3.336 \(\int (e \cos (c+d x))^p (a+a \sin (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=103 \[ -\frac{a^3 2^{\frac{p}{2}+3} (\sin (c+d x)+1)^{-p/2} (e \cos (c+d x))^{p+1} \, _2F_1\left (\frac{1}{2} (-p-4),\frac{p+1}{2};\frac{p+3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{d e (p+1) \sqrt{a \sin (c+d x)+a}} \]

[Out]

-((2^(3 + p/2)*a^3*(e*Cos[c + d*x])^(1 + p)*Hypergeometric2F1[(-4 - 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]]))

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Rubi [A]  time = 0.116803, antiderivative size = 103, 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} \[ -\frac{a^3 2^{\frac{p}{2}+3} (\sin (c+d x)+1)^{-p/2} (e \cos (c+d x))^{p+1} \, _2F_1\left (\frac{1}{2} (-p-4),\frac{p+1}{2};\frac{p+3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{d e (p+1) \sqrt{a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^p*(a + a*Sin[c + d*x])^(5/2),x]

[Out]

-((2^(3 + p/2)*a^3*(e*Cos[c + d*x])^(1 + p)*Hypergeometric2F1[(-4 - 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 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 (e \cos (c+d x))^p (a+a \sin (c+d x))^{5/2} \, 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{5}{2}+\frac{1}{2} (-1+p)} \, dx,x,\sin (c+d x)\right )}{d e}\\ &=\frac{\left (2^{2+\frac{p}{2}} a^4 (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{5}{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^{3+\frac{p}{2}} a^3 (e \cos (c+d x))^{1+p} \, _2F_1\left (\frac{1}{2} (-4-p),\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*}

Mathematica [A]  time = 0.185934, size = 102, normalized size = 0.99 \[ -\frac{a^3 2^{\frac{p}{2}+3} \cos (c+d x) (\sin (c+d x)+1)^{-p/2} (e \cos (c+d x))^p \, _2F_1\left (-\frac{p}{2}-2,\frac{p+1}{2};\frac{p+3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{d (p+1) \sqrt{a (\sin (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^p*(a + a*Sin[c + d*x])^(5/2),x]

[Out]

-((2^(3 + p/2)*a^3*Cos[c + d*x]*(e*Cos[c + d*x])^p*Hypergeometric2F1[-2 - 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 = 0.098, size = 0, normalized size = 0. \begin{align*} \int \left ( e\cos \left ( dx+c \right ) \right ) ^{p} \left ( a+a\sin \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}\, 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))^(5/2),x)

[Out]

int((e*cos(d*x+c))^p*(a+a*sin(d*x+c))^(5/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{5}{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+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^(5/2)*(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 (a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \sqrt{a \sin \left (d x + c\right ) + a} \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+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral(-(a^2*cos(d*x + c)^2 - 2*a^2*sin(d*x + c) - 2*a^2)*sqrt(a*sin(d*x + c) + a)*(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+a*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{5}{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+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)^(5/2)*(e*cos(d*x + c))^p, x)