Integrand size = 23, antiderivative size = 166 \[ \int \frac {\sin ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\left (5 a^3+30 a^2 b+40 a b^2+16 b^3\right ) x}{16 a^4}-\frac {\sqrt {b} (a+b)^{5/2} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a^4 f}-\frac {\left (11 a^2+18 a b+8 b^2\right ) \cos (e+f x) \sin (e+f x)}{16 a^3 f}+\frac {(3 a+2 b) \cos ^3(e+f x) \sin (e+f x)}{8 a^2 f}+\frac {\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f} \] Output:
1/16*(5*a^3+30*a^2*b+40*a*b^2+16*b^3)*x/a^4-b^(1/2)*(a+b)^(5/2)*arctan(b^( 1/2)*tan(f*x+e)/(a+b)^(1/2))/a^4/f-1/16*(11*a^2+18*a*b+8*b^2)*cos(f*x+e)*s in(f*x+e)/a^3/f+1/8*(3*a+2*b)*cos(f*x+e)^3*sin(f*x+e)/a^2/f+1/6*cos(f*x+e) ^3*sin(f*x+e)^3/a/f
Result contains complex when optimal does not.
Time = 3.37 (sec) , antiderivative size = 357, normalized size of antiderivative = 2.15 \[ \int \frac {\sin ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^2(e+f x) \left (3 \sqrt {b} \left (9 a^4+136 a^3 b+384 a^2 b^2+384 a b^3+128 b^4\right ) \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (\cos (2 e)-i \sin (2 e))+\sqrt {b (\cos (e)-i \sin (e))^4} \left (3 a^3 (9 a+8 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )+2 \sqrt {b} \sqrt {a+b} \left (-12 a^3 e+60 a^3 f x+360 a^2 b f x+480 a b^2 f x+192 b^3 f x-3 a \left (15 a^2+32 a b+16 b^2\right ) \sin (2 (e+f x))+3 a^2 (3 a+2 b) \sin (4 (e+f x))-a^3 \sin (6 (e+f x))\right )\right )\right )}{768 a^4 \sqrt {b} \sqrt {a+b} f \left (a+b \sec ^2(e+f x)\right ) \sqrt {b (\cos (e)-i \sin (e))^4}} \] Input:
Integrate[Sin[e + f*x]^6/(a + b*Sec[e + f*x]^2),x]
Output:
((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^2*(3*Sqrt[b]*(9*a^4 + 136*a^3 *b + 384*a^2*b^2 + 384*a*b^3 + 128*b^4)*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]*(3*a^3*(9*a + 8*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]] + 2*Sq rt[b]*Sqrt[a + b]*(-12*a^3*e + 60*a^3*f*x + 360*a^2*b*f*x + 480*a*b^2*f*x + 192*b^3*f*x - 3*a*(15*a^2 + 32*a*b + 16*b^2)*Sin[2*(e + f*x)] + 3*a^2*(3 *a + 2*b)*Sin[4*(e + f*x)] - a^3*Sin[6*(e + f*x)]))))/(768*a^4*Sqrt[b]*Sqr t[a + b]*f*(a + b*Sec[e + f*x]^2)*Sqrt[b*(Cos[e] - I*Sin[e])^4])
Time = 0.44 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.20, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 4620, 372, 27, 440, 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.
