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/(e*sin(d*x+c))^(1/2)+2*e*cos(d*x+c)/a/d/(e*sin(d*x+c))^(1/2)-4*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/si
n(1/2*c+1/4*Pi+1/2*d*x)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*(e*sin(d*x+c))^(1/2)/a/d/sin(d*x+c)^(1/2)

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Rubi [A]  time = 0.21, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, 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 30

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

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 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 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]

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 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 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]

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*}

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Mathematica [C]  time = 0.61, size = 249, normalized size = 2.62 \[ \frac {2 \left (12 e^{2 i c} \sqrt {1-e^{2 i (c+d x)}} \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};e^{2 i (c+d x)}\right )+4 e^{2 i (c+d x)} \sqrt {1-e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};e^{2 i (c+d x)}\right )+6 e^{i (c+d x)}-9 e^{2 i (c+d x)}+3 e^{2 i (2 c+d x)}+6 e^{i (3 c+d x)}-9 e^{2 i c}+3\right ) \sqrt {e \sin (c+d x)}}{3 a \left (1+i e^{i c}\right ) \left (e^{i c}+i\right ) d \left (-1+e^{i (c+d x)}\right ) \left (1+e^{i (c+d x)}\right )} \]

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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fricas [F]  time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e \sin \left (d x + c\right )}}{a \sec \left (d x + c\right ) + a}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {e \sin \left (d x + c\right )}}{a \sec \left (d x + c\right ) + a}\,{d x} \]

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)

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maple [A]  time = 3.89, size = 149, normalized size = 1.57 \[ -\frac {2 e \left (2 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticE \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-\left (\cos ^{2}\left (d x +c \right )\right )+\cos \left (d x +c \right )\right )}{a \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d} \]

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*sin(d*x+c)+2)^(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*sin(d*x+c)+2)^(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {e \sin \left (d x + c\right )}}{a \sec \left (d x + c\right ) + a}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\cos \left (c+d\,x\right )\,\sqrt {e\,\sin \left (c+d\,x\right )}}{a\,\left (\cos \left (c+d\,x\right )+1\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sqrt {e \sin {\left (c + d x \right )}}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \]

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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