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(d*x+c))^(1+p)/d/e/(1+p)-a*(e*cos(d*x+c))^(1+p)*hypergeom([1/2, 1/2+1/2*p],[3/2+1/2*p],cos(d*x+c)^2)*
sin(d*x+c)/d/e/(1+p)/(sin(d*x+c)^2)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 97, normalized size of antiderivative = 1.00, 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 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]

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])

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*}

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Mathematica [C]  time = 0.99, 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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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \sin \left (d x + c\right ) + a\right )} \left (e \cos \left (d x + c\right )\right )^{p}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (d x + c\right ) + a\right )} \left (e \cos \left (d x + c\right )\right )^{p}\,{d x} \]

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)

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maple [F]  time = 1.77, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x +c \right )\right )^{p} \left (a +b \sin \left (d x +c \right )\right )\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (d x + c\right ) + a\right )} \left (e \cos \left (d x + c\right )\right )^{p}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (e\,\cos \left (c+d\,x\right )\right )}^p\,\left (a+b\,\sin \left (c+d\,x\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \cos {\left (c + d x \right )}\right )^{p} \left (a + b \sin {\left (c + d x \right )}\right )\, dx \]

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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