| \(\displaystyle \int \frac {\sin ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)^6}{a+b \sec (e+f x)^2}dx\) |
| \(\Big \downarrow \) 4620 |
| \(\displaystyle \frac {\int \frac {\tan ^6(e+f x)}{\left (\tan ^2(e+f x)+1\right )^4 \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 372 |
| \(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3}-\frac {\int \frac {3 \tan ^2(e+f x) \left (-\left ((2 a+b) \tan ^2(e+f x)\right )+a+b\right )}{\left (\tan ^2(e+f x)+1\right )^3 \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{6 a}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3}-\frac {\int \frac {\tan ^2(e+f x) \left (-\left ((2 a+b) \tan ^2(e+f x)\right )+a+b\right )}{\left (\tan ^2(e+f x)+1\right )^3 \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{2 a}}{f}\) |
| \(\Big \downarrow \) 440 |
| \(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3}-\frac {\frac {\int \frac {(a+b) (3 a+2 b)-\left (8 a^2+13 b a+6 b^2\right ) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right )^2 \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{4 a}-\frac {(3 a+2 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2}}{2 a}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3}-\frac {\frac {\frac {\left (11 a^2+18 a b+8 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {\int \frac {(a+b) (a+2 b) (5 a+4 b)-b \left (11 a^2+18 b a+8 b^2\right ) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{2 a}}{4 a}-\frac {(3 a+2 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2}}{2 a}}{f}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3}-\frac {\frac {\frac {\left (11 a^2+18 a b+8 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {\frac {\left (5 a^3+30 a^2 b+40 a b^2+16 b^3\right ) \int \frac {1}{\tan ^2(e+f x)+1}d\tan (e+f x)}{a}-\frac {16 b (a+b)^3 \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}}{2 a}}{4 a}-\frac {(3 a+2 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2}}{2 a}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3}-\frac {\frac {\frac {\left (11 a^2+18 a b+8 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {\frac {\left (5 a^3+30 a^2 b+40 a b^2+16 b^3\right ) \arctan (\tan (e+f x))}{a}-\frac {16 b (a+b)^3 \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}}{2 a}}{4 a}-\frac {(3 a+2 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2}}{2 a}}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3}-\frac {\frac {\frac {\left (11 a^2+18 a b+8 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {\frac {\left (5 a^3+30 a^2 b+40 a b^2+16 b^3\right ) \arctan (\tan (e+f x))}{a}-\frac {16 \sqrt {b} (a+b)^{5/2} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a}}{2 a}}{4 a}-\frac {(3 a+2 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2}}{2 a}}{f}\) |
Input:
Int[Sin[e + f*x]^6/(a + b*Sec[e + f*x]^2),x]
Output:
(Tan[e + f*x]^3/(6*a*(1 + Tan[e + f*x]^2)^3) - (-1/4*((3*a + 2*b)*Tan[e + f*x])/(a*(1 + Tan[e + f*x]^2)^2) + (-1/2*(((5*a^3 + 30*a^2*b + 40*a*b^2 + 16*b^3)*ArcTan[Tan[e + f*x]])/a - (16*Sqrt[b]*(a + b)^(5/2)*ArcTan[(Sqrt[b ]*Tan[e + f*x])/Sqrt[a + b]])/a)/a + ((11*a^2 + 18*a*b + 8*b^2)*Tan[e + f* x])/(2*a*(1 + Tan[e + f*x]^2)))/(4*a))/(2*a))/f
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[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ )*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[g*(b*e - a*f)*(g*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] - Simp[ g^2/(2*b*(b*c - a*d)*(p + 1)) Int[(g*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f)*(m - 1) + (d*(b*e - a*f)*(m + 2*q + 1) - b*2*(c *f - d*e)*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, q}, x] && LtQ[p, -1] && GtQ[m, 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 = 2.30 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.02
