Integrand size = 23, antiderivative size = 162 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=-\frac {a \sqrt {b} (3 a+4 b) \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{2 (a-b)^{9/2} f}-\frac {a (a+2 b) \cos (e+f x)}{(a-b)^4 f}+\frac {2 a \cos ^3(e+f x)}{3 (a-b)^3 f}-\frac {\cos ^5(e+f x)}{5 (a-b)^2 f}-\frac {a^2 b \sec (e+f x)}{2 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )} \] Output:
-1/2*a*b^(1/2)*(3*a+4*b)*arctan(b^(1/2)*sec(f*x+e)/(a-b)^(1/2))/(a-b)^(9/2 )/f-a*(a+2*b)*cos(f*x+e)/(a-b)^4/f+2/3*a*cos(f*x+e)^3/(a-b)^3/f-1/5*cos(f* x+e)^5/(a-b)^2/f-1/2*a^2*b*sec(f*x+e)/(a-b)^4/f/(a-b+b*sec(f*x+e)^2)
Time = 2.78 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.33 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\frac {\frac {120 a \sqrt {b} (3 a+4 b) \arctan \left (\frac {\sqrt {a-b}-\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{(a-b)^{9/2}}+\frac {120 a \sqrt {b} (3 a+4 b) \arctan \left (\frac {\sqrt {a-b}+\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{(a-b)^{9/2}}+\frac {-30 \cos (e+f x) \left (18 a b+b^2+a^2 \left (5+\frac {8 b}{a+b+(a-b) \cos (2 (e+f x))}\right )\right )+(a-b) (5 (5 a+3 b) \cos (3 (e+f x))+3 (-a+b) \cos (5 (e+f x)))}{(a-b)^4}}{240 f} \] Input:
Integrate[Sin[e + f*x]^5/(a + b*Tan[e + f*x]^2)^2,x]
Output:
((120*a*Sqrt[b]*(3*a + 4*b)*ArcTan[(Sqrt[a - b] - Sqrt[a]*Tan[(e + f*x)/2] )/Sqrt[b]])/(a - b)^(9/2) + (120*a*Sqrt[b]*(3*a + 4*b)*ArcTan[(Sqrt[a - b] + Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]])/(a - b)^(9/2) + (-30*Cos[e + f*x]*( 18*a*b + b^2 + a^2*(5 + (8*b)/(a + b + (a - b)*Cos[2*(e + f*x)]))) + (a - b)*(5*(5*a + 3*b)*Cos[3*(e + f*x)] + 3*(-a + b)*Cos[5*(e + f*x)]))/(a - b) ^4)/(240*f)
Time = 0.86 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.31, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 4147, 365, 25, 361, 25, 1584, 2009}
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.
| \(\displaystyle \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)^5}{\left (a+b \tan (e+f x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 4147 |
| \(\displaystyle \frac {\int \frac {\cos ^6(e+f x) \left (1-\sec ^2(e+f x)\right )^2}{\left (b \sec ^2(e+f x)+a-b\right )^2}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 365 |
| \(\displaystyle \frac {\frac {\int -\frac {\cos ^4(e+f x) \left (-5 (a-b) \sec ^2(e+f x)+10 a-3 b\right )}{\left (b \sec ^2(e+f x)+a-b\right )^2}d\sec (e+f x)}{5 (a-b)}-\frac {\cos ^5(e+f x)}{5 (a-b) \left (a+b \sec ^2(e+f x)-b\right )}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {\int \frac {\cos ^4(e+f x) \left (-5 (a-b) \sec ^2(e+f x)+10 a-3 b\right )}{\left (b \sec ^2(e+f x)+a-b\right )^2}d\sec (e+f x)}{5 (a-b)}-\frac {\cos ^5(e+f x)}{5 (a-b) \left (a+b \sec ^2(e+f x)-b\right )}}{f}\) |
| \(\Big \downarrow \) 361 |
| \(\displaystyle \frac {-\frac {\frac {b \left (5 a^2+2 b^2\right ) \sec (e+f x)}{2 (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )}-\frac {1}{2} b \int -\frac {\cos ^4(e+f x) \left (\frac {\left (5 a^2+2 b^2\right ) \sec ^4(e+f x)}{(a-b)^3}-\frac {2 \left (5 a^2+2 b^2\right ) \sec ^2(e+f x)}{(a-b)^2 b}+\frac {2 (10 a-3 b)}{(a-b) b}\right )}{b \sec ^2(e+f x)+a-b}d\sec (e+f x)}{5 (a-b)}-\frac {\cos ^5(e+f x)}{5 (a-b) \left (a+b \sec ^2(e+f x)-b\right )}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {\frac {1}{2} b \int \frac {\cos ^4(e+f x) \left (\frac {\left (5 a^2+2 b^2\right ) \sec ^4(e+f x)}{(a-b)^3}-\frac {2 \left (5 a^2+2 b^2\right ) \sec ^2(e+f x)}{(a-b)^2 b}+\frac {2 (10 a-3 b)}{(a-b) b}\right )}{b \sec ^2(e+f x)+a-b}d\sec (e+f x)+\frac {b \left (5 a^2+2 b^2\right ) \sec (e+f x)}{2 (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )}}{5 (a-b)}-\frac {\cos ^5(e+f x)}{5 (a-b) \left (a+b \sec ^2(e+f x)-b\right )}}{f}\) |
