Integrand size = 23, antiderivative size = 143 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\sqrt {b} (a+b) (3 a+7 b) \arctan \left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{2 a^{9/2} f}-\frac {(a+b) (a+3 b) \cos (e+f x)}{a^4 f}+\frac {2 (a+b) \cos ^3(e+f x)}{3 a^3 f}-\frac {\cos ^5(e+f x)}{5 a^2 f}-\frac {b (a+b)^2 \cos (e+f x)}{2 a^4 f \left (b+a \cos ^2(e+f x)\right )} \] Output:
1/2*b^(1/2)*(a+b)*(3*a+7*b)*arctan(a^(1/2)*cos(f*x+e)/b^(1/2))/a^(9/2)/f-( a+b)*(a+3*b)*cos(f*x+e)/a^4/f+2/3*(a+b)*cos(f*x+e)^3/a^3/f-1/5*cos(f*x+e)^ 5/a^2/f-1/2*b*(a+b)^2*cos(f*x+e)/a^4/f/(b+a*cos(f*x+e)^2)
Result contains complex when optimal does not.
Time = 5.33 (sec) , antiderivative size = 454, normalized size of antiderivative = 3.17 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\frac {15 \left (3 a^4+384 a^2 b^2+1280 a b^3+896 b^4\right ) \arctan \left (\frac {\left (-\sqrt {a}-i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right ) \sin (e) \tan \left (\frac {f x}{2}\right )+\cos (e) \left (\sqrt {a}-\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right )}{b^{3/2}}+\frac {15 \left (3 a^4+384 a^2 b^2+1280 a b^3+896 b^4\right ) \arctan \left (\frac {\left (-\sqrt {a}+i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right ) \sin (e) \tan \left (\frac {f x}{2}\right )+\cos (e) \left (\sqrt {a}+\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right )}{b^{3/2}}-\frac {45 a^4 \arctan \left (\frac {\sqrt {a}-\sqrt {a+b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{b^{3/2}}-\frac {45 a^4 \arctan \left (\frac {\sqrt {a}+\sqrt {a+b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{b^{3/2}}-\frac {16 \sqrt {a} \cos (e+f x) \left (150 a^3+1436 a^2 b+2960 a b^2+1680 b^3+a \left (125 a^2+688 a b+560 b^2\right ) \cos (2 (e+f x))-2 a^2 (11 a+14 b) \cos (4 (e+f x))+3 a^3 \cos (6 (e+f x))\right )}{a+2 b+a \cos (2 (e+f x))}}{3840 a^{9/2} f} \] Input:
Integrate[Sin[e + f*x]^5/(a + b*Sec[e + f*x]^2)^2,x]
Output:
((15*(3*a^4 + 384*a^2*b^2 + 1280*a*b^3 + 896*b^4)*ArcTan[((-Sqrt[a] - I*Sq rt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2])*Sin[e]*Tan[(f*x)/2] + Cos[e]*(Sqrt[ a] - Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Tan[(f*x)/2]))/Sqrt[b]])/b^(3 /2) + (15*(3*a^4 + 384*a^2*b^2 + 1280*a*b^3 + 896*b^4)*ArcTan[((-Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2])*Sin[e]*Tan[(f*x)/2] + Cos[e]*( Sqrt[a] + Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Tan[(f*x)/2]))/Sqrt[b]]) /b^(3/2) - (45*a^4*ArcTan[(Sqrt[a] - Sqrt[a + b]*Tan[(e + f*x)/2])/Sqrt[b] ])/b^(3/2) - (45*a^4*ArcTan[(Sqrt[a] + Sqrt[a + b]*Tan[(e + f*x)/2])/Sqrt[ b]])/b^(3/2) - (16*Sqrt[a]*Cos[e + f*x]*(150*a^3 + 1436*a^2*b + 2960*a*b^2 + 1680*b^3 + a*(125*a^2 + 688*a*b + 560*b^2)*Cos[2*(e + f*x)] - 2*a^2*(11 *a + 14*b)*Cos[4*(e + f*x)] + 3*a^3*Cos[6*(e + f*x)]))/(a + 2*b + a*Cos[2* (e + f*x)]))/(3840*a^(9/2)*f)
Time = 0.34 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.95, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4621, 366, 25, 363, 254, 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.
