3.300 \(\int \sec ^{10}(e+f x) (a+b \sin ^2(e+f x))^2 \, dx\)

Optimal. Leaf size=106 \[ \frac{\left (6 a^2+6 a b+b^2\right ) \tan ^5(e+f x)}{5 f}+\frac{a^2 \tan (e+f x)}{f}+\frac{(a+b)^2 \tan ^9(e+f x)}{9 f}+\frac{2 (a+b) (2 a+b) \tan ^7(e+f x)}{7 f}+\frac{2 a (2 a+b) \tan ^3(e+f x)}{3 f} \]

[Out]

(a^2*Tan[e + f*x])/f + (2*a*(2*a + b)*Tan[e + f*x]^3)/(3*f) + ((6*a^2 + 6*a*b + b^2)*Tan[e + f*x]^5)/(5*f) + (
2*(a + b)*(2*a + b)*Tan[e + f*x]^7)/(7*f) + ((a + b)^2*Tan[e + f*x]^9)/(9*f)

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Rubi [A]  time = 0.0927193, antiderivative size = 106, 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} \[ \frac{\left (6 a^2+6 a b+b^2\right ) \tan ^5(e+f x)}{5 f}+\frac{a^2 \tan (e+f x)}{f}+\frac{(a+b)^2 \tan ^9(e+f x)}{9 f}+\frac{2 (a+b) (2 a+b) \tan ^7(e+f x)}{7 f}+\frac{2 a (2 a+b) \tan ^3(e+f x)}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^10*(a + b*Sin[e + f*x]^2)^2,x]

[Out]

(a^2*Tan[e + f*x])/f + (2*a*(2*a + b)*Tan[e + f*x]^3)/(3*f) + ((6*a^2 + 6*a*b + b^2)*Tan[e + f*x]^5)/(5*f) + (
2*(a + b)*(2*a + b)*Tan[e + f*x]^7)/(7*f) + ((a + b)^2*Tan[e + f*x]^9)/(9*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 ^{10}(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \left (1+x^2\right )^2 \left (a+(a+b) x^2\right )^2 \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \left (a^2+2 a (2 a+b) x^2+\left (6 a^2+6 a b+b^2\right ) x^4+2 (a+b) (2 a+b) x^6+(a+b)^2 x^8\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{a^2 \tan (e+f x)}{f}+\frac{2 a (2 a+b) \tan ^3(e+f x)}{3 f}+\frac{\left (6 a^2+6 a b+b^2\right ) \tan ^5(e+f x)}{5 f}+\frac{2 (a+b) (2 a+b) \tan ^7(e+f x)}{7 f}+\frac{(a+b)^2 \tan ^9(e+f x)}{9 f}\\ \end{align*}

Mathematica [A]  time = 0.505523, size = 107, normalized size = 1.01 \[ \frac{\sec ^9(e+f x) \left (252 \left (8 a^2+8 a b+3 b^2\right ) \sin (e+f x)+336 \left (4 a^2-a b-b^2\right ) \sin (3 (e+f x))+\left (16 a^2-4 a b+b^2\right ) (36 \sin (5 (e+f x))+9 \sin (7 (e+f x))+\sin (9 (e+f x)))\right )}{10080 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^10*(a + b*Sin[e + f*x]^2)^2,x]

[Out]

(Sec[e + f*x]^9*(252*(8*a^2 + 8*a*b + 3*b^2)*Sin[e + f*x] + 336*(4*a^2 - a*b - b^2)*Sin[3*(e + f*x)] + (16*a^2
- 4*a*b + b^2)*(36*Sin[5*(e + f*x)] + 9*Sin[7*(e + f*x)] + Sin[9*(e + f*x)])))/(10080*f)

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Maple [A]  time = 0.075, size = 195, normalized size = 1.8 \begin{align*}{\frac{1}{f} \left ( -{a}^{2} \left ( -{\frac{128}{315}}-{\frac{ \left ( \sec \left ( fx+e \right ) \right ) ^{8}}{9}}-{\frac{8\, \left ( \sec \left ( fx+e \right ) \right ) ^{6}}{63}}-{\frac{16\, \left ( \sec \left ( fx+e \right ) \right ) ^{4}}{105}}-{\frac{64\, \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{315}} \right ) \tan \left ( fx+e \right ) +2\,ab \left ( 1/9\,{\frac{ \left ( \sin \left ( fx+e \right ) \right ) ^{3}}{ \left ( \cos \left ( fx+e \right ) \right ) ^{9}}}+2/21\,{\frac{ \left ( \sin \left ( fx+e \right ) \right ) ^{3}}{ \left ( \cos \left ( fx+e \right ) \right ) ^{7}}}+{\frac{8\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}}{105\, \left ( \cos \left ( fx+e \right ) \right ) ^{5}}}+{\frac{16\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}}{315\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}}} \right ) +{b}^{2} \left ({\frac{ \left ( \sin \left ( fx+e \right ) \right ) ^{5}}{9\, \left ( \cos \left ( fx+e \right ) \right ) ^{9}}}+{\frac{4\, \left ( \sin \left ( fx+e \right ) \right ) ^{5}}{63\, \left ( \cos \left ( fx+e \right ) \right ) ^{7}}}+{\frac{8\, \left ( \sin \left ( fx+e \right ) \right ) ^{5}}{315\, \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)^10*(a+b*sin(f*x+e)^2)^2,x)

