Integrand size = 23, antiderivative size = 214 \[ \int \frac {\cos ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {b^3 \left (80 a^2+140 a b+63 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{8 a^{11/2} (a+b)^{5/2} f}+\frac {\left (a^2-3 a b+6 b^2\right ) \sin (e+f x)}{a^5 f}-\frac {(2 a-3 b) \sin ^3(e+f x)}{3 a^4 f}+\frac {\sin ^5(e+f x)}{5 a^3 f}-\frac {b^5 \sin (e+f x)}{4 a^5 (a+b) f \left (a+b-a \sin ^2(e+f x)\right )^2}+\frac {b^4 (20 a+17 b) \sin (e+f x)}{8 a^5 (a+b)^2 f \left (a+b-a \sin ^2(e+f x)\right )} \] Output:
-1/8*b^3*(80*a^2+140*a*b+63*b^2)*arctanh(a^(1/2)*sin(f*x+e)/(a+b)^(1/2))/a ^(11/2)/(a+b)^(5/2)/f+(a^2-3*a*b+6*b^2)*sin(f*x+e)/a^5/f-1/3*(2*a-3*b)*sin (f*x+e)^3/a^4/f+1/5*sin(f*x+e)^5/a^3/f-1/4*b^5*sin(f*x+e)/a^5/(a+b)/f/(a+b -a*sin(f*x+e)^2)^2+1/8*b^4*(20*a+17*b)*sin(f*x+e)/a^5/(a+b)^2/f/(a+b-a*sin (f*x+e)^2)
Result contains complex when optimal does not.
Time = 7.26 (sec) , antiderivative size = 2670, normalized size of antiderivative = 12.48 \[ \int \frac {\cos ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Result too large to show} \] Input:
Integrate[Cos[e + f*x]^5/(a + b*Sec[e + f*x]^2)^3,x]
Output:
((5*a^2 - 18*a*b + 48*b^2)*Cos[f*x]*(a + 2*b + a*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*Sin[e])/(64*a^5*f*(a + b*Sec[e + f*x]^2)^3) + ((-80*a^2*b^3 - 14 0*a*b^4 - 63*b^5)*(a + 2*b + a*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*(((I/128 )*ArcTan[((-I)*a*Cos[e] - I*b*Cos[e] + I*a*Cos[3*e] + I*b*Cos[3*e] + a*Sin [e] + b*Sin[e] - Sqrt[a]*Sqrt[a + b]*Cos[e - f*x]*Sqrt[Cos[2*e] - I*Sin[2* e]] + Sqrt[a]*Sqrt[a + b]*Cos[3*e + f*x]*Sqrt[Cos[2*e] - I*Sin[2*e]] + a*S in[3*e] + b*Sin[3*e] - I*Sqrt[a]*Sqrt[a + b]*Sqrt[Cos[2*e] - I*Sin[2*e]]*S in[e - f*x] - (2*I)*Sqrt[a]*Sqrt[a + b]*Sqrt[Cos[2*e] - I*Sin[2*e]]*Sin[e + f*x] + I*Sqrt[a]*Sqrt[a + b]*Sqrt[Cos[2*e] - I*Sin[2*e]]*Sin[3*e + f*x]) /(a*Cos[e] + 3*b*Cos[e] + a*Cos[3*e] + b*Cos[3*e] + a*Cos[e + 2*f*x] + a*C os[3*e + 2*f*x] - (3*I)*a*Sin[e] - I*b*Sin[e] - I*a*Sin[3*e] - I*b*Sin[3*e ] - I*a*Sin[e + 2*f*x] + I*a*Sin[3*e + 2*f*x])]*Cos[e])/(a^(11/2)*Sqrt[a + b]*f*Sqrt[Cos[2*e] - I*Sin[2*e]]) + (ArcTan[((-I)*a*Cos[e] - I*b*Cos[e] + I*a*Cos[3*e] + I*b*Cos[3*e] + a*Sin[e] + b*Sin[e] - Sqrt[a]*Sqrt[a + b]*C os[e - f*x]*Sqrt[Cos[2*e] - I*Sin[2*e]] + Sqrt[a]*Sqrt[a + b]*Cos[3*e + f* x]*Sqrt[Cos[2*e] - I*Sin[2*e]] + a*Sin[3*e] + b*Sin[3*e] - I*Sqrt[a]*Sqrt[ a + b]*Sqrt[Cos[2*e] - I*Sin[2*e]]*Sin[e - f*x] - (2*I)*Sqrt[a]*Sqrt[a + b ]*Sqrt[Cos[2*e] - I*Sin[2*e]]*Sin[e + f*x] + I*Sqrt[a]*Sqrt[a + b]*Sqrt[Co s[2*e] - I*Sin[2*e]]*Sin[3*e + f*x])/(a*Cos[e] + 3*b*Cos[e] + a*Cos[3*e] + b*Cos[3*e] + a*Cos[e + 2*f*x] + a*Cos[3*e + 2*f*x] - (3*I)*a*Sin[e] - ...
