Integrand size = 23, antiderivative size = 117 \[ \int \frac {\sin ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\left (3 a^2+12 a b+8 b^2\right ) x}{8 a^3}-\frac {\sqrt {b} (a+b)^{3/2} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a^3 f}-\frac {(5 a+4 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 a f} \] Output:
1/8*(3*a^2+12*a*b+8*b^2)*x/a^3-b^(1/2)*(a+b)^(3/2)*arctan(b^(1/2)*tan(f*x+ e)/(a+b)^(1/2))/a^3/f-1/8*(5*a+4*b)*cos(f*x+e)*sin(f*x+e)/a^2/f+1/4*cos(f* x+e)^3*sin(f*x+e)/a/f
Result contains complex when optimal does not.
Time = 1.34 (sec) , antiderivative size = 303, normalized size of antiderivative = 2.59 \[ \int \frac {\sin ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^2(e+f x) \left (\sqrt {b} \left (3 a^3+34 a^2 b+64 a b^2+32 b^3\right ) \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (\cos (2 e)-i \sin (2 e))+\sqrt {b (\cos (e)-i \sin (e))^4} \left (a^2 (3 a+2 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )+\sqrt {b} \sqrt {a+b} \left (-2 a^2 e+12 a^2 f x+48 a b f x+32 b^2 f x-8 a (a+b) \sin (2 (e+f x))+a^2 \sin (4 (e+f x))\right )\right )\right )}{64 a^3 \sqrt {b} \sqrt {a+b} f \left (a+b \sec ^2(e+f x)\right ) \sqrt {b (\cos (e)-i \sin (e))^4}} \] Input:
Integrate[Sin[e + f*x]^4/(a + b*Sec[e + f*x]^2),x]
Output:
((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^2*(Sqrt[b]*(3*a^3 + 34*a^2*b + 64*a*b^2 + 32*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*Sin[e])^ 4])]*(Cos[2*e] - I*Sin[2*e]) + Sqrt[b*(Cos[e] - I*Sin[e])^4]*(a^2*(3*a + 2 *b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]] + Sqrt[b]*Sqrt[a + b]*(-2*a ^2*e + 12*a^2*f*x + 48*a*b*f*x + 32*b^2*f*x - 8*a*(a + b)*Sin[2*(e + f*x)] + a^2*Sin[4*(e + f*x)]))))/(64*a^3*Sqrt[b]*Sqrt[a + b]*f*(a + b*Sec[e + f *x]^2)*Sqrt[b*(Cos[e] - I*Sin[e])^4])
Time = 0.34 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.21, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4620, 372, 402, 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.
| \(\displaystyle \int \frac {\sin ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)^4}{a+b \sec (e+f x)^2}dx\) |
| \(\Big \downarrow \) 4620 |
| \(\displaystyle \frac {\int \frac {\tan ^4(e+f x)}{\left (\tan ^2(e+f x)+1\right )^3 \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 372 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2}-\frac {\int \frac {(b-4 (a+b)) \tan ^2(e+f x)+a+b}{\left (\tan ^2(e+f x)+1\right )^2 \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{4 a}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2}-\frac {\frac {(5 a+4 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {\int \frac {(a+b) (3 a+4 b)-b (5 a+4 b) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{2 a}}{4 a}}{f}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2}-\frac {\frac {(5 a+4 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {\frac {\left (3 a^2+12 a b+8 b^2\right ) \int \frac {1}{\tan ^2(e+f x)+1}d\tan (e+f x)}{a}-\frac {8 b (a+b)^2 \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}}{2 a}}{4 a}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2}-\frac {\frac {(5 a+4 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {\frac {\left (3 a^2+12 a b+8 b^2\right ) \arctan (\tan (e+f x))}{a}-\frac {8 b (a+b)^2 \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}}{2 a}}{4 a}}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2}-\frac {\frac {(5 a+4 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {\frac {\left (3 a^2+12 a b+8 b^2\right ) \arctan (\tan (e+f x))}{a}-\frac {8 \sqrt {b} (a+b)^{3/2} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a}}{2 a}}{4 a}}{f}\) |