| \(\Big \downarrow \) 1584 |
| \(\displaystyle \frac {-\frac {\frac {1}{2} b \int \left (\frac {2 (10 a-3 b) \cos ^4(e+f x)}{(a-b)^2 b}-\frac {2 \left (5 a^2+10 b a-b^2\right ) \cos ^2(e+f x)}{(a-b)^3 b}+\frac {5 a (3 a+4 b)}{(a-b)^3 \left (b \sec ^2(e+f x)+a-b\right )}\right )d\sec (e+f x)+\frac {b \left (5 a^2+2 b^2\right ) \sec (e+f x)}{2 (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )}}{5 (a-b)}-\frac {\cos ^5(e+f x)}{5 (a-b) \left (a+b \sec ^2(e+f x)-b\right )}}{f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {\frac {1}{2} b \left (\frac {2 \left (5 a^2+10 a b-b^2\right ) \cos (e+f x)}{b (a-b)^3}+\frac {5 a (3 a+4 b) \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{\sqrt {b} (a-b)^{7/2}}-\frac {2 (10 a-3 b) \cos ^3(e+f x)}{3 b (a-b)^2}\right )+\frac {b \left (5 a^2+2 b^2\right ) \sec (e+f x)}{2 (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )}}{5 (a-b)}-\frac {\cos ^5(e+f x)}{5 (a-b) \left (a+b \sec ^2(e+f x)-b\right )}}{f}\) |
Input:
Int[Sin[e + f*x]^5/(a + b*Tan[e + f*x]^2)^2,x]
Output:
(-1/5*Cos[e + f*x]^5/((a - b)*(a - b + b*Sec[e + f*x]^2)) - ((b*((5*a*(3*a + 4*b)*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/((a - b)^(7/2)*Sqrt[b] ) + (2*(5*a^2 + 10*a*b - b^2)*Cos[e + f*x])/((a - b)^3*b) - (2*(10*a - 3*b )*Cos[e + f*x]^3)/(3*(a - b)^2*b)))/2 + (b*(5*a^2 + 2*b^2)*Sec[e + f*x])/( 2*(a - b)^3*(a - b + b*Sec[e + f*x]^2)))/(5*(a - b)))/f
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : > Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1)) Int[x^m*(a + b*x^2)^(p + 1)*E xpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)] - ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ILtQ[m/ 2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x _Symbol] :> Simp[c^2*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] - Simp[1/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)*(a + b*x^2)^p*Simp[2*b*c^2*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*d^2*(m + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* (a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[ b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Simp[1/(f*ff^ m) Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a - b + b*ff^2*x^2)^p/x^(m + 1 )), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[( m - 1)/2]
Time = 62.51 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.22
| method | result | size |
| derivativedivides | \(\frac {-\frac {\frac {a^{2} \cos \left (f x +e \right )^{5}}{5}-\frac {2 a b \cos \left (f x +e \right )^{5}}{5}+\frac {b^{2} \cos \left (f x +e \right )^{5}}{5}-\frac {2 a^{2} \cos \left (f x +e \right )^{3}}{3}+\frac {2 a b \cos \left (f x +e \right )^{3}}{3}+\cos \left (f x +e \right ) a^{2}+2 a b \cos \left (f x +e \right )}{\left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )^{2}}+\frac {a b \left (-\frac {a \cos \left (f x +e \right )}{2 \left (a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b \right )}+\frac {\left (3 a +4 b \right ) \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {b \left (a -b \right )}}\right )}{2 \sqrt {b \left (a -b \right )}}\right )}{\left (a -b \right )^{4}}}{f}\) | \(197\) |
