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]

-14/15*a^2*(e*cos(d*x+c))^(3/2)/d/e-2/5*(e*cos(d*x+c))^(3/2)*(a^2+a^2*sin(d*x+c))/d/e+14/5*a^2*(cos(1/2*d*x+1/
2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*(e*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, 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 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]

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 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 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]

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*}

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Mathematica [C]  time = 0.05, 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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fricas [F]  time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \sqrt {e \cos \left (d x + c\right )}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {e \cos \left (d x + c\right )} {\left (a \sin \left (d x + c\right ) + a\right )}^{2}\,{d x} \]

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)

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maple [A]  time = 0.79, size = 188, normalized size = 1.79 \[ \frac {2 a^{2} e \left (-24 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+24 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-40 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+21 \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}-6 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+40 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-10 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d} \]

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*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(
sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {e \cos \left (d x + c\right )} {\left (a \sin \left (d x + c\right ) + a\right )}^{2}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {e\,\cos \left (c+d\,x\right )}\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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