3.650 \(\int (e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^m \, dx\)

Optimal. Leaf size=132 \[ \frac{e (1-\sin (c+d x)) (e \cos (c+d x))^{-m-2} \left (-\frac{(a-b) (1-\sin (c+d x))}{(a+b) (\sin (c+d x)+1)}\right )^{m/2} (a+b \sin (c+d x))^{m+1} \, _2F_1\left (m+1,\frac{m+2}{2};m+2;\frac{2 (a+b \sin (c+d x))}{(a+b) (\sin (c+d x)+1)}\right )}{d (m+1) (a+b)} \]

[Out]

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

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Rubi [A]  time = 0.0654538, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.037, Rules used = {2698} \[ \frac{e (1-\sin (c+d x)) (e \cos (c+d x))^{-m-2} \left (-\frac{(a-b) (1-\sin (c+d x))}{(a+b) (\sin (c+d x)+1)}\right )^{m/2} (a+b \sin (c+d x))^{m+1} \, _2F_1\left (m+1,\frac{m+2}{2};m+2;\frac{2 (a+b \sin (c+d x))}{(a+b) (\sin (c+d x)+1)}\right )}{d (m+1) (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2698

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[e + f*x])^(p - 1)*(1 - Sin[e + f*x])*(a + b*Sin[e + f*x])^(m + 1)*(-(((a - b)*(1 - Sin[e + f*x]))/((a + b)
*(1 + Sin[e + f*x]))))^(m/2)*Hypergeometric2F1[m + 1, m/2 + 1, m + 2, (2*(a + b*Sin[e + f*x]))/((a + b)*(1 + S
in[e + f*x]))])/(f*(a + b)*(m + 1)), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && NeQ[a^2 - b^2, 0] && EqQ[m + p +
1, 0]

Rubi steps

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

Mathematica [A]  time = 0.371724, size = 132, normalized size = 1. \[ -\frac{e (\sin (c+d x)+1) (e \cos (c+d x))^{-m-2} \left (\frac{(a+b) (\sin (c+d x)+1)}{(a-b) (\sin (c+d x)-1)}\right )^{m/2} (a+b \sin (c+d x))^{m+1} \, _2F_1\left (m+1,\frac{m+2}{2};m+2;-\frac{2 (a+b \sin (c+d x))}{(a-b) (\sin (c+d x)-1)}\right )}{d (m+1) (a-b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.175, size = 0, normalized size = 0. \begin{align*} \int \left ( e\cos \left ( dx+c \right ) \right ) ^{-1-m} \left ( a+b\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))^(-1-m)*(a+b*sin(d*x+c))^m,x)

[Out]

int((e*cos(d*x+c))^(-1-m)*(a+b*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 )^{-m - 1}{\left (b \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))^(-1-m)*(a+b*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^(-m - 1)*(b*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 )^{-m - 1}{\left (b \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))^(-1-m)*(a+b*sin(d*x+c))^m,x, algorithm="fricas")

[Out]

integral((e*cos(d*x + c))^(-m - 1)*(b*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))**(-1-m)*(a+b*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 )^{-m - 1}{\left (b \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))^(-1-m)*(a+b*sin(d*x+c))^m,x, algorithm="giac")

[Out]

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