\(\int \frac {\sqrt {e \cos (c+d x)}}{a+b \sin (c+d x)} \, dx\) [579]

Optimal result

Integrand size = 25, antiderivative size = 292 \[ \int \frac {\sqrt {e \cos (c+d x)}}{a+b \sin (c+d x)} \, dx=\frac {\sqrt {e} \arctan \left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{\sqrt {b} \sqrt [4]{-a^2+b^2} d}-\frac {\sqrt {e} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{\sqrt {b} \sqrt [4]{-a^2+b^2} d}+\frac {a e \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{b \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}+\frac {a e \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{b \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}} \]

[Out]

arctan(b^(1/2)*(e*cos(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))*e^(1/2)/(-a^2+b^2)^(1/4)/d/b^(1/2)-arctanh(b^(1/
2)*(e*cos(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))*e^(1/2)/(-a^2+b^2)^(1/4)/d/b^(1/2)+a*e*(cos(1/2*d*x+1/2*c)^2
)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(b-(-a^2+b^2)^(1/2)),2^(1/2))*cos(d*x+c)^(1/2)/b/
d/(b-(-a^2+b^2)^(1/2))/(e*cos(d*x+c))^(1/2)+a*e*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin
(1/2*d*x+1/2*c),2*b/(b+(-a^2+b^2)^(1/2)),2^(1/2))*cos(d*x+c)^(1/2)/b/d/(b+(-a^2+b^2)^(1/2))/(e*cos(d*x+c))^(1/
2)

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {2780, 2886, 2884, 335, 304, 211, 214} \[ \int \frac {\sqrt {e \cos (c+d x)}}{a+b \sin (c+d x)} \, dx=\frac {\sqrt {e} \arctan \left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{\sqrt {b} d \sqrt [4]{b^2-a^2}}-\frac {\sqrt {e} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{\sqrt {b} d \sqrt [4]{b^2-a^2}}+\frac {a e \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (c+d x),2\right )}{b d \left (b-\sqrt {b^2-a^2}\right ) \sqrt {e \cos (c+d x)}}+\frac {a e \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (c+d x),2\right )}{b d \left (\sqrt {b^2-a^2}+b\right ) \sqrt {e \cos (c+d x)}} \]

[In]

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

[Out]

(Sqrt[e]*ArcTan[(Sqrt[b]*Sqrt[e*Cos[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(Sqrt[b]*(-a^2 + b^2)^(1/4)*d) -
(Sqrt[e]*ArcTanh[(Sqrt[b]*Sqrt[e*Cos[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(Sqrt[b]*(-a^2 + b^2)^(1/4)*d)
+ (a*e*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c + d*x)/2, 2])/(b*(b - Sqrt[-a^2 + b^2])
*d*Sqrt[e*Cos[c + d*x]]) + (a*e*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c + d*x)/2, 2])/(
b*(b + Sqrt[-a^2 + b^2])*d*Sqrt[e*Cos[c + d*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 214

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

Rule 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2780

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> With[{q = Rt[-a^2
+ b^2, 2]}, Dist[a*(g/(2*b)), Int[1/(Sqrt[g*Cos[e + f*x]]*(q + b*Cos[e + f*x])), x], x] + (-Dist[a*(g/(2*b)),
Int[1/(Sqrt[g*Cos[e + f*x]]*(q - b*Cos[e + f*x])), x], x] + Dist[b*(g/f), Subst[Int[Sqrt[x]/(g^2*(a^2 - b^2)
+ b^2*x^2), x], x, g*Cos[e + f*x]], x])] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0]

Rule 2884

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

Rule 2886

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist
[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt[c + d*Sin[e + f*x]], Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d/
(c + d))*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && N
eQ[c^2 - d^2, 0] &&  !GtQ[c + d, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {(a e) \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {-a^2+b^2}-b \cos (c+d x)\right )} \, dx}{2 b}+\frac {(a e) \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {-a^2+b^2}+b \cos (c+d x)\right )} \, dx}{2 b}+\frac {(b e) \text {Subst}\left (\int \frac {\sqrt {x}}{\left (a^2-b^2\right ) e^2+b^2 x^2} \, dx,x,e \cos (c+d x)\right )}{d} \\ & = \frac {(2 b e) \text {Subst}\left (\int \frac {x^2}{\left (a^2-b^2\right ) e^2+b^2 x^4} \, dx,x,\sqrt {e \cos (c+d x)}\right )}{d}-\frac {\left (a e \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \left (\sqrt {-a^2+b^2}-b \cos (c+d x)\right )} \, dx}{2 b \sqrt {e \cos (c+d x)}}+\frac {\left (a e \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \left (\sqrt {-a^2+b^2}+b \cos (c+d x)\right )} \, dx}{2 b \sqrt {e \cos (c+d x)}} \\ & = \frac {a e \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{b \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}+\frac {a e \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{b \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {e \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e-b x^2} \, dx,x,\sqrt {e \cos (c+d x)}\right )}{d}+\frac {e \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e+b x^2} \, dx,x,\sqrt {e \cos (c+d x)}\right )}{d} \\ & = \frac {\sqrt {e} \arctan \left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{\sqrt {b} \sqrt [4]{-a^2+b^2} d}-\frac {\sqrt {e} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{\sqrt {b} \sqrt [4]{-a^2+b^2} d}+\frac {a e \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{b \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}+\frac {a e \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{b \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 13.01 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.24 \[ \int \frac {\sqrt {e \cos (c+d x)}}{a+b \sin (c+d x)} \, dx=-\frac {2 \sqrt {e \cos (c+d x)} \left (\frac {a \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},1,\frac {7}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(c+d x)}{3 \left (a^2-b^2\right )}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \left (2 \arctan \left (1-\frac {(1+i) \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-\log \left (\sqrt {-a^2+b^2}-(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (c+d x)}+i b \cos (c+d x)\right )+\log \left (\sqrt {-a^2+b^2}+(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (c+d x)}+i b \cos (c+d x)\right )\right )}{\sqrt {b} \sqrt [4]{-a^2+b^2}}\right ) \sin (c+d x) \left (a+b \sqrt {\sin ^2(c+d x)}\right )}{d \sqrt {\cos (c+d x)} \sqrt {\sin ^2(c+d x)} (a+b \sin (c+d x))} \]

