3.566 \(\int \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)} \, dx\)
Optimal. Leaf size=105 \[ -\frac{a \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{f \sqrt{a \sin (e+f x)+a}}-\frac{\sqrt{a} (c+d) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{\sqrt{d} f} \]
[Out]
-((Sqrt[a]*(c + d)*ArcTan[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])])
/(Sqrt[d]*f)) - (a*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(f*Sqrt[a + a*Sin[e + f*x]])
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Rubi [A] time = 0.182373, antiderivative size = 105, normalized size of antiderivative = 1.,
number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used =
{2770, 2775, 205} \[ -\frac{a \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{f \sqrt{a \sin (e+f x)+a}}-\frac{\sqrt{a} (c+d) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{\sqrt{d} f} \]
Antiderivative was successfully verified.
[In]
Int[Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]],x]
[Out]
-((Sqrt[a]*(c + d)*ArcTan[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])])
/(Sqrt[d]*f)) - (a*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(f*Sqrt[a + a*Sin[e + f*x]])
Rule 2770
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(-2*b*Cos[e + f*x]*(c + d*Sin[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(2*n*(b*c + a*d)
)/(b*(2*n + 1)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f}
, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && IntegerQ[2*n]
Rule 2775
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*b)/f, Subst[Int[1/(b + d*x^2), x], x, (b*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])
], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Rule 205
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rubi steps
\begin{align*} \int \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)} \, dx &=-\frac{a \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{f \sqrt{a+a \sin (e+f x)}}+\frac{1}{2} (c+d) \int \frac{\sqrt{a+a \sin (e+f x)}}{\sqrt{c+d \sin (e+f x)}} \, dx\\ &=-\frac{a \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{f \sqrt{a+a \sin (e+f x)}}-\frac{(a (c+d)) \operatorname{Subst}\left (\int \frac{1}{a+d x^2} \, dx,x,\frac{a \cos (e+f x)}{\sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}\right )}{f}\\ &=-\frac{\sqrt{a} (c+d) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}\right )}{\sqrt{d} f}-\frac{a \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{f \sqrt{a+a \sin (e+f x)}}\\ \end{align*}
Mathematica [C] time = 1.29864, size = 350, normalized size = 3.33 \[ \frac{\sqrt{a (\sin (e+f x)+1)} \left (-\frac{2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) (c+d \sin (e+f x))}{f}-\frac{i (c+d) \left (\cos \left (\frac{1}{2} (e+f x)\right )-i \sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\log \left (\frac{e^{-i e} \left (2 \sqrt{d} \sqrt{2 c e^{i (e+f x)}-i d \left (-1+e^{2 i (e+f x)}\right )}+2 \sqrt [4]{-1} c-2 (-1)^{3/4} d e^{i (e+f x)}\right )}{\sqrt{d}}\right )-\log \left (\frac{2 f e^{\frac{1}{2} i (e-2 f x)} \left (i \sqrt{d} \sqrt{2 c e^{i (e+f x)}-i d \left (-1+e^{2 i (e+f x)}\right )}+\sqrt [4]{-1} c e^{i (e+f x)}+(-1)^{3/4} d\right )}{\sqrt{d}}\right )\right ) \sqrt{(\cos (e+f x)+i \sin (e+f x)) (c+d \sin (e+f x))}}{\sqrt{d} f}\right )}{2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \sqrt{c+d \sin (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]],x]
[Out]
(Sqrt[a*(1 + Sin[e + f*x])]*((-2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(c + d*Sin[e + f*x]))/f - (I*(c + d)*(L
