Integrand size = 21, antiderivative size = 48 \[ \int \left (a+b \sec ^2(e+f x)\right ) \tan ^4(e+f x) \, dx=a x-\frac {a \tan (e+f x)}{f}+\frac {a \tan ^3(e+f x)}{3 f}+\frac {b \tan ^5(e+f x)}{5 f} \] Output:
a*x-a*tan(f*x+e)/f+1/3*a*tan(f*x+e)^3/f+1/5*b*tan(f*x+e)^5/f
Time = 0.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.19 \[ \int \left (a+b \sec ^2(e+f x)\right ) \tan ^4(e+f x) \, dx=\frac {a \arctan (\tan (e+f x))}{f}-\frac {a \tan (e+f x)}{f}+\frac {a \tan ^3(e+f x)}{3 f}+\frac {b \tan ^5(e+f x)}{5 f} \] Input:
Integrate[(a + b*Sec[e + f*x]^2)*Tan[e + f*x]^4,x]
Output:
(a*ArcTan[Tan[e + f*x]])/f - (a*Tan[e + f*x])/f + (a*Tan[e + f*x]^3)/(3*f) + (b*Tan[e + f*x]^5)/(5*f)
Time = 0.28 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4629, 2075, 363, 254, 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.
\(\displaystyle \int \tan ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (e+f x)^4 \left (a+b \sec (e+f x)^2\right )dx\) |
\(\Big \downarrow \) 4629 |
\(\displaystyle \frac {\int \frac {\tan ^4(e+f x) \left (a+b \left (\tan ^2(e+f x)+1\right )\right )}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 2075 |
\(\displaystyle \frac {\int \frac {\tan ^4(e+f x) \left (b \tan ^2(e+f x)+a+b\right )}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 363 |
\(\displaystyle \frac {a \int \frac {\tan ^4(e+f x)}{\tan ^2(e+f x)+1}d\tan (e+f x)+\frac {1}{5} b \tan ^5(e+f x)}{f}\) |
\(\Big \downarrow \) 254 |
\(\displaystyle \frac {a \int \left (\tan ^2(e+f x)+\frac {1}{\tan ^2(e+f x)+1}-1\right )d\tan (e+f x)+\frac {1}{5} b \tan ^5(e+f x)}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a \left (\arctan (\tan (e+f x))+\frac {1}{3} \tan ^3(e+f x)-\tan (e+f x)\right )+\frac {1}{5} b \tan ^5(e+f x)}{f}\) |
Input:
Int[(a + b*Sec[e + f*x]^2)*Tan[e + f*x]^4,x]
Output:
((b*Tan[e + f*x]^5)/5 + a*(ArcTan[Tan[e + f*x]] - Tan[e + f*x] + Tan[e + f *x]^3/3))/f
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3)) Int[(e*x)^ m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d , 0] && NeQ[m + 2*p + 3, 0]
Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*Expa ndToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{e, m, p, q}, x] && Binomi alQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] && ! BinomialMatchQ[{u, v}, x]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f _.)*(x_)])^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[ff/f Subst[Int[(d*ff*x)^m*((a + b*(1 + ff^2*x^2)^(n/2))^p/(1 + ff^2*x^2 )), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && Inte gerQ[n/2] && (IntegerQ[m/2] || EqQ[n, 2])
Time = 0.84 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.98
method | result | size |
parts | \(\frac {a \left (\frac {\tan \left (f x +e \right )^{3}}{3}-\tan \left (f x +e \right )+\arctan \left (\tan \left (f x +e \right )\right )\right )}{f}+\frac {b \tan \left (f x +e \right )^{5}}{5 f}\) | \(47\) |
derivativedivides | \(\frac {a \left (\frac {\tan \left (f x +e \right )^{3}}{3}-\tan \left (f x +e \right )+f x +e \right )+\frac {b \sin \left (f x +e \right )^{5}}{5 \cos \left (f x +e \right )^{5}}}{f}\) | \(50\) |
default | \(\frac {a \left (\frac {\tan \left (f x +e \right )^{3}}{3}-\tan \left (f x +e \right )+f x +e \right )+\frac {b \sin \left (f x +e \right )^{5}}{5 \cos \left (f x +e \right )^{5}}}{f}\) | \(50\) |
risch | \(a x +\frac {2 i \left (-30 a \,{\mathrm e}^{8 i \left (f x +e \right )}+15 b \,{\mathrm e}^{8 i \left (f x +e \right )}-90 a \,{\mathrm e}^{6 i \left (f x +e \right )}-110 a \,{\mathrm e}^{4 i \left (f x +e \right )}+30 b \,{\mathrm e}^{4 i \left (f x +e \right )}-70 a \,{\mathrm e}^{2 i \left (f x +e \right )}-20 a +3 b \right )}{15 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{5}}\) | \(103\) |
Input:
int((a+b*sec(f*x+e)^2)*tan(f*x+e)^4,x,method=_RETURNVERBOSE)
Output:
a/f*(1/3*tan(f*x+e)^3-tan(f*x+e)+arctan(tan(f*x+e)))+1/5*b*tan(f*x+e)^5/f
