3.292 \(\int \sec ^8(e+f x) (a+b \sin ^2(e+f x)) \, dx\)

Optimal. Leaf size=72 \[ \frac{(a+b) \tan ^7(e+f x)}{7 f}+\frac{(3 a+2 b) \tan ^5(e+f x)}{5 f}+\frac{(3 a+b) \tan ^3(e+f x)}{3 f}+\frac{a \tan (e+f x)}{f} \]

[Out]

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

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Rubi [A]  time = 0.0552121, antiderivative size = 72, 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 = {3191, 373} \[ \frac{(a+b) \tan ^7(e+f x)}{7 f}+\frac{(3 a+2 b) \tan ^5(e+f x)}{5 f}+\frac{(3 a+b) \tan ^3(e+f x)}{3 f}+\frac{a \tan (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^8*(a + b*Sin[e + f*x]^2),x]

[Out]

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

Rule 3191

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, T
an[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rule 373

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]

Rubi steps

\begin{align*} \int \sec ^8(e+f x) \left (a+b \sin ^2(e+f x)\right ) \, dx &=\frac{\operatorname{Subst}\left (\int \left (1+x^2\right )^2 \left (a+(a+b) x^2\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \left (a+(3 a+b) x^2+(3 a+2 b) x^4+(a+b) x^6\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{a \tan (e+f x)}{f}+\frac{(3 a+b) \tan ^3(e+f x)}{3 f}+\frac{(3 a+2 b) \tan ^5(e+f x)}{5 f}+\frac{(a+b) \tan ^7(e+f x)}{7 f}\\ \end{align*}

Mathematica [A]  time = 0.308562, size = 86, normalized size = 1.19 \[ \frac{\tan (e+f x) \left (15 a \tan ^6(e+f x)+63 a \tan ^4(e+f x)+105 a \tan ^2(e+f x)+105 a+15 b \sec ^6(e+f x)-3 b \sec ^4(e+f x)-4 b \sec ^2(e+f x)-8 b\right )}{105 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^8*(a + b*Sin[e + f*x]^2),x]

[Out]

(Tan[e + f*x]*(105*a - 8*b - 4*b*Sec[e + f*x]^2 - 3*b*Sec[e + f*x]^4 + 15*b*Sec[e + f*x]^6 + 105*a*Tan[e + f*x
]^2 + 63*a*Tan[e + f*x]^4 + 15*a*Tan[e + f*x]^6))/(105*f)

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Maple [A]  time = 0.063, size = 104, normalized size = 1.4 \begin{align*}{\frac{1}{f} \left ( -a \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 ) +b \left ({\frac{ \left ( \sin \left ( fx+e \right ) \right ) ^{3}}{7\, \left ( \cos \left ( fx+e \right ) \right ) ^{7}}}+{\frac{4\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}}{35\, \left ( \cos \left ( fx+e \right ) \right ) ^{5}}}+{\frac{8\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}}{105\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^8*(a+b*sin(f*x+e)^2),x)

[Out]

1/f*(-a*(-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)+b*(1/7*sin(f*x+e)^3/cos(f*x+e
)^7+4/35*sin(f*x+e)^3/cos(f*x+e)^5+8/105*sin(f*x+e)^3/cos(f*x+e)^3))

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Maxima [A]  time = 0.984819, size = 81, normalized size = 1.12 \begin{align*} \frac{15 \,{\left (a + b\right )} \tan \left (f x + e\right )^{7} + 21 \,{\left (3 \, a + 2 \, b\right )} \tan \left (f x + e\right )^{5} + 35 \,{\left (3 \, a + b\right )} \tan \left (f x + e\right )^{3} + 105 \, a \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)^8*(a+b*sin(f*x+e)^2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.8906, size = 189, normalized size = 2.62 \begin{align*} \frac{{\left (8 \,{\left (6 \, a - b\right )} \cos \left (f x + e\right )^{6} + 4 \,{\left (6 \, a - b\right )} \cos \left (f x + e\right )^{4} + 3 \,{\left (6 \, a - b\right )} \cos \left (f x + e\right )^{2} + 15 \, a + 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)^8*(a+b*sin(f*x+e)^2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**8*(a+b*sin(f*x+e)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.1679, size = 119, normalized size = 1.65 \begin{align*} \frac{15 \, a \tan \left (f x + e\right )^{7} + 15 \, b \tan \left (f x + e\right )^{7} + 63 \, a \tan \left (f x + e\right )^{5} + 42 \, b \tan \left (f x + e\right )^{5} + 105 \, a \tan \left (f x + e\right )^{3} + 35 \, b \tan \left (f x + e\right )^{3} + 105 \, a \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)^8*(a+b*sin(f*x+e)^2),x, algorithm="giac")

[Out]

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