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

Optimal. Leaf size=66 \[ \frac{(2 a-b) \cos ^3(e+f x)}{3 f}-\frac{(a-2 b) \cos (e+f x)}{f}-\frac{a \cos ^5(e+f x)}{5 f}+\frac{b \sec (e+f x)}{f} \]

[Out]

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

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Rubi [A]  time = 0.051205, antiderivative size = 66, normalized size of antiderivative = 1., 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 = {4133, 448} \[ \frac{(2 a-b) \cos ^3(e+f x)}{3 f}-\frac{(a-2 b) \cos (e+f x)}{f}-\frac{a \cos ^5(e+f x)}{5 f}+\frac{b \sec (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0445724, size = 88, normalized size = 1.33 \[ -\frac{5 a \cos (e+f x)}{8 f}+\frac{5 a \cos (3 (e+f x))}{48 f}-\frac{a \cos (5 (e+f x))}{80 f}+\frac{7 b \cos (e+f x)}{4 f}-\frac{b \cos (3 (e+f x))}{12 f}+\frac{b \sec (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*a*Cos[e + f*x])/(8*f) + (7*b*Cos[e + f*x])/(4*f) + (5*a*Cos[3*(e + f*x)])/(48*f) - (b*Cos[3*(e + f*x)])/(1
2*f) - (a*Cos[5*(e + f*x)])/(80*f) + (b*Sec[e + f*x])/f

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Maple [A]  time = 0.043, size = 82, normalized size = 1.2 \begin{align*}{\frac{1}{f} \left ( -{\frac{a\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 ) }+b \left ({\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)*sin(f*x+e)^5,x)

[Out]

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

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Maxima [A]  time = 1.01269, size = 78, normalized size = 1.18 \begin{align*} -\frac{3 \, a \cos \left (f x + e\right )^{5} - 5 \,{\left (2 \, a - b\right )} \cos \left (f x + e\right )^{3} + 15 \,{\left (a - 2 \, b\right )} \cos \left (f x + e\right ) - \frac{15 \, b}{\cos \left (f x + e\right )}}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.839111, size = 150, normalized size = 2.27 \begin{align*} -\frac{3 \, a \cos \left (f x + e\right )^{6} - 5 \,{\left (2 \, a - b\right )} \cos \left (f x + e\right )^{4} + 15 \,{\left (a - 2 \, b\right )} \cos \left (f x + e\right )^{2} - 15 \, b}{15 \, f \cos \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)*sin(f*x+e)**5,x)

[Out]

Timed out

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Giac [B]  time = 1.22106, size = 288, normalized size = 4.36 \begin{align*} \frac{2 \,{\left (\frac{15 \, b}{\frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 1} + \frac{8 \, a - 25 \, b - \frac{40 \, a{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac{110 \, b{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac{80 \, a{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{160 \, b{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{90 \, b{\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac{15 \, b{\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)*sin(f*x+e)^5,x, algorithm="giac")

[Out]

2/15*(15*b/((cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 1) + (8*a - 25*b - 40*a*(cos(f*x + e) - 1)/(cos(f*x + e) +
1) + 110*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 80*a*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 160*b*(co
s(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 90*b*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 - 15*b*(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