| default | \(\frac {-\frac {\frac {a^{2} \cos \left (f x +e \right )^{5}}{5}-\frac {2 a b \cos \left (f x +e \right )^{5}}{5}+\frac {b^{2} \cos \left (f x +e \right )^{5}}{5}-\frac {2 a^{2} \cos \left (f x +e \right )^{3}}{3}+\frac {2 a b \cos \left (f x +e \right )^{3}}{3}+\cos \left (f x +e \right ) a^{2}+2 a b \cos \left (f x +e \right )}{\left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )^{2}}+\frac {a b \left (-\frac {a \cos \left (f x +e \right )}{2 \left (a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b \right )}+\frac {\left (3 a +4 b \right ) \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {b \left (a -b \right )}}\right )}{2 \sqrt {b \left (a -b \right )}}\right )}{\left (a -b \right )^{4}}}{f}\) | \(197\) |
| risch | \(-\frac {5 \,{\mathrm e}^{3 i \left (f x +e \right )} a}{96 \left (-a +b \right )^{3} f}-\frac {{\mathrm e}^{3 i \left (f x +e \right )} b}{32 \left (-a +b \right )^{3} f}-\frac {5 \,{\mathrm e}^{i \left (f x +e \right )} a^{2}}{16 f \left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )^{2}}-\frac {9 \,{\mathrm e}^{i \left (f x +e \right )} a b}{8 f \left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )^{2}}-\frac {{\mathrm e}^{i \left (f x +e \right )} b^{2}}{16 f \left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )^{2}}-\frac {5 \,{\mathrm e}^{-i \left (f x +e \right )} a^{2}}{16 \left (a^{4}-4 a^{3} b +6 a^{2} b^{2}-4 a \,b^{3}+b^{4}\right ) f}-\frac {9 \,{\mathrm e}^{-i \left (f x +e \right )} a b}{8 \left (a^{4}-4 a^{3} b +6 a^{2} b^{2}-4 a \,b^{3}+b^{4}\right ) f}-\frac {{\mathrm e}^{-i \left (f x +e \right )} b^{2}}{16 \left (a^{4}-4 a^{3} b +6 a^{2} b^{2}-4 a \,b^{3}+b^{4}\right ) f}-\frac {5 \,{\mathrm e}^{-3 i \left (f x +e \right )} a}{96 \left (-a^{3}+3 a^{2} b -3 a \,b^{2}+b^{3}\right ) f}-\frac {{\mathrm e}^{-3 i \left (f x +e \right )} b}{32 \left (-a^{3}+3 a^{2} b -3 a \,b^{2}+b^{3}\right ) f}-\frac {b \,a^{2} \left ({\mathrm e}^{3 i \left (f x +e \right )}+{\mathrm e}^{i \left (f x +e \right )}\right )}{\left (a^{2}-2 a b +b^{2}\right )^{2} f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}-b \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+2 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a -b \right )}+\frac {3 i \sqrt {b \left (a -b \right )}\, a^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right )}{4 \left (a -b \right )^{5} f}-\frac {i \sqrt {b \left (a -b \right )}\, a \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right ) b}{\left (a -b \right )^{5} f}+\frac {i \sqrt {b \left (a -b \right )}\, a \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right ) b}{\left (a -b \right )^{5} f}-\frac {3 i \sqrt {b \left (a -b \right )}\, a^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right )}{4 \left (a -b \right )^{5} f}-\frac {\cos \left (5 f x +5 e \right )}{80 f \left (a -b \right )^{2}}\) | \(743\) |
Input:
int(sin(f*x+e)^5/(a+b*tan(f*x+e)^2)^2,x,method=_RETURNVERBOSE)
Output:
1/f*(-1/(a^2-2*a*b+b^2)/(a-b)^2*(1/5*a^2*cos(f*x+e)^5-2/5*a*b*cos(f*x+e)^5 +1/5*b^2*cos(f*x+e)^5-2/3*a^2*cos(f*x+e)^3+2/3*a*b*cos(f*x+e)^3+cos(f*x+e) *a^2+2*a*b*cos(f*x+e))+a*b/(a-b)^4*(-1/2*a*cos(f*x+e)/(a*cos(f*x+e)^2-b*co s(f*x+e)^2+b)+1/2*(3*a+4*b)/(b*(a-b))^(1/2)*arctan((a-b)*cos(f*x+e)/(b*(a- b))^(1/2))))
