\(\int \cos (e+f x) (a+b \sec ^2(e+f x))^2 \, dx\) [168]

Optimal result

Integrand size = 21, antiderivative size = 56 \[ \int \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {b (4 a+b) \text {arctanh}(\sin (e+f x))}{2 f}+\frac {a^2 \sin (e+f x)}{f}+\frac {b^2 \sec (e+f x) \tan (e+f x)}{2 f} \]

[Out]

1/2*b*(4*a+b)*arctanh(sin(f*x+e))/f+a^2*sin(f*x+e)/f+1/2*b^2*sec(f*x+e)*tan(f*x+e)/f

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {4232, 398, 393, 212} \[ \int \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {a^2 \sin (e+f x)}{f}+\frac {b (4 a+b) \text {arctanh}(\sin (e+f x))}{2 f}+\frac {b^2 \tan (e+f x) \sec (e+f x)}{2 f} \]

[In]

Int[Cos[e + f*x]*(a + b*Sec[e + f*x]^2)^2,x]

[Out]

(b*(4*a + b)*ArcTanh[Sin[e + f*x]])/(2*f) + (a^2*Sin[e + f*x])/f + (b^2*Sec[e + f*x]*Tan[e + f*x])/(2*f)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 393

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d))*x*((a + b*x^n)^(p
+ 1)/(a*b*n*(p + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x]
/; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 398

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 4232

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fr
eeFactors[Sin[e + f*x], x]}, Dist[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] && IntegerQ[(m - 1)/2] && IntegerQ[n
/2] && IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (a+b-a x^2\right )^2}{\left (1-x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \left (a^2+\frac {b (2 a+b)-2 a b x^2}{\left (1-x^2\right )^2}\right ) \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {a^2 \sin (e+f x)}{f}+\frac {\text {Subst}\left (\int \frac {b (2 a+b)-2 a b x^2}{\left (1-x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {a^2 \sin (e+f x)}{f}+\frac {b^2 \sec (e+f x) \tan (e+f x)}{2 f}+\frac {(b (4 a+b)) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (e+f x)\right )}{2 f} \\ & = \frac {b (4 a+b) \text {arctanh}(\sin (e+f x))}{2 f}+\frac {a^2 \sin (e+f x)}{f}+\frac {b^2 \sec (e+f x) \tan (e+f x)}{2 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.43 \[ \int \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {2 a b \text {arctanh}(\sin (e+f x))}{f}+\frac {b^2 \text {arctanh}(\sin (e+f x))}{2 f}+\frac {a^2 \cos (f x) \sin (e)}{f}+\frac {a^2 \cos (e) \sin (f x)}{f}+\frac {b^2 \sec (e+f x) \tan (e+f x)}{2 f} \]

[In]

Integrate[Cos[e + f*x]*(a + b*Sec[e + f*x]^2)^2,x]

[Out]

(2*a*b*ArcTanh[Sin[e + f*x]])/f + (b^2*ArcTanh[Sin[e + f*x]])/(2*f) + (a^2*Cos[f*x]*Sin[e])/f + (a^2*Cos[e]*Si
n[f*x])/f + (b^2*Sec[e + f*x]*Tan[e + f*x])/(2*f)

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.23

method	result	size
derivativedivides	\(\frac {\sin \left (f x +e \right ) a^{2}+2 a b \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )+b^{2} \left (\frac {\tan \left (f x +e \right ) \sec \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )}{f}\)	\(69\)
default	\(\frac {\sin \left (f x +e \right ) a^{2}+2 a b \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )+b^{2} \left (\frac {\tan \left (f x +e \right ) \sec \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )}{f}\)	\(69\)
parallelrisch	\(\frac {-4 b \left (1+\cos \left (2 f x +2 e \right )\right ) \left (a +\frac {b}{4}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )+4 b \left (1+\cos \left (2 f x +2 e \right )\right ) \left (a +\frac {b}{4}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )+\sin \left (3 f x +3 e \right ) a^{2}+\sin \left (f x +e \right ) \left (a^{2}+2 b^{2}\right )}{2 f \left (1+\cos \left (2 f x +2 e \right )\right )}\)	\(111\)
risch	\(-\frac {i a^{2} {\mathrm e}^{i \left (f x +e \right )}}{2 f}+\frac {i a^{2} {\mathrm e}^{-i \left (f x +e \right )}}{2 f}-\frac {i b^{2} \left ({\mathrm e}^{3 i \left (f x +e \right )}-{\mathrm e}^{i \left (f x +e \right )}\right )}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}-\frac {2 \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) a b}{f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) b^{2}}{2 f}+\frac {2 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) a b}{f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) b^{2}}{2 f}\)	\(163\)
norman	\(\frac {\frac {\left (2 a^{2}+b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{f}+\frac {\left (6 a^{2}-b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{f}-\frac {\left (2 a^{2}+b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}-\frac {\left (6 a^{2}-b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{f}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{3}}-\frac {b \left (4 a +b \right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2 f}+\frac {b \left (4 a +b \right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{2 f}\)	\(180\)