| \(\displaystyle \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)^5}{\left (a+b \sec (e+f x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 4621 |
| \(\displaystyle -\frac {\int \frac {\cos ^4(e+f x) \left (1-\cos ^2(e+f x)\right )^2}{\left (a \cos ^2(e+f x)+b\right )^2}d\cos (e+f x)}{f}\) |
| \(\Big \downarrow \) 366 |
| \(\displaystyle -\frac {\frac {(a+b)^2 \cos ^5(e+f x)}{2 a^2 b \left (a \cos ^2(e+f x)+b\right )}-\frac {\int -\frac {\cos ^4(e+f x) \left (2 a^2+2 b \cos ^2(e+f x) a-5 (a+b)^2\right )}{a \cos ^2(e+f x)+b}d\cos (e+f x)}{2 a^2 b}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\frac {\int \frac {\cos ^4(e+f x) \left (2 a^2+2 b \cos ^2(e+f x) a-5 (a+b)^2\right )}{a \cos ^2(e+f x)+b}d\cos (e+f x)}{2 a^2 b}+\frac {(a+b)^2 \cos ^5(e+f x)}{2 a^2 b \left (a \cos ^2(e+f x)+b\right )}}{f}\) |
| \(\Big \downarrow \) 363 |
| \(\displaystyle -\frac {\frac {\frac {2}{5} b \cos ^5(e+f x)-(a+b) (3 a+7 b) \int \frac {\cos ^4(e+f x)}{a \cos ^2(e+f x)+b}d\cos (e+f x)}{2 a^2 b}+\frac {(a+b)^2 \cos ^5(e+f x)}{2 a^2 b \left (a \cos ^2(e+f x)+b\right )}}{f}\) |
| \(\Big \downarrow \) 254 |
| \(\displaystyle -\frac {\frac {\frac {2}{5} b \cos ^5(e+f x)-(a+b) (3 a+7 b) \int \left (\frac {b^2}{a^2 \left (a \cos ^2(e+f x)+b\right )}-\frac {b}{a^2}+\frac {\cos ^2(e+f x)}{a}\right )d\cos (e+f x)}{2 a^2 b}+\frac {(a+b)^2 \cos ^5(e+f x)}{2 a^2 b \left (a \cos ^2(e+f x)+b\right )}}{f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\frac {(a+b)^2 \cos ^5(e+f x)}{2 a^2 b \left (a \cos ^2(e+f x)+b\right )}+\frac {\frac {2}{5} b \cos ^5(e+f x)-(a+b) (3 a+7 b) \left (\frac {b^{3/2} \arctan \left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{a^{5/2}}-\frac {b \cos (e+f x)}{a^2}+\frac {\cos ^3(e+f x)}{3 a}\right )}{2 a^2 b}}{f}\) |
Input:
Int[Sin[e + f*x]^5/(a + b*Sec[e + f*x]^2)^2,x]
Output:
-((((a + b)^2*Cos[e + f*x]^5)/(2*a^2*b*(b + a*Cos[e + f*x]^2)) + ((2*b*Cos [e + f*x]^5)/5 - (a + b)*(3*a + 7*b)*((b^(3/2)*ArcTan[(Sqrt[a]*Cos[e + f*x ])/Sqrt[b]])/a^(5/2) - (b*Cos[e + f*x])/a^2 + Cos[e + f*x]^3/(3*a)))/(2*a^ 2*b))/f)
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3)) Int[(e*x)^ m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d , 0] && NeQ[m + 2*p + 3, 0]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x_Symbol] :> Simp[(-(b*c - a*d)^2)*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(2*a* b^2*e*(p + 1))), x] + Simp[1/(2*a*b^2*(p + 1)) Int[(e*x)^m*(a + b*x^2)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + 2*b^2*c^2*(p + 1) + 2*a*b*d^2*(p + 1)*x^ 2, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && LtQ[p , -1]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_ )]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*((b + a*(ff*x)^n)^p/(ff*x)^(n*p)), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/ 2] && IntegerQ[n] && IntegerQ[p]
Time = 3.41 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.11
| method | result | size |
| derivativedivides | \(\frac {-\frac {\frac {\cos \left (f x +e \right )^{5} a^{2}}{5}-\frac {2 a^{2} \cos \left (f x +e \right )^{3}}{3}-\frac {2 a \cos \left (f x +e \right )^{3} b}{3}+a^{2} \cos \left (f x +e \right )+4 a b \cos \left (f x +e \right )+3 b^{2} \cos \left (f x +e \right )}{a^{4}}+\frac {b \left (\frac {\left (-\frac {1}{2} a^{2}-a b -\frac {1}{2} b^{2}\right ) \cos \left (f x +e \right )}{b +a \cos \left (f x +e \right )^{2}}+\frac {\left (3 a^{2}+10 a b +7 b^{2}\right ) \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{4}}}{f}\) | \(159\) |
