\(\int \sec ^4(e+f x) (a+b \sec ^2(e+f x))^2 \, dx\) [172]

Optimal result

Integrand size = 23, antiderivative size = 80 \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {(a+b)^2 \tan (e+f x)}{f}+\frac {(a+b) (a+3 b) \tan ^3(e+f x)}{3 f}+\frac {b (2 a+3 b) \tan ^5(e+f x)}{5 f}+\frac {b^2 \tan ^7(e+f x)}{7 f} \]

[Out]

(a+b)^2*tan(f*x+e)/f+1/3*(a+b)*(a+3*b)*tan(f*x+e)^3/f+1/5*b*(2*a+3*b)*tan(f*x+e)^5/f+1/7*b^2*tan(f*x+e)^7/f

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {4231, 380} \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {b (2 a+3 b) \tan ^5(e+f x)}{5 f}+\frac {(a+b) (a+3 b) \tan ^3(e+f x)}{3 f}+\frac {(a+b)^2 \tan (e+f x)}{f}+\frac {b^2 \tan ^7(e+f x)}{7 f} \]

[In]

Int[Sec[e + f*x]^4*(a + b*Sec[e + f*x]^2)^2,x]

[Out]

((a + b)^2*Tan[e + f*x])/f + ((a + b)*(a + 3*b)*Tan[e + f*x]^3)/(3*f) + (b*(2*a + 3*b)*Tan[e + f*x]^5)/(5*f) +
(b^2*Tan[e + f*x]^7)/(7*f)

Rule 380

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

Rule 4231

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fre
eFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/
2), x]^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && IntegerQ[n/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \left (1+x^2\right ) \left (a+b+b x^2\right )^2 \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \left ((a+b)^2+(a+b) (a+3 b) x^2+b (2 a+3 b) x^4+b^2 x^6\right ) \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {(a+b)^2 \tan (e+f x)}{f}+\frac {(a+b) (a+3 b) \tan ^3(e+f x)}{3 f}+\frac {b (2 a+3 b) \tan ^5(e+f x)}{5 f}+\frac {b^2 \tan ^7(e+f x)}{7 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.94 \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {105 (a+b)^2 \tan (e+f x)+35 \left (a^2+4 a b+3 b^2\right ) \tan ^3(e+f x)+21 b (2 a+3 b) \tan ^5(e+f x)+15 b^2 \tan ^7(e+f x)}{105 f} \]

[In]

Integrate[Sec[e + f*x]^4*(a + b*Sec[e + f*x]^2)^2,x]

[Out]

(105*(a + b)^2*Tan[e + f*x] + 35*(a^2 + 4*a*b + 3*b^2)*Tan[e + f*x]^3 + 21*b*(2*a + 3*b)*Tan[e + f*x]^5 + 15*b
^2*Tan[e + f*x]^7)/(105*f)

Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.30

method	result	size
derivativedivides	\(\frac {-a^{2} \left (-\frac {2}{3}-\frac {\sec \left (f x +e \right )^{2}}{3}\right ) \tan \left (f x +e \right )-2 a b \left (-\frac {8}{15}-\frac {\sec \left (f x +e \right )^{4}}{5}-\frac {4 \sec \left (f x +e \right )^{2}}{15}\right ) \tan \left (f x +e \right )-b^{2} \left (-\frac {16}{35}-\frac {\sec \left (f x +e \right )^{6}}{7}-\frac {6 \sec \left (f x +e \right )^{4}}{35}-\frac {8 \sec \left (f x +e \right )^{2}}{35}\right ) \tan \left (f x +e \right )}{f}\)	\(104\)
default	\(\frac {-a^{2} \left (-\frac {2}{3}-\frac {\sec \left (f x +e \right )^{2}}{3}\right ) \tan \left (f x +e \right )-2 a b \left (-\frac {8}{15}-\frac {\sec \left (f x +e \right )^{4}}{5}-\frac {4 \sec \left (f x +e \right )^{2}}{15}\right ) \tan \left (f x +e \right )-b^{2} \left (-\frac {16}{35}-\frac {\sec \left (f x +e \right )^{6}}{7}-\frac {6 \sec \left (f x +e \right )^{4}}{35}-\frac {8 \sec \left (f x +e \right )^{2}}{35}\right ) \tan \left (f x +e \right )}{f}\)	\(104\)
parts	\(-\frac {a^{2} \left (-\frac {2}{3}-\frac {\sec \left (f x +e \right )^{2}}{3}\right ) \tan \left (f x +e \right )}{f}-\frac {b^{2} \left (-\frac {16}{35}-\frac {\sec \left (f x +e \right )^{6}}{7}-\frac {6 \sec \left (f x +e \right )^{4}}{35}-\frac {8 \sec \left (f x +e \right )^{2}}{35}\right ) \tan \left (f x +e \right )}{f}-\frac {2 a b \left (-\frac {8}{15}-\frac {\sec \left (f x +e \right )^{4}}{5}-\frac {4 \sec \left (f x +e \right )^{2}}{15}\right ) \tan \left (f x +e \right )}{f}\)	\(109\)
parallelrisch	\(\frac {\left (1050 a^{2}+2352 a b +1008 b^{2}\right ) \sin \left (3 f x +3 e \right )+\left (490 a^{2}+784 a b +336 b^{2}\right ) \sin \left (5 f x +5 e \right )+\left (70 a^{2}+112 a b +48 b^{2}\right ) \sin \left (7 f x +7 e \right )+630 \sin \left (f x +e \right ) \left (a^{2}+\frac {8}{3} a b +\frac {8}{3} b^{2}\right )}{105 f \left (\cos \left (7 f x +7 e \right )+7 \cos \left (5 f x +5 e \right )+21 \cos \left (3 f x +3 e \right )+35 \cos \left (f x +e \right )\right )}\)	\(145\)
risch	\(\frac {4 i \left (105 a^{2} {\mathrm e}^{10 i \left (f x +e \right )}+455 a^{2} {\mathrm e}^{8 i \left (f x +e \right )}+560 a b \,{\mathrm e}^{8 i \left (f x +e \right )}+770 a^{2} {\mathrm e}^{6 i \left (f x +e \right )}+1400 a b \,{\mathrm e}^{6 i \left (f x +e \right )}+840 b^{2} {\mathrm e}^{6 i \left (f x +e \right )}+630 a^{2} {\mathrm e}^{4 i \left (f x +e \right )}+1176 a b \,{\mathrm e}^{4 i \left (f x +e \right )}+504 b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+245 a^{2} {\mathrm e}^{2 i \left (f x +e \right )}+392 a b \,{\mathrm e}^{2 i \left (f x +e \right )}+168 b^{2} {\mathrm e}^{2 i \left (f x +e \right )}+35 a^{2}+56 a b +24 b^{2}\right )}{105 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{7}}\)	\(199\)
norman	\(\frac {-\frac {2 \left (a^{2}+2 a b +b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}-\frac {2 \left (a^{2}+2 a b +b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{13}}{f}+\frac {4 \left (7 a^{2}+10 a b +3 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3 f}+\frac {4 \left (7 a^{2}+10 a b +3 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}{3 f}+\frac {8 \left (105 a^{2}+182 a b +53 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{35 f}-\frac {2 \left (145 a^{2}+226 a b +129 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{15 f}-\frac {2 \left (145 a^{2}+226 a b +129 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{15 f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{7}}\)	\(225\)

[In]

int(sec(f*x+e)^4*(a+b*sec(f*x+e)^2)^2,x,method=_RETURNVERBOSE)

[Out]

