3.4.70 \(\int (e \cos (c+d x))^{-1-2 m} (a+a \sin (c+d x))^m \, dx\) [370]

Optimal. Leaf size=61 \[ \frac {(e \cos (c+d x))^{-2 m} \, _2F_1\left (1,-m;1-m;\frac {1}{2} (1-\sin (c+d x))\right ) (a+a \sin (c+d x))^m}{2 d e m} \]

[Out]

1/2*hypergeom([1, -m],[1-m],1/2-1/2*sin(d*x+c))*(a+a*sin(d*x+c))^m/d/e/m/((e*cos(d*x+c))^(2*m))

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Rubi [A]
time = 0.05, antiderivative size = 61, normalized size of antiderivative = 1.00, 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 = {2768, 7, 70} \begin {gather*} \frac {(a \sin (c+d x)+a)^m (e \cos (c+d x))^{-2 m} \, _2F_1\left (1,-m;1-m;\frac {1}{2} (1-\sin (c+d x))\right )}{2 d e m} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Hypergeometric2F1[1, -m, 1 - m, (1 - Sin[c + d*x])/2]*(a + a*Sin[c + d*x])^m)/(2*d*e*m*(e*Cos[c + d*x])^(2*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 70

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

Rule 2768

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]

Rubi steps

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

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Mathematica [A]
time = 0.07, size = 61, normalized size = 1.00 \begin {gather*} \frac {(e \cos (c+d x))^{-2 m} \, _2F_1\left (1,-m;1-m;\frac {1}{2} (1-\sin (c+d x))\right ) (a (1+\sin (c+d x)))^m}{2 d e m} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.17, size = 0, normalized size = 0.00 \[\int \left (e \cos \left (d x +c \right )\right )^{-1-2 m} \left (a +a \sin \left (d x +c \right )\right )^{m}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(-1-2*m)*(a+a*sin(d*x+c))^m,x)

[Out]

int((e*cos(d*x+c))^(-1-2*m)*(a+a*sin(d*x+c))^m,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(-1-2*m)*(a+a*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

integrate((cos(d*x + c)*e)^(-2*m - 1)*(a*sin(d*x + c) + a)^m, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(-1-2*m)*(a+a*sin(d*x+c))^m,x, algorithm="fricas")

[Out]

integral((cos(d*x + c)*e)^(-2*m - 1)*(a*sin(d*x + c) + a)^m, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{m} \left (e \cos {\left (c + d x \right )}\right )^{- 2 m - 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))**(-1-2*m)*(a+a*sin(d*x+c))**m,x)

[Out]

Integral((a*(sin(c + d*x) + 1))**m*(e*cos(c + d*x))**(-2*m - 1), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(-1-2*m)*(a+a*sin(d*x+c))^m,x, algorithm="giac")

[Out]

integrate((cos(d*x + c)*e)^(-2*m - 1)*(a*sin(d*x + c) + a)^m, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^m}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{2\,m+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^m/(e*cos(c + d*x))^(2*m + 1),x)

[Out]

int((a + a*sin(c + d*x))^m/(e*cos(c + d*x))^(2*m + 1), x)

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