og[(2*(-1)^(1/4)*c - 2*(-1)^(3/4)*d*E^(I*(e + f*x)) + 2*Sqrt[d]*Sqrt[2*c*E^(I*(e + f*x)) - I*d*(-1 + E^((2*I)*
(e + f*x)))])/(Sqrt[d]*E^(I*e))] - Log[(2*E^((I/2)*(e - 2*f*x))*((-1)^(3/4)*d + (-1)^(1/4)*c*E^(I*(e + f*x)) +
I*Sqrt[d]*Sqrt[2*c*E^(I*(e + f*x)) - I*d*(-1 + E^((2*I)*(e + f*x)))])*f)/Sqrt[d]])*(Cos[(e + f*x)/2] - I*Sin[
(e + f*x)/2])*Sqrt[(Cos[e + f*x] + I*Sin[e + f*x])*(c + d*Sin[e + f*x])])/(Sqrt[d]*f)))/(2*(Cos[(e + f*x)/2] +
Sin[(e + f*x)/2])*Sqrt[c + d*Sin[e + f*x]])
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Maple [F] time = 180., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a+a\sin \left ( fx+e \right ) }\sqrt{c+d\sin \left ( fx+e \right ) }\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sin(f*x+e))^(1/2)*(c+d*sin(f*x+e))^(1/2),x)
[Out]
int((a+a*sin(f*x+e))^(1/2)*(c+d*sin(f*x+e))^(1/2),x)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{d \sin \left (f x + e\right ) + c}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^(1/2)*(c+d*sin(f*x+e))^(1/2),x, algorithm="maxima")
[Out]
integrate(sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c), x)
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Fricas [B] time = 5.20041, size = 2311, normalized size = 22.01 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^(1/2)*(c+d*sin(f*x+e))^(1/2),x, algorithm="fricas")
[Out]
[1/8*(((c + d)*cos(f*x + e) + (c + d)*sin(f*x + e) + c + d)*sqrt(-a/d)*log((128*a*d^4*cos(f*x + e)^5 + a*c^4 +
4*a*c^3*d + 6*a*c^2*d^2 + 4*a*c*d^3 + a*d^4 + 128*(2*a*c*d^3 - a*d^4)*cos(f*x + e)^4 - 32*(5*a*c^2*d^2 - 14*a
*c*d^3 + 13*a*d^4)*cos(f*x + e)^3 - 32*(a*c^3*d - 2*a*c^2*d^2 + 9*a*c*d^3 - 4*a*d^4)*cos(f*x + e)^2 - 8*(16*d^
4*cos(f*x + e)^4 - c^3*d + 17*c^2*d^2 - 59*c*d^3 + 51*d^4 + 24*(c*d^3 - d^4)*cos(f*x + e)^3 - 2*(5*c^2*d^2 - 2
6*c*d^3 + 33*d^4)*cos(f*x + e)^2 - (c^3*d - 7*c^2*d^2 + 31*c*d^3 - 25*d^4)*cos(f*x + e) + (16*d^4*cos(f*x + e)
^3 + c^3*d - 17*c^2*d^2 + 59*c*d^3 - 51*d^4 - 8*(3*c*d^3 - 5*d^4)*cos(f*x + e)^2 - 2*(5*c^2*d^2 - 14*c*d^3 + 1
3*d^4)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)*sqrt(-a/d) + (a*c^4 - 28*
a*c^3*d + 230*a*c^2*d^2 - 476*a*c*d^3 + 289*a*d^4)*cos(f*x + e) + (128*a*d^4*cos(f*x + e)^4 + a*c^4 + 4*a*c^3*
d + 6*a*c^2*d^2 + 4*a*c*d^3 + a*d^4 - 256*(a*c*d^3 - a*d^4)*cos(f*x + e)^3 - 32*(5*a*c^2*d^2 - 6*a*c*d^3 + 5*a
*d^4)*cos(f*x + e)^2 + 32*(a*c^3*d - 7*a*c^2*d^2 + 15*a*c*d^3 - 9*a*d^4)*cos(f*x + e))*sin(f*x + e))/(cos(f*x
+ e) + sin(f*x + e) + 1)) - 8*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)*(cos(f*x + e) - sin(f*x + e) +
1))/(f*cos(f*x + e) + f*sin(f*x + e) + f), 1/4*(((c + d)*cos(f*x + e) + (c + d)*sin(f*x + e) + c + d)*sqrt(a/
d)*arctan(1/4*(8*d^2*cos(f*x + e)^2 - c^2 + 6*c*d - 9*d^2 - 8*(c*d - d^2)*sin(f*x + e))*sqrt(a*sin(f*x + e) +
a)*sqrt(d*sin(f*x + e) + c)*sqrt(a/d)/(2*a*d^2*cos(f*x + e)^3 - (3*a*c*d - a*d^2)*cos(f*x + e)*sin(f*x + e) -
(a*c^2 - a*c*d + 2*a*d^2)*cos(f*x + e))) - 4*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)*(cos(f*x + e) -
sin(f*x + e) + 1))/(f*cos(f*x + e) + f*sin(f*x + e) + f)]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \left (\sin{\left (e + f x \right )} + 1\right )} \sqrt{c + d \sin{\left (e + f x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))**(1/2)*(c+d*sin(f*x+e))**(1/2),x)
[Out]
Integral(sqrt(a*(sin(e + f*x) + 1))*sqrt(c + d*sin(e + f*x)), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{d \sin \left (f x + e\right ) + c}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^(1/2)*(c+d*sin(f*x+e))^(1/2),x, algorithm="giac")
[Out]
integrate(sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c), x)