3.207 \(\int \sqrt{e \cos (c+d x)} (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=105 \[ -\frac{14 a^2 (e \cos (c+d x))^{3/2}}{15 d e}-\frac{2 \left (a^2 \sin (c+d x)+a^2\right ) (e \cos (c+d x))^{3/2}}{5 d e}+\frac{14 a^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{5 d \sqrt{\cos (c+d x)}} \]

[Out]

(-14*a^2*(e*Cos[c + d*x])^(3/2))/(15*d*e) + (14*a^2*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*d*Sqrt[
Cos[c + d*x]]) - (2*(e*Cos[c + d*x])^(3/2)*(a^2 + a^2*Sin[c + d*x]))/(5*d*e)

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Rubi [A]  time = 0.091795, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2678, 2669, 2640, 2639} \[ -\frac{14 a^2 (e \cos (c+d x))^{3/2}}{15 d e}-\frac{2 \left (a^2 \sin (c+d x)+a^2\right ) (e \cos (c+d x))^{3/2}}{5 d e}+\frac{14 a^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{5 d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-14*a^2*(e*Cos[c + d*x])^(3/2))/(15*d*e) + (14*a^2*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*d*Sqrt[
Cos[c + d*x]]) - (2*(e*Cos[c + d*x])^(3/2)*(a^2 + a^2*Sin[c + d*x]))/(5*d*e)

Rule 2678

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

Rule 2669

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

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin{align*} \int \sqrt{e \cos (c+d x)} (a+a \sin (c+d x))^2 \, dx &=-\frac{2 (e \cos (c+d x))^{3/2} \left (a^2+a^2 \sin (c+d x)\right )}{5 d e}+\frac{1}{5} (7 a) \int \sqrt{e \cos (c+d x)} (a+a \sin (c+d x)) \, dx\\ &=-\frac{14 a^2 (e \cos (c+d x))^{3/2}}{15 d e}-\frac{2 (e \cos (c+d x))^{3/2} \left (a^2+a^2 \sin (c+d x)\right )}{5 d e}+\frac{1}{5} \left (7 a^2\right ) \int \sqrt{e \cos (c+d x)} \, dx\\ &=-\frac{14 a^2 (e \cos (c+d x))^{3/2}}{15 d e}-\frac{2 (e \cos (c+d x))^{3/2} \left (a^2+a^2 \sin (c+d x)\right )}{5 d e}+\frac{\left (7 a^2 \sqrt{e \cos (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 \sqrt{\cos (c+d x)}}\\ &=-\frac{14 a^2 (e \cos (c+d x))^{3/2}}{15 d e}+\frac{14 a^2 \sqrt{e \cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d \sqrt{\cos (c+d x)}}-\frac{2 (e \cos (c+d x))^{3/2} \left (a^2+a^2 \sin (c+d x)\right )}{5 d e}\\ \end{align*}

Mathematica [C]  time = 0.0441518, size = 66, normalized size = 0.63 \[ -\frac{8\ 2^{3/4} a^2 (e \cos (c+d x))^{3/2} \, _2F_1\left (-\frac{7}{4},\frac{3}{4};\frac{7}{4};\frac{1}{2} (1-\sin (c+d x))\right )}{3 d e (\sin (c+d x)+1)^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*2^(3/4)*a^2*(e*Cos[c + d*x])^(3/2)*Hypergeometric2F1[-7/4, 3/4, 7/4, (1 - Sin[c + d*x])/2])/(3*d*e*(1 + Si
n[c + d*x])^(3/4))

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Maple [A]  time = 0.388, size = 188, normalized size = 1.8 \begin{align*}{\frac{2\,{a}^{2}e}{15\,d} \left ( -24\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}\cos \left ( 1/2\,dx+c/2 \right ) +24\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}\cos \left ( 1/2\,dx+c/2 \right ) -40\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}+21\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -6\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) +40\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}-10\,\sin \left ( 1/2\,dx+c/2 \right ) \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}e+e}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*a^2*e*(-24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)
+24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-40*sin(1/2*d*x+1/2*c)^5+21*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2
*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-6*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+40*sin(
1/2*d*x+1/2*c)^3-10*sin(1/2*d*x+1/2*c))/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{e \cos \left (d x + c\right )}{\left (a \sin \left (d x + c\right ) + a\right )}^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(e*cos(d*x + c))*(a*sin(d*x + c) + a)^2, 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{e \cos \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(a^2*cos(d*x + c)^2 - 2*a^2*sin(d*x + c) - 2*a^2)*sqrt(e*cos(d*x + c)), 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((a+a*sin(d*x+c))**2*(e*cos(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{e \cos \left (d x + c\right )}{\left (a \sin \left (d x + c\right ) + a\right )}^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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