1/f*(-a^2*(-2/3-1/3*sec(f*x+e)^2)*tan(f*x+e)-2*a*b*(-8/15-1/5*sec(f*x+e)^4-4/15*sec(f*x+e)^2)*tan(f*x+e)-b^2*(
-16/35-1/7*sec(f*x+e)^6-6/35*sec(f*x+e)^4-8/35*sec(f*x+e)^2)*tan(f*x+e))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.18 \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {{\left (2 \, {\left (35 \, a^{2} + 56 \, a b + 24 \, b^{2}\right )} \cos \left (f x + e\right )^{6} + {\left (35 \, a^{2} + 56 \, a b + 24 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 6 \, {\left (7 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 15 \, b^{2}\right )} \sin \left (f x + e\right )}{105 \, f \cos \left (f x + e\right )^{7}} \]

[In]

integrate(sec(f*x+e)^4*(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

1/105*(2*(35*a^2 + 56*a*b + 24*b^2)*cos(f*x + e)^6 + (35*a^2 + 56*a*b + 24*b^2)*cos(f*x + e)^4 + 6*(7*a*b + 3*
b^2)*cos(f*x + e)^2 + 15*b^2)*sin(f*x + e)/(f*cos(f*x + e)^7)

Sympy [F]

\[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2} \sec ^{4}{\left (e + f x \right )}\, dx \]

[In]

integrate(sec(f*x+e)**4*(a+b*sec(f*x+e)**2)**2,x)

[Out]

Integral((a + b*sec(e + f*x)**2)**2*sec(e + f*x)**4, x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.01 \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {15 \, b^{2} \tan \left (f x + e\right )^{7} + 21 \, {\left (2 \, a b + 3 \, b^{2}\right )} \tan \left (f x + e\right )^{5} + 35 \, {\left (a^{2} + 4 \, a b + 3 \, b^{2}\right )} \tan \left (f x + e\right )^{3} + 105 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (f x + e\right )}{105 \, f} \]

[In]

integrate(sec(f*x+e)^4*(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

1/105*(15*b^2*tan(f*x + e)^7 + 21*(2*a*b + 3*b^2)*tan(f*x + e)^5 + 35*(a^2 + 4*a*b + 3*b^2)*tan(f*x + e)^3 + 1
05*(a^2 + 2*a*b + b^2)*tan(f*x + e))/f

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.42 \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {15 \, b^{2} \tan \left (f x + e\right )^{7} + 42 \, a b \tan \left (f x + e\right )^{5} + 63 \, b^{2} \tan \left (f x + e\right )^{5} + 35 \, a^{2} \tan \left (f x + e\right )^{3} + 140 \, a b \tan \left (f x + e\right )^{3} + 105 \, b^{2} \tan \left (f x + e\right )^{3} + 105 \, a^{2} \tan \left (f x + e\right ) + 210 \, a b \tan \left (f x + e\right ) + 105 \, b^{2} \tan \left (f x + e\right )}{105 \, f} \]

[In]

integrate(sec(f*x+e)^4*(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")

[Out]

1/105*(15*b^2*tan(f*x + e)^7 + 42*a*b*tan(f*x + e)^5 + 63*b^2*tan(f*x + e)^5 + 35*a^2*tan(f*x + e)^3 + 140*a*b
*tan(f*x + e)^3 + 105*b^2*tan(f*x + e)^3 + 105*a^2*tan(f*x + e) + 210*a*b*tan(f*x + e) + 105*b^2*tan(f*x + e))
/f

Mupad [B] (verification not implemented)

Time = 18.59 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.88 \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {\mathrm {tan}\left (e+f\,x\right )\,{\left (a+b\right )}^2+{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (\frac {a^2}{3}+\frac {4\,a\,b}{3}+b^2\right )+\frac {b^2\,{\mathrm {tan}\left (e+f\,x\right )}^7}{7}+\frac {b\,{\mathrm {tan}\left (e+f\,x\right )}^5\,\left (2\,a+3\,b\right )}{5}}{f} \]

[In]

int((a + b/cos(e + f*x)^2)^2/cos(e + f*x)^4,x)

[Out]

(tan(e + f*x)*(a + b)^2 + tan(e + f*x)^3*((4*a*b)/3 + a^2/3 + b^2) + (b^2*tan(e + f*x)^7)/7 + (b*tan(e + f*x)^
5*(2*a + 3*b))/5)/f