3.575 \(\int \sqrt {a+b \cos (c+d x) \sin (c+d x)} \, dx\)

Optimal. Leaf size=76 \[ \frac {\sqrt {2 a+b \sin (2 c+2 d x)} E\left (c+d x-\frac {\pi }{4}|\frac {2 b}{2 a+b}\right )}{\sqrt {2} d \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}} \]

[Out]

-1/2*(sin(c+1/4*Pi+d*x)^2)^(1/2)/sin(c+1/4*Pi+d*x)*EllipticE(cos(c+1/4*Pi+d*x),2^(1/2)*(b/(2*a+b))^(1/2))*(2*a
+b*sin(2*d*x+2*c))^(1/2)/d*2^(1/2)/((2*a+b*sin(2*d*x+2*c))/(2*a+b))^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2666, 2655, 2653} \[ \frac {\sqrt {2 a+b \sin (2 c+2 d x)} E\left (c+d x-\frac {\pi }{4}|\frac {2 b}{2 a+b}\right )}{\sqrt {2} d \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Cos[c + d*x]*Sin[c + d*x]],x]

[Out]

(EllipticE[c - Pi/4 + d*x, (2*b)/(2*a + b)]*Sqrt[2*a + b*Sin[2*c + 2*d*x]])/(Sqrt[2]*d*Sqrt[(2*a + b*Sin[2*c +
2*d*x])/(2*a + b)])

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2666

Int[((a_) + cos[(c_.) + (d_.)*(x_)]*(b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Int[(a + (b*Sin[2*c + 2*
d*x])/2)^n, x] /; FreeQ[{a, b, c, d, n}, x]

Rubi steps

\begin {align*} \int \sqrt {a+b \cos (c+d x) \sin (c+d x)} \, dx &=\int \sqrt {a+\frac {1}{2} b \sin (2 c+2 d x)} \, dx\\ &=\frac {\sqrt {a+\frac {1}{2} b \sin (2 c+2 d x)} \int \sqrt {\frac {a}{a+\frac {b}{2}}+\frac {b \sin (2 c+2 d x)}{2 \left (a+\frac {b}{2}\right )}} \, dx}{\sqrt {\frac {a+\frac {1}{2} b \sin (2 c+2 d x)}{a+\frac {b}{2}}}}\\ &=\frac {E\left (c-\frac {\pi }{4}+d x|\frac {2 b}{2 a+b}\right ) \sqrt {2 a+b \sin (2 c+2 d x)}}{\sqrt {2} d \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 75, normalized size = 0.99 \[ \frac {(2 a+b) \sqrt {\frac {2 a+b \sin (2 (c+d x))}{2 a+b}} E\left (c+d x-\frac {\pi }{4}|\frac {2 b}{2 a+b}\right )}{d \sqrt {4 a+2 b \sin (2 (c+d x))}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*Cos[c + d*x]*Sin[c + d*x]],x]

[Out]

((2*a + b)*EllipticE[c - Pi/4 + d*x, (2*b)/(2*a + b)]*Sqrt[(2*a + b*Sin[2*(c + d*x)])/(2*a + b)])/(d*Sqrt[4*a
+ 2*b*Sin[2*(c + d*x)]])

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fricas [F]  time = 2.01, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*cos(d*x + c)*sin(d*x + c) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*cos(d*x + c)*sin(d*x + c) + a), x)

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maple [B]  time = 0.41, size = 312, normalized size = 4.11 \[ -\frac {\sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}\, \sqrt {-\frac {\left (\sin \left (2 d x +2 c \right )-1\right ) b}{2 a +b}}\, \sqrt {-\frac {\left (1+\sin \left (2 d x +2 c \right )\right ) b}{2 a -b}}\, \left (2 a -b \right ) \left (2 \EllipticE \left (\sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}, \sqrt {\frac {2 a -b}{2 a +b}}\right ) a +\EllipticE \left (\sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}, \sqrt {\frac {2 a -b}{2 a +b}}\right ) b -2 a \EllipticF \left (\sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}, \sqrt {\frac {2 a -b}{2 a +b}}\right )-\EllipticF \left (\sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}, \sqrt {\frac {2 a -b}{2 a +b}}\right ) b \right )}{b \cos \left (2 d x +2 c \right ) \sqrt {4 a +2 b \sin \left (2 d x +2 c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2)*(-(sin(2*d*x+2*c)-1)*b/(2*a+b))^(1/2)*(-(1+sin(2*d*x+2*c))*b/(2*a-b))^
(1/2)/b*(2*a-b)*(2*EllipticE(((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))*a+EllipticE(((2*a
+b*sin(2*d*x+2*c))/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))*b-2*a*EllipticF(((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1
/2),((2*a-b)/(2*a+b))^(1/2))-EllipticF(((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))*b)/cos(
2*d*x+2*c)/(4*a+2*b*sin(2*d*x+2*c))^(1/2)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*cos(d*x + c)*sin(d*x + c) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + b \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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