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

Optimal. Leaf size=53 \[ \frac {a^2 \tan (e+f x)}{f}+\frac {(a+b)^2 \tan ^5(e+f x)}{5 f}+\frac {2 a (a+b) \tan ^3(e+f x)}{3 f} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 53, 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, 194} \[ \frac {a^2 \tan (e+f x)}{f}+\frac {(a+b)^2 \tan ^5(e+f x)}{5 f}+\frac {2 a (a+b) \tan ^3(e+f x)}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 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 )^2 \, dx &=\frac {\operatorname {Subst}\left (\int \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 (a+b) x^2+(a+b)^2 x^4\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {a^2 \tan (e+f x)}{f}+\frac {2 a (a+b) \tan ^3(e+f x)}{3 f}+\frac {(a+b)^2 \tan ^5(e+f x)}{5 f}\\ \end {align*}

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Mathematica [A]  time = 0.36, size = 67, normalized size = 1.26 \[ \frac {\tan (e+f x) \left (\left (4 a^2-2 a b-6 b^2\right ) \sec ^2(e+f x)+8 a^2+3 (a+b)^2 \sec ^4(e+f x)-4 a b+3 b^2\right )}{15 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 83, normalized size = 1.57 \[ \frac {{\left ({\left (8 \, a^{2} - 4 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (2 \, a^{2} - a b - 3 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}\right )} \sin \left (f x + e\right )}{15 \, f \cos \left (f x + e\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*((8*a^2 - 4*a*b + 3*b^2)*cos(f*x + e)^4 + 2*(2*a^2 - a*b - 3*b^2)*cos(f*x + e)^2 + 3*a^2 + 6*a*b + 3*b^2)
*sin(f*x + e)/(f*cos(f*x + e)^5)

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giac [A]  time = 0.21, size = 86, normalized size = 1.62 \[ \frac {3 \, a^{2} \tan \left (f x + e\right )^{5} + 6 \, a b \tan \left (f x + e\right )^{5} + 3 \, b^{2} \tan \left (f x + e\right )^{5} + 10 \, a^{2} \tan \left (f x + e\right )^{3} + 10 \, a b \tan \left (f x + e\right )^{3} + 15 \, a^{2} \tan \left (f x + e\right )}{15 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*a^2*tan(f*x + e)^5 + 6*a*b*tan(f*x + e)^5 + 3*b^2*tan(f*x + e)^5 + 10*a^2*tan(f*x + e)^3 + 10*a*b*tan(
f*x + e)^3 + 15*a^2*tan(f*x + e))/f

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maple [B]  time = 0.60, size = 101, normalized size = 1.91 \[ \frac {-a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (f x +e \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (f x +e \right )\right )}{15}\right ) \tan \left (f x +e \right )+2 a b \left (\frac {\sin ^{3}\left (f x +e \right )}{5 \cos \left (f x +e \right )^{5}}+\frac {2 \left (\sin ^{3}\left (f x +e \right )\right )}{15 \cos \left (f x +e \right )^{3}}\right )+\frac {b^{2} \left (\sin ^{5}\left (f x +e \right )\right )}{5 \cos \left (f x +e \right )^{5}}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(-a^2*(-8/15-1/5*sec(f*x+e)^4-4/15*sec(f*x+e)^2)*tan(f*x+e)+2*a*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)+1/5*b^2*sin(f*x+e)^5/cos(f*x+e)^5)

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maxima [A]  time = 0.32, size = 55, normalized size = 1.04 \[ \frac {3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{5} + 10 \, {\left (a^{2} + a b\right )} \tan \left (f x + e\right )^{3} + 15 \, a^{2} \tan \left (f x + e\right )}{15 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*(a^2 + 2*a*b + b^2)*tan(f*x + e)^5 + 10*(a^2 + a*b)*tan(f*x + e)^3 + 15*a^2*tan(f*x + e))/f

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mupad [B]  time = 15.83, size = 44, normalized size = 0.83 \[ \frac {a^2\,\mathrm {tan}\left (e+f\,x\right )+\frac {{\mathrm {tan}\left (e+f\,x\right )}^5\,{\left (a+b\right )}^2}{5}+\frac {2\,a\,{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (a+b\right )}{3}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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