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {\left (-\frac {9}{8} a^{2} b -\frac {1}{2} a \,b^{2}-\frac {11}{16} a^{3}\right ) \tan \left (f x +e \right )^{5}+\left (-2 a^{2} b -a \,b^{2}-\frac {5}{6} a^{3}\right ) \tan \left (f x +e \right )^{3}+\left (-\frac {5}{16} a^{3}-\frac {7}{8} a^{2} b -\frac {1}{2} a \,b^{2}\right ) \tan \left (f x +e \right )}{\left (1+\tan \left (f x +e \right )^{2}\right )^{3}}+\frac {\left (5 a^{3}+30 a^{2} b +40 a \,b^{2}+16 b^{3}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{16}}{a^{4}}-\frac {b \left (a +b \right )^{3} \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{a^{4} \sqrt {\left (a +b \right ) b}}}{f}\) | \(170\) |
| default | \(\frac {\frac {\frac {\left (-\frac {9}{8} a^{2} b -\frac {1}{2} a \,b^{2}-\frac {11}{16} a^{3}\right ) \tan \left (f x +e \right )^{5}+\left (-2 a^{2} b -a \,b^{2}-\frac {5}{6} a^{3}\right ) \tan \left (f x +e \right )^{3}+\left (-\frac {5}{16} a^{3}-\frac {7}{8} a^{2} b -\frac {1}{2} a \,b^{2}\right ) \tan \left (f x +e \right )}{\left (1+\tan \left (f x +e \right )^{2}\right )^{3}}+\frac {\left (5 a^{3}+30 a^{2} b +40 a \,b^{2}+16 b^{3}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{16}}{a^{4}}-\frac {b \left (a +b \right )^{3} \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{a^{4} \sqrt {\left (a +b \right ) b}}}{f}\) | \(170\) |
| risch | \(\frac {5 x}{16 a}+\frac {15 x b}{8 a^{2}}+\frac {5 x \,b^{2}}{2 a^{3}}+\frac {x \,b^{3}}{a^{4}}+\frac {15 i {\mathrm e}^{2 i \left (f x +e \right )}}{128 a f}-\frac {i {\mathrm e}^{-2 i \left (f x +e \right )} b}{4 a^{2} f}-\frac {i {\mathrm e}^{-2 i \left (f x +e \right )} b^{2}}{8 a^{3} f}+\frac {i {\mathrm e}^{2 i \left (f x +e \right )} b}{4 a^{2} f}-\frac {15 i {\mathrm e}^{-2 i \left (f x +e \right )}}{128 a f}+\frac {i {\mathrm e}^{2 i \left (f x +e \right )} b^{2}}{8 a^{3} f}-\frac {\sqrt {-a b -b^{2}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b -b^{2}}-a -2 b}{a}\right )}{2 f \,a^{2}}-\frac {\sqrt {-a b -b^{2}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b -b^{2}}-a -2 b}{a}\right ) b}{f \,a^{3}}-\frac {\sqrt {-a b -b^{2}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b -b^{2}}-a -2 b}{a}\right ) b^{2}}{2 f \,a^{4}}+\frac {\sqrt {-a b -b^{2}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b -b^{2}}+a +2 b}{a}\right )}{2 f \,a^{2}}+\frac {\sqrt {-a b -b^{2}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b -b^{2}}+a +2 b}{a}\right ) b}{f \,a^{3}}+\frac {\sqrt {-a b -b^{2}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b -b^{2}}+a +2 b}{a}\right ) b^{2}}{2 f \,a^{4}}-\frac {\sin \left (6 f x +6 e \right )}{192 a f}+\frac {3 \sin \left (4 f x +4 e \right )}{64 a f}+\frac {\sin \left (4 f x +4 e \right ) b}{32 a^{2} f}\) | \(546\) |
Input:
int(sin(f*x+e)^6/(a+b*sec(f*x+e)^2),x,method=_RETURNVERBOSE)
Output:
1/f*(1/a^4*(((-9/8*a^2*b-1/2*a*b^2-11/16*a^3)*tan(f*x+e)^5+(-2*a^2*b-a*b^2 -5/6*a^3)*tan(f*x+e)^3+(-5/16*a^3-7/8*a^2*b-1/2*a*b^2)*tan(f*x+e))/(1+tan( f*x+e)^2)^3+1/16*(5*a^3+30*a^2*b+40*a*b^2+16*b^3)*arctan(tan(f*x+e)))-b/a^ 4*(a+b)^3/((a+b)*b)^(1/2)*arctan(b*tan(f*x+e)/((a+b)*b)^(1/2)))
Time = 0.12 (sec) , antiderivative size = 428, normalized size of antiderivative = 2.58 \[ \int \frac {\sin ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\left [\frac {3 \, {\left (5 \, a^{3} + 30 \, a^{2} b + 40 \, a b^{2} + 16 \, b^{3}\right )} f x + 12 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-a b - b^{2}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt {-a b - b^{2}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) - {\left (8 \, a^{3} \cos \left (f x + e\right )^{5} - 2 \, {\left (13 \, a^{3} + 6 \, a^{2} b\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (11 \, a^{3} + 18 \, a^{2} b + 8 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \, a^{4} f}, \frac {3 \, {\left (5 \, a^{3} + 30 \, a^{2} b + 40 \, a b^{2} + 16 \, b^{3}\right )} f x + 24 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a b + b^{2}} \arctan \left (\frac {{\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b}{2 \, \sqrt {a b + b^{2}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) - {\left (8 \, a^{3} \cos \left (f x + e\right )^{5} - 2 \, {\left (13 \, a^{3} + 6 \, a^{2} b\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (11 \, a^{3} + 18 \, a^{2} b + 8 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \, a^{4} f}\right ] \] Input:
