Integrand size = 21, antiderivative size = 65 \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {(5 a+4 b) \tan (e+f x)}{5 f}+\frac {b \sec ^4(e+f x) \tan (e+f x)}{5 f}+\frac {(5 a+4 b) \tan ^3(e+f x)}{15 f} \] Output:
1/5*(5*a+4*b)*tan(f*x+e)/f+1/5*b*sec(f*x+e)^4*tan(f*x+e)/f+1/15*(5*a+4*b)* tan(f*x+e)^3/f
Time = 0.16 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.94 \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {a \left (\tan (e+f x)+\frac {1}{3} \tan ^3(e+f x)\right )}{f}+\frac {b \left (\tan (e+f x)+\frac {2}{3} \tan ^3(e+f x)+\frac {1}{5} \tan ^5(e+f x)\right )}{f} \] Input:
Integrate[Sec[e + f*x]^4*(a + b*Sec[e + f*x]^2),x]
Output:
(a*(Tan[e + f*x] + Tan[e + f*x]^3/3))/f + (b*(Tan[e + f*x] + (2*Tan[e + f* x]^3)/3 + Tan[e + f*x]^5/5))/f
Time = 0.31 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.89, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 4534, 3042, 4254, 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.
\(\displaystyle \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (e+f x+\frac {\pi }{2}\right )^4 \left (a+b \csc \left (e+f x+\frac {\pi }{2}\right )^2\right )dx\) |
\(\Big \downarrow \) 4534 |
\(\displaystyle \frac {1}{5} (5 a+4 b) \int \sec ^4(e+f x)dx+\frac {b \tan (e+f x) \sec ^4(e+f x)}{5 f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} (5 a+4 b) \int \csc \left (e+f x+\frac {\pi }{2}\right )^4dx+\frac {b \tan (e+f x) \sec ^4(e+f x)}{5 f}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle \frac {b \tan (e+f x) \sec ^4(e+f x)}{5 f}-\frac {(5 a+4 b) \int \left (\tan ^2(e+f x)+1\right )d(-\tan (e+f x))}{5 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b \tan (e+f x) \sec ^4(e+f x)}{5 f}-\frac {(5 a+4 b) \left (-\frac {1}{3} \tan ^3(e+f x)-\tan (e+f x)\right )}{5 f}\) |
Input:
Int[Sec[e + f*x]^4*(a + b*Sec[e + f*x]^2),x]
Output:
(b*Sec[e + f*x]^4*Tan[e + f*x])/(5*f) - ((5*a + 4*b)*(-Tan[e + f*x] - Tan[ e + f*x]^3/3))/(5*f)
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) )), x] + Simp[(C*m + A*(m + 1))/(m + 1) Int[(b*Csc[e + f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] && !LeQ[m, -1]
Time = 1.52 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\frac {-a \left (-\frac {2}{3}-\frac {\sec \left (f x +e \right )^{2}}{3}\right ) \tan \left (f x +e \right )-b \left (-\frac {8}{15}-\frac {\sec \left (f x +e \right )^{4}}{5}-\frac {4 \sec \left (f x +e \right )^{2}}{15}\right ) \tan \left (f x +e \right )}{f}\) | \(58\) |
default | \(\frac {-a \left (-\frac {2}{3}-\frac {\sec \left (f x +e \right )^{2}}{3}\right ) \tan \left (f x +e \right )-b \left (-\frac {8}{15}-\frac {\sec \left (f x +e \right )^{4}}{5}-\frac {4 \sec \left (f x +e \right )^{2}}{15}\right ) \tan \left (f x +e \right )}{f}\) | \(58\) |
parts | \(-\frac {a \left (-\frac {2}{3}-\frac {\sec \left (f x +e \right )^{2}}{3}\right ) \tan \left (f x +e \right )}{f}-\frac {b \left (-\frac {8}{15}-\frac {\sec \left (f x +e \right )^{4}}{5}-\frac {4 \sec \left (f x +e \right )^{2}}{15}\right ) \tan \left (f x +e \right )}{f}\) | \(60\) |
parallelrisch | \(\frac {\left (50 a +40 b \right ) \sin \left (3 f x +3 e \right )+\left (10 a +8 b \right ) \sin \left (5 f x +5 e \right )+40 \sin \left (f x +e \right ) \left (a +2 b \right )}{15 f \left (\cos \left (5 f x +5 e \right )+5 \cos \left (3 f x +3 e \right )+10 \cos \left (f x +e \right )\right )}\) | \(85\) |
risch | \(\frac {4 i \left (15 a \,{\mathrm e}^{6 i \left (f x +e \right )}+35 a \,{\mathrm e}^{4 i \left (f x +e \right )}+40 b \,{\mathrm e}^{4 i \left (f x +e \right )}+25 a \,{\mathrm e}^{2 i \left (f x +e \right )}+20 b \,{\mathrm e}^{2 i \left (f x +e \right )}+5 a +4 b \right )}{15 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{5}}\) | \(87\) |
