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

Optimal. Leaf size=169 \[ \frac{2 \sqrt{e} \sqrt{\cos (c+d x)+1} \sqrt{a \sin (c+d x)+a} \tan ^{-1}\left (\frac{\sqrt{e} \sin (c+d x)}{\sqrt{\cos (c+d x)+1} \sqrt{e \cos (c+d x)}}\right )}{d (a \sin (c+d x)+a \cos (c+d x)+a)}+\frac{2 \sqrt{e} \sqrt{\cos (c+d x)+1} \sqrt{a \sin (c+d x)+a} \sinh ^{-1}\left (\frac{\sqrt{e \cos (c+d x)}}{\sqrt{e}}\right )}{d (a \sin (c+d x)+a \cos (c+d x)+a)} \]

[Out]

(2*Sqrt[e]*ArcSinh[Sqrt[e*Cos[c + d*x]]/Sqrt[e]]*Sqrt[1 + Cos[c + d*x]]*Sqrt[a + a*Sin[c + d*x]])/(d*(a + a*Co
s[c + d*x] + a*Sin[c + d*x])) + (2*Sqrt[e]*ArcTan[(Sqrt[e]*Sin[c + d*x])/(Sqrt[e*Cos[c + d*x]]*Sqrt[1 + Cos[c
+ d*x]])]*Sqrt[1 + Cos[c + d*x]]*Sqrt[a + a*Sin[c + d*x]])/(d*(a + a*Cos[c + d*x] + a*Sin[c + d*x]))

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Rubi [A]  time = 0.196235, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2684, 2775, 203, 2833, 63, 215} \[ \frac{2 \sqrt{e} \sqrt{\cos (c+d x)+1} \sqrt{a \sin (c+d x)+a} \tan ^{-1}\left (\frac{\sqrt{e} \sin (c+d x)}{\sqrt{\cos (c+d x)+1} \sqrt{e \cos (c+d x)}}\right )}{d (a \sin (c+d x)+a \cos (c+d x)+a)}+\frac{2 \sqrt{e} \sqrt{\cos (c+d x)+1} \sqrt{a \sin (c+d x)+a} \sinh ^{-1}\left (\frac{\sqrt{e \cos (c+d x)}}{\sqrt{e}}\right )}{d (a \sin (c+d x)+a \cos (c+d x)+a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[e]*ArcSinh[Sqrt[e*Cos[c + d*x]]/Sqrt[e]]*Sqrt[1 + Cos[c + d*x]]*Sqrt[a + a*Sin[c + d*x]])/(d*(a + a*Co
s[c + d*x] + a*Sin[c + d*x])) + (2*Sqrt[e]*ArcTan[(Sqrt[e]*Sin[c + d*x])/(Sqrt[e*Cos[c + d*x]]*Sqrt[1 + Cos[c
+ d*x]])]*Sqrt[1 + Cos[c + d*x]]*Sqrt[a + a*Sin[c + d*x]])/(d*(a + a*Cos[c + d*x] + a*Sin[c + d*x]))

Rule 2684

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

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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 2833

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rubi steps

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

Mathematica [C]  time = 0.0829778, size = 77, normalized size = 0.46 \[ -\frac{2 \sqrt [4]{2} (e \cos (c+d x))^{3/2} \, _2F_1\left (\frac{3}{4},\frac{3}{4};\frac{7}{4};\frac{1}{2} (1-\sin (c+d x))\right )}{3 d e \sqrt [4]{\sin (c+d x)+1} \sqrt{a (\sin (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*2^(1/4)*(e*Cos[c + d*x])^(3/2)*Hypergeometric2F1[3/4, 3/4, 7/4, (1 - Sin[c + d*x])/2])/(3*d*e*(1 + Sin[c +
d*x])^(1/4)*Sqrt[a*(1 + Sin[c + d*x])])

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Maple [A]  time = 0.095, size = 141, normalized size = 0.8 \begin{align*}{\frac{\sqrt{2} \left ( 1-\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) \right ) }{d\sin \left ( dx+c \right ) }\sqrt{e\cos \left ( dx+c \right ) } \left ( \arctan \left ({\frac{\sqrt{2}}{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}} \right ) +{\it Artanh} \left ({\frac{\sqrt{2}\sin \left ( dx+c \right ) }{2\,\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}} \right ) \right ){\frac{1}{\sqrt{a \left ( 1+\sin \left ( dx+c \right ) \right ) }}}{\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*2^(1/2)*(e*cos(d*x+c))^(1/2)*(1-cos(d*x+c)+sin(d*x+c))*(arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^
(1/2))+arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)))/(a*(1+sin(d*x+c)))^(1/
2)/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(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 )}}{\sqrt{a \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+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )}}{\sqrt{a \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+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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