3.43 \(\int \sin ^5(e+f x) (a+b \tan ^2(e+f x))^2 \, dx\)

Optimal. Leaf size=107 \[ -\frac{\left (a^2-6 a b+6 b^2\right ) \cos (e+f x)}{f}-\frac{(a-b)^2 \cos ^5(e+f x)}{5 f}+\frac{2 (a-2 b) (a-b) \cos ^3(e+f x)}{3 f}+\frac{2 b (a-2 b) \sec (e+f x)}{f}+\frac{b^2 \sec ^3(e+f x)}{3 f} \]

[Out]

-(((a^2 - 6*a*b + 6*b^2)*Cos[e + f*x])/f) + (2*(a - 2*b)*(a - b)*Cos[e + f*x]^3)/(3*f) - ((a - b)^2*Cos[e + f*
x]^5)/(5*f) + (2*(a - 2*b)*b*Sec[e + f*x])/f + (b^2*Sec[e + f*x]^3)/(3*f)

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Rubi [A]  time = 0.107058, antiderivative size = 107, 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 = {3664, 448} \[ -\frac{\left (a^2-6 a b+6 b^2\right ) \cos (e+f x)}{f}-\frac{(a-b)^2 \cos ^5(e+f x)}{5 f}+\frac{2 (a-2 b) (a-b) \cos ^3(e+f x)}{3 f}+\frac{2 b (a-2 b) \sec (e+f x)}{f}+\frac{b^2 \sec ^3(e+f x)}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^5*(a + b*Tan[e + f*x]^2)^2,x]

[Out]

-(((a^2 - 6*a*b + 6*b^2)*Cos[e + f*x])/f) + (2*(a - 2*b)*(a - b)*Cos[e + f*x]^3)/(3*f) - ((a - b)^2*Cos[e + f*
x]^5)/(5*f) + (2*(a - 2*b)*b*Sec[e + f*x])/f + (b^2*Sec[e + f*x]^3)/(3*f)

Rule 3664

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a - b + b*ff^2*x^2)^p)/x^(
m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 448

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 0]

Rubi steps

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

Mathematica [A]  time = 0.719437, size = 97, normalized size = 0.91 \[ \frac{-30 \left (5 a^2-38 a b+41 b^2\right ) \cos (e+f x)+5 (5 a-13 b) (a-b) \cos (3 (e+f x))-3 (a-b)^2 \cos (5 (e+f x))+480 b (a-2 b) \sec (e+f x)+80 b^2 \sec ^3(e+f x)}{240 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^5*(a + b*Tan[e + f*x]^2)^2,x]

[Out]

(-30*(5*a^2 - 38*a*b + 41*b^2)*Cos[e + f*x] + 5*(5*a - 13*b)*(a - b)*Cos[3*(e + f*x)] - 3*(a - b)^2*Cos[5*(e +
f*x)] + 480*(a - 2*b)*b*Sec[e + f*x] + 80*b^2*Sec[e + f*x]^3)/(240*f)

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Maple [A]  time = 0.084, size = 185, normalized size = 1.7 \begin{align*}{\frac{1}{f} \left ( -{\frac{{a}^{2}\cos \left ( fx+e \right ) }{5} \left ({\frac{8}{3}}+ \left ( \sin \left ( fx+e \right ) \right ) ^{4}+{\frac{4\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{3}} \right ) }+2\,ab \left ({\frac{ \left ( \sin \left ( fx+e \right ) \right ) ^{8}}{\cos \left ( fx+e \right ) }}+ \left ({\frac{16}{5}}+ \left ( \sin \left ( fx+e \right ) \right ) ^{6}+6/5\, \left ( \sin \left ( fx+e \right ) \right ) ^{4}+8/5\, \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) \right ) +{b}^{2} \left ({\frac{ \left ( \sin \left ( fx+e \right ) \right ) ^{10}}{3\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}}}-{\frac{7\, \left ( \sin \left ( fx+e \right ) \right ) ^{10}}{3\,\cos \left ( fx+e \right ) }}-{\frac{7\,\cos \left ( fx+e \right ) }{3} \left ({\frac{128}{35}}+ \left ( \sin \left ( fx+e \right ) \right ) ^{8}+{\frac{8\, \left ( \sin \left ( fx+e \right ) \right ) ^{6}}{7}}+{\frac{48\, \left ( \sin \left ( fx+e \right ) \right ) ^{4}}{35}}+{\frac{64\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{35}} \right ) } \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^5*(a+b*tan(f*x+e)^2)^2,x)

[Out]

1/f*(-1/5*a^2*(8/3+sin(f*x+e)^4+4/3*sin(f*x+e)^2)*cos(f*x+e)+2*a*b*(sin(f*x+e)^8/cos(f*x+e)+(16/5+sin(f*x+e)^6
+6/5*sin(f*x+e)^4+8/5*sin(f*x+e)^2)*cos(f*x+e))+b^2*(1/3*sin(f*x+e)^10/cos(f*x+e)^3-7/3*sin(f*x+e)^10/cos(f*x+
e)-7/3*(128/35+sin(f*x+e)^8+8/7*sin(f*x+e)^6+48/35*sin(f*x+e)^4+64/35*sin(f*x+e)^2)*cos(f*x+e)))

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Maxima [A]  time = 0.983669, size = 140, normalized size = 1.31 \begin{align*} -\frac{3 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - 10 \,{\left (a^{2} - 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + 15 \,{\left (a^{2} - 6 \, a b + 6 \, b^{2}\right )} \cos \left (f x + e\right ) - \frac{5 \,{\left (6 \,{\left (a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}\right )}}{\cos \left (f x + e\right )^{3}}}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5*(a+b*tan(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.05938, size = 258, normalized size = 2.41 \begin{align*} -\frac{3 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{8} - 10 \,{\left (a^{2} - 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{6} + 15 \,{\left (a^{2} - 6 \, a b + 6 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 30 \,{\left (a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 5 \, b^{2}}{15 \, f \cos \left (f x + e\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5*(a+b*tan(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

-1/15*(3*(a^2 - 2*a*b + b^2)*cos(f*x + e)^8 - 10*(a^2 - 3*a*b + 2*b^2)*cos(f*x + e)^6 + 15*(a^2 - 6*a*b + 6*b^
2)*cos(f*x + e)^4 - 30*(a*b - 2*b^2)*cos(f*x + e)^2 - 5*b^2)/(f*cos(f*x + e)^3)

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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(sin(f*x+e)**5*(a+b*tan(f*x+e)**2)**2,x)

[Out]

Timed out

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Giac [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(sin(f*x+e)^5*(a+b*tan(f*x+e)^2)^2,x, algorithm="giac")

[Out]

Timed out