norman | \(\frac {-\frac {2 \left (a +b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}-\frac {2 \left (a +b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{f}+\frac {8 \left (2 a +b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3 f}+\frac {8 \left (2 a +b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{3 f}-\frac {4 \left (25 a +29 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{15 f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{5}}\) | \(119\) |
Input:
int(sec(f*x+e)^4*(a+b*sec(f*x+e)^2),x,method=_RETURNVERBOSE)
Output:
1/f*(-a*(-2/3-1/3*sec(f*x+e)^2)*tan(f*x+e)-b*(-8/15-1/5*sec(f*x+e)^4-4/15* sec(f*x+e)^2)*tan(f*x+e))
Time = 0.07 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.86 \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {{\left (2 \, {\left (5 \, a + 4 \, b\right )} \cos \left (f x + e\right )^{4} + {\left (5 \, a + 4 \, b\right )} \cos \left (f x + e\right )^{2} + 3 \, b\right )} \sin \left (f x + e\right )}{15 \, f \cos \left (f x + e\right )^{5}} \] Input:
integrate(sec(f*x+e)^4*(a+b*sec(f*x+e)^2),x, algorithm="fricas")
Output:
1/15*(2*(5*a + 4*b)*cos(f*x + e)^4 + (5*a + 4*b)*cos(f*x + e)^2 + 3*b)*sin (f*x + e)/(f*cos(f*x + e)^5)
\[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right ) \sec ^{4}{\left (e + f x \right )}\, dx \] Input:
integrate(sec(f*x+e)**4*(a+b*sec(f*x+e)**2),x)
Output:
Integral((a + b*sec(e + f*x)**2)*sec(e + f*x)**4, x)
Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.66 \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {3 \, b \tan \left (f x + e\right )^{5} + 5 \, {\left (a + 2 \, b\right )} \tan \left (f x + e\right )^{3} + 15 \, {\left (a + b\right )} \tan \left (f x + e\right )}{15 \, f} \] Input:
integrate(sec(f*x+e)^4*(a+b*sec(f*x+e)^2),x, algorithm="maxima")
Output:
1/15*(3*b*tan(f*x + e)^5 + 5*(a + 2*b)*tan(f*x + e)^3 + 15*(a + b)*tan(f*x + e))/f
Time = 0.13 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.88 \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {3 \, b \tan \left (f x + e\right )^{5} + 5 \, a \tan \left (f x + e\right )^{3} + 10 \, b \tan \left (f x + e\right )^{3} + 15 \, a \tan \left (f x + e\right ) + 15 \, b \tan \left (f x + e\right )}{15 \, f} \] Input:
integrate(sec(f*x+e)^4*(a+b*sec(f*x+e)^2),x, algorithm="giac")
Output:
1/15*(3*b*tan(f*x + e)^5 + 5*a*tan(f*x + e)^3 + 10*b*tan(f*x + e)^3 + 15*a *tan(f*x + e) + 15*b*tan(f*x + e))/f
Time = 15.74 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.65 \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {\frac {b\,{\mathrm {tan}\left (e+f\,x\right )}^5}{5}+\left (\frac {a}{3}+\frac {2\,b}{3}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^3+\left (a+b\right )\,\mathrm {tan}\left (e+f\,x\right )}{f} \] Input:
int((a + b/cos(e + f*x)^2)/cos(e + f*x)^4,x)
Output:
(tan(e + f*x)^3*(a/3 + (2*b)/3) + (b*tan(e + f*x)^5)/5 + tan(e + f*x)*(a + b))/f
Time = 0.17 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.42 \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {\sin \left (f x +e \right ) \left (10 \sin \left (f x +e \right )^{4} a +8 \sin \left (f x +e \right )^{4} b -25 \sin \left (f x +e \right )^{2} a -20 \sin \left (f x +e \right )^{2} b +15 a +15 b \right )}{15 \cos \left (f x +e \right ) f \left (\sin \left (f x +e \right )^{4}-2 \sin \left (f x +e \right )^{2}+1\right )} \] Input:
int(sec(f*x+e)^4*(a+b*sec(f*x+e)^2),x)
Output:
(sin(e + f*x)*(10*sin(e + f*x)**4*a + 8*sin(e + f*x)**4*b - 25*sin(e + f*x )**2*a - 20*sin(e + f*x)**2*b + 15*a + 15*b))/(15*cos(e + f*x)*f*(sin(e + f*x)**4 - 2*sin(e + f*x)**2 + 1))