f} \]
-1/2*((a + b)*tan(f*x + e)/(a*b^2*tan(f*x + e)^2 + a^2*b + a*b^2) - 2*(f*x + e)/a^2 - (a^2 - a*b - 2*b^2)*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/(sq rt((a + b)*b)*a^2*b))/f
Time = 1.02 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.33 \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\frac {2 \, {\left (f x + e\right )}}{a^{2}} + \frac {{\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} {\left (a^{2} - a b - 2 \, b^{2}\right )}}{\sqrt {a b + b^{2}} a^{2} b} - \frac {a \tan \left (f x + e\right ) + b \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )} a b}}{2 \, f} \]
1/2*(2*(f*x + e)/a^2 + (pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan (f*x + e)/sqrt(a*b + b^2)))*(a^2 - a*b - 2*b^2)/(sqrt(a*b + b^2)*a^2*b) - (a*tan(f*x + e) + b*tan(f*x + e))/((b*tan(f*x + e)^2 + a + b)*a*b))/f
Time = 19.79 (sec) , antiderivative size = 285, normalized size of antiderivative = 3.17 \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (e+f\,x\right )}{\frac {3\,b}{2\,a}-\frac {a}{2\,b}+1}-\frac {\mathrm {tan}\left (e+f\,x\right )}{2\,\left (\frac {b}{a}+\frac {3\,b^2}{2\,a^2}-\frac {1}{2}\right )}+\frac {3\,b\,\mathrm {tan}\left (e+f\,x\right )}{2\,\left (a+\frac {3\,b}{2}-\frac {a^2}{2\,b}\right )}\right )}{a^2\,f}-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (a+b\right )}{2\,a\,b\,f\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )}-\frac {\mathrm {atanh}\left (\frac {3\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-b^4-a\,b^3}}{2\,\left (\frac {a\,b}{4}-a^2+\frac {3\,b^2}{2}+\frac {a^3}{4\,b}\right )}-\frac {5\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-b^4-a\,b^3}}{4\,\left (\frac {a^2}{4}-a\,b+\frac {b^2}{4}+\frac {3\,b^3}{2\,a}\right )}+\frac {\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-b^4-a\,b^3}}{4\,\left (\frac {a\,b}{4}-b^2+\frac {b^3}{4\,a}+\frac {3\,b^4}{2\,a^2}\right )}\right )\,\sqrt {-b^3\,\left (a+b\right )}\,\left (a-2\,b\right )}{2\,a^2\,b^3\,f} \]
atan(tan(e + f*x)/((3*b)/(2*a) - a/(2*b) + 1) - tan(e + f*x)/(2*(b/a + (3* b^2)/(2*a^2) - 1/2)) + (3*b*tan(e + f*x))/(2*(a + (3*b)/2 - a^2/(2*b))))/( a^2*f) - (tan(e + f*x)*(a + b))/(2*a*b*f*(a + b + b*tan(e + f*x)^2)) - (at anh((3*tan(e + f*x)*(- a*b^3 - b^4)^(1/2))/(2*((a*b)/4 - a^2 + (3*b^2)/2 + a^3/(4*b))) - (5*tan(e + f*x)*(- a*b^3 - b^4)^(1/2))/(4*(a^2/4 - a*b + b^ 2/4 + (3*b^3)/(2*a))) + (tan(e + f*x)*(- a*b^3 - b^4)^(1/2))/(4*((a*b)/4 - b^2 + b^3/(4*a) + (3*b^4)/(2*a^2))))*(-b^3*(a + b))^(1/2)*(a - 2*b))/(2*a ^2*b^3*f)