Time = 0.14 (sec) , antiderivative size = 593, normalized size of antiderivative = 3.66 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\left [-\frac {12 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{7} - 4 \, {\left (10 \, a^{3} - 23 \, a^{2} b + 16 \, a b^{2} - 3 \, b^{3}\right )} \cos \left (f x + e\right )^{5} + 20 \, {\left (3 \, a^{3} + a^{2} b - 4 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left (3 \, a^{2} b + 4 \, a b^{2} + {\left (3 \, a^{3} + a^{2} b - 4 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-\frac {b}{a - b}} \log \left (\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (a - b\right )} \sqrt {-\frac {b}{a - b}} \cos \left (f x + e\right ) - b}{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}\right ) + 30 \, {\left (3 \, a^{2} b + 4 \, a b^{2}\right )} \cos \left (f x + e\right )}{60 \, {\left ({\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} b - 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - 4 \, a b^{4} + b^{5}\right )} f\right )}}, -\frac {6 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{7} - 2 \, {\left (10 \, a^{3} - 23 \, a^{2} b + 16 \, a b^{2} - 3 \, b^{3}\right )} \cos \left (f x + e\right )^{5} + 10 \, {\left (3 \, a^{3} + a^{2} b - 4 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} + 15 \, {\left (3 \, a^{2} b + 4 \, a b^{2} + {\left (3 \, a^{3} + a^{2} b - 4 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{a - b}} \arctan \left (-\frac {{\left (a - b\right )} \sqrt {\frac {b}{a - b}} \cos \left (f x + e\right )}{b}\right ) + 15 \, {\left (3 \, a^{2} b + 4 \, a b^{2}\right )} \cos \left (f x + e\right )}{30 \, {\left ({\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} b - 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - 4 \, a b^{4} + b^{5}\right )} f\right )}}\right ] \] Input:
integrate(sin(f*x+e)^5/(a+b*tan(f*x+e)^2)^2,x, algorithm="fricas")
Output:
[-1/60*(12*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^7 - 4*(10*a^3 - 23 *a^2*b + 16*a*b^2 - 3*b^3)*cos(f*x + e)^5 + 20*(3*a^3 + a^2*b - 4*a*b^2)*c os(f*x + e)^3 - 15*(3*a^2*b + 4*a*b^2 + (3*a^3 + a^2*b - 4*a*b^2)*cos(f*x + e)^2)*sqrt(-b/(a - b))*log(((a - b)*cos(f*x + e)^2 + 2*(a - b)*sqrt(-b/( a - b))*cos(f*x + e) - b)/((a - b)*cos(f*x + e)^2 + b)) + 30*(3*a^2*b + 4* a*b^2)*cos(f*x + e))/((a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*f*cos(f*x + e)^2 + (a^4*b - 4*a^3*b^2 + 6*a^2*b^3 - 4*a*b^4 + b^5)*f ), -1/30*(6*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^7 - 2*(10*a^3 - 2 3*a^2*b + 16*a*b^2 - 3*b^3)*cos(f*x + e)^5 + 10*(3*a^3 + a^2*b - 4*a*b^2)* cos(f*x + e)^3 + 15*(3*a^2*b + 4*a*b^2 + (3*a^3 + a^2*b - 4*a*b^2)*cos(f*x + e)^2)*sqrt(b/(a - b))*arctan(-(a - b)*sqrt(b/(a - b))*cos(f*x + e)/b) + 15*(3*a^2*b + 4*a*b^2)*cos(f*x + e))/((a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^ 2*b^3 + 5*a*b^4 - b^5)*f*cos(f*x + e)^2 + (a^4*b - 4*a^3*b^2 + 6*a^2*b^3 - 4*a*b^4 + b^5)*f)]
Timed out. \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\text {Timed out} \] Input:
integrate(sin(f*x+e)**5/(a+b*tan(f*x+e)**2)**2,x)
Output:
Timed out
Exception generated. \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(sin(f*x+e)^5/(a+b*tan(f*x+e)^2)^2,x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for more details)Is
Leaf count of result is larger than twice the leaf count of optimal. 530 vs. \(2 (146) = 292\).
