3.1028 \(\int \cos ^4(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx\)

Optimal. Leaf size=129 \[ -\frac{a^2 2^{m+\frac{5}{2}} (A (m+5)+B m) \cos ^5(e+f x) (\sin (e+f x)+1)^{-m-\frac{1}{2}} (a \sin (e+f x)+a)^{m-2} \, _2F_1\left (\frac{5}{2},-m-\frac{3}{2};\frac{7}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{5 f (m+5)}-\frac{B \cos ^5(e+f x) (a \sin (e+f x)+a)^m}{f (m+5)} \]

[Out]

-(2^(5/2 + m)*a^2*(B*m + A*(5 + m))*Cos[e + f*x]^5*Hypergeometric2F1[5/2, -3/2 - m, 7/2, (1 - Sin[e + f*x])/2]
*(1 + Sin[e + f*x])^(-1/2 - m)*(a + a*Sin[e + f*x])^(-2 + m))/(5*f*(5 + m)) - (B*Cos[e + f*x]^5*(a + a*Sin[e +
f*x])^m)/(f*(5 + m))

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Rubi [A]  time = 0.197929, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {2860, 2689, 70, 69} \[ -\frac{a^2 2^{m+\frac{5}{2}} (A (m+5)+B m) \cos ^5(e+f x) (\sin (e+f x)+1)^{-m-\frac{1}{2}} (a \sin (e+f x)+a)^{m-2} \, _2F_1\left (\frac{5}{2},-m-\frac{3}{2};\frac{7}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{5 f (m+5)}-\frac{B \cos ^5(e+f x) (a \sin (e+f x)+a)^m}{f (m+5)} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^4*(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]),x]

[Out]

-(2^(5/2 + m)*a^2*(B*m + A*(5 + m))*Cos[e + f*x]^5*Hypergeometric2F1[5/2, -3/2 - m, 7/2, (1 - Sin[e + f*x])/2]
*(1 + Sin[e + f*x])^(-1/2 - m)*(a + a*Sin[e + f*x])^(-2 + m))/(5*f*(5 + m)) - (B*Cos[e + f*x]^5*(a + a*Sin[e +
f*x])^m)/(f*(5 + m))

Rule 2860

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[(d*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(f*g*(m + p + 1)), x]
+ Dist[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; Fre
eQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[m + p + 1, 0]

Rule 2689

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

Rule 70

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

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

\begin{align*} \int \cos ^4(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx &=-\frac{B \cos ^5(e+f x) (a+a \sin (e+f x))^m}{f (5+m)}+\left (A+\frac{B m}{5+m}\right ) \int \cos ^4(e+f x) (a+a \sin (e+f x))^m \, dx\\ &=-\frac{B \cos ^5(e+f x) (a+a \sin (e+f x))^m}{f (5+m)}+\frac{\left (a^2 \left (A+\frac{B m}{5+m}\right ) \cos ^5(e+f x)\right ) \operatorname{Subst}\left (\int (a-a x)^{3/2} (a+a x)^{\frac{3}{2}+m} \, dx,x,\sin (e+f x)\right )}{f (a-a \sin (e+f x))^{5/2} (a+a \sin (e+f x))^{5/2}}\\ &=-\frac{B \cos ^5(e+f x) (a+a \sin (e+f x))^m}{f (5+m)}+\frac{\left (2^{\frac{3}{2}+m} a^3 \left (A+\frac{B m}{5+m}\right ) \cos ^5(e+f x) (a+a \sin (e+f x))^{-2+m} \left (\frac{a+a \sin (e+f x)}{a}\right )^{-\frac{1}{2}-m}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}+\frac{x}{2}\right )^{\frac{3}{2}+m} (a-a x)^{3/2} \, dx,x,\sin (e+f x)\right )}{f (a-a \sin (e+f x))^{5/2}}\\ &=-\frac{2^{\frac{5}{2}+m} a^2 \left (A+\frac{B m}{5+m}\right ) \cos ^5(e+f x) \, _2F_1\left (\frac{5}{2},-\frac{3}{2}-m;\frac{7}{2};\frac{1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac{1}{2}-m} (a+a \sin (e+f x))^{-2+m}}{5 f}-\frac{B \cos ^5(e+f x) (a+a \sin (e+f x))^m}{f (5+m)}\\ \end{align*}

Mathematica [A]  time = 0.463207, size = 111, normalized size = 0.86 \[ -\frac{\cos ^5(e+f x) (\sin (e+f x)+1)^{-m-\frac{5}{2}} (a (\sin (e+f x)+1))^m \left (2^{m+\frac{5}{2}} (A (m+5)+B m) \, _2F_1\left (\frac{5}{2},-m-\frac{3}{2};\frac{7}{2};\frac{1}{2} (1-\sin (e+f x))\right )+5 B (\sin (e+f x)+1)^{m+\frac{5}{2}}\right )}{5 f (m+5)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^4*(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]),x]

[Out]

-(Cos[e + f*x]^5*(1 + Sin[e + f*x])^(-5/2 - m)*(a*(1 + Sin[e + f*x]))^m*(2^(5/2 + m)*(B*m + A*(5 + m))*Hyperge
ometric2F1[5/2, -3/2 - m, 7/2, (1 - Sin[e + f*x])/2] + 5*B*(1 + Sin[e + f*x])^(5/2 + m)))/(5*f*(5 + m))

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Maple [F]  time = 3.684, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( fx+e \right ) \right ) ^{4} \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(cos(f*x+e)^4*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x)

[Out]

int(cos(f*x+e)^4*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x)

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Maxima [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(cos(f*x+e)^4*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (B \cos \left (f x + e\right )^{4} \sin \left (f x + e\right ) + A \cos \left (f x + e\right )^{4}\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(cos(f*x+e)^4*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x, algorithm="fricas")

[Out]

integral((B*cos(f*x + e)^4*sin(f*x + e) + A*cos(f*x + e)^4)*(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(cos(f*x+e)**4*(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} \cos \left (f x + e\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^4*(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*cos(f*x + e)^4, x)