3.338 \(\int (e \cos (c+d x))^p \sqrt {a+a \sin (c+d x)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2689, 70, 69} \[ -\frac {a 2^{\frac {p}{2}+1} (\sin (c+d x)+1)^{-p/2} (e \cos (c+d x))^{p+1} \, _2F_1\left (-\frac {p}{2},\frac {p+1}{2};\frac {p+3}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{d e (p+1) \sqrt {a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 70

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

Rule 2689

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

Rubi steps

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

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Mathematica [C]  time = 3.89, size = 310, normalized size = 3.20 \[ \frac {(1+i) 2^{-p} e^{-\frac {1}{2} i d x} \sqrt {a (\sin (c+d x)+1)} \cos ^{-p}(c+d x) (e \cos (c+d x))^p \left (e^{-i d x} \left (i \sin (c) \left (-1+e^{2 i d x}\right )+\cos (c) \left (1+e^{2 i d x}\right )\right )\right )^p \left (i \sin (2 c) e^{2 i d x}+\cos (2 c) e^{2 i d x}+1\right )^{-p} \left ((2 p+1) e^{i d x} \left (\cos \left (\frac {c}{2}\right )+i \sin \left (\frac {c}{2}\right )\right ) \, _2F_1\left (\frac {1}{4} (1-2 p),-p;\frac {1}{4} (5-2 p);-e^{2 i d x} (\cos (c)+i \sin (c))^2\right )+(2 p-1) \left (\sin \left (\frac {c}{2}\right )+i \cos \left (\frac {c}{2}\right )\right ) \, _2F_1\left (\frac {1}{4} (-2 p-1),-p;\frac {1}{4} (3-2 p);-e^{2 i d x} (\cos (c)+i \sin (c))^2\right )\right )}{d (2 p-1) (2 p+1) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + I)*(e*Cos[c + d*x])^p*(E^(I*d*x)*(1 + 2*p)*Hypergeometric2F1[(1 - 2*p)/4, -p, (5 - 2*p)/4, -(E^((2*I)*d*
x)*(Cos[c] + I*Sin[c])^2)]*(Cos[c/2] + I*Sin[c/2]) + (-1 + 2*p)*Hypergeometric2F1[(-1 - 2*p)/4, -p, (3 - 2*p)/
4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*(I*Cos[c/2] + Sin[c/2]))*(((1 + E^((2*I)*d*x))*Cos[c] + I*(-1 + E^(
(2*I)*d*x))*Sin[c])/E^(I*d*x))^p*Sqrt[a*(1 + Sin[c + d*x])])/(2^p*d*E^((I/2)*d*x)*(-1 + 2*p)*(1 + 2*p)*Cos[c +
d*x]^p*(1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c])^p*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {a \sin \left (d x + c\right ) + a} \left (e \cos \left (d x + c\right )\right )^{p}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^p*(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^p*(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(cos((d*x+c)/2-pi/4))]Simplification assuming c near 0Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to c
heck sign: (2*pi/x/2)>(-2*pi/x/2)Unable to divide, perhaps due to rounding error%%%{64,[0,2,0,2,2,1,1,1]%%%} /
%%%{128*i,[0,2,0,2,2,0,0,0]%%%} Error: Bad Argument Value

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maple [F]  time = 0.20, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x +c \right )\right )^{p} \sqrt {a +a \sin \left (d x +c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^p*(a+a*sin(d*x+c))^(1/2),x)

[Out]

int((e*cos(d*x+c))^p*(a+a*sin(d*x+c))^(1/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^p*(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(a*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 \[ \int {\left (e\,\cos \left (c+d\,x\right )\right )}^p\,\sqrt {a+a\,\sin \left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(c + d*x))^p*(a + a*sin(c + d*x))^(1/2),x)

[Out]

int((e*cos(c + d*x))^p*(a + a*sin(c + d*x))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )} \left (e \cos {\left (c + d x \right )}\right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))**p*(a+a*sin(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(a*(sin(c + d*x) + 1))*(e*cos(c + d*x))**p, x)

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