Time = 0.67 (sec) , antiderivative size = 530, normalized size of antiderivative = 3.27 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx =\text {Too large to display} \] Input:
integrate(sin(f*x+e)^5/(a+b*tan(f*x+e)^2)^2,x, algorithm="giac")
Output:
-1/30*(15*(3*a^2*b + 4*a*b^2)*arctan(-(a*cos(f*x + e) - b*cos(f*x + e) - b )/(sqrt(a*b - b^2)*cos(f*x + e) + sqrt(a*b - b^2)))/((a^4 - 4*a^3*b + 6*a^ 2*b^2 - 4*a*b^3 + b^4)*sqrt(a*b - b^2)) + 30*(a^2*b + a^2*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 2*a*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1))/( (a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*(a + 2*a*(cos(f*x + e) - 1)/(c os(f*x + e) + 1) - 4*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + a*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)) - 4*(8*a^2 + 34*a*b + 3*b^2 - 40*a^2*(c os(f*x + e) - 1)/(cos(f*x + e) + 1) - 140*a*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 80*a^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 160*a*b*(co s(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 30*b^2*(cos(f*x + e) - 1)^2/(cos( f*x + e) + 1)^2 - 180*a*b*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 30*a *b*(cos(f*x + e) - 1)^4/(cos(f*x + e) + 1)^4 + 15*b^2*(cos(f*x + e) - 1)^4 /(cos(f*x + e) + 1)^4)/((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*((cos( f*x + e) - 1)/(cos(f*x + e) + 1) - 1)^5))/f
Time = 11.93 (sec) , antiderivative size = 1049, normalized size of antiderivative = 6.48 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx =\text {Too large to display} \] Input:
int(sin(e + f*x)^5/(a + b*tan(e + f*x)^2)^2,x)
Output:
- ((6*a*b^2 + 83*a^2*b + 16*a^3)/(15*(a - b)*(3*a*b^2 - 3*a^2*b + a^3 - b^ 3)) + (tan(e/2 + (f*x)/2)^8*(366*a*b^2 - 83*a^2*b + 32*a^3))/(3*(a - b)*(3 *a*b^2 - 3*a^2*b + a^3 - b^3)) + (tan(e/2 + (f*x)/2)^4*(1336*a*b^2 + 223*a ^2*b + 16*a^3))/(15*(a - b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)) + (2*tan(e/2 + (f*x)/2)^10*(11*a*b^2 + 6*a^2*b + 4*b^3))/((a - b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)) + (4*tan(e/2 + (f*x)/2)^6*(73*a*b^2 + 32*a^2*b - 12*a^3 + 12*b ^3))/(3*(a - b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)) + (2*tan(e/2 + (f*x)/2)^2 *(145*a*b^2 + 134*a^2*b + 24*a^3 + 12*b^3))/(15*(a - b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)) + (a*tan(e/2 + (f*x)/2)^12*(3*a*b + 4*b^2))/((a - b)*(3*a*b ^2 - 3*a^2*b + a^3 - b^3)))/(f*(a + tan(e/2 + (f*x)/2)^4*(a + 20*b) + tan( e/2 + (f*x)/2)^10*(a + 20*b) + tan(e/2 + (f*x)/2)^2*(3*a + 4*b) + tan(e/2 + (f*x)/2)^12*(3*a + 4*b) - tan(e/2 + (f*x)/2)^6*(5*a - 40*b) - tan(e/2 + (f*x)/2)^8*(5*a - 40*b) + a*tan(e/2 + (f*x)/2)^14)) - (a*b^(1/2)*atan(((a - b)^9*(tan(e/2 + (f*x)/2)^2*((b^(1/2)*(3*a + 4*b)*(24*a^12*b + 32*a^3*b^1 0 - 232*a^4*b^9 + 704*a^5*b^8 - 1120*a^6*b^7 + 896*a^7*b^6 - 112*a^8*b^5 - 448*a^9*b^4 + 416*a^10*b^3 - 160*a^11*b^2))/(4*(a - b)^(17/2)) - (a*b^(1/ 2)*(a - 2*b)*(3*a + 4*b)^2*(224*a^14*b - 16*a^15 + 32*a^2*b^13 - 400*a^3*b ^12 + 2304*a^4*b^11 - 8096*a^5*b^10 + 19360*a^6*b^9 - 33264*a^7*b^8 + 4224 0*a^8*b^7 - 40128*a^9*b^6 + 28512*a^10*b^5 - 14960*a^11*b^4 + 5632*a^12*b^ 3 - 1440*a^13*b^2))/(32*(a - b)^(27/2))) - (a*b^(1/2)*(a - 2*b)*(3*a + ...
