Integrand size = 23, antiderivative size = 181 \[ \int \frac {\cos ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {b^2 \left (48 a^2+80 a b+35 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{8 a^{9/2} (a+b)^{5/2} f}+\frac {(a-3 b) \sin (e+f x)}{a^4 f}-\frac {\sin ^3(e+f x)}{3 a^3 f}+\frac {b^4 \sin (e+f x)}{4 a^4 (a+b) f \left (a+b-a \sin ^2(e+f x)\right )^2}-\frac {b^3 (16 a+13 b) \sin (e+f x)}{8 a^4 (a+b)^2 f \left (a+b-a \sin ^2(e+f x)\right )} \] Output:
1/8*b^2*(48*a^2+80*a*b+35*b^2)*arctanh(a^(1/2)*sin(f*x+e)/(a+b)^(1/2))/a^( 9/2)/(a+b)^(5/2)/f+(a-3*b)*sin(f*x+e)/a^4/f-1/3*sin(f*x+e)^3/a^3/f+1/4*b^4 *sin(f*x+e)/a^4/(a+b)/f/(a+b-a*sin(f*x+e)^2)^2-1/8*b^3*(16*a+13*b)*sin(f*x +e)/a^4/(a+b)^2/f/(a+b-a*sin(f*x+e)^2)
Time = 3.11 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.07 \[ \int \frac {\cos ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {-\frac {3 b^2 \left (48 a^2+80 a b+35 b^2\right ) \left (\log \left (\sqrt {a+b}-\sqrt {a} \sin (e+f x)\right )-\log \left (\sqrt {a+b}+\sqrt {a} \sin (e+f x)\right )\right )}{(a+b)^{5/2}}+12 \sqrt {a} \left (-12 b-\frac {b^4 (9 a+22 b+13 a \cos (2 (e+f x)))}{(a+b)^2 (a+2 b+a \cos (2 (e+f x)))^2}+a \left (3-\frac {16 b^3}{(a+b)^2 (a+2 b+a \cos (2 (e+f x)))}\right )\right ) \sin (e+f x)+4 a^{3/2} \sin (3 (e+f x))}{48 a^{9/2} f} \] Input:
Integrate[Cos[e + f*x]^3/(a + b*Sec[e + f*x]^2)^3,x]
Output:
((-3*b^2*(48*a^2 + 80*a*b + 35*b^2)*(Log[Sqrt[a + b] - Sqrt[a]*Sin[e + f*x ]] - Log[Sqrt[a + b] + Sqrt[a]*Sin[e + f*x]]))/(a + b)^(5/2) + 12*Sqrt[a]* (-12*b - (b^4*(9*a + 22*b + 13*a*Cos[2*(e + f*x)]))/((a + b)^2*(a + 2*b + a*Cos[2*(e + f*x)])^2) + a*(3 - (16*b^3)/((a + b)^2*(a + 2*b + a*Cos[2*(e + f*x)]))))*Sin[e + f*x] + 4*a^(3/2)*Sin[3*(e + f*x)])/(48*a^(9/2)*f)
Time = 0.41 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.94, 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.
