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

Optimal. Leaf size=50 \[ \frac {(a+b) \tan ^5(e+f x)}{5 f}+\frac {(2 a+b) \tan ^3(e+f x)}{3 f}+\frac {a \tan (e+f x)}{f} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 50, normalized size of antiderivative = 1.00, 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} \[ \frac {(a+b) \tan ^5(e+f x)}{5 f}+\frac {(2 a+b) \tan ^3(e+f x)}{3 f}+\frac {a \tan (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.18, size = 64, normalized size = 1.28 \[ \frac {\tan (e+f x) \left (3 a \tan ^4(e+f x)+10 a \tan ^2(e+f x)+15 a+3 b \sec ^4(e+f x)-b \sec ^2(e+f x)-2 b\right )}{15 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Tan[e + f*x]*(15*a - 2*b - b*Sec[e + f*x]^2 + 3*b*Sec[e + f*x]^4 + 10*a*Tan[e + f*x]^2 + 3*a*Tan[e + f*x]^4))
/(15*f)

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fricas [A]  time = 0.42, size = 59, normalized size = 1.18 \[ \frac {{\left (2 \, {\left (4 \, a - b\right )} \cos \left (f x + e\right )^{4} + {\left (4 \, a - b\right )} \cos \left (f x + e\right )^{2} + 3 \, a + 3 \, b\right )} \sin \left (f x + e\right )}{15 \, f \cos \left (f x + e\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(2*(4*a - b)*cos(f*x + e)^4 + (4*a - b)*cos(f*x + e)^2 + 3*a + 3*b)*sin(f*x + e)/(f*cos(f*x + e)^5)

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giac [A]  time = 0.20, size = 64, normalized size = 1.28 \[ \frac {3 \, a \tan \left (f x + e\right )^{5} + 3 \, b \tan \left (f x + e\right )^{5} + 10 \, a \tan \left (f x + e\right )^{3} + 5 \, b \tan \left (f x + e\right )^{3} + 15 \, a \tan \left (f x + e\right )}{15 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*a*tan(f*x + e)^5 + 3*b*tan(f*x + e)^5 + 10*a*tan(f*x + e)^3 + 5*b*tan(f*x + e)^3 + 15*a*tan(f*x + e))/
f

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maple [A]  time = 0.54, size = 76, normalized size = 1.52 \[ \frac {-a \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (f x +e \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (f x +e \right )\right )}{15}\right ) \tan \left (f x +e \right )+b \left (\frac {\sin ^{3}\left (f x +e \right )}{5 \cos \left (f x +e \right )^{5}}+\frac {2 \left (\sin ^{3}\left (f x +e \right )\right )}{15 \cos \left (f x +e \right )^{3}}\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^6*(a+b*sin(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*(1/5*sin(f*x+e)^3/cos(f*x+e)^5+2/15*sin(f*x+e)
^3/cos(f*x+e)^3))

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maxima [A]  time = 0.35, size = 43, normalized size = 0.86 \[ \frac {3 \, {\left (a + b\right )} \tan \left (f x + e\right )^{5} + 5 \, {\left (2 \, a + b\right )} \tan \left (f x + e\right )^{3} + 15 \, a \tan \left (f x + e\right )}{15 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 13.90, size = 45, normalized size = 0.90 \[ \frac {\left (\frac {a}{5}+\frac {b}{5}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^5+\left (\frac {2\,a}{3}+\frac {b}{3}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^3+a\,\mathrm {tan}\left (e+f\,x\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(e + f*x)^2)/cos(e + f*x)^6,x)

[Out]

(tan(e + f*x)^3*((2*a)/3 + b/3) + tan(e + f*x)^5*(a/5 + b/5) + a*tan(e + f*x))/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)**6*(a+b*sin(f*x+e)**2),x)

[Out]

Timed out

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