Time = 0.25 (sec) , antiderivative size = 1032, normalized size of antiderivative = 6.37 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx =\text {Too large to display} \] Input:
int(sin(f*x+e)^5/(a+b*tan(f*x+e)^2)^2,x)
Output:
(45*sqrt(b)*sqrt(a - b)*atan((sqrt(a - b) - sqrt(a)*tan((e + f*x)/2))/sqrt (b))*sin(e + f*x)**2*a**3 + 15*sqrt(b)*sqrt(a - b)*atan((sqrt(a - b) - sqr t(a)*tan((e + f*x)/2))/sqrt(b))*sin(e + f*x)**2*a**2*b - 60*sqrt(b)*sqrt(a - b)*atan((sqrt(a - b) - sqrt(a)*tan((e + f*x)/2))/sqrt(b))*sin(e + f*x)* *2*a*b**2 - 45*sqrt(b)*sqrt(a - b)*atan((sqrt(a - b) - sqrt(a)*tan((e + f* x)/2))/sqrt(b))*a**3 - 60*sqrt(b)*sqrt(a - b)*atan((sqrt(a - b) - sqrt(a)* tan((e + f*x)/2))/sqrt(b))*a**2*b + 45*sqrt(b)*sqrt(a - b)*atan((sqrt(a - b) + sqrt(a)*tan((e + f*x)/2))/sqrt(b))*sin(e + f*x)**2*a**3 + 15*sqrt(b)* sqrt(a - b)*atan((sqrt(a - b) + sqrt(a)*tan((e + f*x)/2))/sqrt(b))*sin(e + f*x)**2*a**2*b - 60*sqrt(b)*sqrt(a - b)*atan((sqrt(a - b) + sqrt(a)*tan(( e + f*x)/2))/sqrt(b))*sin(e + f*x)**2*a*b**2 - 45*sqrt(b)*sqrt(a - b)*atan ((sqrt(a - b) + sqrt(a)*tan((e + f*x)/2))/sqrt(b))*a**3 - 60*sqrt(b)*sqrt( a - b)*atan((sqrt(a - b) + sqrt(a)*tan((e + f*x)/2))/sqrt(b))*a**2*b - 6*c os(e + f*x)*sin(e + f*x)**6*a**4 + 24*cos(e + f*x)*sin(e + f*x)**6*a**3*b - 36*cos(e + f*x)*sin(e + f*x)**6*a**2*b**2 + 24*cos(e + f*x)*sin(e + f*x) **6*a*b**3 - 6*cos(e + f*x)*sin(e + f*x)**6*b**4 - 2*cos(e + f*x)*sin(e + f*x)**4*a**4 - 6*cos(e + f*x)*sin(e + f*x)**4*a**3*b + 30*cos(e + f*x)*sin (e + f*x)**4*a**2*b**2 - 34*cos(e + f*x)*sin(e + f*x)**4*a*b**3 + 12*cos(e + f*x)*sin(e + f*x)**4*b**4 - 8*cos(e + f*x)*sin(e + f*x)**2*a**4 - 40*co s(e + f*x)*sin(e + f*x)**2*a**3*b + 98*cos(e + f*x)*sin(e + f*x)**2*a**...