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

Optimal. Leaf size=86 \[ \frac{2^{\frac{1}{2}-m} (1-\sin (c+d x))^{m-\frac{1}{2}} (a \sin (c+d x)+a)^m (e \cos (c+d x))^{1-2 m} \, _2F_1\left (\frac{1}{2},\frac{1}{2} (2 m+1);\frac{3}{2};\frac{1}{2} (\sin (c+d x)+1)\right )}{d e} \]

[Out]

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

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Rubi [A]  time = 0.0904185, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2689, 7, 70, 69} \[ \frac{2^{\frac{1}{2}-m} (1-\sin (c+d x))^{m-\frac{1}{2}} (a \sin (c+d x)+a)^m (e \cos (c+d x))^{1-2 m} \, _2F_1\left (\frac{1}{2},\frac{1}{2} (2 m+1);\frac{3}{2};\frac{1}{2} (\sin (c+d x)+1)\right )}{d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0854901, size = 90, normalized size = 1.05 \[ \frac{\sqrt{2} \cos (c+d x) (a (\sin (c+d x)+1))^m (e \cos (c+d x))^{-2 m} \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2}-m;\frac{1}{2} (1-\sin (c+d x))\right )}{d (2 m-1) \sqrt{\sin (c+d x)+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.42, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+a\sin \left ( dx+c \right ) \right ) ^{m}}{ \left ( e\cos \left ( dx+c \right ) \right ) ^{2\,m}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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