3.371 \(\int (e \cos (c+d x))^{-3-2 m} (a+a \sin (c+d x))^m \, dx\)

Optimal. Leaf size=70 \[ \frac{(a \sin (c+d x)+a)^{m+1} (e \cos (c+d x))^{-2 (m+1)} \, _2F_1\left (2,-m-1;-m;\frac{1}{2} (1-\sin (c+d x))\right )}{4 a d e (m+1)} \]

[Out]

(Hypergeometric2F1[2, -1 - m, -m, (1 - Sin[c + d*x])/2]*(a + a*Sin[c + d*x])^(1 + m))/(4*a*d*e*(1 + m)*(e*Cos[
c + d*x])^(2*(1 + m)))

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Rubi [A]  time = 0.0751064, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2689, 7, 68} \[ \frac{(a \sin (c+d x)+a)^{m+1} (e \cos (c+d x))^{-2 (m+1)} \, _2F_1\left (2,-m-1;-m;\frac{1}{2} (1-\sin (c+d x))\right )}{4 a d e (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(-3 - 2*m)*(a + a*Sin[c + d*x])^m,x]

[Out]

(Hypergeometric2F1[2, -1 - m, -m, (1 - Sin[c + d*x])/2]*(a + a*Sin[c + d*x])^(1 + m))/(4*a*d*e*(1 + m)*(e*Cos[
c + d*x])^(2*(1 + m)))

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 7

Int[(u_.)*(Px_)^(p_), x_Symbol] :> Int[u*Px^Simplify[p], x] /; PolyQ[Px, x] &&  !RationalQ[p] && FreeQ[p, x] &
& RationalQ[Simplify[p]]

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 (e \cos (c+d x))^{-3-2 m} (a+a \sin (c+d x))^m \, dx &=\frac{\left (a^2 (e \cos (c+d x))^{-2-2 m} (a-a \sin (c+d x))^{\frac{1}{2} (2+2 m)} (a+a \sin (c+d x))^{\frac{1}{2} (2+2 m)}\right ) \operatorname{Subst}\left (\int (a-a x)^{\frac{1}{2} (-4-2 m)} (a+a x)^{\frac{1}{2} (-4-2 m)+m} \, dx,x,\sin (c+d x)\right )}{d e}\\ &=\frac{\left (a^2 (e \cos (c+d x))^{-2-2 m} (a-a \sin (c+d x))^{\frac{1}{2} (2+2 m)} (a+a \sin (c+d x))^{\frac{1}{2} (2+2 m)}\right ) \operatorname{Subst}\left (\int \frac{(a-a x)^{\frac{1}{2} (-4-2 m)}}{(a+a x)^2} \, dx,x,\sin (c+d x)\right )}{d e}\\ &=\frac{(e \cos (c+d x))^{-2 (1+m)} \, _2F_1\left (2,-1-m;-m;\frac{1}{2} (1-\sin (c+d x))\right ) (a+a \sin (c+d x))^{1+m}}{4 a d e (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.136757, size = 76, normalized size = 1.09 \[ \frac{\sec ^2(c+d x) (a (\sin (c+d x)+1))^{m+1} (e \cos (c+d x))^{-2 m} \, _2F_1\left (2,-m-1;-m;\frac{1}{2} (1-\sin (c+d x))\right )}{4 a d e^3 (m+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(-3 - 2*m)*(a + a*Sin[c + d*x])^m,x]

[Out]

(Hypergeometric2F1[2, -1 - m, -m, (1 - Sin[c + d*x])/2]*Sec[c + d*x]^2*(a*(1 + Sin[c + d*x]))^(1 + m))/(4*a*d*
e^3*(1 + m)*(e*Cos[c + d*x])^(2*m))

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Maple [F]  time = 0.802, size = 0, normalized size = 0. \begin{align*} \int \left ( e\cos \left ( dx+c \right ) \right ) ^{-3-2\,m} \left ( a+a\sin \left ( dx+c \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(-3-2*m)*(a+a*sin(d*x+c))^m,x)

[Out]

int((e*cos(d*x+c))^(-3-2*m)*(a+a*sin(d*x+c))^m,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e \cos \left (d x + c\right )\right )^{-2 \, m - 3}{\left (a \sin \left (d x + c\right ) + a\right )}^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(-3-2*m)*(a+a*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^(-2*m - 3)*(a*sin(d*x + c) + a)^m, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (e \cos \left (d x + c\right )\right )^{-2 \, m - 3}{\left (a \sin \left (d x + c\right ) + a\right )}^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(-3-2*m)*(a+a*sin(d*x+c))^m,x, algorithm="fricas")

[Out]

integral((e*cos(d*x + c))^(-2*m - 3)*(a*sin(d*x + c) + 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((e*cos(d*x+c))**(-3-2*m)*(a+a*sin(d*x+c))**m,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e \cos \left (d x + c\right )\right )^{-2 \, m - 3}{\left (a \sin \left (d x + c\right ) + a\right )}^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(-3-2*m)*(a+a*sin(d*x+c))^m,x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^(-2*m - 3)*(a*sin(d*x + c) + a)^m, x)