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

Optimal. Leaf size=97 \[ -\frac{\left (a^2-4 a b+b^2\right ) \cos (e+f x)}{f}-\frac{a^2 \cos ^5(e+f x)}{5 f}+\frac{2 a (a-b) \cos ^3(e+f x)}{3 f}+\frac{2 b (a-b) \sec (e+f x)}{f}+\frac{b^2 \sec ^3(e+f x)}{3 f} \]

[Out]

-(((a^2 - 4*a*b + b^2)*Cos[e + f*x])/f) + (2*a*(a - b)*Cos[e + f*x]^3)/(3*f) - (a^2*Cos[e + f*x]^5)/(5*f) + (2
*(a - b)*b*Sec[e + f*x])/f + (b^2*Sec[e + f*x]^3)/(3*f)

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Rubi [A]  time = 0.0897856, antiderivative size = 97, 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 = {4133, 448} \[ -\frac{\left (a^2-4 a b+b^2\right ) \cos (e+f x)}{f}-\frac{a^2 \cos ^5(e+f x)}{5 f}+\frac{2 a (a-b) \cos ^3(e+f x)}{3 f}+\frac{2 b (a-b) \sec (e+f x)}{f}+\frac{b^2 \sec ^3(e+f x)}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a^2 - 4*a*b + b^2)*Cos[e + f*x])/f) + (2*a*(a - b)*Cos[e + f*x]^3)/(3*f) - (a^2*Cos[e + f*x]^5)/(5*f) + (2
*(a - b)*b*Sec[e + f*x])/f + (b^2*Sec[e + f*x]^3)/(3*f)

Rule 4133

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

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

Mathematica [A]  time = 0.632687, size = 118, normalized size = 1.22 \[ -\frac{\sec ^3(e+f x) \left (24 \left (22 a^2-215 a b+120 b^2\right ) \cos (2 (e+f x))+12 \left (7 a^2-60 a b+20 b^2\right ) \cos (4 (e+f x))-16 a^2 \cos (6 (e+f x))+3 a^2 \cos (8 (e+f x))+425 a^2+40 a b \cos (6 (e+f x))-4400 a b+2000 b^2\right )}{1920 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((425*a^2 - 4400*a*b + 2000*b^2 + 24*(22*a^2 - 215*a*b + 120*b^2)*Cos[2*(e + f*x)] + 12*(7*a^2 - 60*a*b + 20*
b^2)*Cos[4*(e + f*x)] - 16*a^2*Cos[6*(e + f*x)] + 40*a*b*Cos[6*(e + f*x)] + 3*a^2*Cos[8*(e + f*x)])*Sec[e + f*
x]^3)/(1920*f)

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Maple [A]  time = 0.053, size = 155, normalized size = 1.6 \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 ) ^{6}}{\cos \left ( fx+e \right ) }}+ \left ( 8/3+ \left ( \sin \left ( fx+e \right ) \right ) ^{4}+4/3\, \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 ) ^{6}}{3\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}}}-{\frac{ \left ( \sin \left ( fx+e \right ) \right ) ^{6}}{\cos \left ( fx+e \right ) }}- \left ({\frac{8}{3}}+ \left ( \sin \left ( fx+e \right ) \right ) ^{4}+{\frac{4\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{3}} \right ) \cos \left ( fx+e \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Maxima [A]  time = 1.00513, size = 120, normalized size = 1.24 \begin{align*} -\frac{3 \, a^{2} \cos \left (f x + e\right )^{5} - 10 \,{\left (a^{2} - a b\right )} \cos \left (f x + e\right )^{3} + 15 \,{\left (a^{2} - 4 \, a b + b^{2}\right )} \cos \left (f x + e\right ) - \frac{5 \,{\left (6 \,{\left (a b - 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((a+b*sec(f*x+e)^2)^2*sin(f*x+e)^5,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.511764, size = 217, normalized size = 2.24 \begin{align*} -\frac{3 \, a^{2} \cos \left (f x + e\right )^{8} - 10 \,{\left (a^{2} - a b\right )} \cos \left (f x + e\right )^{6} + 15 \,{\left (a^{2} - 4 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 30 \,{\left (a b - 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((a+b*sec(f*x+e)^2)^2*sin(f*x+e)^5,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.3028, size = 602, normalized size = 6.21 \begin{align*} \frac{2 \,{\left (\frac{5 \,{\left (6 \, a b - 5 \, b^{2} + \frac{12 \, a b{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac{12 \, b^{2}{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac{6 \, a b{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{3 \, b^{2}{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}}{{\left (\frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 1\right )}^{3}} + \frac{8 \, a^{2} - 50 \, a b + 15 \, b^{2} - \frac{40 \, a^{2}{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac{220 \, a b{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac{60 \, b^{2}{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac{80 \, a^{2}{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{320 \, a b{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{90 \, b^{2}{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{180 \, a b{\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac{60 \, b^{2}{\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac{30 \, a b{\left (\cos \left (f x + e\right ) - 1\right )}^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{15 \, b^{2}{\left (\cos \left (f x + e\right ) - 1\right )}^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}}{{\left (\frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} - 1\right )}^{5}}\right )}}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(5*(6*a*b - 5*b^2 + 12*a*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 12*b^2*(cos(f*x + e) - 1)/(cos(f*x + e
) + 1) + 6*a*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 3*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)/((
cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 1)^3 + (8*a^2 - 50*a*b + 15*b^2 - 40*a^2*(cos(f*x + e) - 1)/(cos(f*x +
e) + 1) + 220*a*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 60*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 80*a^
2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 320*a*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 90*b^2*(cos(
f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 180*a*b*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 - 60*b^2*(cos(f*x + e
) - 1)^3/(cos(f*x + e) + 1)^3 - 30*a*b*(cos(f*x + e) - 1)^4/(cos(f*x + e) + 1)^4 + 15*b^2*(cos(f*x + e) - 1)^4
/(cos(f*x + e) + 1)^4)/((cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 1)^5)/f