Time = 0.10 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.50 \[ \int \left (a+b \sec ^2(e+f x)\right ) \tan ^4(e+f x) \, dx=\frac {15 \, a f x \cos \left (f x + e\right )^{5} - {\left ({\left (20 \, a - 3 \, b\right )} \cos \left (f x + e\right )^{4} - {\left (5 \, a - 6 \, b\right )} \cos \left (f x + e\right )^{2} - 3 \, b\right )} \sin \left (f x + e\right )}{15 \, f \cos \left (f x + e\right )^{5}} \] Input:
integrate((a+b*sec(f*x+e)^2)*tan(f*x+e)^4,x, algorithm="fricas")
Output:
1/15*(15*a*f*x*cos(f*x + e)^5 - ((20*a - 3*b)*cos(f*x + e)^4 - (5*a - 6*b) *cos(f*x + e)^2 - 3*b)*sin(f*x + e))/(f*cos(f*x + e)^5)
Time = 0.83 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.12 \[ \int \left (a+b \sec ^2(e+f x)\right ) \tan ^4(e+f x) \, dx=a \left (\begin {cases} x + \frac {\tan ^{3}{\left (e + f x \right )}}{3 f} - \frac {\tan {\left (e + f x \right )}}{f} & \text {for}\: f \neq 0 \\x \tan ^{4}{\left (e \right )} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} x \tan ^{4}{\left (e \right )} \sec ^{2}{\left (e \right )} & \text {for}\: f = 0 \\\frac {\tan ^{5}{\left (e + f x \right )}}{5 f} & \text {otherwise} \end {cases}\right ) \] Input:
integrate((a+b*sec(f*x+e)**2)*tan(f*x+e)**4,x)
Output:
a*Piecewise((x + tan(e + f*x)**3/(3*f) - tan(e + f*x)/f, Ne(f, 0)), (x*tan (e)**4, True)) + b*Piecewise((x*tan(e)**4*sec(e)**2, Eq(f, 0)), (tan(e + f *x)**5/(5*f), True))
Time = 0.11 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.94 \[ \int \left (a+b \sec ^2(e+f x)\right ) \tan ^4(e+f x) \, dx=\frac {3 \, b \tan \left (f x + e\right )^{5} + 5 \, a \tan \left (f x + e\right )^{3} + 15 \, {\left (f x + e\right )} a - 15 \, a \tan \left (f x + e\right )}{15 \, f} \] Input:
integrate((a+b*sec(f*x+e)^2)*tan(f*x+e)^4,x, algorithm="maxima")
Output:
1/15*(3*b*tan(f*x + e)^5 + 5*a*tan(f*x + e)^3 + 15*(f*x + e)*a - 15*a*tan( f*x + e))/f
Time = 0.35 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.19 \[ \int \left (a+b \sec ^2(e+f x)\right ) \tan ^4(e+f x) \, dx=\frac {{\left (f x + e\right )} a}{f} + \frac {3 \, b f^{4} \tan \left (f x + e\right )^{5} + 5 \, a f^{4} \tan \left (f x + e\right )^{3} - 15 \, a f^{4} \tan \left (f x + e\right )}{15 \, f^{5}} \] Input:
integrate((a+b*sec(f*x+e)^2)*tan(f*x+e)^4,x, algorithm="giac")
Output:
(f*x + e)*a/f + 1/15*(3*b*f^4*tan(f*x + e)^5 + 5*a*f^4*tan(f*x + e)^3 - 15 *a*f^4*tan(f*x + e))/f^5
Time = 15.57 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.83 \[ \int \left (a+b \sec ^2(e+f x)\right ) \tan ^4(e+f x) \, dx=\frac {\frac {b\,{\mathrm {tan}\left (e+f\,x\right )}^5}{5}+\frac {a\,{\mathrm {tan}\left (e+f\,x\right )}^3}{3}-a\,\mathrm {tan}\left (e+f\,x\right )+a\,f\,x}{f} \] Input:
int(tan(e + f*x)^4*(a + b/cos(e + f*x)^2),x)
Output:
((a*tan(e + f*x)^3)/3 - a*tan(e + f*x) + (b*tan(e + f*x)^5)/5 + a*f*x)/f
Time = 0.15 (sec) , antiderivative size = 224, normalized size of antiderivative = 4.67 \[ \int \left (a+b \sec ^2(e+f x)\right ) \tan ^4(e+f x) \, dx=\frac {5 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} \tan \left (f x +e \right )^{3} a -15 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} \tan \left (f x +e \right ) a +15 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} a f x -10 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} \tan \left (f x +e \right )^{3} a +30 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} \tan \left (f x +e \right ) a -30 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a f x +5 \cos \left (f x +e \right ) \tan \left (f x +e \right )^{3} a -15 \cos \left (f x +e \right ) \tan \left (f x +e \right ) a +15 \cos \left (f x +e \right ) a f x +3 \sin \left (f x +e \right )^{5} b}{15 \cos \left (f x +e \right ) f \left (\sin \left (f x +e \right )^{4}-2 \sin \left (f x +e \right )^{2}+1\right )} \] Input:
int((a+b*sec(f*x+e)^2)*tan(f*x+e)^4,x)
Output:
(5*cos(e + f*x)*sin(e + f*x)**4*tan(e + f*x)**3*a - 15*cos(e + f*x)*sin(e + f*x)**4*tan(e + f*x)*a + 15*cos(e + f*x)*sin(e + f*x)**4*a*f*x - 10*cos( e + f*x)*sin(e + f*x)**2*tan(e + f*x)**3*a + 30*cos(e + f*x)*sin(e + f*x)* *2*tan(e + f*x)*a - 30*cos(e + f*x)*sin(e + f*x)**2*a*f*x + 5*cos(e + f*x) *tan(e + f*x)**3*a - 15*cos(e + f*x)*tan(e + f*x)*a + 15*cos(e + f*x)*a*f* x + 3*sin(e + f*x)**5*b)/(15*cos(e + f*x)*f*(sin(e + f*x)**4 - 2*sin(e + f *x)**2 + 1))