| default | \(\frac {-\frac {\frac {\cos \left (f x +e \right )^{5} a^{2}}{5}-\frac {2 a^{2} \cos \left (f x +e \right )^{3}}{3}-\frac {2 a \cos \left (f x +e \right )^{3} b}{3}+a^{2} \cos \left (f x +e \right )+4 a b \cos \left (f x +e \right )+3 b^{2} \cos \left (f x +e \right )}{a^{4}}+\frac {b \left (\frac {\left (-\frac {1}{2} a^{2}-a b -\frac {1}{2} b^{2}\right ) \cos \left (f x +e \right )}{b +a \cos \left (f x +e \right )^{2}}+\frac {\left (3 a^{2}+10 a b +7 b^{2}\right ) \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{4}}}{f}\) | \(159\) |
| risch | \(\frac {5 \,{\mathrm e}^{3 i \left (f x +e \right )}}{96 a^{2} f}+\frac {{\mathrm e}^{3 i \left (f x +e \right )} b}{12 a^{3} f}-\frac {5 \,{\mathrm e}^{i \left (f x +e \right )}}{16 f \,a^{2}}-\frac {7 \,{\mathrm e}^{i \left (f x +e \right )} b}{4 f \,a^{3}}-\frac {3 \,{\mathrm e}^{i \left (f x +e \right )} b^{2}}{2 f \,a^{4}}-\frac {5 \,{\mathrm e}^{-i \left (f x +e \right )}}{16 f \,a^{2}}-\frac {7 \,{\mathrm e}^{-i \left (f x +e \right )} b}{4 f \,a^{3}}-\frac {3 \,{\mathrm e}^{-i \left (f x +e \right )} b^{2}}{2 f \,a^{4}}+\frac {5 \,{\mathrm e}^{-3 i \left (f x +e \right )}}{96 a^{2} f}+\frac {{\mathrm e}^{-3 i \left (f x +e \right )} b}{12 a^{3} f}-\frac {\left (a^{2}+2 a b +b^{2}\right ) b \left ({\mathrm e}^{3 i \left (f x +e \right )}+{\mathrm e}^{i \left (f x +e \right )}\right )}{f \,a^{4} \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 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right )}{4 a^{3} f}+\frac {5 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right ) b}{2 a^{4} f}-\frac {7 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right ) b^{2}}{4 a^{5} f}+\frac {7 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right ) b^{2}}{4 a^{5} f}+\frac {3 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right )}{4 a^{3} f}-\frac {5 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right ) b}{2 a^{4} f}-\frac {\cos \left (5 f x +5 e \right )}{80 f \,a^{2}}\) | \(561\) |
Input:
int(sin(f*x+e)^5/(a+b*sec(f*x+e)^2)^2,x,method=_RETURNVERBOSE)
Output:
1/f*(-1/a^4*(1/5*cos(f*x+e)^5*a^2-2/3*a^2*cos(f*x+e)^3-2/3*a*cos(f*x+e)^3* b+a^2*cos(f*x+e)+4*a*b*cos(f*x+e)+3*b^2*cos(f*x+e))+b/a^4*((-1/2*a^2-a*b-1 /2*b^2)*cos(f*x+e)/(b+a*cos(f*x+e)^2)+1/2*(3*a^2+10*a*b+7*b^2)/(a*b)^(1/2) *arctan(a*cos(f*x+e)/(a*b)^(1/2))))
Time = 0.11 (sec) , antiderivative size = 405, normalized size of antiderivative = 2.83 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\left [-\frac {12 \, a^{3} \cos \left (f x + e\right )^{7} - 4 \, {\left (10 \, a^{3} + 7 \, a^{2} b\right )} \cos \left (f x + e\right )^{5} + 20 \, {\left (3 \, a^{3} + 10 \, a^{2} b + 7 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left (3 \, a^{2} b + 10 \, a b^{2} + 7 \, b^{3} + {\left (3 \, a^{3} + 10 \, a^{2} b + 7 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, a \sqrt {-\frac {b}{a}} \cos \left (f x + e\right ) - b}{a \cos \left (f x + e\right )^{2} + b}\right ) + 30 \, {\left (3 \, a^{2} b + 10 \, a b^{2} + 7 \, b^{3}\right )} \cos \left (f x + e\right )}{60 \, {\left (a^{5} f \cos \left (f x + e\right )^{2} + a^{4} b f\right )}}, -\frac {6 \, a^{3} \cos \left (f x + e\right )^{7} - 2 \, {\left (10 \, a^{3} + 7 \, a^{2} b\right )} \cos \left (f x + e\right )^{5} + 10 \, {\left (3 \, a^{3} + 10 \, a^{2} b + 7 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left (3 \, a^{2} b + 10 \, a b^{2} + 7 \, b^{3} + {\left (3 \, a^{3} + 10 \, a^{2} b + 7 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}} \cos \left (f x + e\right )}{b}\right ) + 15 \, {\left (3 \, a^{2} b + 10 \, a b^{2} + 7 \, b^{3}\right )} \cos \left (f x + e\right )}{30 \, {\left (a^{5} f \cos \left (f x + e\right )^{2} + a^{4} b f\right )}}\right ] \] Input:
integrate(sin(f*x+e)^5/(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")
Output:
[-1/60*(12*a^3*cos(f*x + e)^7 - 4*(10*a^3 + 7*a^2*b)*cos(f*x + e)^5 + 20*( 3*a^3 + 10*a^2*b + 7*a*b^2)*cos(f*x + e)^3 - 15*(3*a^2*b + 10*a*b^2 + 7*b^ 3 + (3*a^3 + 10*a^2*b + 7*a*b^2)*cos(f*x + e)^2)*sqrt(-b/a)*log(-(a*cos(f* x + e)^2 + 2*a*sqrt(-b/a)*cos(f*x + e) - b)/(a*cos(f*x + e)^2 + b)) + 30*( 3*a^2*b + 10*a*b^2 + 7*b^3)*cos(f*x + e))/(a^5*f*cos(f*x + e)^2 + a^4*b*f) , -1/30*(6*a^3*cos(f*x + e)^7 - 2*(10*a^3 + 7*a^2*b)*cos(f*x + e)^5 + 10*( 3*a^3 + 10*a^2*b + 7*a*b^2)*cos(f*x + e)^3 - 15*(3*a^2*b + 10*a*b^2 + 7*b^ 3 + (3*a^3 + 10*a^2*b + 7*a*b^2)*cos(f*x + e)^2)*sqrt(b/a)*arctan(a*sqrt(b /a)*cos(f*x + e)/b) + 15*(3*a^2*b + 10*a*b^2 + 7*b^3)*cos(f*x + e))/(a^5*f *cos(f*x + e)^2 + a^4*b*f)]
Timed out. \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\text {Timed out} \] Input:
integrate(sin(f*x+e)**5/(a+b*sec(f*x+e)**2)**2,x)
Output:
Timed out
Time = 0.11 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.03 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=-\frac {\frac {15 \, {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )}{a^{5} \cos \left (f x + e\right )^{2} + a^{4} b} - \frac {15 \, {\left (3 \, a^{2} b + 10 \, a b^{2} + 7 \, b^{3}\right )} \arctan \left (\frac {a \cos \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {a b} a^{4}} + \frac {2 \, {\left (3 \, a^{2} \cos \left (f x + e\right )^{5} - 10 \, {\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{3} + 15 \, {\left (a^{2} + 4 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )\right )}}{a^{4}}}{30 \, f} \] Input:
integrate(sin(f*x+e)^5/(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")
Output:
-1/30*(15*(a^2*b + 2*a*b^2 + b^3)*cos(f*x + e)/(a^5*cos(f*x + e)^2 + a^4*b ) - 15*(3*a^2*b + 10*a*b^2 + 7*b^3)*arctan(a*cos(f*x + e)/sqrt(a*b))/(sqrt (a*b)*a^4) + 2*(3*a^2*cos(f*x + e)^5 - 10*(a^2 + a*b)*cos(f*x + e)^3 + 15* (a^2 + 4*a*b + 3*b^2)*cos(f*x + e))/a^4)/f
Time = 0.12 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.44 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {{\left (3 \, a^{2} b + 10 \, a b^{2} + 7 \, b^{3}\right )} \arctan \left (\frac {a \cos \left (f x + e\right )}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{4} f} - \frac {a^{2} b \cos \left (f x + e\right ) + 2 \, a b^{2} \cos \left (f x + e\right ) + b^{3} \cos \left (f x + e\right )}{2 \, {\left (a \cos \left (f x + e\right )^{2} + b\right )} a^{4} f} - \frac {3 \, a^{8} f^{4} \cos \left (f x + e\right )^{5} - 10 \, a^{8} f^{4} \cos \left (f x + e\right )^{3} - 10 \, a^{7} b f^{4} \cos \left (f x + e\right )^{3} + 15 \, a^{8} f^{4} \cos \left (f x + e\right ) + 60 \, a^{7} b f^{4} \cos \left (f x + e\right ) + 45 \, a^{6} b^{2} f^{4} \cos \left (f x + e\right )}{15 \, a^{10} f^{5}} \] Input:
