3.7.18 \(\int (e \cos (c+d x))^p (a+b \sin (c+d x)) \, dx\) [618]

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

[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 = 0.04, 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 = {2748, 2722} \begin {gather*} -\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)} \end {gather*}

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 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[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] Result contains complex when optimal does not.
time = 0.65, size = 240, normalized size = 2.47 \begin {gather*} -\frac {(e \cos (c+d x))^p \left (2^{-1-p} b \left (e^{-i (c+d x)}+e^{i (c+d x)}\right )^p \left (1+e^{2 i (c+d x)}\right ) \cos ^{-p}(c+d x) \left (-e^{-i (c+d x)} (-1+p) \, _2F_1\left (1,\frac {1+p}{2};\frac {1-p}{2};-e^{2 i (c+d x)}\right )+e^{i (c+d x)} (1+p) \, _2F_1\left (1,\frac {3+p}{2};\frac {3-p}{2};-e^{2 i (c+d x)}\right )\right ) \sqrt {\sin ^2(c+d x)}-\frac {1}{2} a (-1+p) \, _2F_1\left (\frac {1}{2},\frac {1+p}{2};\frac {3+p}{2};\cos ^2(c+d x)\right ) \sin (2 (c+d x))\right )}{\left (d-d p^2\right ) \sqrt {\sin ^2(c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[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 = 0.15, 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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)*(cos(d*x + c)*e)^p, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)*(cos(d*x + c)*e)^p, x)

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

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)*(cos(d*x + c)*e)^p, x)

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

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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