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

Optimal. Leaf size=80 \[ \frac{a^2 \tan (e+f x)}{f}+\frac{(a+b)^2 \tan ^7(e+f x)}{7 f}+\frac{(a+b) (3 a+b) \tan ^5(e+f x)}{5 f}+\frac{a (3 a+2 b) \tan ^3(e+f x)}{3 f} \]

[Out]

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

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Rubi [A]  time = 0.0757315, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {3191, 373} \[ \frac{a^2 \tan (e+f x)}{f}+\frac{(a+b)^2 \tan ^7(e+f x)}{7 f}+\frac{(a+b) (3 a+b) \tan ^5(e+f x)}{5 f}+\frac{a (3 a+2 b) \tan ^3(e+f x)}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.478616, size = 92, normalized size = 1.15 \[ \frac{\tan (e+f x) \left (6 \left (3 a^2-a b-4 b^2\right ) \sec ^4(e+f x)+\left (24 a^2-8 a b+3 b^2\right ) \sec ^2(e+f x)+48 a^2+15 (a+b)^2 \sec ^6(e+f x)-16 a b+6 b^2\right )}{105 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((48*a^2 - 16*a*b + 6*b^2 + (24*a^2 - 8*a*b + 3*b^2)*Sec[e + f*x]^2 + 6*(3*a^2 - a*b - 4*b^2)*Sec[e + f*x]^4 +
15*(a + b)^2*Sec[e + f*x]^6)*Tan[e + f*x])/(105*f)

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Maple [A]  time = 0.075, size = 149, normalized size = 1.9 \begin{align*}{\frac{1}{f} \left ( -{a}^{2} \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 ) +2\,ab \left ( 1/7\,{\frac{ \left ( \sin \left ( fx+e \right ) \right ) ^{3}}{ \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 ) +{b}^{2} \left ({\frac{ \left ( \sin \left ( fx+e \right ) \right ) ^{5}}{7\, \left ( \cos \left ( fx+e \right ) \right ) ^{7}}}+{\frac{2\, \left ( \sin \left ( fx+e \right ) \right ) ^{5}}{35\, \left ( \cos \left ( fx+e \right ) \right ) ^{5}}} \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)^2,x)

[Out]

1/f*(-a^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)+2*a*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)+b^2*(1/7*sin(f*x+e)^5/cos(f*x+e)^7+2
/35*sin(f*x+e)^5/cos(f*x+e)^5))

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Maxima [A]  time = 0.97737, size = 109, normalized size = 1.36 \begin{align*} \frac{15 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{7} + 21 \,{\left (3 \, a^{2} + 4 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{5} + 35 \,{\left (3 \, a^{2} + 2 \, a b\right )} \tan \left (f x + e\right )^{3} + 105 \, a^{2} \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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.95275, size = 261, normalized size = 3.26 \begin{align*} \frac{{\left (2 \,{\left (24 \, a^{2} - 8 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{6} +{\left (24 \, a^{2} - 8 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 6 \,{\left (3 \, a^{2} - a b - 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 15 \, a^{2} + 30 \, a b + 15 \, b^{2}\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)^2,x, algorithm="fricas")

[Out]

1/105*(2*(24*a^2 - 8*a*b + 3*b^2)*cos(f*x + e)^6 + (24*a^2 - 8*a*b + 3*b^2)*cos(f*x + e)^4 + 6*(3*a^2 - a*b -
4*b^2)*cos(f*x + e)^2 + 15*a^2 + 30*a*b + 15*b^2)*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)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.15234, size = 171, normalized size = 2.14 \begin{align*} \frac{15 \, a^{2} \tan \left (f x + e\right )^{7} + 30 \, a b \tan \left (f x + e\right )^{7} + 15 \, b^{2} \tan \left (f x + e\right )^{7} + 63 \, a^{2} \tan \left (f x + e\right )^{5} + 84 \, a b \tan \left (f x + e\right )^{5} + 21 \, b^{2} \tan \left (f x + e\right )^{5} + 105 \, a^{2} \tan \left (f x + e\right )^{3} + 70 \, a b \tan \left (f x + e\right )^{3} + 105 \, a^{2} \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)^2,x, algorithm="giac")

[Out]

1/105*(15*a^2*tan(f*x + e)^7 + 30*a*b*tan(f*x + e)^7 + 15*b^2*tan(f*x + e)^7 + 63*a^2*tan(f*x + e)^5 + 84*a*b*
tan(f*x + e)^5 + 21*b^2*tan(f*x + e)^5 + 105*a^2*tan(f*x + e)^3 + 70*a*b*tan(f*x + e)^3 + 105*a^2*tan(f*x + e)
)/f