Integrand size = 23, antiderivative size = 191 \[ \int \frac {\sin ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {3 \left (a^2+8 a b+8 b^2\right ) x}{8 a^4}-\frac {3 \sqrt {b} \sqrt {a+b} (a+2 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 a^4 f}-\frac {(5 a+6 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 a f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {3 b (3 a+4 b) \tan (e+f x)}{8 a^3 f \left (a+b+b \tan ^2(e+f x)\right )} \]
3/8*(a^2+8*a*b+8*b^2)*x/a^4-3/2*(a+2*b)*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1 /2))*b^(1/2)*(a+b)^(1/2)/a^4/f-1/8*(5*a+6*b)*cos(f*x+e)*sin(f*x+e)/a^2/f/( a+b+b*tan(f*x+e)^2)+1/4*cos(f*x+e)^3*sin(f*x+e)/a/f/(a+b+b*tan(f*x+e)^2)-3 /8*b*(3*a+4*b)*tan(f*x+e)/a^3/f/(a+b+b*tan(f*x+e)^2)
Result contains complex when optimal does not.
Time = 11.56 (sec) , antiderivative size = 1105, normalized size of antiderivative = 5.79 \[ \int \frac {\sin ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=-\frac {(a+2 b+a \cos (2 e+2 f x))^2 \sec ^4(e+f x) \left (16 x+\frac {\left (-a^3+6 a^2 b+24 a b^2+16 b^3\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 ) (\cos (2 e)-i \sin (2 e))}{b (a+b)^{3/2} f \sqrt {b (\cos (e)-i \sin (e))^4}}+\frac {\left (a^2+8 a b+8 b^2\right ) ((a+2 b) \sin (2 e)-a \sin (2 f x))}{b (a+b) f (a+2 b+a \cos (2 (e+f x))) (\cos (e)-\sin (e)) (\cos (e)+\sin (e))}\right )}{256 a^2 \left (a+b \sec ^2(e+f x)\right )^2}+\frac {3 (a+2 b+a \cos (2 e+2 f x))^2 \sec ^4(e+f x) \left (\frac {(a+2 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}-\frac {a \sqrt {b} \sin (2 (e+f x))}{(a+b) (a+2 b+a \cos (2 (e+f x)))}\right )}{1024 b^{3/2} f \left (a+b \sec ^2(e+f x)\right )^2}+\frac {(a+2 b+a \cos (2 e+2 f x))^2 \sec ^4(e+f x) \left (-\frac {\left (a^5-30 a^4 b-480 a^3 b^2-1600 a^2 b^3-1920 a b^4-768 b^5\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 ) (\cos (2 e)-i \sin (2 e))}{\sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}+\frac {\sec (2 e) \left (32 b \left (5 a^4+39 a^3 b+106 a^2 b^2+120 a b^3+48 b^4\right ) f x \cos (2 e)+16 a b \left (5 a^3+29 a^2 b+48 a b^2+24 b^3\right ) f x \cos (2 f x)+80 a^4 b f x \cos (4 e+2 f x)+464 a^3 b^2 f x \cos (4 e+2 f x)+768 a^2 b^3 f x \cos (4 e+2 f x)+384 a b^4 f x \cos (4 e+2 f x)+a^5 \sin (2 e)+34 a^4 b \sin (2 e)+224 a^3 b^2 \sin (2 e)+576 a^2 b^3 \sin (2 e)+640 a b^4 \sin (2 e)+256 b^5 \sin (2 e)-a^5 \sin (2 f x)-62 a^4 b \sin (2 f x)-318 a^3 b^2 \sin (2 f x)-512 a^2 b^3 \sin (2 f x)-256 a b^4 \sin (2 f x)-12 a^4 b \sin (2 (e+2 f x))-36 a^3 b^2 \sin (2 (e+2 f x))-24 a^2 b^3 \sin (2 (e+2 f x))-30 a^4 b \sin (4 e+2 f x)-158 a^3 b^2 \sin (4 e+2 f x)-256 a^2 b^3 \sin (4 e+2 f x)-128 a b^4 \sin (4 e+2 f x)-12 a^4 b \sin (6 e+4 f x)-36 a^3 b^2 \sin (6 e+4 f x)-24 a^2 b^3 \sin (6 e+4 f x)+2 a^4 b \sin (4 e+6 f x)+2 a^3 b^2 \sin (4 e+6 f x)+2 a^4 b \sin (8 e+6 f x)+2 a^3 b^2 \sin (8 e+6 f x)\right )}{a+2 b+a \cos (2 (e+f x))}\right )}{1024 a^4 b (a+b) f \left (a+b \sec ^2(e+f x)\right )^2} \]