Time = 0.43 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4635, 300, 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.
| \(\displaystyle \int \frac {\cos ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sec (e+f x)^5 \left (a+b \sec (e+f x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 4635 |
| \(\displaystyle \frac {\int \frac {\left (1-\sin ^2(e+f x)\right )^5}{\left (-a \sin ^2(e+f x)+a+b\right )^3}d\sin (e+f x)}{f}\) |
| \(\Big \downarrow \) 300 |
| \(\displaystyle \frac {\int \left (\frac {\sin ^4(e+f x)}{a^3}-\frac {(2 a-3 b) \sin ^2(e+f x)}{a^4}+\frac {a^2-3 b a+6 b^2}{a^5}-\frac {10 a^2 b^3 \sin ^4(e+f x)-5 a b^3 (4 a+3 b) \sin ^2(e+f x)+b^3 \left (10 a^2+15 b a+6 b^2\right )}{a^5 \left (-a \sin ^2(e+f x)+a+b\right )^3}\right )d\sin (e+f x)}{f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {b^5 \sin (e+f x)}{4 a^5 (a+b) \left (-a \sin ^2(e+f x)+a+b\right )^2}+\frac {b^4 (20 a+17 b) \sin (e+f x)}{8 a^5 (a+b)^2 \left (-a \sin ^2(e+f x)+a+b\right )}-\frac {(2 a-3 b) \sin ^3(e+f x)}{3 a^4}+\frac {\sin ^5(e+f x)}{5 a^3}-\frac {b^3 \left (80 a^2+140 a b+63 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{8 a^{11/2} (a+b)^{5/2}}+\frac {\left (a^2-3 a b+6 b^2\right ) \sin (e+f x)}{a^5}}{f}\) |
Input:
Int[Cos[e + f*x]^5/(a + b*Sec[e + f*x]^2)^3,x]
Output:
(-1/8*(b^3*(80*a^2 + 140*a*b + 63*b^2)*ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt [a + b]])/(a^(11/2)*(a + b)^(5/2)) + ((a^2 - 3*a*b + 6*b^2)*Sin[e + f*x])/ a^5 - ((2*a - 3*b)*Sin[e + f*x]^3)/(3*a^4) + Sin[e + f*x]^5/(5*a^3) - (b^5 *Sin[e + f*x])/(4*a^5*(a + b)*(a + b - a*Sin[e + f*x]^2)^2) + (b^4*(20*a + 17*b)*Sin[e + f*x])/(8*a^5*(a + b)^2*(a + b - a*Sin[e + f*x]^2)))/f
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int [PolynomialDivide[(a + b*x^2)^p, (c + d*x^2)^(-q), x], x] /; FreeQ[{a, b, c , d}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_ ))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f Subst[Int[ExpandToSum[b + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 - ff^2*x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && In tegerQ[(m - 1)/2] && IntegerQ[n/2] && IntegerQ[p]
Time = 6.95 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.00
