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

Optimal. Leaf size=292 \[ \frac{\sqrt{e} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{\sqrt{b} d \sqrt [4]{b^2-a^2}}-\frac{\sqrt{e} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{\sqrt{b} d \sqrt [4]{b^2-a^2}}+\frac{a e \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{b d \left (b-\sqrt{b^2-a^2}\right ) \sqrt{e \cos (c+d x)}}+\frac{a e \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{b d \left (\sqrt{b^2-a^2}+b\right ) \sqrt{e \cos (c+d x)}} \]

[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]])

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Rubi [A]  time = 0.582375, antiderivative size = 292, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {2701, 2807, 2805, 329, 298, 205, 208} \[ \frac{\sqrt{e} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{\sqrt{b} d \sqrt [4]{b^2-a^2}}-\frac{\sqrt{e} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{\sqrt{b} d \sqrt [4]{b^2-a^2}}+\frac{a e \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{b d \left (b-\sqrt{b^2-a^2}\right ) \sqrt{e \cos (c+d x)}}+\frac{a e \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{b d \left (\sqrt{b^2-a^2}+b\right ) \sqrt{e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[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 2701

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 2807

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*
Sin[e + f*x])/(c + d)]), 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]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[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 329

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 298

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 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]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{\sqrt{e \cos (c+d x)}}{a+b \sin (c+d x)} \, dx &=-\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) \operatorname{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) \operatorname{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)} \Pi \left (\frac{2 b}{b-\sqrt{-a^2+b^2}};\left .\frac{1}{2} (c+d x)\right |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)} \Pi \left (\frac{2 b}{b+\sqrt{-a^2+b^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{b \left (b+\sqrt{-a^2+b^2}\right ) d \sqrt{e \cos (c+d x)}}-\frac{e \operatorname{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 \operatorname{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} \tan ^{-1}\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} \tanh ^{-1}\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)} \Pi \left (\frac{2 b}{b-\sqrt{-a^2+b^2}};\left .\frac{1}{2} (c+d x)\right |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)} \Pi \left (\frac{2 b}{b+\sqrt{-a^2+b^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{b \left (b+\sqrt{-a^2+b^2}\right ) d \sqrt{e \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 15.9687, size = 361, normalized size = 1.24 \[ -\frac{2 \sin (c+d x) \sqrt{e \cos (c+d x)} \left (a+b \sqrt{\sin ^2(c+d x)}\right ) \left (\frac{a \cos ^{\frac{3}{2}}(c+d x) F_1\left (\frac{3}{4};\frac{1}{2},1;\frac{7}{4};\cos ^2(c+d x),\frac{b^2 \cos ^2(c+d x)}{b^2-a^2}\right )}{3 \left (a^2-b^2\right )}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) \left (-\log \left (-(1+i) \sqrt{b} \sqrt [4]{b^2-a^2} \sqrt{\cos (c+d x)}+\sqrt{b^2-a^2}+i b \cos (c+d x)\right )+\log \left ((1+i) \sqrt{b} \sqrt [4]{b^2-a^2} \sqrt{\cos (c+d x)}+\sqrt{b^2-a^2}+i b \cos (c+d x)\right )+2 \tan ^{-1}\left (1-\frac{(1+i) \sqrt{b} \sqrt{\cos (c+d x)}}{\sqrt [4]{b^2-a^2}}\right )-2 \tan ^{-1}\left (1+\frac{(1+i) \sqrt{b} \sqrt{\cos (c+d x)}}{\sqrt [4]{b^2-a^2}}\right )\right )}{\sqrt{b} \sqrt [4]{b^2-a^2}}\right )}{d \sqrt{\sin ^2(c+d x)} \sqrt{\cos (c+d x)} (a+b \sin (c+d x))} \]

Warning: Unable to verify antiderivative.

[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]))

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Maple [C]  time = 1.838, size = 682, normalized size = 2.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*e*b*sum((_R^6-_R^4*e-_R^2*e^2+e^3)/(_R^7*b^2-3*_R^5*b^2*e+8*_R^3*a^2*e^2-5*_R^3*b^2*e^2-_R*b^2*e^3)*ln((
-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)-e^(1/2)*cos(1/2*d*x+1/2*c)*2^(1/2)-_R),_R=RootOf(b^2*_Z^8-4*b^2*e*_Z^6+(16*
a^2*e^2-10*b^2*e^2)*_Z^4-4*b^2*e^3*_Z^2+b^2*e^4))-1/8/d*(e*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1
/2)*e/a/b^2*sum(1/_alpha*(8*(e*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)*(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))*_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*(_alp
ha^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*_alpha^2-3)/(4*a^2-3*b^
2)*(4*cos(1/2*d*x+1/2*c)^2*a^2-3*b^2*cos(1/2*d*x+1/2*c)^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))*(-sin(1/2*d*x+1/2*c)^2*e*(2*s
in(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)/(-sin(1/2*d*x+1/2*c)^2*e*(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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{e \cos \left (d x + c\right )}}{b \sin \left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{e \cos \left (d x + c\right )}}{b \sin \left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)