3.3.95 \(\int \frac {\sqrt {e \csc (c+d x)}}{a+a \sec (c+d x)} \, dx\) [295]

Optimal. Leaf size=105 \[ \frac {2 \cot (c+d x) \sqrt {e \csc (c+d x)}}{3 a d}-\frac {2 \csc (c+d x) \sqrt {e \csc (c+d x)}}{3 a d}+\frac {4 \sqrt {e \csc (c+d x)} F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{3 a d} \]

[Out]

2/3*cot(d*x+c)*(e*csc(d*x+c))^(1/2)/a/d-2/3*csc(d*x+c)*(e*csc(d*x+c))^(1/2)/a/d-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))*(e*csc(d*x+c))^(1/2)*sin(d*x+
c)^(1/2)/a/d

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Rubi [A]
time = 0.14, antiderivative size = 105, 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 = {3963, 3957, 2918, 2644, 30, 2647, 2720} \begin {gather*} -\frac {2 \csc (c+d x) \sqrt {e \csc (c+d x)}}{3 a d}+\frac {2 \cot (c+d x) \sqrt {e \csc (c+d x)}}{3 a d}+\frac {4 \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \csc (c+d x)}}{3 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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]

Rule 3963

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*((g_.)*sec[(e_.) + (f_.)*(x_)])^(p_), x_Symbol] :> Dist[g^Int
Part[p]*(g*Sec[e + f*x])^FracPart[p]*Cos[e + f*x]^FracPart[p], Int[(a + b*Csc[e + f*x])^m/Cos[e + f*x]^p, x],
x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&  !IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {\sqrt {e \csc (c+d x)}}{a+a \sec (c+d x)} \, dx &=\left (\sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {1}{(a+a \sec (c+d x)) \sqrt {\sin (c+d x)}} \, dx\\ &=-\left (\left (\sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {\cos (c+d x)}{(-a-a \cos (c+d x)) \sqrt {\sin (c+d x)}} \, dx\right )\\ &=\frac {\left (\sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {\cos (c+d x)}{\sin ^{\frac {5}{2}}(c+d x)} \, dx}{a}-\frac {\left (\sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {\cos ^2(c+d x)}{\sin ^{\frac {5}{2}}(c+d x)} \, dx}{a}\\ &=\frac {2 \cot (c+d x) \sqrt {e \csc (c+d x)}}{3 a d}+\frac {\left (2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx}{3 a}+\frac {\left (\sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{x^{5/2}} \, dx,x,\sin (c+d x)\right )}{a d}\\ &=\frac {2 \cot (c+d x) \sqrt {e \csc (c+d x)}}{3 a d}-\frac {2 \csc (c+d x) \sqrt {e \csc (c+d x)}}{3 a d}+\frac {4 \sqrt {e \csc (c+d x)} 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}\\ \end {align*}

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Mathematica [A]
time = 0.25, size = 60, normalized size = 0.57 \begin {gather*} \frac {2 (e \csc (c+d x))^{3/2} \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 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains complex when optimal does not.
time = 0.18, size = 326, normalized size = 3.10

method	result	size
default	\(\frac {\sqrt {\frac {e}{\sin \left (d x +c \right )}}\, \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (2 i \cos \left (d x +c \right ) \sin \left (d x +c \right ) \EllipticF \left (\sqrt {\frac {i \cos \left (d x +c \right )+\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {i \cos \left (d x +c \right )+\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \cos \left (d x +c \right )-\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}+2 i \EllipticF \left (\sqrt {\frac {i \cos \left (d x +c \right )+\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {i \cos \left (d x +c \right )+\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \cos \left (d x +c \right )-\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}\, \sin \left (d x +c \right )+\sqrt {2}\, \cos \left (d x +c \right )-\sqrt {2}\right ) \sqrt {2}}{3 a d \sin \left (d x +c \right )^{5}}\)	\(326\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/a/d*(e/sin(d*x+c))^(1/2)*(1+cos(d*x+c))^2*(-1+cos(d*x+c))^2*(2*I*cos(d*x+c)*sin(d*x+c)*((I*cos(d*x+c)+sin(
d*x+c)-I)/sin(d*x+c))^(1/2)*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-sin(d*x+c)-I)/sin(d*x+c))^(1
/2)*EllipticF(((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2),1/2*2^(1/2))+2*I*EllipticF(((I*cos(d*x+c)+sin(d*x
+c)-I)/sin(d*x+c))^(1/2),1/2*2^(1/2))*((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2)*(-I*(-1+cos(d*x+c))/sin(d
*x+c))^(1/2)*(-(I*cos(d*x+c)-sin(d*x+c)-I)/sin(d*x+c))^(1/2)*sin(d*x+c)+2^(1/2)*cos(d*x+c)-2^(1/2))/sin(d*x+c)
^5*2^(1/2)

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

[Out]

e^(1/2)*integrate(sqrt(csc(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 = 0.80, size = 103, normalized size = 0.98 \begin {gather*} -\frac {2 \, {\left (\sqrt {2 i} {\left (i \, \cos \left (d x + c\right ) e^{\frac {1}{2}} + i \, e^{\frac {1}{2}}\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {-2 i} {\left (-i \, \cos \left (d x + c\right ) e^{\frac {1}{2}} - i \, e^{\frac {1}{2}}\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + e^{\frac {1}{2}} \sqrt {\sin \left (d x + c\right )}\right )}}{3 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

integrate(sqrt(e*csc(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 {\frac {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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