3.3.8 \(\int \frac {(a+a \sin (c+d x))^2}{\sqrt {e \cos (c+d x)}} \, dx\) [208]

Optimal. Leaf size=105 \[ -\frac {10 a^2 \sqrt {e \cos (c+d x)}}{3 d e}+\frac {10 a^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d \sqrt {e \cos (c+d x)}}-\frac {2 \sqrt {e \cos (c+d x)} \left (a^2+a^2 \sin (c+d x)\right )}{3 d e} \]

[Out]

10/3*a^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2
)/d/(e*cos(d*x+c))^(1/2)-10/3*a^2*(e*cos(d*x+c))^(1/2)/d/e-2/3*(a^2+a^2*sin(d*x+c))*(e*cos(d*x+c))^(1/2)/d/e

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Rubi [A]
time = 0.07, 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 = {2757, 2748, 2721, 2720} \begin {gather*} -\frac {10 a^2 \sqrt {e \cos (c+d x)}}{3 d e}-\frac {2 \left (a^2 \sin (c+d x)+a^2\right ) \sqrt {e \cos (c+d x)}}{3 d e}+\frac {10 a^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d \sqrt {e \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-10*a^2*Sqrt[e*Cos[c + d*x]])/(3*d*e) + (10*a^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(3*d*Sqrt[e*Cos
[c + d*x]]) - (2*Sqrt[e*Cos[c + d*x]]*(a^2 + a^2*Sin[c + d*x]))/(3*d*e)

Rule 2720

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

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rule 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[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 2757

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 \frac {(a+a \sin (c+d x))^2}{\sqrt {e \cos (c+d x)}} \, dx &=-\frac {2 \sqrt {e \cos (c+d x)} \left (a^2+a^2 \sin (c+d x)\right )}{3 d e}+\frac {1}{3} (5 a) \int \frac {a+a \sin (c+d x)}{\sqrt {e \cos (c+d x)}} \, dx\\ &=-\frac {10 a^2 \sqrt {e \cos (c+d x)}}{3 d e}-\frac {2 \sqrt {e \cos (c+d x)} \left (a^2+a^2 \sin (c+d x)\right )}{3 d e}+\frac {1}{3} \left (5 a^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx\\ &=-\frac {10 a^2 \sqrt {e \cos (c+d x)}}{3 d e}-\frac {2 \sqrt {e \cos (c+d x)} \left (a^2+a^2 \sin (c+d x)\right )}{3 d e}+\frac {\left (5 a^2 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{3 \sqrt {e \cos (c+d x)}}\\ &=-\frac {10 a^2 \sqrt {e \cos (c+d x)}}{3 d e}+\frac {10 a^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d \sqrt {e \cos (c+d x)}}-\frac {2 \sqrt {e \cos (c+d x)} \left (a^2+a^2 \sin (c+d x)\right )}{3 d e}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.02, size = 64, normalized size = 0.61 \begin {gather*} -\frac {8 \sqrt [4]{2} a^2 \sqrt {e \cos (c+d x)} \, _2F_1\left (-\frac {5}{4},\frac {1}{4};\frac {5}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{d e \sqrt [4]{1+\sin (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*2^(1/4)*a^2*Sqrt[e*Cos[c + d*x]]*Hypergeometric2F1[-5/4, 1/4, 5/4, (1 - Sin[c + d*x])/2])/(d*e*(1 + Sin[c
+ d*x])^(1/4))

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Maple [A]
time = 1.41, size = 152, normalized size = 1.45

method	result	size
default	\(-\frac {2 a^{2} \left (-4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-12 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\)	\(152\)

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,method=_RETURNVERBOSE)

[Out]

-2/3/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*a^2*(-4*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+5*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+2*sin(1/2*
d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-12*sin(1/2*d*x+1/2*c)^3+6*sin(1/2*d*x+1/2*c))/d

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 86, normalized size = 0.82 \begin {gather*} \frac {{\left (-5 i \, \sqrt {2} a^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} a^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 2 \, {\left (a^{2} \sin \left (d x + c\right ) + 6 \, a^{2}\right )} \sqrt {\cos \left (d x + c\right )}\right )} e^{\left (-\frac {1}{2}\right )}}{3 \, d} \end {gather*}

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]

1/3*(-5*I*sqrt(2)*a^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*I*sqrt(2)*a^2*weierstrassP
Inverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 2*(a^2*sin(d*x + c) + 6*a^2)*sqrt(cos(d*x + c)))*e^(-1/2)/d

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{2} \left (\int \frac {1}{\sqrt {e \cos {\left (c + d x \right )}}}\, dx + \int \frac {2 \sin {\left (c + d x \right )}}{\sqrt {e \cos {\left (c + d x \right )}}}\, dx + \int \frac {\sin ^{2}{\left (c + d x \right )}}{\sqrt {e \cos {\left (c + d x \right )}}}\, dx\right ) \end {gather*}

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]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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