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

Optimal. Leaf size=97 \[ \frac{2^{1-\frac{m}{2}} (1-\sin (c+d x))^{\frac{m}{2}-1} (a \sin (c+d x)+a)^m (e \cos (c+d x))^{2-m} \, _2F_1\left (\frac{m}{2},\frac{m+2}{2};\frac{m+4}{2};\frac{1}{2} (\sin (c+d x)+1)\right )}{d e (m+2)} \]

[Out]

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

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Rubi [A]  time = 0.106128, antiderivative size = 97, 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, 70, 69} \[ \frac{2^{1-\frac{m}{2}} (1-\sin (c+d x))^{\frac{m}{2}-1} (a \sin (c+d x)+a)^m (e \cos (c+d x))^{2-m} \, _2F_1\left (\frac{m}{2},\frac{m+2}{2};\frac{m+4}{2};\frac{1}{2} (\sin (c+d x)+1)\right )}{d e (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2^(1 - m/2)*(e*Cos[c + d*x])^(2 - m)*Hypergeometric2F1[m/2, (2 + m)/2, (4 + m)/2, (1 + Sin[c + d*x])/2]*(1 -
Sin[c + d*x])^(-1 + m/2)*(a + a*Sin[c + d*x])^m)/(d*e*(2 + 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 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))^{1-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))^{\frac{1}{2} (-2+m)} (a+a \sin (c+d x))^{\frac{1}{2} (-2+m)}\right ) \operatorname{Subst}\left (\int (a-a x)^{-m/2} (a+a x)^{m/2} \, dx,x,\sin (c+d x)\right )}{d e}\\ &=\frac{\left (2^{-m/2} a^2 (e \cos (c+d x))^{2-m} (a-a \sin (c+d x))^{\frac{1}{2} (-2+m)-\frac{m}{2}} \left (\frac{a-a \sin (c+d x)}{a}\right )^{m/2} (a+a \sin (c+d x))^{\frac{1}{2} (-2+m)}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}-\frac{x}{2}\right )^{-m/2} (a+a x)^{m/2} \, dx,x,\sin (c+d x)\right )}{d e}\\ &=\frac{2^{1-\frac{m}{2}} (e \cos (c+d x))^{2-m} \, _2F_1\left (\frac{m}{2},\frac{2+m}{2};\frac{4+m}{2};\frac{1}{2} (1+\sin (c+d x))\right ) (1-\sin (c+d x))^{-1+\frac{m}{2}} (a+a \sin (c+d x))^m}{d e (2+m)}\\ \end{align*}

Mathematica [A]  time = 0.254181, size = 97, normalized size = 1. \[ \frac{2^{\frac{m}{2}+1} (\sin (c+d x)+1)^{-\frac{m}{2}-1} (a (\sin (c+d x)+1))^m (e \cos (c+d x))^{2-m} \, _2F_1\left (1-\frac{m}{2},-\frac{m}{2};2-\frac{m}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{d e (m-2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.15, size = 0, normalized size = 0. \begin{align*} \int \left ( e\cos \left ( dx+c \right ) \right ) ^{1-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))^(1-m)*(a+a*sin(d*x+c))^m,x)

[Out]

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

[Out]

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

[Out]

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

[Out]

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