[In]

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

[Out]

(-2*Sqrt[e*Cos[c + d*x]]*((a*AppellF1[3/4, 1/2, 1, 7/4, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)]*Cos
[c + d*x]^(3/2))/(3*(a^2 - b^2)) + ((1/8 + I/8)*(2*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Cos[c + d*x]])/(-a^2 + b^2
)^(1/4)] - 2*ArcTan[1 + ((1 + I)*Sqrt[b]*Sqrt[Cos[c + d*x]])/(-a^2 + b^2)^(1/4)] - Log[Sqrt[-a^2 + b^2] - (1 +
I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[c + d*x]] + I*b*Cos[c + d*x]] + Log[Sqrt[-a^2 + b^2] + (1 + I)*Sqrt[b]
*(-a^2 + b^2)^(1/4)*Sqrt[Cos[c + d*x]] + I*b*Cos[c + d*x]]))/(Sqrt[b]*(-a^2 + b^2)^(1/4)))*Sin[c + d*x]*(a + b
*Sqrt[Sin[c + d*x]^2]))/(d*Sqrt[Cos[c + d*x]]*Sqrt[Sin[c + d*x]^2]*(a + b*Sin[c + d*x]))

Maple [C] (warning: unable to verify)

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

Time = 2.39 (sec) , antiderivative size = 825, normalized size of antiderivative = 2.83

method	result	size
default	\(\text {Expression too large to display}\)	\(825\)

[In]

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

[Out]

(1/4*e/b/(e^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*(ln((2*e*cos(1/2*d*x+1/2*c)^2-e-(e^2*(a^2-b^2)/b^2)^(1/4)*(2*e*cos(
1/2*d*x+1/2*c)^2-e)^(1/2)*2^(1/2)+(e^2*(a^2-b^2)/b^2)^(1/2))/(2*e*cos(1/2*d*x+1/2*c)^2-e+(e^2*(a^2-b^2)/b^2)^(
1/4)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)*2^(1/2)+(e^2*(a^2-b^2)/b^2)^(1/2)))+2*arctan((2^(1/2)*(2*e*cos(1/2*d*x
+1/2*c)^2-e)^(1/2)+(e^2*(a^2-b^2)/b^2)^(1/4))/(e^2*(a^2-b^2)/b^2)^(1/4))+2*arctan((2^(1/2)*(2*e*cos(1/2*d*x+1/
2*c)^2-e)^(1/2)-(e^2*(a^2-b^2)/b^2)^(1/4))/(e^2*(a^2-b^2)/b^2)^(1/4)))-1/8*(e*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1
/2*d*x+1/2*c)^2)^(1/2)/a*e/b^2*sum(1/_alpha*(8*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*
EllipticPi(cos(1/2*d*x+1/2*c),-4*b^2/a^2*(_alpha^2-1),2^(1/2))*(e*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)*_alpha
^3*b^2-8*b^2*_alpha*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*
c),-4*b^2/a^2*(_alpha^2-1),2^(1/2))*(e*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)+a^2*2^(1/2)*arctanh(1/2*e*(4*_alp
ha^2-3)/(4*a^2-3*b^2)*(4*a^2*cos(1/2*d*x+1/2*c)^2-3*cos(1/2*d*x+1/2*c)^2*b^2+b^2*_alpha^2-3*a^2+2*b^2)*2^(1/2)
/(e*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)/(-e*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2))*(-e*sin(1/
2*d*x+1/2*c)^2*(2*sin(1/2*d*x+1/2*c)^2-1))^(1/2))/(e*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)/(-e*sin(1/2*d*x+1/2
*c)^2*(2*sin(1/2*d*x+1/2*c)^2-1))^(1/2),_alpha=RootOf(4*_Z^4*b^2-4*_Z^2*b^2+a^2))/sin(1/2*d*x+1/2*c)/(e*(2*cos
(1/2*d*x+1/2*c)^2-1))^(1/2))/d

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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