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {a^{2} \sin \left (f x +e \right )^{5}}{5}-\frac {2 a^{2} \sin \left (f x +e \right )^{3}}{3}+a b \sin \left (f x +e \right )^{3}+\sin \left (f x +e \right ) a^{2}-3 \sin \left (f x +e \right ) a b +6 \sin \left (f x +e \right ) b^{2}}{a^{5}}+\frac {b^{3} \left (\frac {-\frac {a b \left (20 a +17 b \right ) \sin \left (f x +e \right )^{3}}{8 \left (a^{2}+2 a b +b^{2}\right )}+\frac {5 \left (4 a +3 b \right ) b \sin \left (f x +e \right )}{8 \left (a +b \right )}}{\left (-a -b +a \sin \left (f x +e \right )^{2}\right )^{2}}-\frac {\left (80 a^{2}+140 a b +63 b^{2}\right ) \operatorname {arctanh}\left (\frac {a \sin \left (f x +e \right )}{\sqrt {a \left (a +b \right )}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) \sqrt {a \left (a +b \right )}}\right )}{a^{5}}}{f}\) | \(214\) |
| default | \(\frac {\frac {\frac {a^{2} \sin \left (f x +e \right )^{5}}{5}-\frac {2 a^{2} \sin \left (f x +e \right )^{3}}{3}+a b \sin \left (f x +e \right )^{3}+\sin \left (f x +e \right ) a^{2}-3 \sin \left (f x +e \right ) a b +6 \sin \left (f x +e \right ) b^{2}}{a^{5}}+\frac {b^{3} \left (\frac {-\frac {a b \left (20 a +17 b \right ) \sin \left (f x +e \right )^{3}}{8 \left (a^{2}+2 a b +b^{2}\right )}+\frac {5 \left (4 a +3 b \right ) b \sin \left (f x +e \right )}{8 \left (a +b \right )}}{\left (-a -b +a \sin \left (f x +e \right )^{2}\right )^{2}}-\frac {\left (80 a^{2}+140 a b +63 b^{2}\right ) \operatorname {arctanh}\left (\frac {a \sin \left (f x +e \right )}{\sqrt {a \left (a +b \right )}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) \sqrt {a \left (a +b \right )}}\right )}{a^{5}}}{f}\) | \(214\) |
| risch | \(\frac {5 i {\mathrm e}^{-i \left (f x +e \right )}}{16 a^{3} f}-\frac {i {\mathrm e}^{5 i \left (f x +e \right )}}{160 a^{3} f}+\frac {5 i {\mathrm e}^{-3 i \left (f x +e \right )}}{96 a^{3} f}+\frac {3 i {\mathrm e}^{-i \left (f x +e \right )} b^{2}}{f \,a^{5}}-\frac {5 i {\mathrm e}^{3 i \left (f x +e \right )}}{96 a^{3} f}+\frac {9 i {\mathrm e}^{i \left (f x +e \right )} b}{8 a^{4} f}-\frac {i b^{4} \left (20 a^{2} {\mathrm e}^{7 i \left (f x +e \right )}+17 a b \,{\mathrm e}^{7 i \left (f x +e \right )}+20 a^{2} {\mathrm e}^{5 i \left (f x +e \right )}+89 a b \,{\mathrm e}^{5 i \left (f x +e \right )}+60 b^{2} {\mathrm e}^{5 i \left (f x +e \right )}-20 a^{2} {\mathrm e}^{3 i \left (f x +e \right )}-89 a b \,{\mathrm e}^{3 i \left (f x +e \right )}-60 b^{2} {\mathrm e}^{3 i \left (f x +e \right )}-20 a^{2} {\mathrm e}^{i \left (f x +e \right )}-17 a b \,{\mathrm e}^{i \left (f x +e \right )}\right )}{4 a^{5} \left (a +b \right )^{2} 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 )^{2}}-\frac {3 i {\mathrm e}^{i \left (f x +e \right )} b^{2}}{f \,a^{5}}+\frac {i {\mathrm e}^{-5 i \left (f x +e \right )}}{160 a^{3} f}+\frac {i {\mathrm e}^{3 i \left (f x +e \right )} b}{8 a^{4} f}-\frac {5 i {\mathrm e}^{i \left (f x +e \right )}}{16 a^{3} f}-\frac {i {\mathrm e}^{-3 i \left (f x +e \right )} b}{8 a^{4} f}-\frac {9 i {\mathrm e}^{-i \left (f x +e \right )} b}{8 a^{4} f}+\frac {5 b^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right )}{\sqrt {a^{2}+a b}\, \left (a +b \right )^{2} f \,a^{3}}+\frac {35 b^{4} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right )}{4 \sqrt {a^{2}+a b}\, \left (a +b \right )^{2} f \,a^{4}}+\frac {63 b^{5} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right )}{16 \sqrt {a^{2}+a b}\, \left (a +b \right )^{2} f \,a^{5}}-\frac {5 b^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right )}{\sqrt {a^{2}+a b}\, \left (a +b \right )^{2} f \,a^{3}}-\frac {35 b^{4} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right )}{4 \sqrt {a^{2}+a b}\, \left (a +b \right )^{2} f \,a^{4}}-\frac {63 b^{5} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right )}{16 \sqrt {a^{2}+a b}\, \left (a +b \right )^{2} f \,a^{5}}\) | \(787\) |
Input:
int(cos(f*x+e)^5/(a+b*sec(f*x+e)^2)^3,x,method=_RETURNVERBOSE)
Output:
1/f*(1/a^5*(1/5*a^2*sin(f*x+e)^5-2/3*a^2*sin(f*x+e)^3+a*b*sin(f*x+e)^3+sin (f*x+e)*a^2-3*sin(f*x+e)*a*b+6*sin(f*x+e)*b^2)+b^3/a^5*((-1/8*a*b*(20*a+17 *b)/(a^2+2*a*b+b^2)*sin(f*x+e)^3+5/8*(4*a+3*b)*b/(a+b)*sin(f*x+e))/(-a-b+a *sin(f*x+e)^2)^2-1/8*(80*a^2+140*a*b+63*b^2)/(a^2+2*a*b+b^2)/(a*(a+b))^(1/ 2)*arctanh(a*sin(f*x+e)/(a*(a+b))^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 485 vs. \(2 (202) = 404\).
Time = 0.18 (sec) , antiderivative size = 995, normalized size of antiderivative = 4.65 \[ \int \frac {\cos ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:
integrate(cos(f*x+e)^5/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")
Output:
[1/240*(15*(80*a^2*b^5 + 140*a*b^6 + 63*b^7 + (80*a^4*b^3 + 140*a^3*b^4 + 63*a^2*b^5)*cos(f*x + e)^4 + 2*(80*a^3*b^4 + 140*a^2*b^5 + 63*a*b^6)*cos(f *x + e)^2)*sqrt(a^2 + a*b)*log(-(a*cos(f*x + e)^2 + 2*sqrt(a^2 + a*b)*sin( f*x + e) - 2*a - b)/(a*cos(f*x + e)^2 + b)) + 2*(24*(a^8 + 3*a^7*b + 3*a^6 *b^2 + a^5*b^3)*cos(f*x + e)^8 + 64*a^6*b^2 - 48*a^5*b^3 + 192*a^4*b^4 + 1 774*a^3*b^5 + 2415*a^2*b^6 + 945*a*b^7 + 8*(4*a^8 + 3*a^7*b - 15*a^6*b^2 - 23*a^5*b^3 - 9*a^4*b^4)*cos(f*x + e)^6 + 8*(8*a^8 + 2*a^7*b + 21*a^6*b^2 + 131*a^5*b^3 + 167*a^4*b^4 + 63*a^3*b^5)*cos(f*x + e)^4 + (128*a^7*b - 64 *a^6*b^2 + 360*a^5*b^3 + 3044*a^4*b^4 + 4067*a^3*b^5 + 1575*a^2*b^6)*cos(f *x + e)^2)*sin(f*x + e))/((a^11 + 3*a^10*b + 3*a^9*b^2 + a^8*b^3)*f*cos(f* x + e)^4 + 2*(a^10*b + 3*a^9*b^2 + 3*a^8*b^3 + a^7*b^4)*f*cos(f*x + e)^2 + (a^9*b^2 + 3*a^8*b^3 + 3*a^7*b^4 + a^6*b^5)*f), 1/120*(15*(80*a^2*b^5 + 1 40*a*b^6 + 63*b^7 + (80*a^4*b^3 + 140*a^3*b^4 + 63*a^2*b^5)*cos(f*x + e)^4 + 2*(80*a^3*b^4 + 140*a^2*b^5 + 63*a*b^6)*cos(f*x + e)^2)*sqrt(-a^2 - a*b )*arctan(sqrt(-a^2 - a*b)*sin(f*x + e)/(a + b)) + (24*(a^8 + 3*a^7*b + 3*a ^6*b^2 + a^5*b^3)*cos(f*x + e)^8 + 64*a^6*b^2 - 48*a^5*b^3 + 192*a^4*b^4 + 1774*a^3*b^5 + 2415*a^2*b^6 + 945*a*b^7 + 8*(4*a^8 + 3*a^7*b - 15*a^6*b^2 - 23*a^5*b^3 - 9*a^4*b^4)*cos(f*x + e)^6 + 8*(8*a^8 + 2*a^7*b + 21*a^6*b^ 2 + 131*a^5*b^3 + 167*a^4*b^4 + 63*a^3*b^5)*cos(f*x + e)^4 + (128*a^7*b - 64*a^6*b^2 + 360*a^5*b^3 + 3044*a^4*b^4 + 4067*a^3*b^5 + 1575*a^2*b^6)*...
Timed out. \[ \int \frac {\cos ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Timed out} \] Input:
integrate(cos(f*x+e)**5/(a+b*sec(f*x+e)**2)**3,x)
Output:
Timed out
Time = 0.11 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.42 \[ \int \frac {\cos ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\frac {15 \, {\left (80 \, a^{2} b^{3} + 140 \, a b^{4} + 63 \, b^{5}\right )} \log \left (\frac {a \sin \left (f x + e\right ) - \sqrt {{\left (a + b\right )} a}}{a \sin \left (f x + e\right ) + \sqrt {{\left (a + b\right )} a}}\right )}{{\left (a^{7} + 2 \, a^{6} b + a^{5} b^{2}\right )} \sqrt {{\left (a + b\right )} a}} - \frac {30 \, {\left ({\left (20 \, a^{2} b^{4} + 17 \, a b^{5}\right )} \sin \left (f x + e\right )^{3} - 5 \, {\left (4 \, a^{2} b^{4} + 7 \, a b^{5} + 3 \, b^{6}\right )} \sin \left (f x + e\right )\right )}}{a^{9} + 4 \, a^{8} b + 6 \, a^{7} b^{2} + 4 \, a^{6} b^{3} + a^{5} b^{4} + {\left (a^{9} + 2 \, a^{8} b + a^{7} b^{2}\right )} \sin \left (f x + e\right )^{4} - 2 \, {\left (a^{9} + 3 \, a^{8} b + 3 \, a^{7} b^{2} + a^{6} b^{3}\right )} \sin \left (f x + e\right )^{2}} + \frac {16 \, {\left (3 \, a^{2} \sin \left (f x + e\right )^{5} - 5 \, {\left (2 \, a^{2} - 3 \, a b\right )} \sin \left (f x + e\right )^{3} + 15 \, {\left (a^{2} - 3 \, a b + 6 \, b^{2}\right )} \sin \left (f x + e\right )\right )}}{a^{5}}}{240 \, f} \] Input:
integrate(cos(f*x+e)^5/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")
Output:
1/240*(15*(80*a^2*b^3 + 140*a*b^4 + 63*b^5)*log((a*sin(f*x + e) - sqrt((a + b)*a))/(a*sin(f*x + e) + sqrt((a + b)*a)))/((a^7 + 2*a^6*b + a^5*b^2)*sq rt((a + b)*a)) - 30*((20*a^2*b^4 + 17*a*b^5)*sin(f*x + e)^3 - 5*(4*a^2*b^4 + 7*a*b^5 + 3*b^6)*sin(f*x + e))/(a^9 + 4*a^8*b + 6*a^7*b^2 + 4*a^6*b^3 + a^5*b^4 + (a^9 + 2*a^8*b + a^7*b^2)*sin(f*x + e)^4 - 2*(a^9 + 3*a^8*b + 3 *a^7*b^2 + a^6*b^3)*sin(f*x + e)^2) + 16*(3*a^2*sin(f*x + e)^5 - 5*(2*a^2 - 3*a*b)*sin(f*x + e)^3 + 15*(a^2 - 3*a*b + 6*b^2)*sin(f*x + e))/a^5)/f