| \(\displaystyle \int \frac {\cos ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sec (e+f x)^3 \left (a+b \sec (e+f x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 4635 |
| \(\displaystyle \frac {\int \frac {\left (1-\sin ^2(e+f x)\right )^4}{\left (-a \sin ^2(e+f x)+a+b\right )^3}d\sin (e+f x)}{f}\) |
| \(\Big \downarrow \) 300 |
| \(\displaystyle \frac {\int \left (-\frac {\sin ^2(e+f x)}{a^3}+\frac {a-3 b}{a^4}+\frac {6 a^2 b^2 \sin ^4(e+f x)-4 a b^2 (3 a+2 b) \sin ^2(e+f x)+b^2 \left (6 a^2+8 b a+3 b^2\right )}{a^4 \left (-a \sin ^2(e+f x)+a+b\right )^3}\right )d\sin (e+f x)}{f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {b^4 \sin (e+f x)}{4 a^4 (a+b) \left (-a \sin ^2(e+f x)+a+b\right )^2}-\frac {b^3 (16 a+13 b) \sin (e+f x)}{8 a^4 (a+b)^2 \left (-a \sin ^2(e+f x)+a+b\right )}+\frac {(a-3 b) \sin (e+f x)}{a^4}-\frac {\sin ^3(e+f x)}{3 a^3}+\frac {b^2 \left (48 a^2+80 a b+35 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{8 a^{9/2} (a+b)^{5/2}}}{f}\) |
Input:
Int[Cos[e + f*x]^3/(a + b*Sec[e + f*x]^2)^3,x]
Output:
((b^2*(48*a^2 + 80*a*b + 35*b^2)*ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b ]])/(8*a^(9/2)*(a + b)^(5/2)) + ((a - 3*b)*Sin[e + f*x])/a^4 - Sin[e + f*x ]^3/(3*a^3) + (b^4*Sin[e + f*x])/(4*a^4*(a + b)*(a + b - a*Sin[e + f*x]^2) ^2) - (b^3*(16*a + 13*b)*Sin[e + f*x])/(8*a^4*(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 = 3.97 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.98
| method | result | size |
| derivativedivides | \(\frac {-\frac {\frac {a \sin \left (f x +e \right )^{3}}{3}-\sin \left (f x +e \right ) a +3 \sin \left (f x +e \right ) b}{a^{4}}-\frac {b^{2} \left (\frac {-\frac {a b \left (16 a +13 b \right ) \sin \left (f x +e \right )^{3}}{8 \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (16 a +11 b \right ) b \sin \left (f x +e \right )}{8 a +8 b}}{\left (-a -b +a \sin \left (f x +e \right )^{2}\right )^{2}}-\frac {\left (48 a^{2}+80 a b +35 b^{2}\right ) \operatorname {arctanh}\left (\frac {a \sin \left (f x +e \right )}{\sqrt {a \left (a +b \right )}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) \sqrt {a \left (a +b \right )}}\right )}{a^{4}}}{f}\) | \(177\) |
| default | \(\frac {-\frac {\frac {a \sin \left (f x +e \right )^{3}}{3}-\sin \left (f x +e \right ) a +3 \sin \left (f x +e \right ) b}{a^{4}}-\frac {b^{2} \left (\frac {-\frac {a b \left (16 a +13 b \right ) \sin \left (f x +e \right )^{3}}{8 \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (16 a +11 b \right ) b \sin \left (f x +e \right )}{8 a +8 b}}{\left (-a -b +a \sin \left (f x +e \right )^{2}\right )^{2}}-\frac {\left (48 a^{2}+80 a b +35 b^{2}\right ) \operatorname {arctanh}\left (\frac {a \sin \left (f x +e \right )}{\sqrt {a \left (a +b \right )}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) \sqrt {a \left (a +b \right )}}\right )}{a^{4}}}{f}\) | \(177\) |
| risch | \(-\frac {i {\mathrm e}^{3 i \left (f x +e \right )}}{24 a^{3} f}-\frac {3 i {\mathrm e}^{i \left (f x +e \right )}}{8 a^{3} f}+\frac {3 i {\mathrm e}^{i \left (f x +e \right )} b}{2 a^{4} f}+\frac {3 i {\mathrm e}^{-i \left (f x +e \right )}}{8 a^{3} f}-\frac {3 i {\mathrm e}^{-i \left (f x +e \right )} b}{2 a^{4} f}+\frac {i {\mathrm e}^{-3 i \left (f x +e \right )}}{24 a^{3} f}+\frac {i b^{3} \left (16 a^{2} {\mathrm e}^{7 i \left (f x +e \right )}+13 a b \,{\mathrm e}^{7 i \left (f x +e \right )}+16 a^{2} {\mathrm e}^{5 i \left (f x +e \right )}+69 a b \,{\mathrm e}^{5 i \left (f x +e \right )}+44 b^{2} {\mathrm e}^{5 i \left (f x +e \right )}-16 a^{2} {\mathrm e}^{3 i \left (f x +e \right )}-69 a b \,{\mathrm e}^{3 i \left (f x +e \right )}-44 b^{2} {\mathrm e}^{3 i \left (f x +e \right )}-16 a^{2} {\mathrm e}^{i \left (f x +e \right )}-13 a b \,{\mathrm e}^{i \left (f x +e \right )}\right )}{4 a^{4} \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 \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 ) b^{2}}{\sqrt {a^{2}+a b}\, \left (a +b \right )^{2} f \,a^{2}}+\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 )}{16 \sqrt {a^{2}+a b}\, \left (a +b \right )^{2} f \,a^{4}}-\frac {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 ) b^{2}}{\sqrt {a^{2}+a b}\, \left (a +b \right )^{2} f \,a^{2}}-\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 )}{16 \sqrt {a^{2}+a b}\, \left (a +b \right )^{2} f \,a^{4}}\) | \(671\) |
Input:
int(cos(f*x+e)^3/(a+b*sec(f*x+e)^2)^3,x,method=_RETURNVERBOSE)
Output:
1/f*(-1/a^4*(1/3*a*sin(f*x+e)^3-sin(f*x+e)*a+3*sin(f*x+e)*b)-b^2/a^4*((-1/ 8*a*b*(16*a+13*b)/(a^2+2*a*b+b^2)*sin(f*x+e)^3+1/8*(16*a+11*b)*b/(a+b)*sin (f*x+e))/(-a-b+a*sin(f*x+e)^2)^2-1/8*(48*a^2+80*a*b+35*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. 416 vs. \(2 (171) = 342\).
