3.4.75 \(\int (e \cos (c+d x))^{-2-2 m} (a+a \sin (c+d x))^m \, dx\) [375]

Optimal. Leaf size=87 \[ -\frac {2^{-\frac {1}{2}-m} (e \cos (c+d x))^{-1-2 m} \, _2F_1\left (-\frac {1}{2},\frac {1}{2} (3+2 m);\frac {1}{2};\frac {1}{2} (1+\sin (c+d x))\right ) (1-\sin (c+d x))^{\frac {1}{2}+m} (a+a \sin (c+d x))^m}{d e} \]

[Out]

-2^(-1/2-m)*(e*cos(d*x+c))^(-1-2*m)*hypergeom([-1/2, 3/2+m],[1/2],1/2+1/2*sin(d*x+c))*(1-sin(d*x+c))^(1/2+m)*(
a+a*sin(d*x+c))^m/d/e

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Rubi [A]
time = 0.07, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2768, 7, 72, 71} \begin {gather*} -\frac {2^{-m-\frac {1}{2}} (1-\sin (c+d x))^{m+\frac {1}{2}} (a \sin (c+d x)+a)^m (e \cos (c+d x))^{-2 m-1} \, _2F_1\left (-\frac {1}{2},\frac {1}{2} (2 m+3);\frac {1}{2};\frac {1}{2} (\sin (c+d x)+1)\right )}{d e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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]))

Rule 72

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

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Mathematica [A]
time = 0.22, size = 87, normalized size = 1.00 \begin {gather*} \frac {(e \cos (c+d x))^{-1-2 m} \, _2F_1\left (\frac {3}{2},-\frac {1}{2}-m;\frac {1}{2}-m;\frac {1}{2} (1-\sin (c+d x))\right ) \sqrt {1+\sin (c+d x)} (a (1+\sin (c+d x)))^m}{\sqrt {2} e (d+2 d m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.16, size = 0, normalized size = 0.00 \[\int \left (e \cos \left (d x +c \right )\right )^{-2-2 m} \left (a +a \sin \left (d x +c \right )\right )^{m}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \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+2}} \,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 + 2),x)

[Out]

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

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