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

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

[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))

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Rubi [A]  time = 0.149241, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {2692, 2669, 2643} \[ -\frac{\left (a^2 (p+2)+b^2\right ) \sin (c+d x) (e \cos (c+d x))^{p+1} \, _2F_1\left (\frac{1}{2},\frac{p+1}{2};\frac{p+3}{2};\cos ^2(c+d x)\right )}{d e (p+1) (p+2) \sqrt{\sin ^2(c+d x)}}-\frac{a b (p+3) (e \cos (c+d x))^{p+1}}{d e (p+1) (p+2)}-\frac{b (a+b \sin (c+d x)) (e \cos (c+d x))^{p+1}}{d e (p+2)} \]

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 2692

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
])

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))^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*}

Mathematica [C]  time = 1.02257, size = 285, normalized size = 1.82 \[ -\frac{(e \cos (c+d x))^p \left (-\frac{1}{2} a^2 (p-1) \sin (2 (c+d x)) \, _2F_1\left (\frac{1}{2},\frac{p+1}{2};\frac{p+3}{2};\cos ^2(c+d x)\right )+a b 2^{-p} \left (1+e^{2 i (c+d x)}\right ) \left (e^{-i (c+d x)}+e^{i (c+d x)}\right )^p \sqrt{\sin ^2(c+d x)} \left ((p+1) e^{i (c+d x)} \, _2F_1\left (1,\frac{p+3}{2};\frac{3-p}{2};-e^{2 i (c+d x)}\right )-(p-1) e^{-i (c+d x)} \, _2F_1\left (1,\frac{p+1}{2};\frac{1-p}{2};-e^{2 i (c+d x)}\right )\right ) \cos ^{-p}(c+d x)-\frac{1}{2} b^2 (p-1) \sin (2 (c+d x)) \, _2F_1\left (-\frac{1}{2},\frac{p+1}{2};\frac{p+3}{2};\cos ^2(c+d x)\right )\right )}{\left (d-d p^2\right ) \sqrt{\sin ^2(c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(e*Cos[c + d*x])^p*(a + b*Sin[c + d*x])^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 = 2.566, 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 ) ^{2}\, 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))^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., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (d x + c\right ) + a\right )}^{2} \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))^2,x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c) + a)^2*(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^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}\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))^2,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))**2,x)

[Out]

Timed out

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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 )}^{2} \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))^2,x, algorithm="giac")

[Out]

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