-1/256*((a + 2*b + a*Cos[2*e + 2*f*x])^2*Sec[e + f*x]^4*(16*x + ((-a^3 + 6 *a^2*b + 24*a*b^2 + 16*b^3)*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*S in[e])^4])]*(Cos[2*e] - I*Sin[2*e]))/(b*(a + b)^(3/2)*f*Sqrt[b*(Cos[e] - I *Sin[e])^4]) + ((a^2 + 8*a*b + 8*b^2)*((a + 2*b)*Sin[2*e] - a*Sin[2*f*x])) /(b*(a + b)*f*(a + 2*b + a*Cos[2*(e + f*x)])*(Cos[e] - Sin[e])*(Cos[e] + S in[e]))))/(a^2*(a + b*Sec[e + f*x]^2)^2) + (3*(a + 2*b + a*Cos[2*e + 2*f*x ])^2*Sec[e + f*x]^4*(((a + 2*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]] )/(a + b)^(3/2) - (a*Sqrt[b]*Sin[2*(e + f*x)])/((a + b)*(a + 2*b + a*Cos[2 *(e + f*x)]))))/(1024*b^(3/2)*f*(a + b*Sec[e + f*x]^2)^2) + ((a + 2*b + a* Cos[2*e + 2*f*x])^2*Sec[e + f*x]^4*(-(((a^5 - 30*a^4*b - 480*a^3*b^2 - 160 0*a^2*b^3 - 1920*a*b^4 - 768*b^5)*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])]*(Cos[2*e] - I*Sin[2*e]))/(Sqrt[a + b]*Sqrt[b*(Cos[e] - I *Sin[e])^4])) + (Sec[2*e]*(32*b*(5*a^4 + 39*a^3*b + 106*a^2*b^2 + 120*a*b^ 3 + 48*b^4)*f*x*Cos[2*e] + 16*a*b*(5*a^3 + 29*a^2*b + 48*a*b^2 + 24*b^3)*f *x*Cos[2*f*x] + 80*a^4*b*f*x*Cos[4*e + 2*f*x] + 464*a^3*b^2*f*x*Cos[4*e + 2*f*x] + 768*a^2*b^3*f*x*Cos[4*e + 2*f*x] + 384*a*b^4*f*x*Cos[4*e + 2*f*x] + a^5*Sin[2*e] + 34*a^4*b*Sin[2*e] + 224*a^3*b^2*Sin[2*e] + 576*a^2*b^3*S in[2*e] + 640*a*b^4*Sin[2*e] + 256*b^5*Sin[2*e] - a^5*Sin[2*f*x] - 62*a...
Time = 0.40 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.12, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3042, 4620, 372, 402, 27, 402, 27, 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.
| \(\displaystyle \int \frac {\sin ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)^4}{\left (a+b \sec (e+f x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 4620 |
| \(\displaystyle \frac {\int \frac {\tan ^4(e+f x)}{\left (\tan ^2(e+f x)+1\right )^3 \left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 372 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )}-\frac {\int \frac {-\left ((4 a+5 b) \tan ^2(e+f x)\right )+a+b}{\left (\tan ^2(e+f x)+1\right )^2 \left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{4 a}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )}-\frac {\frac {(5 a+6 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}-\frac {\int \frac {3 \left ((a+b) (a+2 b)-b (5 a+6 b) \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{2 a}}{4 a}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )}-\frac {\frac {(5 a+6 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}-\frac {3 \int \frac {(a+b) (a+2 b)-b (5 a+6 b) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{2 a}}{4 a}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )}-\frac {\frac {(5 a+6 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}-\frac {3 \left (\frac {\int \frac {2 (a+b) \left ((a+b) (a+4 b)-b (3 a+4 b) \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{2 a (a+b)}-\frac {b (3 a+4 b) \tan (e+f x)}{a \left (a+b \tan ^2(e+f x)+b\right )}\right )}{2 a}}{4 a}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )}-\frac {\frac {(5 a+6 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}-\frac {3 \left (\frac {\int \frac {(a+b) (a+4 b)-b (3 a+4 b) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{a}-\frac {b (3 a+4 b) \tan (e+f x)}{a \left (a+b \tan ^2(e+f x)+b\right )}\right )}{2 a}}{4 