\(\int \sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)} \, dx\) [566]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [F(-1)]
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 29, antiderivative size = 104 \[ \int \sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)} \, dx=-\frac {\sqrt {3} (c+d) \arctan \left (\frac {\sqrt {3} \sqrt {d} \cos (e+f x)}{\sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{\sqrt {d} f}-\frac {3 \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {3+3 \sin (e+f x)}} \]

[Out]

-(c+d)*arctan(cos(f*x+e)*a^(1/2)*d^(1/2)/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(1/2))*a^(1/2)/f/d^(1/2)-a*co
s(f*x+e)*(c+d*sin(f*x+e))^(1/2)/f/(a+a*sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.01, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2849, 2854, 211} \[ \int \sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)} \, dx=-\frac {\sqrt {a} (c+d) \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{\sqrt {d} f}-\frac {a \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}} \]

[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 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2849

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 2854

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]

Rubi steps \begin{align*} \text {integral}& = -\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)) \text {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) \arctan \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] (verified)

Result contains complex when optimal does not.

Time = 0.93 (sec) , antiderivative size = 345, normalized size of antiderivative = 3.32 \[ \int \sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)} \, dx=\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) \sqrt {3} \sqrt {1+\sin (e+f x)} \left (\frac {i (c+d) e^{\frac {1}{2} i (e+f x)} \sqrt {2 c-i d e^{-i (e+f x)} \left (-1+e^{2 i (e+f x)}\right )} \left ((-1)^{3/4} \sqrt {2} \arctan \left (\frac {\sqrt [4]{-1} \left (d-i c e^{i (e+f x)}\right )}{\sqrt {d} \sqrt {-2 c e^{i (e+f x)}+i d \left (-1+e^{2 i (e+f x)}\right )}}\right )-(1+i) \text {arctanh}\left (\frac {(-1)^{3/4} \left (c-i d e^{i (e+f x)}\right )}{\sqrt {d} \sqrt {-2 c e^{i (e+f x)}+i d \left (-1+e^{2 i (e+f x)}\right )}}\right )\right )}{\sqrt {d} \sqrt {-2 c e^{i (e+f x)}+i d \left (-1+e^{2 i (e+f x)}\right )} f}-\frac {(2-2 i) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c+d \sin (e+f x)}}{f}\right )}{\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )} \]

[In]

Integrate[Sqrt[3 + 3*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]],x]

[Out]

((1/4 + I/4)*Sqrt[3]*Sqrt[1 + Sin[e + f*x]]*((I*(c + d)*E^((I/2)*(e + f*x))*Sqrt[2*c - (I*d*(-1 + E^((2*I)*(e
+ f*x))))/E^(I*(e + f*x))]*((-1)^(3/4)*Sqrt[2]*ArcTan[((-1)^(1/4)*(d - I*c*E^(I*(e + f*x))))/(Sqrt[d]*Sqrt[-2*
c*E^(I*(e + f*x)) + I*d*(-1 + E^((2*I)*(e + f*x)))])] - (1 + I)*ArcTanh[((-1)^(3/4)*(c - I*d*E^(I*(e + f*x))))
/(Sqrt[d]*Sqrt[-2*c*E^(I*(e + f*x)) + I*d*(-1 + E^((2*I)*(e + f*x)))])]))/(Sqrt[d]*Sqrt[-2*c*E^(I*(e + f*x)) +
 I*d*(-1 + E^((2*I)*(e + f*x)))]*f) - ((2 - 2*I)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[c + d*Sin[e + f*x]
])/f))/(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])

Maple [F(-1)]

Timed out.

\[\int \sqrt {a +a \sin \left (f x +e \right )}\, \sqrt {c +d \sin \left (f x +e \right )}d x\]

[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)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 243 vs. \(2 (89) = 178\).

Time = 0.59 (sec) , antiderivative size = 945, normalized size of antiderivative = 9.09 \[ \int \sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)} \, dx=\text {Too large to display} \]

[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)]

Sympy [F]

\[ \int \sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)} \, dx=\int \sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )} \sqrt {c + d \sin {\left (e + f x \right )}}\, dx \]

[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)

Maxima [F]

\[ \int \sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)} \, dx=\int { \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {d \sin \left (f x + e\right ) + c} \,d x } \]

[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)

Giac [F]

\[ \int \sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)} \, dx=\int { \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {d \sin \left (f x + e\right ) + c} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)} \, dx=\int \sqrt {a+a\,\sin \left (e+f\,x\right )}\,\sqrt {c+d\,\sin \left (e+f\,x\right )} \,d x \]

[In]

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

[Out]

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