Integrand size = 23, antiderivative size = 49 \[ \int \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {b^2 \text {arctanh}(\sin (e+f x))}{f}+\frac {a (a+2 b) \sin (e+f x)}{f}-\frac {a^2 \sin ^3(e+f x)}{3 f} \] Output:
b^2*arctanh(sin(f*x+e))/f+a*(a+2*b)*sin(f*x+e)/f-1/3*a^2*sin(f*x+e)^3/f
Time = 0.04 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.47 \[ \int \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {b^2 \coth ^{-1}(\sin (e+f x))}{f}+\frac {2 a b \cos (f x) \sin (e)}{f}+\frac {2 a b \cos (e) \sin (f x)}{f}+\frac {a^2 \sin (e+f x)}{f}-\frac {a^2 \sin ^3(e+f x)}{3 f} \] Input:
Integrate[Cos[e + f*x]^3*(a + b*Sec[e + f*x]^2)^2,x]
Output:
(b^2*ArcCoth[Sin[e + f*x]])/f + (2*a*b*Cos[f*x]*Sin[e])/f + (2*a*b*Cos[e]* Sin[f*x])/f + (a^2*Sin[e + f*x])/f - (a^2*Sin[e + f*x]^3)/(3*f)
Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.90, 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.
\(\displaystyle \int \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \sec (e+f x)^2\right )^2}{\sec (e+f x)^3}dx\) |
\(\Big \downarrow \) 4635 |
\(\displaystyle \frac {\int \frac {\left (-a \sin ^2(e+f x)+a+b\right )^2}{1-\sin ^2(e+f x)}d\sin (e+f x)}{f}\) |
\(\Big \downarrow \) 300 |
\(\displaystyle \frac {\int \left (\frac {b^2}{1-\sin ^2(e+f x)}-a^2 \sin ^2(e+f x)+a (a+2 b)\right )d\sin (e+f x)}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{3} a^2 \sin ^3(e+f x)+a (a+2 b) \sin (e+f x)+b^2 \text {arctanh}(\sin (e+f x))}{f}\) |
Input:
Int[Cos[e + f*x]^3*(a + b*Sec[e + f*x]^2)^2,x]
Output:
(b^2*ArcTanh[Sin[e + f*x]] + a*(a + 2*b)*Sin[e + f*x] - (a^2*Sin[e + f*x]^ 3)/3)/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 = 1.01 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.12
method | result | size |
derivativedivides | \(\frac {\frac {a^{2} \left (2+\cos \left (f x +e \right )^{2}\right ) \sin \left (f x +e \right )}{3}+2 \sin \left (f x +e \right ) a b +b^{2} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}\) | \(55\) |
default | \(\frac {\frac {a^{2} \left (2+\cos \left (f x +e \right )^{2}\right ) \sin \left (f x +e \right )}{3}+2 \sin \left (f x +e \right ) a b +b^{2} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}\) | \(55\) |
parallelrisch | \(\frac {-12 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) b^{2}+12 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) b^{2}+\sin \left (3 f x +3 e \right ) a^{2}+9 \left (a +\frac {8 b}{3}\right ) \sin \left (f x +e \right ) a}{12 f}\) | \(68\) |
risch | \(-\frac {3 i a^{2} {\mathrm e}^{i \left (f x +e \right )}}{8 f}-\frac {i {\mathrm e}^{i \left (f x +e \right )} a b}{f}+\frac {3 i a^{2} {\mathrm e}^{-i \left (f x +e \right )}}{8 f}+\frac {i {\mathrm e}^{-i \left (f x +e \right )} a b}{f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) b^{2}}{f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) b^{2}}{f}+\frac {a^{2} \sin \left (3 f x +3 e \right )}{12 f}\) | \(130\) |
norman | \(\frac {-\frac {4 a \left (a -2 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{f}+\frac {4 a \left (a -2 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{f}-\frac {2 a \left (a +2 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}+\frac {2 a \left (a +2 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}{f}+\frac {2 a \left (7 a +6 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3 f}-\frac {2 a \left (7 a +6 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{3 f}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{3}}+\frac {b^{2} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{f}-\frac {b^{2} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{f}\) | \(207\) |
