3.330 \(\int (e \cos (c+d x))^p (a+a \sin (c+d x)) \, dx\)

Optimal. Leaf size=93 \[ -\frac{a 2^{\frac{p}{2}+\frac{3}{2}} (\sin (c+d x)+1)^{\frac{1}{2} (-p-1)} (e \cos (c+d x))^{p+1} \, _2F_1\left (\frac{1}{2} (-p-1),\frac{p+1}{2};\frac{p+3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{d e (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.0568303, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2688, 69} \[ -\frac{a 2^{\frac{p}{2}+\frac{3}{2}} (\sin (c+d x)+1)^{\frac{1}{2} (-p-1)} (e \cos (c+d x))^{p+1} \, _2F_1\left (\frac{1}{2} (-p-1),\frac{p+1}{2};\frac{p+3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{d e (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2688

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

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

Mathematica [C]  time = 1.34029, size = 245, normalized size = 2.63 \[ -\frac{i a 2^{-p-1} \left (e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )\right )^{p+1} (\sin (c+d x)+1) \left ((p+1) e^{i (c+d x)} \left (i p e^{i (c+d x)} \, _2F_1\left (1,\frac{p+3}{2};\frac{3-p}{2};-e^{2 i (c+d x)}\right )-2 (p-1) \, _2F_1\left (1,\frac{p+2}{2};1-\frac{p}{2};-e^{2 i (c+d x)}\right )\right )-i (p-1) p \, _2F_1\left (1,\frac{p+1}{2};\frac{1-p}{2};-e^{2 i (c+d x)}\right )\right ) \cos ^{-p}(c+d x) (e \cos (c+d x))^p}{d (p-1) p (p+1) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I)*2^(-1 - p)*a*((1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x)))^(1 + p)*(e*Cos[c + d*x])^p*((-I)*(-1 + p)*p*Hyp
ergeometric2F1[1, (1 + p)/2, (1 - p)/2, -E^((2*I)*(c + d*x))] + E^(I*(c + d*x))*(1 + p)*(-2*(-1 + p)*Hypergeom
etric2F1[1, (2 + p)/2, 1 - p/2, -E^((2*I)*(c + d*x))] + I*E^(I*(c + d*x))*p*Hypergeometric2F1[1, (3 + p)/2, (3
- p)/2, -E^((2*I)*(c + d*x))]))*(1 + Sin[c + d*x]))/(d*(-1 + p)*p*(1 + p)*Cos[c + d*x]^p*(Cos[(c + d*x)/2] +
Sin[(c + d*x)/2])^2)

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Maple [F]  time = 0.907, size = 0, normalized size = 0. \begin{align*} \int \left ( e\cos \left ( dx+c \right ) \right ) ^{p} \left ( a+a\sin \left ( dx+c \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^p*(a+a*sin(d*x+c)),x)

[Out]

int((e*cos(d*x+c))^p*(a+a*sin(d*x+c)),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sin(d*x + c) + a)*(e*cos(d*x + c))^p, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*sin(d*x + c) + a)*(e*cos(d*x + c))^p, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))**p*(a+a*sin(d*x+c)),x)

[Out]

a*(Integral((e*cos(c + d*x))**p, x) + Integral((e*cos(c + d*x))**p*sin(c + d*x), x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sin(d*x + c) + a)*(e*cos(d*x + c))^p, x)