3.1.64 \(\int \frac {1}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}} \, dx\) [64]

Optimal. Leaf size=307 \[ \frac {\sqrt {b} \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{\left (-a^2+b^2\right )^{3/4} d \sqrt {e}}+\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{\left (-a^2+b^2\right )^{3/4} d \sqrt {e}}+\frac {a \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{\left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}}+\frac {a \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{\left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}} \]

[Out]

arctan(b^(1/2)*(e*sin(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))*b^(1/2)/(-a^2+b^2)^(3/4)/d/e^(1/2)+arctanh(b^(1/
2)*(e*sin(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))*b^(1/2)/(-a^2+b^2)^(3/4)/d/e^(1/2)-a*(sin(1/2*c+1/4*Pi+1/2*d
*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticPi(cos(1/2*c+1/4*Pi+1/2*d*x),2*b/(b-(-a^2+b^2)^(1/2)),2^(1/2))*
sin(d*x+c)^(1/2)/d/(a^2-b*(b-(-a^2+b^2)^(1/2)))/(e*sin(d*x+c))^(1/2)-a*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin
(1/2*c+1/4*Pi+1/2*d*x)*EllipticPi(cos(1/2*c+1/4*Pi+1/2*d*x),2*b/(b+(-a^2+b^2)^(1/2)),2^(1/2))*sin(d*x+c)^(1/2)
/d/(a^2-b*(b+(-a^2+b^2)^(1/2)))/(e*sin(d*x+c))^(1/2)

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Rubi [A]
time = 0.37, antiderivative size = 307, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {2781, 2886, 2884, 335, 218, 214, 211} \begin {gather*} \frac {\sqrt {b} \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d \sqrt {e} \left (b^2-a^2\right )^{3/4}}+\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d \sqrt {e} \left (b^2-a^2\right )^{3/4}}+\frac {a \sqrt {\sin (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {e \sin (c+d x)}}+\frac {a \sqrt {\sin (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {e \sin (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*Cos[c + d*x])*Sqrt[e*Sin[c + d*x]]),x]

[Out]

(Sqrt[b]*ArcTan[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/((-a^2 + b^2)^(3/4)*d*Sqrt[e]) +
(Sqrt[b]*ArcTanh[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/((-a^2 + b^2)^(3/4)*d*Sqrt[e])
+ (a*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/((a^2 - b*(b - Sqrt[
-a^2 + b^2]))*d*Sqrt[e*Sin[c + d*x]]) + (a*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c - Pi/2 + d*x)/2, 2]*Sqr
t[Sin[c + d*x]])/((a^2 - b*(b + Sqrt[-a^2 + b^2]))*d*Sqrt[e*Sin[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 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[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 2781

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> With[{q = Rt[
-a^2 + b^2, 2]}, Dist[-a/(2*q), Int[1/(Sqrt[g*Cos[e + f*x]]*(q + b*Cos[e + f*x])), x], x] + (Dist[b*(g/f), Sub
st[Int[1/(Sqrt[x]*(g^2*(a^2 - b^2) + b^2*x^2)), x], x, g*Cos[e + f*x]], x] - Dist[a/(2*q), Int[1/(Sqrt[g*Cos[e
+ f*x]]*(q - b*Cos[e + f*x])), 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*} \int \frac {1}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}} \, dx &=-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{2 \sqrt {-a^2+b^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{2 \sqrt {-a^2+b^2}}-\frac {(b e) \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (\left (a^2-b^2\right ) e^2+b^2 x^2\right )} \, dx,x,e \sin (c+d x)\right )}{d}\\ &=-\frac {(2 b e) \text {Subst}\left (\int \frac {1}{\left (a^2-b^2\right ) e^2+b^2 x^4} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{d}-\frac {\left (a \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{2 \sqrt {-a^2+b^2} \sqrt {e \sin (c+d x)}}-\frac {\left (a \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{2 \sqrt {-a^2+b^2} \sqrt {e \sin (c+d x)}}\\ &=\frac {a \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{\left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}}+\frac {a \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{\left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}}+\frac {b \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e-b x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{\sqrt {-a^2+b^2} d}+\frac {b \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e+b x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{\sqrt {-a^2+b^2} d}\\ &=\frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{\left (-a^2+b^2\right )^{3/4} d \sqrt {e}}+\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{\left (-a^2+b^2\right )^{3/4} d \sqrt {e}}+\frac {a \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{\left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}}+\frac {a \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{\left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
time = 12.03, size = 261, normalized size = 0.85 \begin {gather*} \frac {10 (a+b) F_1\left (\frac {1}{4};-\frac {1}{2},1;\frac {5}{4};-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) \sqrt {e \sin (c+d x)}}{d e (a+b \cos (c+d x)) \left (5 (a+b) F_1\left (\frac {1}{4};-\frac {1}{2},1;\frac {5}{4};-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )+2 \left (-2 (a-b) F_1\left (\frac {5}{4};-\frac {1}{2},2;\frac {9}{4};-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )+(a+b) F_1\left (\frac {5}{4};\frac {1}{2},1;\frac {9}{4};-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + b*Cos[c + d*x])*Sqrt[e*Sin[c + d*x]]),x]

