3.7.17 \(\int (e \cos (c+d x))^p (a+b \sin (c+d x))^2 \, dx\) [617]

Optimal. Leaf size=157 \[ -\frac {a b (3+p) (e \cos (c+d x))^{1+p}}{d e (1+p) (2+p)}-\frac {\left (b^2+a^2 (2+p)\right ) (e \cos (c+d x))^{1+p} \, _2F_1\left (\frac {1}{2},\frac {1+p}{2};\frac {3+p}{2};\cos ^2(c+d x)\right ) \sin (c+d x)}{d e (1+p) (2+p) \sqrt {\sin ^2(c+d x)}}-\frac {b (e \cos (c+d x))^{1+p} (a+b \sin (c+d x))}{d e (2+p)} \]

[Out]

-a*b*(3+p)*(e*cos(d*x+c))^(1+p)/d/e/(1+p)/(2+p)-b*(e*cos(d*x+c))^(1+p)*(a+b*sin(d*x+c))/d/e/(2+p)-(b^2+a^2*(2+
p))*(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)/(2+p)/(sin(
d*x+c)^2)^(1/2)

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Rubi [A]
time = 0.11, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2771, 2748, 2722} \begin {gather*} -\frac {\left (a^2 (p+2)+b^2\right ) \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) (p+2) \sqrt {\sin ^2(c+d x)}}-\frac {a b (p+3) (e \cos (c+d x))^{p+1}}{d e (p+1) (p+2)}-\frac {b (a+b \sin (c+d x)) (e \cos (c+d x))^{p+1}}{d e (p+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^p*(a + b*Sin[c + d*x])^2,x]

[Out]

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

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])

Rule 2771

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

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 0.70, size = 285, normalized size = 1.82 \begin {gather*} -\frac {(e \cos (c+d x))^p \left (2^{-p} a 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} b^2 (-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))-\frac {1}{2} a^2 (-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])^2,x]

[Out]

-(((e*Cos[c + d*x])^p*((a*b*(E^((-I)*(c + d*x)) + E^(I*(c + d*x)))^p*(1 + E^((2*I)*(c + d*x)))*(-(((-1 + p)*Hy
pergeometric2F1[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)*Hyp
ergeometric2F1[1, (3 + p)/2, (3 - p)/2, -E^((2*I)*(c + d*x))])*Sqrt[Sin[c + d*x]^2])/(2^p*Cos[c + d*x]^p) - (b
^2*(-1 + p)*Hypergeometric2F1[-1/2, (1 + p)/2, (3 + p)/2, Cos[c + d*x]^2]*Sin[2*(c + d*x)])/2 - (a^2*(-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)*Sqrt[Sin[c + d
*x]^2]))

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Maple [F]
time = 0.70, size = 0, normalized size = 0.00 \[\int \left (e \cos \left (d x +c \right )\right )^{p} \left (a +b \sin \left (d x +c \right )\right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^p*(a+b*sin(d*x+c))^2,x)

[Out]

int((e*cos(d*x+c))^p*(a+b*sin(d*x+c))^2,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))^2,x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c) + a)^2*(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))^2,x, algorithm="fricas")

[Out]

integral(-(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)*(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 )^{2}\, 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))**2,x)

[Out]

Integral((e*cos(c + d*x))**p*(a + b*sin(c + d*x))**2, 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))^2,x, algorithm="giac")

[Out]

integrate((b*sin(d*x + c) + a)^2*(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 )}^2 \,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))^2,x)

[Out]

int((e*cos(c + d*x))^p*(a + b*sin(c + d*x))^2, x)

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