3.6.50 \(\int \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^2 \, dx\) [550]

Optimal. Leaf size=109 \[ -\frac {14 a b (e \cos (c+d x))^{3/2}}{15 d e}+\frac {2 \left (5 a^2+2 b^2\right ) \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)}}-\frac {2 b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{5 d e} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.08, antiderivative size = 109, 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 = {2771, 2748, 2721, 2719} \begin {gather*} \frac {2 \left (5 a^2+2 b^2\right ) 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)}}-\frac {14 a b (e \cos (c+d x))^{3/2}}{15 d e}-\frac {2 b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{5 d e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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 2771

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[1/(m + p), Int[(g*Cos[e + f*x]
)^p*(a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + a^2*(m + p) + a*b*(2*m + p - 1)*Sin[e + f*x]), x], x] /; FreeQ
[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && NeQ[m + p, 0] && (IntegersQ[2*m, 2*p] || IntegerQ
[m])

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.34, size = 80, normalized size = 0.73 \begin {gather*} \frac {\sqrt {e \cos (c+d x)} \left (6 \left (5 a^2+2 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )-2 b \cos ^{\frac {3}{2}}(c+d x) (10 a+3 b \sin (c+d x))\right )}{15 d \sqrt {\cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[e*Cos[c + d*x]]*(6*(5*a^2 + 2*b^2)*EllipticE[(c + d*x)/2, 2] - 2*b*Cos[c + d*x]^(3/2)*(10*a + 3*b*Sin[c
+ d*x])))/(15*d*Sqrt[Cos[c + d*x]])

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(250\) vs. \(2(121)=242\).
time = 4.52, size = 251, normalized size = 2.30

method	result	size
default	\(\frac {2 e \left (-24 b^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-40 a b \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+24 b^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2}+6 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{2}+40 a b \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 b^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-10 a b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\)	\(251\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.12, size = 126, normalized size = 1.16 \begin {gather*} \frac {3 i \, \sqrt {2} {\left (5 \, a^{2} + 2 \, b^{2}\right )} e^{\frac {1}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} {\left (5 \, a^{2} + 2 \, b^{2}\right )} e^{\frac {1}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (3 \, b^{2} \cos \left (d x + c\right ) e^{\frac {1}{2}} \sin \left (d x + c\right ) + 10 \, a b \cos \left (d x + c\right ) e^{\frac {1}{2}}\right )} \sqrt {\cos \left (d x + c\right )}}{15 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*I*sqrt(2)*(5*a^2 + 2*b^2)*e^(1/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*s
in(d*x + c))) - 3*I*sqrt(2)*(5*a^2 + 2*b^2)*e^(1/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x
+ c) - I*sin(d*x + c))) - 2*(3*b^2*cos(d*x + c)*e^(1/2)*sin(d*x + c) + 10*a*b*cos(d*x + c)*e^(1/2))*sqrt(cos(d
*x + c)))/d

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________