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

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Rubi [A]  time = 0.08, antiderivative size = 80, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.48, 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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fricas [A]  time = 0.41, size = 108, normalized size = 1.35 \[ \frac {{\left (2 \, {\left (24 \, a^{2} - 8 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{6} + {\left (24 \, a^{2} - 8 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 6 \, {\left (3 \, a^{2} - a b - 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 15 \, a^{2} + 30 \, a b + 15 \, b^{2}\right )} \sin \left (f x + e\right )}{105 \, f \cos \left (f x + e\right )^{7}} \]

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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giac [A]  time = 0.25, size = 127, normalized size = 1.59 \[ \frac {15 \, a^{2} \tan \left (f x + e\right )^{7} + 30 \, a b \tan \left (f x + e\right )^{7} + 15 \, b^{2} \tan \left (f x + e\right )^{7} + 63 \, a^{2} \tan \left (f x + e\right )^{5} + 84 \, a b \tan \left (f x + e\right )^{5} + 21 \, b^{2} \tan \left (f x + e\right )^{5} + 105 \, a^{2} \tan \left (f x + e\right )^{3} + 70 \, a b \tan \left (f x + e\right )^{3} + 105 \, a^{2} \tan \left (f x + e\right )}{105 \, f} \]

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

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maple [A]  time = 0.60, size = 149, normalized size = 1.86 \[ \frac {-a^{2} \left (-\frac {16}{35}-\frac {\left (\sec ^{6}\left (f x +e \right )\right )}{7}-\frac {6 \left (\sec ^{4}\left (f x +e \right )\right )}{35}-\frac {8 \left (\sec ^{2}\left (f x +e \right )\right )}{35}\right ) \tan \left (f x +e \right )+2 a b \left (\frac {\sin ^{3}\left (f x +e \right )}{7 \cos \left (f x +e \right )^{7}}+\frac {4 \left (\sin ^{3}\left (f x +e \right )\right )}{35 \cos \left (f x +e \right )^{5}}+\frac {8 \left (\sin ^{3}\left (f x +e \right )\right )}{105 \cos \left (f x +e \right )^{3}}\right )+b^{2} \left (\frac {\sin ^{5}\left (f x +e \right )}{7 \cos \left (f x +e \right )^{7}}+\frac {2 \left (\sin ^{5}\left (f x +e \right )\right )}{35 \cos \left (f x +e \right )^{5}}\right )}{f} \]

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 = 0.35, size = 81, normalized size = 1.01 \[ \frac {15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{7} + 21 \, {\left (3 \, a^{2} + 4 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{5} + 35 \, {\left (3 \, a^{2} + 2 \, a b\right )} \tan \left (f x + e\right )^{3} + 105 \, a^{2} \tan \left (f x + e\right )}{105 \, f} \]

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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mupad [B]  time = 15.14, size = 72, normalized size = 0.90 \[ \frac {a^2\,\mathrm {tan}\left (e+f\,x\right )+\frac {{\mathrm {tan}\left (e+f\,x\right )}^7\,{\left (a+b\right )}^2}{7}+{\mathrm {tan}\left (e+f\,x\right )}^5\,\left (\frac {3\,a^2}{5}+\frac {4\,a\,b}{5}+\frac {b^2}{5}\right )+\frac {a\,{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (3\,a+2\,b\right )}{3}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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