a}}{f}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )}-\frac {\frac {(5 a+6 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}-\frac {3 \left (\frac {\frac {\left (a^2+8 a b+8 b^2\right ) \int \frac {1}{\tan ^2(e+f x)+1}d\tan (e+f x)}{a}-\frac {4 b (a+b) (a+2 b) \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}}{a}-\frac {b (3 a+4 b) \tan (e+f x)}{a \left (a+b \tan ^2(e+f x)+b\right )}\right )}{2 a}}{4 a}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )}-\frac {\frac {(5 a+6 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}-\frac {3 \left (\frac {\frac {\left (a^2+8 a b+8 b^2\right ) \arctan (\tan (e+f x))}{a}-\frac {4 b (a+b) (a+2 b) \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}}{a}-\frac {b (3 a+4 b) \tan (e+f x)}{a \left (a+b \tan ^2(e+f x)+b\right )}\right )}{2 a}}{4 a}}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )}-\frac {\frac {(5 a+6 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}-\frac {3 \left (\frac {\frac {\left (a^2+8 a b+8 b^2\right ) \arctan (\tan (e+f x))}{a}-\frac {4 \sqrt {b} \sqrt {a+b} (a+2 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a}}{a}-\frac {b (3 a+4 b) \tan (e+f x)}{a \left (a+b \tan ^2(e+f x)+b\right )}\right )}{2 a}}{4 a}}{f}\) |
(Tan[e + f*x]/(4*a*(1 + Tan[e + f*x]^2)^2*(a + b + b*Tan[e + f*x]^2)) - (( (5*a + 6*b)*Tan[e + f*x])/(2*a*(1 + Tan[e + f*x]^2)*(a + b + b*Tan[e + f*x ]^2)) - (3*((((a^2 + 8*a*b + 8*b^2)*ArcTan[Tan[e + f*x]])/a - (4*Sqrt[b]*S qrt[a + b]*(a + 2*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/a)/a - (b *(3*a + 4*b)*Tan[e + f*x])/(a*(a + b + b*Tan[e + f*x]^2))))/(2*a))/(4*a))/ f
3.1.48.3.1 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[(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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_ )]^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m + 1)/f Subst[Int[x^m*(ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p/(1 + f f^2*x^2)^(m/2 + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && IntegerQ[n/2]
Time = 3.17 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.77
| method | result | size |
| derivativedivides | \(\frac {-\frac {\left (a +b \right ) b \left (\frac {a \tan \left (f x +e \right )}{2 a +2 b +2 b \tan \left (f x +e \right )^{2}}+\frac {3 \left (a +2 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \sqrt {\left (a +b \right ) b}}\right )}{a^{4}}+\frac {\frac {\left (-a b -\frac {5}{8} a^{2}\right ) \tan \left (f x +e \right )^{3}+\left (-\frac {3}{8} a^{2}-a b \right ) \tan \left (f x +e \right )}{\left (1+\tan \left (f x +e \right )^{2}\right )^{2}}+\frac {3 \left (a^{2}+8 a b +8 b^{2}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{8}}{a^{4}}}{f}\) | \(147\) |
| default | \(\frac {-\frac {\left (a +b \right ) b \left (\frac {a \tan \left (f x +e \right )}{2 a +2 b +2 b \tan \left (f x +e \right )^{2}}+\frac {3 \left (a +2 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \sqrt {\left (a +b \right ) b}}\right )}{a^{4}}+\frac {\frac {\left (-a b -\frac {5}{8} a^{2}\right ) \tan \left (f x +e \right )^{3}+\left (-\frac {3}{8} a^{2}-a b \right ) \tan \left (f x +e \right )}{\left (1+\tan \left (f x +e \right )^{2}\right )^{2}}+\frac {3 \left (a^{2}+8 a b +8 b^{2}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{8}}{a^{4}}}{f}\) | \(147\) |