integrate(sin(f*x+e)^6/(a+b*sec(f*x+e)^2),x, algorithm="fricas")
Output:
[1/48*(3*(5*a^3 + 30*a^2*b + 40*a*b^2 + 16*b^3)*f*x + 12*(a^2 + 2*a*b + 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))*sqr t(-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)) - (8*a^3*cos(f*x + e)^5 - 2*(13*a^3 + 6*a^2*b)*cos(f*x + e)^3 + 3*(11*a^3 + 18*a^2*b + 8*a*b^2)*cos(f*x + e))*sin(f*x + e))/(a^4*f), 1/48 *(3*(5*a^3 + 30*a^2*b + 40*a*b^2 + 16*b^3)*f*x + 24*(a^2 + 2*a*b + b^2)*sq rt(a*b + b^2)*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)/(sqrt(a*b + b^2)*c os(f*x + e)*sin(f*x + e))) - (8*a^3*cos(f*x + e)^5 - 2*(13*a^3 + 6*a^2*b)* cos(f*x + e)^3 + 3*(11*a^3 + 18*a^2*b + 8*a*b^2)*cos(f*x + e))*sin(f*x + e ))/(a^4*f)]
Timed out. \[ \int \frac {\sin ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\text {Timed out} \] Input:
integrate(sin(f*x+e)**6/(a+b*sec(f*x+e)**2),x)
Output:
Timed out
Time = 0.11 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.26 \[ \int \frac {\sin ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {3 \, {\left (11 \, a^{2} + 18 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )^{5} + 8 \, {\left (5 \, a^{2} + 12 \, a b + 6 \, b^{2}\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (5 \, a^{2} + 14 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )}{a^{3} \tan \left (f x + e\right )^{6} + 3 \, a^{3} \tan \left (f x + e\right )^{4} + 3 \, a^{3} \tan \left (f x + e\right )^{2} + a^{3}} - \frac {3 \, {\left (5 \, a^{3} + 30 \, a^{2} b + 40 \, a b^{2} + 16 \, b^{3}\right )} {\left (f x + e\right )}}{a^{4}} + \frac {48 \, {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a^{4}}}{48 \, f} \] Input:
integrate(sin(f*x+e)^6/(a+b*sec(f*x+e)^2),x, algorithm="maxima")
Output:
-1/48*((3*(11*a^2 + 18*a*b + 8*b^2)*tan(f*x + e)^5 + 8*(5*a^2 + 12*a*b + 6 *b^2)*tan(f*x + e)^3 + 3*(5*a^2 + 14*a*b + 8*b^2)*tan(f*x + e))/(a^3*tan(f *x + e)^6 + 3*a^3*tan(f*x + e)^4 + 3*a^3*tan(f*x + e)^2 + a^3) - 3*(5*a^3 + 30*a^2*b + 40*a*b^2 + 16*b^3)*(f*x + e)/a^4 + 48*(a^3*b + 3*a^2*b^2 + 3* a*b^3 + b^4)*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/(sqrt((a + b)*b)*a^4)) /f
Time = 0.17 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.43 \[ \int \frac {\sin ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\frac {3 \, {\left (5 \, a^{3} + 30 \, a^{2} b + 40 \, a b^{2} + 16 \, b^{3}\right )} {\left (f x + e\right )}}{a^{4}} - \frac {48 \, {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )}}{\sqrt {a b + b^{2}} a^{4}} - \frac {33 \, a^{2} \tan \left (f x + e\right )^{5} + 54 \, a b \tan \left (f x + e\right )^{5} + 24 \, b^{2} \tan \left (f x + e\right )^{5} + 40 \, a^{2} \tan \left (f x + e\right )^{3} + 96 \, a b \tan \left (f x + e\right )^{3} + 48 \, b^{2} \tan \left (f x + e\right )^{3} + 15 \, a^{2} \tan \left (f x + e\right ) + 42 \, a b \tan \left (f x + e\right ) + 24 \, b^{2} \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{3} a^{3}}}{48 \, f} \] Input:
integrate(sin(f*x+e)^6/(a+b*sec(f*x+e)^2),x, algorithm="giac")
Output:
1/48*(3*(5*a^3 + 30*a^2*b + 40*a*b^2 + 16*b^3)*(f*x + e)/a^4 - 48*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*(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) - (33*a^2*tan(f*x + e)^5 + 54*a*b*tan(f*x + e)^5 + 24*b^2*tan(f*x + e)^5 + 40*a^2*tan(f*x + e )^3 + 96*a*b*tan(f*x + e)^3 + 48*b^2*tan(f*x + e)^3 + 15*a^2*tan(f*x + e) + 42*a*b*tan(f*x + e) + 24*b^2*tan(f*x + e))/((tan(f*x + e)^2 + 1)^3*a^3)) /f