Time = 0.19 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.27 \[ \int \frac {\cos ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\frac {15 \, {\left (80 \, a^{2} b^{3} + 140 \, a b^{4} + 63 \, b^{5}\right )} \arctan \left (\frac {a \sin \left (f x + e\right )}{\sqrt {-a^{2} - a b}}\right )}{{\left (a^{7} + 2 \, a^{6} b + a^{5} b^{2}\right )} \sqrt {-a^{2} - a b}} - \frac {15 \, {\left (20 \, a^{2} b^{4} \sin \left (f x + e\right )^{3} + 17 \, a b^{5} \sin \left (f x + e\right )^{3} - 20 \, a^{2} b^{4} \sin \left (f x + e\right ) - 35 \, a b^{5} \sin \left (f x + e\right ) - 15 \, b^{6} \sin \left (f x + e\right )\right )}}{{\left (a^{7} + 2 \, a^{6} b + a^{5} b^{2}\right )} {\left (a \sin \left (f x + e\right )^{2} - a - b\right )}^{2}} + \frac {8 \, {\left (3 \, a^{12} \sin \left (f x + e\right )^{5} - 10 \, a^{12} \sin \left (f x + e\right )^{3} + 15 \, a^{11} b \sin \left (f x + e\right )^{3} + 15 \, a^{12} \sin \left (f x + e\right ) - 45 \, a^{11} b \sin \left (f x + e\right ) + 90 \, a^{10} b^{2} \sin \left (f x + e\right )\right )}}{a^{15}}}{120 \, f} \] Input:
integrate(cos(f*x+e)^5/(a+b*sec(f*x+e)^2)^3,x, algorithm="giac")
Output:
1/120*(15*(80*a^2*b^3 + 140*a*b^4 + 63*b^5)*arctan(a*sin(f*x + e)/sqrt(-a^ 2 - a*b))/((a^7 + 2*a^6*b + a^5*b^2)*sqrt(-a^2 - a*b)) - 15*(20*a^2*b^4*si n(f*x + e)^3 + 17*a*b^5*sin(f*x + e)^3 - 20*a^2*b^4*sin(f*x + e) - 35*a*b^ 5*sin(f*x + e) - 15*b^6*sin(f*x + e))/((a^7 + 2*a^6*b + a^5*b^2)*(a*sin(f* x + e)^2 - a - b)^2) + 8*(3*a^12*sin(f*x + e)^5 - 10*a^12*sin(f*x + e)^3 + 15*a^11*b*sin(f*x + e)^3 + 15*a^12*sin(f*x + e) - 45*a^11*b*sin(f*x + e) + 90*a^10*b^2*sin(f*x + e))/a^15)/f
Time = 0.26 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.20 \[ \int \frac {\cos ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\frac {5\,\sin \left (e+f\,x\right )\,\left (3\,b^5+4\,a\,b^4\right )}{8\,\left (a+b\right )}-\frac {{\sin \left (e+f\,x\right )}^3\,\left (20\,a^2\,b^4+17\,a\,b^5\right )}{8\,{\left (a+b\right )}^2}}{f\,\left (2\,a^6\,b-{\sin \left (e+f\,x\right )}^2\,\left (2\,a^7+2\,b\,a^6\right )+a^7+a^5\,b^2+a^7\,{\sin \left (e+f\,x\right )}^4\right )}+\frac {{\sin \left (e+f\,x\right )}^5}{5\,a^3\,f}+\frac {{\sin \left (e+f\,x\right )}^3\,\left (\frac {a+b}{a^4}-\frac {5}{3\,a^3}\right )}{f}+\frac {\sin \left (e+f\,x\right )\,\left (\frac {10}{a^3}-\frac {3\,{\left (a+b\right )}^2}{a^5}+\frac {3\,\left (a+b\right )\,\left (\frac {3\,\left (a+b\right )}{a^4}-\frac {5}{a^3}\right )}{a}\right )}{f}+\frac {b^3\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sin \left (e+f\,x\right )\,1{}\mathrm {i}}{\sqrt {a+b}}\right )\,\left (80\,a^2+140\,a\,b+63\,b^2\right )\,1{}\mathrm {i}}{8\,a^{11/2}\,f\,{\left (a+b\right )}^{5/2}} \] Input:
int(cos(e + f*x)^5/(a + b/cos(e + f*x)^2)^3,x)
Output:
((5*sin(e + f*x)*(4*a*b^4 + 3*b^5))/(8*(a + b)) - (sin(e + f*x)^3*(17*a*b^ 5 + 20*a^2*b^4))/(8*(a + b)^2))/(f*(2*a^6*b - sin(e + f*x)^2*(2*a^6*b + 2* a^7) + a^7 + a^5*b^2 + a^7*sin(e + f*x)^4)) + sin(e + f*x)^5/(5*a^3*f) + ( sin(e + f*x)^3*((a + b)/a^4 - 5/(3*a^3)))/f + (sin(e + f*x)*(10/a^3 - (3*( a + b)^2)/a^5 + (3*(a + b)*((3*(a + b))/a^4 - 5/a^3))/a))/f + (b^3*atan((a ^(1/2)*sin(e + f*x)*1i)/(a + b)^(1/2))*(140*a*b + 80*a^2 + 63*b^2)*1i)/(8* a^(11/2)*f*(a + b)^(5/2))
Time = 0.49 (sec) , antiderivative size = 1887, normalized size of antiderivative = 8.82 \[ \int \frac {\cos ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:
int(cos(f*x+e)^5/(a+b*sec(f*x+e)^2)^3,x)
Output:
(1200*sqrt(a)*sqrt(a + b)*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b ) - 2*sqrt(a)*tan((e + f*x)/2))*sin(e + f*x)**4*a**4*b**3 + 2100*sqrt(a)*s qrt(a + b)*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt(a)*t an((e + f*x)/2))*sin(e + f*x)**4*a**3*b**4 + 945*sqrt(a)*sqrt(a + b)*log(s qrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt(a)*tan((e + f*x)/2)) *sin(e + f*x)**4*a**2*b**5 - 2400*sqrt(a)*sqrt(a + b)*log(sqrt(a + b)*tan( (e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt(a)*tan((e + f*x)/2))*sin(e + f*x)** 2*a**4*b**3 - 6600*sqrt(a)*sqrt(a + b)*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt(a)*tan((e + f*x)/2))*sin(e + f*x)**2*a**3*b**4 - 6 090*sqrt(a)*sqrt(a + b)*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt(a)*tan((e + f*x)/2))*sin(e + f*x)**2*a**2*b**5 - 1890*sqrt(a)*sqr t(a + b)*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt(a)*tan ((e + f*x)/2))*sin(e + f*x)**2*a*b**6 + 1200*sqrt(a)*sqrt(a + b)*log(sqrt( a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt(a)*tan((e + f*x)/2))*a** 4*b**3 + 4500*sqrt(a)*sqrt(a + b)*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sq rt(a + b) - 2*sqrt(a)*tan((e + f*x)/2))*a**3*b**4 + 6345*sqrt(a)*sqrt(a + b)*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt(a)*tan((e + f*x)/2))*a**2*b**5 + 3990*sqrt(a)*sqrt(a + b)*log(sqrt(a + b)*tan((e + f*x )/2)**2 + sqrt(a + b) - 2*sqrt(a)*tan((e + f*x)/2))*a*b**6 + 945*sqrt(a)*s qrt(a + b)*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt(a...