| risch | \(\frac {3 x}{8 a^{2}}+\frac {3 x b}{a^{3}}+\frac {3 x \,b^{2}}{a^{4}}-\frac {i {\mathrm e}^{4 i \left (f x +e \right )}}{64 a^{2} f}+\frac {i {\mathrm e}^{2 i \left (f x +e \right )}}{8 a^{2} f}+\frac {i {\mathrm e}^{2 i \left (f x +e \right )} b}{4 a^{3} f}-\frac {i {\mathrm e}^{-2 i \left (f x +e \right )}}{8 a^{2} f}-\frac {i {\mathrm e}^{-2 i \left (f x +e \right )} b}{4 a^{3} f}+\frac {i {\mathrm e}^{-4 i \left (f x +e \right )}}{64 a^{2} f}-\frac {i b \left (a^{2} {\mathrm e}^{2 i \left (f x +e \right )}+3 a b \,{\mathrm e}^{2 i \left (f x +e \right )}+2 b^{2} {\mathrm e}^{2 i \left (f x +e \right )}+a^{2}+a b \right )}{a^{4} f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )}+\frac {3 \sqrt {-a b -b^{2}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b -b^{2}}+a +2 b}{a}\right )}{4 f \,a^{3}}+\frac {3 \sqrt {-a b -b^{2}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b -b^{2}}+a +2 b}{a}\right ) b}{2 f \,a^{4}}-\frac {3 \sqrt {-a b -b^{2}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b -b^{2}}-a -2 b}{a}\right )}{4 f \,a^{3}}-\frac {3 \sqrt {-a b -b^{2}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b -b^{2}}-a -2 b}{a}\right ) b}{2 f \,a^{4}}\) | \(458\) |
1/f*(-(a+b)*b/a^4*(1/2*a*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)+3/2*(a+2*b)/((a+b )*b)^(1/2)*arctan(b*tan(f*x+e)/((a+b)*b)^(1/2)))+1/a^4*(((-a*b-5/8*a^2)*ta n(f*x+e)^3+(-3/8*a^2-a*b)*tan(f*x+e))/(1+tan(f*x+e)^2)^2+3/8*(a^2+8*a*b+8* b^2)*arctan(tan(f*x+e))))
Time = 0.31 (sec) , antiderivative size = 522, normalized size of antiderivative = 2.73 \[ \int \frac {\sin ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\left [\frac {3 \, {\left (a^{3} + 8 \, a^{2} b + 8 \, a b^{2}\right )} f x \cos \left (f x + e\right )^{2} + 3 \, {\left (a^{2} b + 8 \, a b^{2} + 8 \, b^{3}\right )} f x + 3 \, {\left ({\left (a^{2} + 2 \, a b\right )} \cos \left (f x + e\right )^{2} + a b + 2 \, b^{2}\right )} \sqrt {-a b - b^{2}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt {-a b - b^{2}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) + {\left (2 \, a^{3} \cos \left (f x + e\right )^{5} - {\left (5 \, a^{3} + 6 \, a^{2} b\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left (3 \, a^{2} b + 4 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \, {\left (a^{5} f \cos \left (f x + e\right )^{2} + a^{4} b f\right )}}, \frac {3 \, {\left (a^{3} + 8 \, a^{2} b + 8 \, a b^{2}\right )} f x \cos \left (f x + e\right )^{2} + 3 \, {\left (a^{2} b + 8 \, a b^{2} + 8 \, b^{3}\right )} f x + 6 \, {\left ({\left (a^{2} + 2 \, a b\right )} \cos \left (f x + e\right )^{2} + a b + 2 \, b^{2}\right )} \sqrt {a b + b^{2}} \arctan \left (\frac {{\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b}{2 \, \sqrt {a b + b^{2}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) + {\left (2 \, a^{3} \cos \left (f x + e\right )^{5} - {\left (5 \, a^{3} + 6 \, a^{2} b\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left (3 \, a^{2} b + 4 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \, {\left (a^{5} f \cos \left (f x + e\right )^{2} + a^{4} b f\right )}}\right ] \]
