\(\int \frac {(a+b \sin (c+d x))^2}{\sqrt {e \cos (c+d x)}} \, dx\) [551]

Optimal result

Integrand size = 25, antiderivative size = 109 \[ \int \frac {(a+b \sin (c+d x))^2}{\sqrt {e \cos (c+d x)}} \, dx=-\frac {10 a b \sqrt {e \cos (c+d x)}}{3 d e}+\frac {2 \left (3 a^2+2 b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {e \cos (c+d x)}}-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{3 d e} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2771, 2748, 2721, 2720} \[ \int \frac {(a+b \sin (c+d x))^2}{\sqrt {e \cos (c+d x)}} \, dx=\frac {2 \left (3 a^2+2 b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {e \cos (c+d x)}}-\frac {10 a b \sqrt {e \cos (c+d x)}}{3 d e}-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{3 d e} \]

[In]

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

[Out]

(-10*a*b*Sqrt[e*Cos[c + d*x]])/(3*d*e) + (2*(3*a^2 + 2*b^2)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(3*d
*Sqrt[e*Cos[c + d*x]]) - (2*b*Sqrt[e*Cos[c + d*x]]*(a + b*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 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*} \text {integral}& = -\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{3 d e}+\frac {2}{3} \int \frac {\frac {3 a^2}{2}+b^2+\frac {5}{2} a b \sin (c+d x)}{\sqrt {e \cos (c+d x)}} \, dx \\ & = -\frac {10 a b \sqrt {e \cos (c+d x)}}{3 d e}-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{3 d e}+\frac {1}{3} \left (3 a^2+2 b^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx \\ & = -\frac {10 a b \sqrt {e \cos (c+d x)}}{3 d e}-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{3 d e}+\frac {\left (\left (3 a^2+2 b^2\right ) \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 b \sqrt {e \cos (c+d x)}}{3 d e}+\frac {2 \left (3 a^2+2 b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {e \cos (c+d x)}}-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{3 d e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.79 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.69 \[ \int \frac {(a+b \sin (c+d x))^2}{\sqrt {e \cos (c+d x)}} \, dx=\frac {2 \left (3 a^2+2 b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-2 b \cos (c+d x) (6 a+b \sin (c+d x))}{3 d \sqrt {e \cos (c+d x)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.52 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.93

method	result	size
default	\(\frac {\frac {8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}}{3}-\frac {4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}}{3}-2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2}-\frac {4 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{2}}{3}+8 a b \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) a b}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\)	\(210\)
parts	\(\frac {2 a^{2} \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \operatorname {am}^{-1}\left (\frac {d x}{2}+\frac {c}{2}| \sqrt {2}\right )}{d \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}}+\frac {4 b^{2} \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-e \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\, d}-\frac {4 a b \sqrt {e \cos \left (d x +c \right )}}{d e}\)	\(268\)

[In]

int((a+b*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)*(4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*b^2-2*co
s(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*b^2-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Elli
pticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2-2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Elliptic
F(cos(1/2*d*x+1/2*c),2^(1/2))*b^2+12*a*b*sin(1/2*d*x+1/2*c)^3-6*sin(1/2*d*x+1/2*c)*a*b)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.09 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.99 \[ \int \frac {(a+b \sin (c+d x))^2}{\sqrt {e \cos (c+d x)}} \, dx=\frac {\sqrt {2} {\left (-3 i \, a^{2} - 2 i \, b^{2}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (3 i \, a^{2} + 2 i \, b^{2}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 2 \, {\left (b^{2} \sin \left (d x + c\right ) + 6 \, a b\right )} \sqrt {e \cos \left (d x + c\right )}}{3 \, d e} \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+b \sin (c+d x))^2}{\sqrt {e \cos (c+d x)}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {(a+b \sin (c+d x))^2}{\sqrt {e \cos (c+d x)}} \, dx=\int { \frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{2}}{\sqrt {e \cos \left (d x + c\right )}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {(a+b \sin (c+d x))^2}{\sqrt {e \cos (c+d x)}} \, dx=\int { \frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{2}}{\sqrt {e \cos \left (d x + c\right )}} \,d x } \]

[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(e*cos(d*x + c)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \sin (c+d x))^2}{\sqrt {e \cos (c+d x)}} \, dx=\int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^2}{\sqrt {e\,\cos \left (c+d\,x\right )}} \,d x \]

[In]

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

[Out]

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