Input:
int(cos(f*x+e)^3*(a+b*sec(f*x+e)^2)^2,x,method=_RETURNVERBOSE)
Output:
1/f*(1/3*a^2*(2+cos(f*x+e)^2)*sin(f*x+e)+2*sin(f*x+e)*a*b+b^2*ln(sec(f*x+e )+tan(f*x+e)))
Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.35 \[ \int \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {3 \, b^{2} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \, b^{2} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (a^{2} \cos \left (f x + e\right )^{2} + 2 \, a^{2} + 6 \, a b\right )} \sin \left (f x + e\right )}{6 \, f} \] Input:
integrate(cos(f*x+e)^3*(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")
Output:
1/6*(3*b^2*log(sin(f*x + e) + 1) - 3*b^2*log(-sin(f*x + e) + 1) + 2*(a^2*c os(f*x + e)^2 + 2*a^2 + 6*a*b)*sin(f*x + e))/f
\[ \int \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2} \cos ^{3}{\left (e + f x \right )}\, dx \] Input:
integrate(cos(f*x+e)**3*(a+b*sec(f*x+e)**2)**2,x)
Output:
Integral((a + b*sec(e + f*x)**2)**2*cos(e + f*x)**3, x)
Time = 0.03 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.29 \[ \int \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=-\frac {2 \, a^{2} \sin \left (f x + e\right )^{3} - 3 \, b^{2} \log \left (\sin \left (f x + e\right ) + 1\right ) + 3 \, b^{2} \log \left (\sin \left (f x + e\right ) - 1\right ) - 6 \, {\left (a^{2} + 2 \, a b\right )} \sin \left (f x + e\right )}{6 \, f} \] Input:
integrate(cos(f*x+e)^3*(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")
Output:
-1/6*(2*a^2*sin(f*x + e)^3 - 3*b^2*log(sin(f*x + e) + 1) + 3*b^2*log(sin(f *x + e) - 1) - 6*(a^2 + 2*a*b)*sin(f*x + e))/f
Time = 0.14 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.43 \[ \int \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=-\frac {2 \, a^{2} \sin \left (f x + e\right )^{3} - 3 \, b^{2} \log \left ({\left | \sin \left (f x + e\right ) + 1 \right |}\right ) + 3 \, b^{2} \log \left ({\left | \sin \left (f x + e\right ) - 1 \right |}\right ) - 6 \, a^{2} \sin \left (f x + e\right ) - 12 \, a b \sin \left (f x + e\right )}{6 \, f} \] Input:
integrate(cos(f*x+e)^3*(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")
Output:
-1/6*(2*a^2*sin(f*x + e)^3 - 3*b^2*log(abs(sin(f*x + e) + 1)) + 3*b^2*log( abs(sin(f*x + e) - 1)) - 6*a^2*sin(f*x + e) - 12*a*b*sin(f*x + e))/f
Time = 14.88 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.98 \[ \int \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=-\frac {\sin \left (e+f\,x\right )\,\left (a^2-2\,a\,\left (a+b\right )\right )+\frac {a^2\,{\sin \left (e+f\,x\right )}^3}{3}-b^2\,\mathrm {atanh}\left (\sin \left (e+f\,x\right )\right )}{f} \] Input:
int(cos(e + f*x)^3*(a + b/cos(e + f*x)^2)^2,x)
Output:
-(sin(e + f*x)*(a^2 - 2*a*(a + b)) + (a^2*sin(e + f*x)^3)/3 - b^2*atanh(si n(e + f*x)))/f
Time = 0.15 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.51 \[ \int \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {-3 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) b^{2}+3 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) b^{2}-\sin \left (f x +e \right )^{3} a^{2}+3 \sin \left (f x +e \right ) a^{2}+6 \sin \left (f x +e \right ) a b}{3 f} \] Input:
int(cos(f*x+e)^3*(a+b*sec(f*x+e)^2)^2,x)
Output:
( - 3*log(tan((e + f*x)/2) - 1)*b**2 + 3*log(tan((e + f*x)/2) + 1)*b**2 - sin(e + f*x)**3*a**2 + 3*sin(e + f*x)*a**2 + 6*sin(e + f*x)*a*b)/(3*f)