[1/8*(3*(a^3 + 8*a^2*b + 8*a*b^2)*f*x*cos(f*x + e)^2 + 3*(a^2*b + 8*a*b^2 + 8*b^3)*f*x + 3*((a^2 + 2*a*b)*cos(f*x + e)^2 + a*b + 2*b^2)*sqrt(-a*b - b^2)*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 + 2*b)*cos(f*x + e)^3 - b*cos(f*x + e))*sqrt(-a*b - b^2)*s in(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)) + (2 *a^3*cos(f*x + e)^5 - (5*a^3 + 6*a^2*b)*cos(f*x + e)^3 - 3*(3*a^2*b + 4*a* b^2)*cos(f*x + e))*sin(f*x + e))/(a^5*f*cos(f*x + e)^2 + a^4*b*f), 1/8*(3* (a^3 + 8*a^2*b + 8*a*b^2)*f*x*cos(f*x + e)^2 + 3*(a^2*b + 8*a*b^2 + 8*b^3) *f*x + 6*((a^2 + 2*a*b)*cos(f*x + e)^2 + a*b + 2*b^2)*sqrt(a*b + b^2)*arct an(1/2*((a + 2*b)*cos(f*x + e)^2 - b)/(sqrt(a*b + b^2)*cos(f*x + e)*sin(f* x + e))) + (2*a^3*cos(f*x + e)^5 - (5*a^3 + 6*a^2*b)*cos(f*x + e)^3 - 3*(3 *a^2*b + 4*a*b^2)*cos(f*x + e))*sin(f*x + e))/(a^5*f*cos(f*x + e)^2 + a^4* b*f)]
\[ \int \frac {\sin ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\int \frac {\sin ^{4}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2}}\, dx \]
Time = 0.26 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.07 \[ \int \frac {\sin ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=-\frac {\frac {3 \, {\left (3 \, a b + 4 \, b^{2}\right )} \tan \left (f x + e\right )^{5} + {\left (5 \, a^{2} + 24 \, a b + 24 \, b^{2}\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (a^{2} + 5 \, a b + 4 \, b^{2}\right )} \tan \left (f x + e\right )}{a^{3} b \tan \left (f x + e\right )^{6} + {\left (a^{4} + 3 \, a^{3} b\right )} \tan \left (f x + e\right )^{4} + a^{4} + a^{3} b + {\left (2 \, a^{4} + 3 \, a^{3} b\right )} \tan \left (f x + e\right )^{2}} - \frac {3 \, {\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} {\left (f x + e\right )}}{a^{4}} + \frac {12 \, {\left (a^{2} b + 3 \, a b^{2} + 2 \, b^{3}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a^{4}}}{8 \, f} \]
-1/8*((3*(3*a*b + 4*b^2)*tan(f*x + e)^5 + (5*a^2 + 24*a*b + 24*b^2)*tan(f* x + e)^3 + 3*(a^2 + 5*a*b + 4*b^2)*tan(f*x + e))/(a^3*b*tan(f*x + e)^6 + ( a^4 + 3*a^3*b)*tan(f*x + e)^4 + a^4 + a^3*b + (2*a^4 + 3*a^3*b)*tan(f*x + e)^2) - 3*(a^2 + 8*a*b + 8*b^2)*(f*x + e)/a^4 + 12*(a^2*b + 3*a*b^2 + 2*b^ 3)*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/(sqrt((a + b)*b)*a^4))/f
Time = 0.34 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.01 \[ \int \frac {\sin ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\frac {3 \, {\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} {\left (f x + e\right )}}{a^{4}} - \frac {12 \, {\left (a^{2} b + 3 \, a b^{2} + 2 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )}}{\sqrt {a b + b^{2}} a^{4}} - \frac {4 \, {\left (a b \tan \left (f x + e\right ) + b^{2} \tan \left (f x + e\right )\right )}}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )} a^{3}} - \frac {5 \, a \tan \left (f x + e\right )^{3} + 8 \, b \tan \left (f x + e\right )^{3} + 3 \, a \tan \left (f x + e\right ) + 8 \, b \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{2} a^{3}}}{8 \, f} \]
