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

Optimal. Leaf size=95 \[ -\frac{2 e}{a d \sqrt{e \sin (c+d x)}}+\frac{2 e \cos (c+d x)}{a d \sqrt{e \sin (c+d x)}}+\frac{4 E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{a d \sqrt{\sin (c+d x)}} \]

[Out]

(-2*e)/(a*d*Sqrt[e*Sin[c + d*x]]) + (2*e*Cos[c + d*x])/(a*d*Sqrt[e*Sin[c + d*x]]) + (4*EllipticE[(c - Pi/2 + d
*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(a*d*Sqrt[Sin[c + d*x]])

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Rubi [A]  time = 0.207713, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {3872, 2839, 2564, 30, 2567, 2640, 2639} \[ -\frac{2 e}{a d \sqrt{e \sin (c+d x)}}+\frac{2 e \cos (c+d x)}{a d \sqrt{e \sin (c+d x)}}+\frac{4 E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{a d \sqrt{\sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*e)/(a*d*Sqrt[e*Sin[c + d*x]]) + (2*e*Cos[c + d*x])/(a*d*Sqrt[e*Sin[c + d*x]]) + (4*EllipticE[(c - Pi/2 + d
*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(a*d*Sqrt[Sin[c + d*x]])

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2839

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

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2567

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(a*Cos[e +
f*x])^(m - 1)*(b*Sin[e + f*x])^(n + 1))/(b*f*(n + 1)), x] + Dist[(a^2*(m - 1))/(b^2*(n + 1)), Int[(a*Cos[e +
f*x])^(m - 2)*(b*Sin[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[m, 1] && LtQ[n, -1] && (Intege
rsQ[2*m, 2*n] || EqQ[m + n, 0])

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin{align*} \int \frac{\sqrt{e \sin (c+d x)}}{a+a \sec (c+d x)} \, dx &=-\int \frac{\cos (c+d x) \sqrt{e \sin (c+d x)}}{-a-a \cos (c+d x)} \, dx\\ &=\frac{e^2 \int \frac{\cos (c+d x)}{(e \sin (c+d x))^{3/2}} \, dx}{a}-\frac{e^2 \int \frac{\cos ^2(c+d x)}{(e \sin (c+d x))^{3/2}} \, dx}{a}\\ &=\frac{2 e \cos (c+d x)}{a d \sqrt{e \sin (c+d x)}}+\frac{2 \int \sqrt{e \sin (c+d x)} \, dx}{a}+\frac{e \operatorname{Subst}\left (\int \frac{1}{x^{3/2}} \, dx,x,e \sin (c+d x)\right )}{a d}\\ &=-\frac{2 e}{a d \sqrt{e \sin (c+d x)}}+\frac{2 e \cos (c+d x)}{a d \sqrt{e \sin (c+d x)}}+\frac{\left (2 \sqrt{e \sin (c+d x)}\right ) \int \sqrt{\sin (c+d x)} \, dx}{a \sqrt{\sin (c+d x)}}\\ &=-\frac{2 e}{a d \sqrt{e \sin (c+d x)}}+\frac{2 e \cos (c+d x)}{a d \sqrt{e \sin (c+d x)}}+\frac{4 E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{a d \sqrt{\sin (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 0.587984, size = 249, normalized size = 2.62 \[ \frac{2 \left (12 e^{2 i c} \sqrt{1-e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},e^{2 i (c+d x)}\right )+4 e^{2 i (c+d x)} \sqrt{1-e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},e^{2 i (c+d x)}\right )+6 e^{i (c+d x)}-9 e^{2 i (c+d x)}+3 e^{2 i (2 c+d x)}+6 e^{i (3 c+d x)}-9 e^{2 i c}+3\right ) \sqrt{e \sin (c+d x)}}{3 a \left (1+i e^{i c}\right ) \left (e^{i c}+i\right ) d \left (-1+e^{i (c+d x)}\right ) \left (1+e^{i (c+d x)}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(3 - 9*E^((2*I)*c) + 6*E^(I*(c + d*x)) - 9*E^((2*I)*(c + d*x)) + 3*E^((2*I)*(2*c + d*x)) + 6*E^(I*(3*c + d*
x)) + 12*E^((2*I)*c)*Sqrt[1 - E^((2*I)*(c + d*x))]*Hypergeometric2F1[-1/4, 1/2, 3/4, E^((2*I)*(c + d*x))] + 4*
E^((2*I)*(c + d*x))*Sqrt[1 - E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, E^((2*I)*(c + d*x))])*Sqrt[
e*Sin[c + d*x]])/(3*a*d*(1 + I*E^(I*c))*(I + E^(I*c))*(-1 + E^(I*(c + d*x)))*(1 + E^(I*(c + d*x))))

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Maple [A]  time = 1.421, size = 149, normalized size = 1.6 \begin{align*} -2\,{\frac{e \left ( 2\,\sqrt{-\sin \left ( dx+c \right ) +1}\sqrt{2+2\,\sin \left ( dx+c \right ) }\sqrt{\sin \left ( dx+c \right ) }{\it EllipticE} \left ( \sqrt{-\sin \left ( dx+c \right ) +1},1/2\,\sqrt{2} \right ) -\sqrt{-\sin \left ( dx+c \right ) +1}\sqrt{2+2\,\sin \left ( dx+c \right ) }\sqrt{\sin \left ( dx+c \right ) }{\it EllipticF} \left ( \sqrt{-\sin \left ( dx+c \right ) +1},1/2\,\sqrt{2} \right ) - \left ( \cos \left ( dx+c \right ) \right ) ^{2}+\cos \left ( dx+c \right ) \right ) }{a\cos \left ( dx+c \right ) \sqrt{e\sin \left ( dx+c \right ) }d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/a/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*e*(2*(-sin(d*x+c)+1)^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*Ellipt
icE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-(-sin(d*x+c)+1)^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*EllipticF
((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-cos(d*x+c)^2+cos(d*x+c))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{e \sin \left (d x + c\right )}}{a \sec \left (d x + c\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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