3.2.24 \(\int \frac {1}{(a+a \sec (c+d x)) \sqrt {e \sin (c+d x)}} \, dx\) [124]

Optimal. Leaf size=101 \[ -\frac {2 e}{3 a d (e \sin (c+d x))^{3/2}}+\frac {2 e \cos (c+d x)}{3 a d (e \sin (c+d x))^{3/2}}+\frac {4 F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{3 a d \sqrt {e \sin (c+d x)}} \]

[Out]

-2/3*e/a/d/(e*sin(d*x+c))^(3/2)+2/3*e*cos(d*x+c)/a/d/(e*sin(d*x+c))^(3/2)-4/3*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1
/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticF(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*sin(d*x+c)^(1/2)/a/d/(e*sin(d*x+c))
^(1/2)

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Rubi [A]
time = 0.16, antiderivative size = 101, 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, 2720} \begin {gather*} -\frac {2 e}{3 a d (e \sin (c+d x))^{3/2}}+\frac {2 e \cos (c+d x)}{3 a d (e \sin (c+d x))^{3/2}}+\frac {4 \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{3 a d \sqrt {e \sin (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sec[c + d*x])*Sqrt[e*Sin[c + d*x]]),x]

[Out]

(-2*e)/(3*a*d*(e*Sin[c + d*x])^(3/2)) + (2*e*Cos[c + d*x])/(3*a*d*(e*Sin[c + d*x])^(3/2)) + (4*EllipticF[(c -
Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(3*a*d*Sqrt[e*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 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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 {1}{(a+a \sec (c+d x)) \sqrt {e \sin (c+d x)}} \, dx &=-\int \frac {\cos (c+d x)}{(-a-a \cos (c+d x)) \sqrt {e \sin (c+d x)}} \, dx\\ &=\frac {e^2 \int \frac {\cos (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx}{a}-\frac {e^2 \int \frac {\cos ^2(c+d x)}{(e \sin (c+d x))^{5/2}} \, dx}{a}\\ &=\frac {2 e \cos (c+d x)}{3 a d (e \sin (c+d x))^{3/2}}+\frac {2 \int \frac {1}{\sqrt {e \sin (c+d x)}} \, dx}{3 a}+\frac {e \text {Subst}\left (\int \frac {1}{x^{5/2}} \, dx,x,e \sin (c+d x)\right )}{a d}\\ &=-\frac {2 e}{3 a d (e \sin (c+d x))^{3/2}}+\frac {2 e \cos (c+d x)}{3 a d (e \sin (c+d x))^{3/2}}+\frac {\left (2 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx}{3 a \sqrt {e \sin (c+d x)}}\\ &=-\frac {2 e}{3 a d (e \sin (c+d x))^{3/2}}+\frac {2 e \cos (c+d x)}{3 a d (e \sin (c+d x))^{3/2}}+\frac {4 F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{3 a d \sqrt {e \sin (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.36, size = 77, normalized size = 0.76 \begin {gather*} \frac {2 \cot \left (\frac {1}{2} (c+d x)\right ) \left (-1+\cos (c+d x)-2 F\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right ) \sin ^{\frac {3}{2}}(c+d x)\right )}{3 a d (1+\cos (c+d x)) \sqrt {e \sin (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sec[c + d*x])*Sqrt[e*Sin[c + d*x]]),x]

[Out]

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

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Maple [A]
time = 0.18, size = 121, normalized size = 1.20

method	result	size
default	\(\frac {-\frac {2 e}{3 a \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {2 \left (\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {5}{2}}\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )+\sin ^{3}\left (d x +c \right )-\sin \left (d x +c \right )\right )}{3 a \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\)	\(121\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-2/3/a*e/(e*sin(d*x+c))^(3/2)-2/3*((-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(5/2)*EllipticF((-
sin(d*x+c)+1)^(1/2),1/2*2^(1/2))+sin(d*x+c)^3-sin(d*x+c))/a/sin(d*x+c)^2/cos(d*x+c)/(e*sin(d*x+c))^(1/2))/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(1/(a+a*sec(d*x+c))/(e*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.69, size = 104, normalized size = 1.03 \begin {gather*} \frac {2 \, {\left (\sqrt {-i} {\left (\sqrt {2} \cos \left (d x + c\right ) + \sqrt {2}\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {i} {\left (\sqrt {2} \cos \left (d x + c\right ) + \sqrt {2}\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - \sqrt {\sin \left (d x + c\right )}\right )}}{3 \, {\left (a d \cos \left (d x + c\right ) e^{\frac {1}{2}} + a d e^{\frac {1}{2}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate(1/((a*sec(d*x + c) + a)*sqrt(e*sin(d*x + c))), 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 )}{a\,\sqrt {e\,\sin \left (c+d\,x\right )}\,\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(1/((e*sin(c + d*x))^(1/2)*(a + a/cos(c + d*x))),x)

[Out]

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

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