[Out]

1/f*(-a^2*(-128/315-1/9*sec(f*x+e)^8-8/63*sec(f*x+e)^6-16/105*sec(f*x+e)^4-64/315*sec(f*x+e)^2)*tan(f*x+e)+2*a
*b*(1/9*sin(f*x+e)^3/cos(f*x+e)^9+2/21*sin(f*x+e)^3/cos(f*x+e)^7+8/105*sin(f*x+e)^3/cos(f*x+e)^5+16/315*sin(f*
x+e)^3/cos(f*x+e)^3)+b^2*(1/9*sin(f*x+e)^5/cos(f*x+e)^9+4/63*sin(f*x+e)^5/cos(f*x+e)^7+8/315*sin(f*x+e)^5/cos(
f*x+e)^5))

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Maxima [A]  time = 0.988156, size = 139, normalized size = 1.31 \begin{align*} \frac{35 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{9} + 90 \,{\left (2 \, a^{2} + 3 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{7} + 63 \,{\left (6 \, a^{2} + 6 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{5} + 210 \,{\left (2 \, a^{2} + a b\right )} \tan \left (f x + e\right )^{3} + 315 \, a^{2} \tan \left (f x + e\right )}{315 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^10*(a+b*sin(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

1/315*(35*(a^2 + 2*a*b + b^2)*tan(f*x + e)^9 + 90*(2*a^2 + 3*a*b + b^2)*tan(f*x + e)^7 + 63*(6*a^2 + 6*a*b + b
^2)*tan(f*x + e)^5 + 210*(2*a^2 + a*b)*tan(f*x + e)^3 + 315*a^2*tan(f*x + e))/f

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Fricas [A]  time = 1.90386, size = 316, normalized size = 2.98 \begin{align*} \frac{{\left (8 \,{\left (16 \, a^{2} - 4 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{8} + 4 \,{\left (16 \, a^{2} - 4 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{6} + 3 \,{\left (16 \, a^{2} - 4 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 10 \,{\left (4 \, a^{2} - a b - 5 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 35 \, a^{2} + 70 \, a b + 35 \, b^{2}\right )} \sin \left (f x + e\right )}{315 \, f \cos \left (f x + e\right )^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^10*(a+b*sin(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

1/315*(8*(16*a^2 - 4*a*b + b^2)*cos(f*x + e)^8 + 4*(16*a^2 - 4*a*b + b^2)*cos(f*x + e)^6 + 3*(16*a^2 - 4*a*b +
b^2)*cos(f*x + e)^4 + 10*(4*a^2 - a*b - 5*b^2)*cos(f*x + e)^2 + 35*a^2 + 70*a*b + 35*b^2)*sin(f*x + e)/(f*cos
(f*x + e)^9)

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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)**10*(a+b*sin(f*x+e)**2)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.17024, size = 227, normalized size = 2.14 \begin{align*} \frac{35 \, a^{2} \tan \left (f x + e\right )^{9} + 70 \, a b \tan \left (f x + e\right )^{9} + 35 \, b^{2} \tan \left (f x + e\right )^{9} + 180 \, a^{2} \tan \left (f x + e\right )^{7} + 270 \, a b \tan \left (f x + e\right )^{7} + 90 \, b^{2} \tan \left (f x + e\right )^{7} + 378 \, a^{2} \tan \left (f x + e\right )^{5} + 378 \, a b \tan \left (f x + e\right )^{5} + 63 \, b^{2} \tan \left (f x + e\right )^{5} + 420 \, a^{2} \tan \left (f x + e\right )^{3} + 210 \, a b \tan \left (f x + e\right )^{3} + 315 \, a^{2} \tan \left (f x + e\right )}{315 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^10*(a+b*sin(f*x+e)^2)^2,x, algorithm="giac")

[Out]

1/315*(35*a^2*tan(f*x + e)^9 + 70*a*b*tan(f*x + e)^9 + 35*b^2*tan(f*x + e)^9 + 180*a^2*tan(f*x + e)^7 + 270*a*
b*tan(f*x + e)^7 + 90*b^2*tan(f*x + e)^7 + 378*a^2*tan(f*x + e)^5 + 378*a*b*tan(f*x + e)^5 + 63*b^2*tan(f*x +
e)^5 + 420*a^2*tan(f*x + e)^3 + 210*a*b*tan(f*x + e)^3 + 315*a^2*tan(f*x + e))/f