[In]

int(cos(f*x+e)*(a+b*sec(f*x+e)^2)^2,x,method=_RETURNVERBOSE)

[Out]

1/f*(sin(f*x+e)*a^2+2*a*b*ln(sec(f*x+e)+tan(f*x+e))+b^2*(1/2*tan(f*x+e)*sec(f*x+e)+1/2*ln(sec(f*x+e)+tan(f*x+e
))))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.68 \[ \int \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {{\left (4 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} \log \left (\sin \left (f x + e\right ) + 1\right ) - {\left (4 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (2 \, a^{2} \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}{4 \, f \cos \left (f x + e\right )^{2}} \]

[In]

integrate(cos(f*x+e)*(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

1/4*((4*a*b + b^2)*cos(f*x + e)^2*log(sin(f*x + e) + 1) - (4*a*b + b^2)*cos(f*x + e)^2*log(-sin(f*x + e) + 1)
+ 2*(2*a^2*cos(f*x + e)^2 + b^2)*sin(f*x + e))/(f*cos(f*x + e)^2)

Sympy [F]

\[ \int \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2} \cos {\left (e + f x \right )}\, dx \]

[In]

integrate(cos(f*x+e)*(a+b*sec(f*x+e)**2)**2,x)

[Out]

Integral((a + b*sec(e + f*x)**2)**2*cos(e + f*x), x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.55 \[ \int \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=-\frac {b^{2} {\left (\frac {2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 4 \, a b {\left (\log \left (\sin \left (f x + e\right ) + 1\right ) - \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 4 \, a^{2} \sin \left (f x + e\right )}{4 \, f} \]

[In]

integrate(cos(f*x+e)*(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

-1/4*(b^2*(2*sin(f*x + e)/(sin(f*x + e)^2 - 1) - log(sin(f*x + e) + 1) + log(sin(f*x + e) - 1)) - 4*a*b*(log(s
in(f*x + e) + 1) - log(sin(f*x + e) - 1)) - 4*a^2*sin(f*x + e))/f

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.41 \[ \int \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {4 \, a^{2} \sin \left (f x + e\right ) + {\left (4 \, a b + b^{2}\right )} \log \left ({\left | \sin \left (f x + e\right ) + 1 \right |}\right ) - {\left (4 \, a b + b^{2}\right )} \log \left ({\left | \sin \left (f x + e\right ) - 1 \right |}\right ) - \frac {2 \, b^{2} \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1}}{4 \, f} \]

[In]

integrate(cos(f*x+e)*(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")

[Out]

1/4*(4*a^2*sin(f*x + e) + (4*a*b + b^2)*log(abs(sin(f*x + e) + 1)) - (4*a*b + b^2)*log(abs(sin(f*x + e) - 1))
- 2*b^2*sin(f*x + e)/(sin(f*x + e)^2 - 1))/f

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.98 \[ \int \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {a^2\,\sin \left (e+f\,x\right )+\frac {b\,\mathrm {atanh}\left (\sin \left (e+f\,x\right )\right )\,\left (4\,a+b\right )}{2}-\frac {b^2\,\sin \left (e+f\,x\right )}{2\,\left ({\sin \left (e+f\,x\right )}^2-1\right )}}{f} \]

[In]

int(cos(e + f*x)*(a + b/cos(e + f*x)^2)^2,x)

[Out]

(a^2*sin(e + f*x) + (b*atanh(sin(e + f*x))*(4*a + b))/2 - (b^2*sin(e + f*x))/(2*(sin(e + f*x)^2 - 1)))/f