3.4.38 \(\int (e \cos (c+d x))^p \sqrt {a+a \sin (c+d x)} \, dx\) [338]

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

[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 = 0.07, 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 = {2768, 72, 71} \begin {gather*} -\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}} \end {gather*}

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[-1/2*p, (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 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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 72

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 2768

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 ) \text {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 ) \text {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] Result contains complex when optimal does not.
time = 3.94, size = 310, normalized size = 3.20 \begin {gather*} \frac {(1+i) 2^{-p} e^{-\frac {1}{2} i d x} \cos ^{-p}(c+d x) (e \cos (c+d x))^p \left (e^{i d x} (1+2 p) \, _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 ) \left (\cos \left (\frac {c}{2}\right )+i \sin \left (\frac {c}{2}\right )\right )+(-1+2 p) \, _2F_1\left (\frac {1}{4} (-1-2 p),-p;\frac {1}{4} (3-2 p);-e^{2 i d x} (\cos (c)+i \sin (c))^2\right ) \left (i \cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right )\right ) \left (e^{-i d x} \left (\left (1+e^{2 i d x}\right ) \cos (c)+i \left (-1+e^{2 i d x}\right ) \sin (c)\right )\right )^p \left (1+e^{2 i d x} \cos (2 c)+i e^{2 i d x} \sin (2 c)\right )^{-p} \sqrt {a (1+\sin (c+d x))}}{d (-1+2 p) (1+2 p) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \end {gather*}

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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Maple [F]
time = 0.15, 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 \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+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

[Out]

integral(sqrt(a*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 \sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )} \left (e \cos {\left (c + d x \right )}\right )^{p}\, dx \end {gather*}

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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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(co

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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\,\sqrt {a+a\,\sin \left (c+d\,x\right )} \,d x \end {gather*}

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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