1/8*(3*(a^2 + 8*a*b + 8*b^2)*(f*x + e)/a^4 - 12*(a^2*b + 3*a*b^2 + 2*b^3)* (pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^ 2)))/(sqrt(a*b + b^2)*a^4) - 4*(a*b*tan(f*x + e) + b^2*tan(f*x + e))/((b*t an(f*x + e)^2 + a + b)*a^3) - (5*a*tan(f*x + e)^3 + 8*b*tan(f*x + e)^3 + 3 *a*tan(f*x + e) + 8*b*tan(f*x + e))/((tan(f*x + e)^2 + 1)^2*a^3))/f
Time = 19.42 (sec) , antiderivative size = 435, normalized size of antiderivative = 2.28 \[ \int \frac {\sin ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {3\,\mathrm {atanh}\left (\frac {27\,b^3\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-b^2-a\,b}}{64\,\left (\frac {27\,a\,b^3}{64}+\frac {81\,b^4}{64}+\frac {27\,b^5}{32\,a}\right )}+\frac {27\,b^4\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-b^2-a\,b}}{32\,\left (\frac {27\,a^2\,b^3}{64}+\frac {81\,a\,b^4}{64}+\frac {27\,b^5}{32}\right )}\right )\,\left (a+2\,b\right )\,\sqrt {-b\,\left (a+b\right )}}{2\,a^4\,f}-\frac {\frac {3\,{\mathrm {tan}\left (e+f\,x\right )}^5\,\left (4\,b^2+3\,a\,b\right )}{8\,a^3}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (5\,a^2+24\,a\,b+24\,b^2\right )}{8\,a^3}+\frac {3\,\mathrm {tan}\left (e+f\,x\right )\,\left (a^2+5\,a\,b+4\,b^2\right )}{8\,a^3}}{f\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^6+\left (a+3\,b\right )\,{\mathrm {tan}\left (e+f\,x\right )}^4+\left (2\,a+3\,b\right )\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )}-\frac {\mathrm {atan}\left (\frac {27\,b^2\,\mathrm {tan}\left (e+f\,x\right )}{256\,\left (\frac {27\,b^2}{256}+\frac {243\,b^3}{256\,a}+\frac {27\,b^4}{16\,a^2}+\frac {27\,b^5}{32\,a^3}\right )}+\frac {243\,b^3\,\mathrm {tan}\left (e+f\,x\right )}{256\,\left (\frac {27\,a\,b^2}{256}+\frac {243\,b^3}{256}+\frac {27\,b^4}{16\,a}+\frac {27\,b^5}{32\,a^2}\right )}+\frac {27\,b^4\,\mathrm {tan}\left (e+f\,x\right )}{16\,\left (\frac {243\,a\,b^3}{256}+\frac {27\,b^4}{16}+\frac {27\,a^2\,b^2}{256}+\frac {27\,b^5}{32\,a}\right )}+\frac {27\,b^5\,\mathrm {tan}\left (e+f\,x\right )}{32\,\left (\frac {27\,a^3\,b^2}{256}+\frac {243\,a^2\,b^3}{256}+\frac {27\,a\,b^4}{16}+\frac {27\,b^5}{32}\right )}\right )\,\left (a^2\,1{}\mathrm {i}+a\,b\,8{}\mathrm {i}+b^2\,8{}\mathrm {i}\right )\,3{}\mathrm {i}}{8\,a^4\,f} \]
(3*atanh((27*b^3*tan(e + f*x)*(- a*b - b^2)^(1/2))/(64*((27*a*b^3)/64 + (8 1*b^4)/64 + (27*b^5)/(32*a))) + (27*b^4*tan(e + f*x)*(- a*b - b^2)^(1/2))/ (32*((81*a*b^4)/64 + (27*b^5)/32 + (27*a^2*b^3)/64)))*(a + 2*b)*(-b*(a + b ))^(1/2))/(2*a^4*f) - (atan((27*b^2*tan(e + f*x))/(256*((27*b^2)/256 + (24 3*b^3)/(256*a) + (27*b^4)/(16*a^2) + (27*b^5)/(32*a^3))) + (243*b^3*tan(e + f*x))/(256*((27*a*b^2)/256 + (243*b^3)/256 + (27*b^4)/(16*a) + (27*b^5)/ (32*a^2))) + (27*b^4*tan(e + f*x))/(16*((243*a*b^3)/256 + (27*b^4)/16 + (2 7*a^2*b^2)/256 + (27*b^5)/(32*a))) + (27*b^5*tan(e + f*x))/(32*((27*a*b^4) /16 + (27*b^5)/32 + (243*a^2*b^3)/256 + (27*a^3*b^2)/256)))*(a*b*8i + a^2* 1i + b^2*8i)*3i)/(8*a^4*f) - ((3*tan(e + f*x)^5*(3*a*b + 4*b^2))/(8*a^3) + (tan(e + f*x)^3*(24*a*b + 5*a^2 + 24*b^2))/(8*a^3) + (3*tan(e + f*x)*(5*a *b + a^2 + 4*b^2))/(8*a^3))/(f*(a + b + tan(e + f*x)^2*(2*a + 3*b) + b*tan (e + f*x)^6 + tan(e + f*x)^4*(a + 3*b)))