3.2.23 \(\int \frac {\sqrt {e \sin (c+d x)}}{a+a \sec (c+d x)} \, dx\) [123]

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

[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 = 0.15, 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 = {3957, 2918, 2644, 30, 2647, 2721, 2719} \begin {gather*} -\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)}} \end {gather*}

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 2644

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 2647

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 2719

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

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rule 2918

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 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[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 \text {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] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.42, size = 249, normalized size = 2.62 \begin {gather*} \frac {2 \left (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)}} \, _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 )\right ) \sqrt {e \sin (c+d x)}}{3 a d \left (1+i e^{i c}\right ) \left (i+e^{i c}\right ) \left (-1+e^{i (c+d x)}\right ) \left (1+e^{i (c+d x)}\right )} \end {gather*}

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 = 0.17, size = 148, normalized size = 1.56

method	result	size
default	\(\frac {2 e \left (\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticE \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )+\cos ^{2}\left (d x +c \right )-\cos \left (d x +c \right )\right )}{a \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}\)	\(148\)

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,method=_RETURNVERBOSE)

[Out]

2/a/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*e*((-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))-2*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticE(
(-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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.28, size = 121, normalized size = 1.27 \begin {gather*} -\frac {2 \, {\left (e^{\frac {1}{2}} \sin \left (d x + c\right )^{\frac {3}{2}} + \sqrt {-i} {\left (-i \, \sqrt {2} \cos \left (d x + c\right ) e^{\frac {1}{2}} - i \, \sqrt {2} e^{\frac {1}{2}}\right )} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + \sqrt {i} {\left (i \, \sqrt {2} \cos \left (d x + c\right ) e^{\frac {1}{2}} + i \, \sqrt {2} e^{\frac {1}{2}}\right )} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )\right )}}{a d \cos \left (d x + c\right ) + a d} \end {gather*}

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]

-2*(e^(1/2)*sin(d*x + c)^(3/2) + sqrt(-I)*(-I*sqrt(2)*cos(d*x + c)*e^(1/2) - I*sqrt(2)*e^(1/2))*weierstrassZet
a(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c))) + sqrt(I)*(I*sqrt(2)*cos(d*x + c)*e^(1/2) +
I*sqrt(2)*e^(1/2))*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a*d*cos(d
*x + c) + a*d)

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

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 = 0.00, 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((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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

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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