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

Optimal. Leaf size=101 \[ -\frac{a 2^{p/2} (\sin (c+d x)+1)^{1-\frac{p}{2}} (e \cos (c+d x))^{p+1} \, _2F_1\left (\frac{2-p}{2},\frac{p+1}{2};\frac{p+3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{d e (p+1) (a \sin (c+d x)+a)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.108896, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2689, 70, 69} \[ -\frac{a 2^{p/2} (\sin (c+d x)+1)^{1-\frac{p}{2}} (e \cos (c+d x))^{p+1} \, _2F_1\left (\frac{2-p}{2},\frac{p+1}{2};\frac{p+3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{d e (p+1) (a \sin (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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]))

Rubi steps

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

Mathematica [A]  time = 0.110652, size = 97, normalized size = 0.96 \[ -\frac{2^{p/2} \cos (c+d x) (\sin (c+d x)+1)^{-p/2} (e \cos (c+d x))^p \, _2F_1\left (1-\frac{p}{2},\frac{p+1}{2};\frac{p+3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{d (p+1) \sqrt{a (\sin (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.091, size = 0, normalized size = 0. \begin{align*} \int{ \left ( e\cos \left ( dx+c \right ) \right ) ^{p}{\frac{1}{\sqrt{a+a\sin \left ( dx+c \right ) }}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \cos \left (d x + c\right )\right )^{p}}{\sqrt{a \sin \left (d x + c\right ) + a}}\,{d x} \end{align*}

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\left (e \cos \left (d x + c\right )\right )^{p}}{\sqrt{a \sin \left (d x + c\right ) + a}}, x\right ) \end{align*}

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \cos{\left (c + d x \right )}\right )^{p}}{\sqrt{a \left (\sin{\left (c + d x \right )} + 1\right )}}\, dx \end{align*}

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \cos \left (d x + c\right )\right )^{p}}{\sqrt{a \sin \left (d x + c\right ) + a}}\,{d x} \end{align*}

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]

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