Input:
Int[Sin[e + f*x]^4/(a + b*Sec[e + f*x]^2),x]
Output:
(Tan[e + f*x]/(4*a*(1 + Tan[e + f*x]^2)^2) - (-1/2*(((3*a^2 + 12*a*b + 8*b ^2)*ArcTan[Tan[e + f*x]])/a - (8*Sqrt[b]*(a + b)^(3/2)*ArcTan[(Sqrt[b]*Tan [e + f*x])/Sqrt[a + b]])/a)/a + ((5*a + 4*b)*Tan[e + f*x])/(2*a*(1 + Tan[e + f*x]^2)))/(4*a))/f
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 = 1.21 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.02
| method | result | size |
| derivativedivides | \(\frac {-\frac {b \left (a +b \right )^{2} \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{a^{3} \sqrt {\left (a +b \right ) b}}+\frac {\frac {\left (-\frac {1}{2} a b -\frac {5}{8} a^{2}\right ) \tan \left (f x +e \right )^{3}+\left (-\frac {3}{8} a^{2}-\frac {1}{2} a b \right ) \tan \left (f x +e \right )}{\left (1+\tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (3 a^{2}+12 a b +8 b^{2}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{8}}{a^{3}}}{f}\) | \(119\) |
| default | \(\frac {-\frac {b \left (a +b \right )^{2} \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{a^{3} \sqrt {\left (a +b \right ) b}}+\frac {\frac {\left (-\frac {1}{2} a b -\frac {5}{8} a^{2}\right ) \tan \left (f x +e \right )^{3}+\left (-\frac {3}{8} a^{2}-\frac {1}{2} a b \right ) \tan \left (f x +e \right )}{\left (1+\tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (3 a^{2}+12 a b +8 b^{2}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{8}}{a^{3}}}{f}\) | \(119\) |
| risch | \(\frac {3 x}{8 a}+\frac {3 x b}{2 a^{2}}+\frac {x \,b^{2}}{a^{3}}+\frac {i {\mathrm e}^{2 i \left (f x +e \right )}}{8 a f}+\frac {i {\mathrm e}^{2 i \left (f x +e \right )} b}{8 a^{2} f}-\frac {i {\mathrm e}^{-2 i \left (f x +e \right )}}{8 a f}-\frac {i {\mathrm e}^{-2 i \left (f x +e \right )} b}{8 a^{2} f}+\frac {\sqrt {-a b -b^{2}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b -b^{2}}+a +2 b}{a}\right )}{2 f \,a^{2}}+\frac {\sqrt {-a b -b^{2}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b -b^{2}}+a +2 b}{a}\right ) b}{2 f \,a^{3}}-\frac {\sqrt {-a b -b^{2}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b -b^{2}}-a -2 b}{a}\right )}{2 f \,a^{2}}-\frac {\sqrt {-a b -b^{2}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b -b^{2}}-a -2 b}{a}\right ) b}{2 f \,a^{3}}+\frac {\sin \left (4 f x +4 e \right )}{32 a f}\) | \(342\) |
Input:
int(sin(f*x+e)^4/(a+b*sec(f*x+e)^2),x,method=_RETURNVERBOSE)
Output:
1/f*(-b/a^3*(a+b)^2/((a+b)*b)^(1/2)*arctan(b*tan(f*x+e)/((a+b)*b)^(1/2))+1 /a^3*(((-1/2*a*b-5/8*a^2)*tan(f*x+e)^3+(-3/8*a^2-1/2*a*b)*tan(f*x+e))/(1+t an(f*x+e)^2)^2+1/8*(3*a^2+12*a*b+8*b^2)*arctan(tan(f*x+e))))
Time = 0.11 (sec) , antiderivative size = 332, normalized size of antiderivative = 2.84 \[ \int \frac {\sin ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\left [\frac {{\left (3 \, a^{2} + 12 \, a b + 8 \, b^{2}\right )} f x + 2 \, \sqrt {-a b - b^{2}} {\left (a + b\right )} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt {-a b - b^{2}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) + {\left (2 \, a^{2} \cos \left (f x + e\right )^{3} - {\left (5 \, a^{2} + 4 \, a b\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \, a^{3} f}, \frac {{\left (3 \, a^{2} + 12 \, a b + 8 \, b^{2}\right )} f x + 4 \, \sqrt {a b + b^{2}} {\left (a + b\right )} \arctan \left (\frac {{\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b}{2 \, \sqrt {a b + b^{2}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) + {\left (2 \, a^{2} \cos \left (f x + e\right )^{3} - {\left (5 \, a^{2} + 4 \, a b\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \, a^{3} f}\right ] \] Input:
integrate(sin(f*x+e)^4/(a+b*sec(f*x+e)^2),x, algorithm="fricas")
Output:
[1/8*((3*a^2 + 12*a*b + 8*b^2)*f*x + 2*sqrt(-a*b - b^2)*(a + 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 + 2*b)*cos(f*x + e)^3 - b*cos(f*x + e))*sqrt(-a*b - b^2)*sin(f*x + e) + b ^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)) + (2*a^2*cos(f*x + e)^3 - (5*a^2 + 4*a*b)*cos(f*x + e))*sin(f*x + e))/(a^3*f), 1/8*((3*a^2 + 12*a*b + 8*b^2)*f*x + 4*sqrt(a*b + b^2)*(a + b)*arctan(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^2*cos( f*x + e)^3 - (5*a^2 + 4*a*b)*cos(f*x + e))*sin(f*x + e))/(a^3*f)]
\[ \int \frac {\sin ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\int \frac {\sin ^{4}{\left (e + f x \right )}}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \] Input:
integrate(sin(f*x+e)**4/(a+b*sec(f*x+e)**2),x)
Output:
Integral(sin(e + f*x)**4/(a + b*sec(e + f*x)**2), x)
Time = 0.12 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.17 \[ \int \frac {\sin ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {{\left (5 \, a + 4 \, b\right )} \tan \left (f x + e\right )^{3} + {\left (3 \, a + 4 \, b\right )} \tan \left (f x + e\right )}{a^{2} \tan \left (f x + e\right )^{4} + 2 \, a^{2} \tan \left (f x + e\right )^{2} + a^{2}} - \frac {{\left (3 \, a^{2} + 12 \, a b + 8 \, b^{2}\right )} {\left (f x + e\right )}}{a^{3}} + \frac {8 \, {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a^{3}}}{8 \, f} \] Input:
integrate(sin(f*x+e)^4/(a+b*sec(f*x+e)^2),x, algorithm="maxima")
Output:
-1/8*(((5*a + 4*b)*tan(f*x + e)^3 + (3*a + 4*b)*tan(f*x + e))/(a^2*tan(f*x + e)^4 + 2*a^2*tan(f*x + e)^2 + a^2) - (3*a^2 + 12*a*b + 8*b^2)*(f*x + e) /a^3 + 8*(a^2*b + 2*a*b^2 + b^3)*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/(s qrt((a + b)*b)*a^3))/f
Time = 0.16 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.30 \[ \int \frac {\sin ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\frac {{\left (3 \, a^{2} + 12 \, a b + 8 \, b^{2}\right )} {\left (f x + e\right )}}{a^{3}} - \frac {8 \, {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )}}{\sqrt {a b + b^{2}} a^{3}} - \frac {5 \, a \tan \left (f x + e\right )^{3} + 4 \, b \tan \left (f x + e\right )^{3} + 3 \, a \tan \left (f x + e\right ) + 4 \, b \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{2} a^{2}}}{8 \, f} \] Input:
integrate(sin(f*x+e)^4/(a+b*sec(f*x+e)^2),x, algorithm="giac")
Output:
1/8*((3*a^2 + 12*a*b + 8*b^2)*(f*x + e)/a^3 - 8*(a^2*b + 2*a*b^2 + b^3)*(p i*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^3) - (5*a*tan(f*x + e)^3 + 4*b*tan(f*x + e)^3 + 3*a* tan(f*x + e) + 4*b*tan(f*x + e))/((tan(f*x + e)^2 + 1)^2*a^2))/f
Time = 12.75 (sec) , antiderivative size = 494, normalized size of antiderivative = 4.22 \[ \int \frac {\sin ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\mathrm {atanh}\left (\frac {9\,b^3\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-a^3\,b-3\,a^2\,b^2-3\,a\,b^3-b^4}}{32\,\left (\frac {13\,a\,b^4}{16}+\frac {25\,b^5}{32}+\frac {9\,a^2\,b^3}{32}+\frac {b^6}{4\,a}\right )}+\frac {b^4\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-a^3\,b-3\,a^2\,b^2-3\,a\,b^3-b^4}}{4\,\left (\frac {9\,a^3\,b^3}{32}+\frac {13\,a^2\,b^4}{16}+\frac {25\,a\,b^5}{32}+\frac {b^6}{4}\right )}\right )\,\sqrt {-b\,{\left (a+b\right )}^3}}{a^3\,f}-\frac {\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (3\,a+4\,b\right )}{8\,a^2}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (5\,a+4\,b\right )}{8\,a^2}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^4+2\,{\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}-\frac {\mathrm {atan}\left (\frac {159\,b^3\,\mathrm {tan}\left (e+f\,x\right )}{256\,\left (\frac {27\,a\,b^2}{256}+\frac {159\,b^3}{256}+\frac {75\,b^4}{64\,a}+\frac {29\,b^5}{32\,a^2}+\frac {b^6}{4\,a^3}\right )}+\frac {75\,b^4\,\mathrm {tan}\left (e+f\,x\right )}{64\,\left (\frac {159\,a\,b^3}{256}+\frac {75\,b^4}{64}+\frac {27\,a^2\,b^2}{256}+\frac {29\,b^5}{32\,a}+\frac {b^6}{4\,a^2}\right )}+\frac {29\,b^5\,\mathrm {tan}\left (e+f\,x\right )}{32\,\left (\frac {75\,a\,b^4}{64}+\frac {29\,b^5}{32}+\frac {159\,a^2\,b^3}{256}+\frac {27\,a^3\,b^2}{256}+\frac {b^6}{4\,a}\right )}+\frac {b^6\,\mathrm {tan}\left (e+f\,x\right )}{4\,\left (\frac {27\,a^4\,b^2}{256}+\frac {159\,a^3\,b^3}{256}+\frac {75\,a^2\,b^4}{64}+\frac {29\,a\,b^5}{32}+\frac {b^6}{4}\right )}+\frac {27\,b^2\,\mathrm {tan}\left (e+f\,x\right )}{256\,\left (\frac {27\,b^2}{256}+\frac {159\,b^3}{256\,a}+\frac {75\,b^4}{64\,a^2}+\frac {29\,b^5}{32\,a^3}+\frac {b^6}{4\,a^4}\right )}\right )\,\left (a^2\,3{}\mathrm {i}+a\,b\,12{}\mathrm {i}+b^2\,8{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,a^3\,f} \] Input:
int(sin(e + f*x)^4/(a + b/cos(e + f*x)^2),x)
Output:
(atanh((9*b^3*tan(e + f*x)*(- 3*a*b^3 - a^3*b - b^4 - 3*a^2*b^2)^(1/2))/(3 2*((13*a*b^4)/16 + (25*b^5)/32 + (9*a^2*b^3)/32 + b^6/(4*a))) + (b^4*tan(e + f*x)*(- 3*a*b^3 - a^3*b - b^4 - 3*a^2*b^2)^(1/2))/(4*((25*a*b^5)/32 + b ^6/4 + (13*a^2*b^4)/16 + (9*a^3*b^3)/32)))*(-b*(a + b)^3)^(1/2))/(a^3*f) - ((tan(e + f*x)*(3*a + 4*b))/(8*a^2) + (tan(e + f*x)^3*(5*a + 4*b))/(8*a^2 ))/(f*(2*tan(e + f*x)^2 + tan(e + f*x)^4 + 1)) - (atan((159*b^3*tan(e + f* x))/(256*((27*a*b^2)/256 + (159*b^3)/256 + (75*b^4)/(64*a) + (29*b^5)/(32* a^2) + b^6/(4*a^3))) + (75*b^4*tan(e + f*x))/(64*((159*a*b^3)/256 + (75*b^ 4)/64 + (27*a^2*b^2)/256 + (29*b^5)/(32*a) + b^6/(4*a^2))) + (29*b^5*tan(e + f*x))/(32*((75*a*b^4)/64 + (29*b^5)/32 + (159*a^2*b^3)/256 + (27*a^3*b^ 2)/256 + b^6/(4*a))) + (b^6*tan(e + f*x))/(4*((29*a*b^5)/32 + b^6/4 + (75* a^2*b^4)/64 + (159*a^3*b^3)/256 + (27*a^4*b^2)/256)) + (27*b^2*tan(e + f*x ))/(256*((27*b^2)/256 + (159*b^3)/(256*a) + (75*b^4)/(64*a^2) + (29*b^5)/( 32*a^3) + b^6/(4*a^4))))*(a*b*12i + a^2*3i + b^2*8i)*1i)/(8*a^3*f)
Time = 0.18 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.97 \[ \int \frac {\sin ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {-8 \sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\sqrt {a}}{\sqrt {b}}\right ) a -8 \sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\sqrt {a}}{\sqrt {b}}\right ) b -8 \sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\sqrt {a}}{\sqrt {b}}\right ) a -8 \sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\sqrt {a}}{\sqrt {b}}\right ) b -2 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} a^{2}-3 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a^{2}-4 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a b +3 a^{2} e +3 a^{2} f x +12 a b e +12 a b f x +8 b^{2} e +8 b^{2} f x}{8 a^{3} f} \] Input:
int(sin(f*x+e)^4/(a+b*sec(f*x+e)^2),x)
Output:
( - 8*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sq rt(b))*a - 8*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt (a))/sqrt(b))*b - 8*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*a - 8*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*b - 2*cos(e + f*x)*sin(e + f*x)**3*a**2 - 3*co s(e + f*x)*sin(e + f*x)*a**2 - 4*cos(e + f*x)*sin(e + f*x)*a*b + 3*a**2*e + 3*a**2*f*x + 12*a*b*e + 12*a*b*f*x + 8*b**2*e + 8*b**2*f*x)/(8*a**3*f)