3.158 \(\int \sec ^6(e+f x) (a+b \sec ^2(e+f x)) \, dx\)

Optimal. Leaf size=87 \[ \frac{(7 a+6 b) \tan ^5(e+f x)}{35 f}+\frac{2 (7 a+6 b) \tan ^3(e+f x)}{21 f}+\frac{(7 a+6 b) \tan (e+f x)}{7 f}+\frac{b \tan (e+f x) \sec ^6(e+f x)}{7 f} \]

[Out]

((7*a + 6*b)*Tan[e + f*x])/(7*f) + (b*Sec[e + f*x]^6*Tan[e + f*x])/(7*f) + (2*(7*a + 6*b)*Tan[e + f*x]^3)/(21*
f) + ((7*a + 6*b)*Tan[e + f*x]^5)/(35*f)

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Rubi [A]  time = 0.049371, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {4046, 3767} \[ \frac{(7 a+6 b) \tan ^5(e+f x)}{35 f}+\frac{2 (7 a+6 b) \tan ^3(e+f x)}{21 f}+\frac{(7 a+6 b) \tan (e+f x)}{7 f}+\frac{b \tan (e+f x) \sec ^6(e+f x)}{7 f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^6*(a + b*Sec[e + f*x]^2),x]

[Out]

((7*a + 6*b)*Tan[e + f*x])/(7*f) + (b*Sec[e + f*x]^6*Tan[e + f*x])/(7*f) + (2*(7*a + 6*b)*Tan[e + f*x]^3)/(21*
f) + ((7*a + 6*b)*Tan[e + f*x]^5)/(35*f)

Rule 4046

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] + Dist[(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]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin{align*} \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx &=\frac{b \sec ^6(e+f x) \tan (e+f x)}{7 f}+\frac{1}{7} (7 a+6 b) \int \sec ^6(e+f x) \, dx\\ &=\frac{b \sec ^6(e+f x) \tan (e+f x)}{7 f}-\frac{(7 a+6 b) \operatorname{Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (e+f x)\right )}{7 f}\\ &=\frac{(7 a+6 b) \tan (e+f x)}{7 f}+\frac{b \sec ^6(e+f x) \tan (e+f x)}{7 f}+\frac{2 (7 a+6 b) \tan ^3(e+f x)}{21 f}+\frac{(7 a+6 b) \tan ^5(e+f x)}{35 f}\\ \end{align*}

Mathematica [A]  time = 0.311456, size = 81, normalized size = 0.93 \[ \frac{a \left (\frac{1}{5} \tan ^5(e+f x)+\frac{2}{3} \tan ^3(e+f x)+\tan (e+f x)\right )}{f}+\frac{b \left (\frac{1}{7} \tan ^7(e+f x)+\frac{3}{5} \tan ^5(e+f x)+\tan ^3(e+f x)+\tan (e+f x)\right )}{f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^6*(a + b*Sec[e + f*x]^2),x]

[Out]

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

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Maple [A]  time = 0.03, size = 78, normalized size = 0.9 \begin{align*}{\frac{1}{f} \left ( -a \left ( -{\frac{8}{15}}-{\frac{ \left ( \sec \left ( fx+e \right ) \right ) ^{4}}{5}}-{\frac{4\, \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{15}} \right ) \tan \left ( fx+e \right ) -b \left ( -{\frac{16}{35}}-{\frac{ \left ( \sec \left ( fx+e \right ) \right ) ^{6}}{7}}-{\frac{6\, \left ( \sec \left ( fx+e \right ) \right ) ^{4}}{35}}-{\frac{8\, \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{35}} \right ) \tan \left ( fx+e \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^6*(a+b*sec(f*x+e)^2),x)

[Out]

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

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Maxima [A]  time = 0.983035, size = 81, normalized size = 0.93 \begin{align*} \frac{15 \, b \tan \left (f x + e\right )^{7} + 21 \,{\left (a + 3 \, b\right )} \tan \left (f x + e\right )^{5} + 35 \,{\left (2 \, a + 3 \, b\right )} \tan \left (f x + e\right )^{3} + 105 \,{\left (a + b\right )} \tan \left (f x + e\right )}{105 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^6*(a+b*sec(f*x+e)^2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.475483, size = 188, normalized size = 2.16 \begin{align*} \frac{{\left (8 \,{\left (7 \, a + 6 \, b\right )} \cos \left (f x + e\right )^{6} + 4 \,{\left (7 \, a + 6 \, b\right )} \cos \left (f x + e\right )^{4} + 3 \,{\left (7 \, a + 6 \, b\right )} \cos \left (f x + e\right )^{2} + 15 \, b\right )} \sin \left (f x + e\right )}{105 \, f \cos \left (f x + e\right )^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^6*(a+b*sec(f*x+e)^2),x, algorithm="fricas")

[Out]

1/105*(8*(7*a + 6*b)*cos(f*x + e)^6 + 4*(7*a + 6*b)*cos(f*x + e)^4 + 3*(7*a + 6*b)*cos(f*x + e)^2 + 15*b)*sin(
f*x + e)/(f*cos(f*x + e)^7)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \sec ^{2}{\left (e + f x \right )}\right ) \sec ^{6}{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**6*(a+b*sec(f*x+e)**2),x)

[Out]

Integral((a + b*sec(e + f*x)**2)*sec(e + f*x)**6, x)

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Giac [A]  time = 1.32198, size = 116, normalized size = 1.33 \begin{align*} \frac{15 \, b \tan \left (f x + e\right )^{7} + 21 \, a \tan \left (f x + e\right )^{5} + 63 \, b \tan \left (f x + e\right )^{5} + 70 \, a \tan \left (f x + e\right )^{3} + 105 \, b \tan \left (f x + e\right )^{3} + 105 \, a \tan \left (f x + e\right ) + 105 \, b \tan \left (f x + e\right )}{105 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^6*(a+b*sec(f*x+e)^2),x, algorithm="giac")

[Out]

1/105*(15*b*tan(f*x + e)^7 + 21*a*tan(f*x + e)^5 + 63*b*tan(f*x + e)^5 + 70*a*tan(f*x + e)^3 + 105*b*tan(f*x +
e)^3 + 105*a*tan(f*x + e) + 105*b*tan(f*x + e))/f