Time = 0.15 (sec) , antiderivative size = 856, normalized size of antiderivative = 4.73 \[ \int \frac {\cos ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:
integrate(cos(f*x+e)^3/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")
Output:
[1/48*(3*(48*a^2*b^4 + 80*a*b^5 + 35*b^6 + (48*a^4*b^2 + 80*a^3*b^3 + 35*a ^2*b^4)*cos(f*x + e)^4 + 2*(48*a^3*b^3 + 80*a^2*b^4 + 35*a*b^5)*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*(16*a^5*b^2 - 24*a^4*b^3 - 210* a^3*b^4 - 275*a^2*b^5 - 105*a*b^6 + 8*(a^7 + 3*a^6*b + 3*a^5*b^2 + a^4*b^3 )*cos(f*x + e)^6 + 8*(2*a^7 - a^6*b - 15*a^5*b^2 - 19*a^4*b^3 - 7*a^3*b^4) *cos(f*x + e)^4 + (32*a^6*b - 40*a^5*b^2 - 360*a^4*b^3 - 463*a^3*b^4 - 175 *a^2*b^5)*cos(f*x + e)^2)*sin(f*x + e))/((a^10 + 3*a^9*b + 3*a^8*b^2 + a^7 *b^3)*f*cos(f*x + e)^4 + 2*(a^9*b + 3*a^8*b^2 + 3*a^7*b^3 + a^6*b^4)*f*cos (f*x + e)^2 + (a^8*b^2 + 3*a^7*b^3 + 3*a^6*b^4 + a^5*b^5)*f), -1/24*(3*(48 *a^2*b^4 + 80*a*b^5 + 35*b^6 + (48*a^4*b^2 + 80*a^3*b^3 + 35*a^2*b^4)*cos( f*x + e)^4 + 2*(48*a^3*b^3 + 80*a^2*b^4 + 35*a*b^5)*cos(f*x + e)^2)*sqrt(- a^2 - a*b)*arctan(sqrt(-a^2 - a*b)*sin(f*x + e)/(a + b)) - (16*a^5*b^2 - 2 4*a^4*b^3 - 210*a^3*b^4 - 275*a^2*b^5 - 105*a*b^6 + 8*(a^7 + 3*a^6*b + 3*a ^5*b^2 + a^4*b^3)*cos(f*x + e)^6 + 8*(2*a^7 - a^6*b - 15*a^5*b^2 - 19*a^4* b^3 - 7*a^3*b^4)*cos(f*x + e)^4 + (32*a^6*b - 40*a^5*b^2 - 360*a^4*b^3 - 4 63*a^3*b^4 - 175*a^2*b^5)*cos(f*x + e)^2)*sin(f*x + e))/((a^10 + 3*a^9*b + 3*a^8*b^2 + a^7*b^3)*f*cos(f*x + e)^4 + 2*(a^9*b + 3*a^8*b^2 + 3*a^7*b^3 + a^6*b^4)*f*cos(f*x + e)^2 + (a^8*b^2 + 3*a^7*b^3 + 3*a^6*b^4 + a^5*b^5)* f)]
Timed out. \[ \int \frac {\cos ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Timed out} \] Input:
integrate(cos(f*x+e)**3/(a+b*sec(f*x+e)**2)**3,x)
Output:
Timed out
Time = 0.11 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.50 \[ \int \frac {\cos ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {3 \, {\left (48 \, a^{2} b^{2} + 80 \, a b^{3} + 35 \, b^{4}\right )} \log \left (\frac {a \sin \left (f x + e\right ) - \sqrt {{\left (a + b\right )} a}}{a \sin \left (f x + e\right ) + \sqrt {{\left (a + b\right )} a}}\right )}{{\left (a^{6} + 2 \, a^{5} b + a^{4} b^{2}\right )} \sqrt {{\left (a + b\right )} a}} - \frac {6 \, {\left ({\left (16 \, a^{2} b^{3} + 13 \, a b^{4}\right )} \sin \left (f x + e\right )^{3} - {\left (16 \, a^{2} b^{3} + 27 \, a b^{4} + 11 \, b^{5}\right )} \sin \left (f x + e\right )\right )}}{a^{8} + 4 \, a^{7} b + 6 \, a^{6} b^{2} + 4 \, a^{5} b^{3} + a^{4} b^{4} + {\left (a^{8} + 2 \, a^{7} b + a^{6} b^{2}\right )} \sin \left (f x + e\right )^{4} - 2 \, {\left (a^{8} + 3 \, a^{7} b + 3 \, a^{6} b^{2} + a^{5} b^{3}\right )} \sin \left (f x + e\right )^{2}} + \frac {16 \, {\left (a \sin \left (f x + e\right )^{3} - 3 \, {\left (a - 3 \, b\right )} \sin \left (f x + e\right )\right )}}{a^{4}}}{48 \, f} \] Input:
integrate(cos(f*x+e)^3/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")
Output:
-1/48*(3*(48*a^2*b^2 + 80*a*b^3 + 35*b^4)*log((a*sin(f*x + e) - sqrt((a + b)*a))/(a*sin(f*x + e) + sqrt((a + b)*a)))/((a^6 + 2*a^5*b + a^4*b^2)*sqrt ((a + b)*a)) - 6*((16*a^2*b^3 + 13*a*b^4)*sin(f*x + e)^3 - (16*a^2*b^3 + 2 7*a*b^4 + 11*b^5)*sin(f*x + e))/(a^8 + 4*a^7*b + 6*a^6*b^2 + 4*a^5*b^3 + a ^4*b^4 + (a^8 + 2*a^7*b + a^6*b^2)*sin(f*x + e)^4 - 2*(a^8 + 3*a^7*b + 3*a ^6*b^2 + a^5*b^3)*sin(f*x + e)^2) + 16*(a*sin(f*x + e)^3 - 3*(a - 3*b)*sin (f*x + e))/a^4)/f
Time = 0.19 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.27 \[ \int \frac {\cos ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {3 \, {\left (48 \, a^{2} b^{2} + 80 \, a b^{3} + 35 \, b^{4}\right )} \arctan \left (\frac {a \sin \left (f x + e\right )}{\sqrt {-a^{2} - a b}}\right )}{{\left (a^{6} + 2 \, a^{5} b + a^{4} b^{2}\right )} \sqrt {-a^{2} - a b}} - \frac {3 \, {\left (16 \, a^{2} b^{3} \sin \left (f x + e\right )^{3} + 13 \, a b^{4} \sin \left (f x + e\right )^{3} - 16 \, a^{2} b^{3} \sin \left (f x + e\right ) - 27 \, a b^{4} \sin \left (f x + e\right ) - 11 \, b^{5} \sin \left (f x + e\right )\right )}}{{\left (a^{6} + 2 \, a^{5} b + a^{4} b^{2}\right )} {\left (a \sin \left (f x + e\right )^{2} - a - b\right )}^{2}} + \frac {8 \, {\left (a^{6} \sin \left (f x + e\right )^{3} - 3 \, a^{6} \sin \left (f x + e\right ) + 9 \, a^{5} b \sin \left (f x + e\right )\right )}}{a^{9}}}{24 \, f} \] Input:
integrate(cos(f*x+e)^3/(a+b*sec(f*x+e)^2)^3,x, algorithm="giac")
Output:
-1/24*(3*(48*a^2*b^2 + 80*a*b^3 + 35*b^4)*arctan(a*sin(f*x + e)/sqrt(-a^2 - a*b))/((a^6 + 2*a^5*b + a^4*b^2)*sqrt(-a^2 - a*b)) - 3*(16*a^2*b^3*sin(f *x + e)^3 + 13*a*b^4*sin(f*x + e)^3 - 16*a^2*b^3*sin(f*x + e) - 27*a*b^4*s in(f*x + e) - 11*b^5*sin(f*x + e))/((a^6 + 2*a^5*b + a^4*b^2)*(a*sin(f*x + e)^2 - a - b)^2) + 8*(a^6*sin(f*x + e)^3 - 3*a^6*sin(f*x + e) + 9*a^5*b*s in(f*x + e))/a^9)/f
