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

Optimal. Leaf size=95 \[ -\frac {a^2 2^{\frac {p}{2}+\frac {5}{2}} (\sin (c+d x)+1)^{\frac {1}{2} (-p-1)} (e \cos (c+d x))^{p+1} \, _2F_1\left (\frac {1}{2} (-p-3),\frac {p+1}{2};\frac {p+3}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{d e (p+1)} \]

[Out]

-2^(5/2+1/2*p)*a^2*(e*cos(d*x+c))^(1+p)*hypergeom([1/2+1/2*p, -3/2-1/2*p],[3/2+1/2*p],1/2-1/2*sin(d*x+c))*(1+s
in(d*x+c))^(-1/2-1/2*p)/d/e/(1+p)

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Rubi [A]  time = 0.08, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2688, 69} \[ -\frac {a^2 2^{\frac {p}{2}+\frac {5}{2}} (\sin (c+d x)+1)^{\frac {1}{2} (-p-1)} (e \cos (c+d x))^{p+1} \, _2F_1\left (\frac {1}{2} (-p-3),\frac {p+1}{2};\frac {p+3}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{d e (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2^(5/2 + p/2)*a^2*(e*Cos[c + d*x])^(1 + p)*Hypergeometric2F1[(-3 - p)/2, (1 + p)/2, (3 + p)/2, (1 - Sin[c +
d*x])/2]*(1 + Sin[c + d*x])^((-1 - p)/2))/(d*e*(1 + p)))

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rule 2688

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

Rubi steps

\begin {align*} \int (e \cos (c+d x))^p (a+a \sin (c+d x))^2 \, dx &=\frac {\left (a^2 (e \cos (c+d x))^{1+p} (1-\sin (c+d x))^{\frac {1}{2} (-1-p)} (1+\sin (c+d x))^{\frac {1}{2} (-1-p)}\right ) \operatorname {Subst}\left (\int (1-x)^{\frac {1}{2} (-1+p)} (1+x)^{2+\frac {1}{2} (-1+p)} \, dx,x,\sin (c+d x)\right )}{d e}\\ &=-\frac {2^{\frac {5}{2}+\frac {p}{2}} a^2 (e \cos (c+d x))^{1+p} \, _2F_1\left (\frac {1}{2} (-3-p),\frac {1+p}{2};\frac {3+p}{2};\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{\frac {1}{2} (-1-p)}}{d e (1+p)}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 94, normalized size = 0.99 \[ -\frac {a^2 2^{\frac {p+5}{2}} \cos (c+d x) (\sin (c+d x)+1)^{\frac {1}{2} (-p-1)} (e \cos (c+d x))^p \, _2F_1\left (\frac {1}{2} (-p-3),\frac {p+1}{2};\frac {p+3}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{d (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2^((5 + p)/2)*a^2*Cos[c + d*x]*(e*Cos[c + d*x])^p*Hypergeometric2F1[(-3 - p)/2, (1 + p)/2, (3 + p)/2, (1 -
Sin[c + d*x])/2]*(1 + Sin[c + d*x])^((-1 - p)/2))/(d*(1 + p)))

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \left (e \cos \left (d x + c\right )\right )^{p}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 4.30, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x +c \right )\right )^{p} \left (a +a \sin \left (d x +c \right )\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (e\,\cos \left (c+d\,x\right )\right )}^p\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral((e*cos(c + d*x))**p, x) + Integral(2*(e*cos(c + d*x))**p*sin(c + d*x), x) + Integral((e*cos(c +
d*x))**p*sin(c + d*x)**2, x))

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