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

Optimal. Leaf size=97 \[ -\frac{a \sin (c+d x) (e \cos (c+d x))^{p+1} \, _2F_1\left (\frac{1}{2},\frac{p+1}{2};\frac{p+3}{2};\cos ^2(c+d x)\right )}{d e (p+1) \sqrt{\sin ^2(c+d x)}}-\frac{b (e \cos (c+d x))^{p+1}}{d e (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.0520769, antiderivative size = 97, 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 = {2669, 2643} \[ -\frac{a \sin (c+d x) (e \cos (c+d x))^{p+1} \, _2F_1\left (\frac{1}{2},\frac{p+1}{2};\frac{p+3}{2};\cos ^2(c+d x)\right )}{d e (p+1) \sqrt{\sin ^2(c+d x)}}-\frac{b (e \cos (c+d x))^{p+1}}{d e (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2669

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

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rubi steps

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

Mathematica [C]  time = 0.934504, size = 240, normalized size = 2.47 \[ -\frac{(e \cos (c+d x))^p \left (-\frac{1}{2} a (p-1) \sin (2 (c+d x)) \, _2F_1\left (\frac{1}{2},\frac{p+1}{2};\frac{p+3}{2};\cos ^2(c+d x)\right )+b 2^{-p-1} \left (1+e^{2 i (c+d x)}\right ) \left (e^{-i (c+d x)}+e^{i (c+d x)}\right )^p \sqrt{\sin ^2(c+d x)} \left ((p+1) e^{i (c+d x)} \, _2F_1\left (1,\frac{p+3}{2};\frac{3-p}{2};-e^{2 i (c+d x)}\right )-(p-1) e^{-i (c+d x)} \, _2F_1\left (1,\frac{p+1}{2};\frac{1-p}{2};-e^{2 i (c+d x)}\right )\right ) \cos ^{-p}(c+d x)\right )}{\left (d-d p^2\right ) \sqrt{\sin ^2(c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(((e*Cos[c + d*x])^p*((2^(-1 - p)*b*(E^((-I)*(c + d*x)) + E^(I*(c + d*x)))^p*(1 + E^((2*I)*(c + d*x)))*(-(((-
1 + p)*Hypergeometric2F1[1, (1 + p)/2, (1 - p)/2, -E^((2*I)*(c + d*x))])/E^(I*(c + d*x))) + E^(I*(c + d*x))*(1
+ p)*Hypergeometric2F1[1, (3 + p)/2, (3 - p)/2, -E^((2*I)*(c + d*x))])*Sqrt[Sin[c + d*x]^2])/Cos[c + d*x]^p -
(a*(-1 + p)*Hypergeometric2F1[1/2, (1 + p)/2, (3 + p)/2, Cos[c + d*x]^2]*Sin[2*(c + d*x)])/2))/((d - d*p^2)*S
qrt[Sin[c + d*x]^2]))

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Maple [F]  time = 0.875, size = 0, normalized size = 0. \begin{align*} \int \left ( e\cos \left ( dx+c \right ) \right ) ^{p} \left ( a+b\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+b*sin(d*x+c)),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \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+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

integrate((b*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 (b \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+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

integral((b*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*} \int \left (e \cos{\left (c + d x \right )}\right )^{p} \left (a + b \sin{\left (c + d x \right )}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \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+b*sin(d*x+c)),x, algorithm="giac")

[Out]

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