3.1026 \(\int \sec ^5(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx\)

Optimal. Leaf size=104 \[ \frac{a^4 (A+B) (a \sin (e+f x)+a)^{m-2}}{4 f (a-a \sin (e+f x))^2}-\frac{a^2 (A (4-m)-B m) (a \sin (e+f x)+a)^{m-2} \, _2F_1\left (2,m-2;m-1;\frac{1}{2} (\sin (e+f x)+1)\right )}{16 f (2-m)} \]

[Out]

-(a^2*(A*(4 - m) - B*m)*Hypergeometric2F1[2, -2 + m, -1 + m, (1 + Sin[e + f*x])/2]*(a + a*Sin[e + f*x])^(-2 +
m))/(16*f*(2 - m)) + (a^4*(A + B)*(a + a*Sin[e + f*x])^(-2 + m))/(4*f*(a - a*Sin[e + f*x])^2)

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Rubi [A]  time = 0.139399, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {2836, 78, 68} \[ \frac{a^4 (A+B) (a \sin (e+f x)+a)^{m-2}}{4 f (a-a \sin (e+f x))^2}-\frac{a^2 (A (4-m)-B m) (a \sin (e+f x)+a)^{m-2} \, _2F_1\left (2,m-2;m-1;\frac{1}{2} (\sin (e+f x)+1)\right )}{16 f (2-m)} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^5*(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]),x]

[Out]

-(a^2*(A*(4 - m) - B*m)*Hypergeometric2F1[2, -2 + m, -1 + m, (1 + Sin[e + f*x])/2]*(a + a*Sin[e + f*x])^(-2 +
m))/(16*f*(2 - m)) + (a^4*(A + B)*(a + a*Sin[e + f*x])^(-2 + m))/(4*f*(a - a*Sin[e + f*x])^2)

Rule 2836

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d*x)/b
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

\begin{align*} \int \sec ^5(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx &=\frac{a^5 \operatorname{Subst}\left (\int \frac{(a+x)^{-3+m} \left (A+\frac{B x}{a}\right )}{(a-x)^3} \, dx,x,a \sin (e+f x)\right )}{f}\\ &=\frac{a^4 (A+B) (a+a \sin (e+f x))^{-2+m}}{4 f (a-a \sin (e+f x))^2}+\frac{\left (a^4 (A (4-m)-B m)\right ) \operatorname{Subst}\left (\int \frac{(a+x)^{-3+m}}{(a-x)^2} \, dx,x,a \sin (e+f x)\right )}{4 f}\\ &=-\frac{a^2 (A (4-m)-B m) \, _2F_1\left (2,-2+m;-1+m;\frac{1}{2} (1+\sin (e+f x))\right ) (a+a \sin (e+f x))^{-2+m}}{16 f (2-m)}+\frac{a^4 (A+B) (a+a \sin (e+f x))^{-2+m}}{4 f (a-a \sin (e+f x))^2}\\ \end{align*}

Mathematica [A]  time = 0.16677, size = 76, normalized size = 0.73 \[ \frac{a^2 (a (\sin (e+f x)+1))^{m-2} \left (\frac{4 (A+B)}{(\sin (e+f x)-1)^2}-\frac{(A (m-4)+B m) \, _2F_1\left (2,m-2;m-1;\frac{1}{2} (\sin (e+f x)+1)\right )}{m-2}\right )}{16 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^5*(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]),x]

[Out]

(a^2*(-(((A*(-4 + m) + B*m)*Hypergeometric2F1[2, -2 + m, -1 + m, (1 + Sin[e + f*x])/2])/(-2 + m)) + (4*(A + B)
)/(-1 + Sin[e + f*x])^2)*(a*(1 + Sin[e + f*x]))^(-2 + m))/(16*f)

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Maple [F]  time = 0.341, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( fx+e \right ) \right ) ^{5} \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( A+B\sin \left ( fx+e \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^5*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x)

[Out]

int(sec(f*x+e)^5*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^5*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m*sec(f*x + e)^5, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (B \sec \left (f x + e\right )^{5} \sin \left (f x + e\right ) + A \sec \left (f x + e\right )^{5}\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^5*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x, algorithm="fricas")

[Out]

integral((B*sec(f*x + e)^5*sin(f*x + e) + A*sec(f*x + e)^5)*(a*sin(f*x + e) + a)^m, x)

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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(sec(f*x+e)**5*(a+a*sin(f*x+e))**m*(A+B*sin(f*x+e)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^5*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x, algorithm="giac")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m*sec(f*x + e)^5, x)