Time = 13.84 (sec) , antiderivative size = 1448, normalized size of antiderivative = 8.72 \[ \int \frac {\sin ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\text {Too large to display} \] Input:
int(sin(e + f*x)^6/(a + b/cos(e + f*x)^2),x)
Output:
(atanh((25*b^3*tan(e + f*x)*(- 5*a*b^5 - a^5*b - b^6 - 10*a^2*b^4 - 10*a^3 *b^3 - 5*a^4*b^2)^(1/2))/(128*((227*a*b^5)/128 + (217*b^6)/128 + (119*a^2* b^4)/128 + (25*a^3*b^3)/128 + (13*b^7)/(16*a) + (5*b^8)/(32*a^2))) + (11*b ^4*tan(e + f*x)*(- 5*a*b^5 - a^5*b - b^6 - 10*a^2*b^4 - 10*a^3*b^3 - 5*a^4 *b^2)^(1/2))/(32*((217*a*b^6)/128 + (13*b^7)/16 + (227*a^2*b^5)/128 + (119 *a^3*b^4)/128 + (25*a^4*b^3)/128 + (5*b^8)/(32*a))) + (5*b^5*tan(e + f*x)* (- 5*a*b^5 - a^5*b - b^6 - 10*a^2*b^4 - 10*a^3*b^3 - 5*a^4*b^2)^(1/2))/(32 *((13*a*b^7)/16 + (5*b^8)/32 + (217*a^2*b^6)/128 + (227*a^3*b^5)/128 + (11 9*a^4*b^4)/128 + (25*a^5*b^3)/128)))*(-b*(a + b)^5)^(1/2))/(a^4*f) - (atan (((((((512*a^8*b^5 + 1408*a^9*b^4 + 1216*a^10*b^3 + 320*a^11*b^2)/(256*a^9 ) - (tan(e + f*x)*(2048*a^8*b^3 + 1024*a^9*b^2)*(a*b^2*40i + a^2*b*30i + a ^3*5i + b^3*16i))/(4096*a^10))*(a*b^2*40i + a^2*b*30i + a^3*5i + b^3*16i)) /(32*a^4) - (tan(e + f*x)*(2816*a*b^8 + 512*b^9 + 6400*a^2*b^7 + 7680*a^3* b^6 + 5140*a^4*b^5 + 1836*a^5*b^4 + 281*a^6*b^3))/(128*a^6))*(a*b^2*40i + a^2*b*30i + a^3*5i + b^3*16i)*1i)/(32*a^4) - (((((512*a^8*b^5 + 1408*a^9*b ^4 + 1216*a^10*b^3 + 320*a^11*b^2)/(256*a^9) + (tan(e + f*x)*(2048*a^8*b^3 + 1024*a^9*b^2)*(a*b^2*40i + a^2*b*30i + a^3*5i + b^3*16i))/(4096*a^10))* (a*b^2*40i + a^2*b*30i + a^3*5i + b^3*16i))/(32*a^4) + (tan(e + f*x)*(2816 *a*b^8 + 512*b^9 + 6400*a^2*b^7 + 7680*a^3*b^6 + 5140*a^4*b^5 + 1836*a^5*b ^4 + 281*a^6*b^3))/(128*a^6))*(a*b^2*40i + a^2*b*30i + a^3*5i + b^3*16i...
Time = 0.20 (sec) , antiderivative size = 384, normalized size of antiderivative = 2.31 \[ \int \frac {\sin ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {-48 \sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\sqrt {a}}{\sqrt {b}}\right ) a^{2}-96 \sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\sqrt {a}}{\sqrt {b}}\right ) a b -48 \sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\sqrt {a}}{\sqrt {b}}\right ) b^{2}-48 \sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\sqrt {a}}{\sqrt {b}}\right ) a^{2}-96 \sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\sqrt {a}}{\sqrt {b}}\right ) a b -48 \sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\sqrt {a}}{\sqrt {b}}\right ) b^{2}-8 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{5} a^{3}-10 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} a^{3}-12 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} a^{2} b -15 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a^{3}-42 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a^{2} b -24 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a \,b^{2}+15 a^{3} e +15 a^{3} f x +90 a^{2} b e +90 a^{2} b f x +120 a \,b^{2} e +120 a \,b^{2} f x +48 b^{3} e +48 b^{3} f x}{48 a^{4} f} \] Input:
int(sin(f*x+e)^6/(a+b*sec(f*x+e)^2),x)
Output:
( - 48*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/s qrt(b))*a**2 - 96*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*a*b - 48*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*b**2 - 48*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*a**2 - 96*sqrt(b)*sqrt(a + b)*atan ((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*a*b - 48*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*b**2 - 8*cos( e + f*x)*sin(e + f*x)**5*a**3 - 10*cos(e + f*x)*sin(e + f*x)**3*a**3 - 12* cos(e + f*x)*sin(e + f*x)**3*a**2*b - 15*cos(e + f*x)*sin(e + f*x)*a**3 - 42*cos(e + f*x)*sin(e + f*x)*a**2*b - 24*cos(e + f*x)*sin(e + f*x)*a*b**2 + 15*a**3*e + 15*a**3*f*x + 90*a**2*b*e + 90*a**2*b*f*x + 120*a*b**2*e + 1 20*a*b**2*f*x + 48*b**3*e + 48*b**3*f*x)/(48*a**4*f)