[Out]

(10*(a + b)*AppellF1[1/4, -1/2, 1, 5/4, -Tan[(c + d*x)/2]^2, ((-a + b)*Tan[(c + d*x)/2]^2)/(a + b)]*Sqrt[e*Sin
[c + d*x]])/(d*e*(a + b*Cos[c + d*x])*(5*(a + b)*AppellF1[1/4, -1/2, 1, 5/4, -Tan[(c + d*x)/2]^2, ((-a + b)*Ta
n[(c + d*x)/2]^2)/(a + b)] + 2*(-2*(a - b)*AppellF1[5/4, -1/2, 2, 9/4, -Tan[(c + d*x)/2]^2, ((-a + b)*Tan[(c +
d*x)/2]^2)/(a + b)] + (a + b)*AppellF1[5/4, 1/2, 1, 9/4, -Tan[(c + d*x)/2]^2, ((-a + b)*Tan[(c + d*x)/2]^2)/(
a + b)])*Tan[(c + d*x)/2]^2))

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Maple [A]
time = 0.17, size = 600, normalized size = 1.95

method	result	size
default	\(\frac {-\frac {b e \left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {e \sin \left (d x +c \right )+\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}} \sqrt {e \sin \left (d x +c \right )}\, \sqrt {2}+\sqrt {\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}}}{e \sin \left (d x +c \right )-\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}} \sqrt {e \sin \left (d x +c \right )}\, \sqrt {2}+\sqrt {\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}}}\right )}{4 \left (a^{2} e^{2}-b^{2} e^{2}\right )}-\frac {b e \left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \sin \left (d x +c \right )}}{\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}}}+1\right )}{2 \left (a^{2} e^{2}-b^{2} e^{2}\right )}-\frac {b e \left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \sin \left (d x +c \right )}}{\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}}}-1\right )}{2 \left (a^{2} e^{2}-b^{2} e^{2}\right )}+\frac {a \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \left (\EllipticPi \left (\sqrt {-\sin \left (d x +c \right )+1}, -\frac {b}{\sqrt {-a^{2}+b^{2}}-b}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a^{2}+b^{2}}+\EllipticPi \left (\sqrt {-\sin \left (d x +c \right )+1}, -\frac {b}{\sqrt {-a^{2}+b^{2}}-b}, \frac {\sqrt {2}}{2}\right ) b +\EllipticPi \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {b}{b +\sqrt {-a^{2}+b^{2}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a^{2}+b^{2}}-\EllipticPi \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {b}{b +\sqrt {-a^{2}+b^{2}}}, \frac {\sqrt {2}}{2}\right ) b \right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (\sqrt {-a^{2}+b^{2}}-b \right ) \left (b +\sqrt {-a^{2}+b^{2}}\right ) \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\)	\(600\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {e \sin {\left (c + d x \right )}} \left (a + b \cos {\left (c + d x \right )}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt(e*sin(c + d*x))*(a + b*cos(c + d*x))), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{\sqrt {e\,\sin \left (c+d\,x\right )}\,\left (a+b\,\cos \left (c+d\,x\right )\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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