Time = 0.36 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.41 \[ \int \frac {\cos ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {b^2\,\ln \left (\sqrt {a+b}+\sqrt {a}\,\sin \left (e+f\,x\right )\right )\,\left (3\,a^2+5\,a\,b+\frac {35\,b^2}{16}\right )}{a^{9/2}\,f\,{\left (a+b\right )}^{5/2}}-\frac {\frac {\sin \left (e+f\,x\right )\,\left (11\,b^4+16\,a\,b^3\right )}{8\,\left (a+b\right )}-\frac {{\sin \left (e+f\,x\right )}^3\,\left (16\,a^2\,b^3+13\,a\,b^4\right )}{8\,{\left (a+b\right )}^2}}{f\,\left (2\,a^5\,b-{\sin \left (e+f\,x\right )}^2\,\left (2\,a^6+2\,b\,a^5\right )+a^6+a^4\,b^2+a^6\,{\sin \left (e+f\,x\right )}^4\right )}-\frac {{\sin \left (e+f\,x\right )}^3}{3\,a^3\,f}-\frac {b^2\,\ln \left (\sqrt {a}\,\sin \left (e+f\,x\right )-\sqrt {a+b}\right )\,\left (48\,a^2+80\,a\,b+35\,b^2\right )}{16\,a^{9/2}\,f\,{\left (a+b\right )}^{5/2}}-\frac {\sin \left (e+f\,x\right )\,\left (\frac {3\,\left (a+b\right )}{a^4}-\frac {4}{a^3}\right )}{f} \] Input:
int(cos(e + f*x)^3/(a + b/cos(e + f*x)^2)^3,x)
Output:
(b^2*log((a + b)^(1/2) + a^(1/2)*sin(e + f*x))*(5*a*b + 3*a^2 + (35*b^2)/1 6))/(a^(9/2)*f*(a + b)^(5/2)) - ((sin(e + f*x)*(16*a*b^3 + 11*b^4))/(8*(a + b)) - (sin(e + f*x)^3*(13*a*b^4 + 16*a^2*b^3))/(8*(a + b)^2))/(f*(2*a^5* b - sin(e + f*x)^2*(2*a^5*b + 2*a^6) + a^6 + a^4*b^2 + a^6*sin(e + f*x)^4) ) - sin(e + f*x)^3/(3*a^3*f) - (b^2*log(a^(1/2)*sin(e + f*x) - (a + b)^(1/ 2))*(80*a*b + 48*a^2 + 35*b^2))/(16*a^(9/2)*f*(a + b)^(5/2)) - (sin(e + f* x)*((3*(a + b))/a^4 - 4/a^3))/f
Time = 0.34 (sec) , antiderivative size = 1766, normalized size of antiderivative = 9.76 \[ \int \frac {\cos ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:
int(cos(f*x+e)^3/(a+b*sec(f*x+e)^2)^3,x)
Output:
( - 144*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**2 - 240*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**3*b**3 - 105*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**2*b**4 + 288*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**2 + 768*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**3 + 69 0*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**4 + 210*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*b**5 - 144*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* *2 - 528*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**3*b**3 - 729*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**4 - 450*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**5 - 105*sqrt(a)*sqrt(a + b)*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt(a)*tan((...