integrate(sin(f*x+e)^5/(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")
Output:
1/2*(3*a^2*b + 10*a*b^2 + 7*b^3)*arctan(a*cos(f*x + e)/sqrt(a*b))/(sqrt(a* b)*a^4*f) - 1/2*(a^2*b*cos(f*x + e) + 2*a*b^2*cos(f*x + e) + b^3*cos(f*x + e))/((a*cos(f*x + e)^2 + b)*a^4*f) - 1/15*(3*a^8*f^4*cos(f*x + e)^5 - 10* a^8*f^4*cos(f*x + e)^3 - 10*a^7*b*f^4*cos(f*x + e)^3 + 15*a^8*f^4*cos(f*x + e) + 60*a^7*b*f^4*cos(f*x + e) + 45*a^6*b^2*f^4*cos(f*x + e))/(a^10*f^5)
Time = 0.15 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.36 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {{\cos \left (e+f\,x\right )}^3\,\left (\frac {2\,b}{3\,a^3}+\frac {2}{3\,a^2}\right )}{f}-\frac {{\cos \left (e+f\,x\right )}^5}{5\,a^2\,f}-\frac {\cos \left (e+f\,x\right )\,\left (\frac {1}{a^2}-\frac {b^2}{a^4}+\frac {2\,b\,\left (\frac {2\,b}{a^3}+\frac {2}{a^2}\right )}{a}\right )}{f}-\frac {\cos \left (e+f\,x\right )\,\left (\frac {a^2\,b}{2}+a\,b^2+\frac {b^3}{2}\right )}{f\,\left (a^5\,{\cos \left (e+f\,x\right )}^2+b\,a^4\right )}+\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {b}\,\cos \left (e+f\,x\right )\,\left (a+b\right )\,\left (3\,a+7\,b\right )}{3\,a^2\,b+10\,a\,b^2+7\,b^3}\right )\,\left (a+b\right )\,\left (3\,a+7\,b\right )}{2\,a^{9/2}\,f} \] Input:
int(sin(e + f*x)^5/(a + b/cos(e + f*x)^2)^2,x)
Output:
(cos(e + f*x)^3*((2*b)/(3*a^3) + 2/(3*a^2)))/f - cos(e + f*x)^5/(5*a^2*f) - (cos(e + f*x)*(1/a^2 - b^2/a^4 + (2*b*((2*b)/a^3 + 2/a^2))/a))/f - (cos( e + f*x)*(a*b^2 + (a^2*b)/2 + b^3/2))/(f*(a^4*b + a^5*cos(e + f*x)^2)) + ( b^(1/2)*atan((a^(1/2)*b^(1/2)*cos(e + f*x)*(a + b)*(3*a + 7*b))/(10*a*b^2 + 3*a^2*b + 7*b^3))*(a + b)*(3*a + 7*b))/(2*a^(9/2)*f)
Time = 0.25 (sec) , antiderivative size = 782, normalized size of antiderivative = 5.47 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx =\text {Too large to display} \] Input:
int(sin(f*x+e)^5/(a+b*sec(f*x+e)^2)^2,x)
Output:
( - 45*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt( b))*sin(e + f*x)**2*a**3 - 150*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**2*a**2*b - 105*sqrt(b)*sqrt(a)*a tan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**2*a*b* *2 + 45*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt (b))*a**3 + 195*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt( a))/sqrt(b))*a**2*b + 255*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/ 2) - sqrt(a))/sqrt(b))*a*b**2 + 105*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan( (e + f*x)/2) - sqrt(a))/sqrt(b))*b**3 + 45*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*sin(e + f*x)**2*a**3 + 150*sqrt(b) *sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*sin(e + f* x)**2*a**2*b + 105*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) + sq rt(a))/sqrt(b))*sin(e + f*x)**2*a*b**2 - 45*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*a**3 - 195*sqrt(b)*sqrt(a)*atan(( sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*a**2*b - 255*sqrt(b)*sqrt (a)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*a*b**2 - 105*sq rt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*b**3 - 6*cos(e + f*x)*sin(e + f*x)**6*a**4 - 2*cos(e + f*x)*sin(e + f*x)**4*a** 4 - 14*cos(e + f*x)*sin(e + f*x)**4*a**3*b - 8*cos(e + f*x)*sin(e + f*x)** 2*a**4 - 72*cos(e + f*x)*sin(e + f*x)**2*a**3*b - 70*cos(e + f*x)*sin(e...