3.297 \(\int \frac {1}{(e \csc (c+d x))^{3/2} (a+a \sec (c+d x))} \, dx\)

Optimal. Leaf size=106 \[ \frac {2}{a d e \sqrt {e \csc (c+d x)}}-\frac {2 \cos (c+d x)}{3 a d e \sqrt {e \csc (c+d x)}}-\frac {4 F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{3 a d e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.23, antiderivative size = 106, 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 = {3878, 3872, 2839, 2564, 30, 2569, 2641} \[ \frac {2}{a d e \sqrt {e \csc (c+d x)}}-\frac {2 \cos (c+d x)}{3 a d e \sqrt {e \csc (c+d x)}}-\frac {4 F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{3 a d e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/((e*Csc[c + d*x])^(3/2)*(a + a*Sec[c + d*x])),x]

[Out]

2/(a*d*e*Sqrt[e*Csc[c + d*x]]) - (2*Cos[c + d*x])/(3*a*d*e*Sqrt[e*Csc[c + d*x]]) - (4*EllipticF[(c - Pi/2 + d*
x)/2, 2])/(3*a*d*e*Sqrt[e*Csc[c + d*x]]*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 2569

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

Rule 2641

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

Rule 3878

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

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Mathematica [A]  time = 0.40, size = 70, normalized size = 0.66 \[ \frac {4 F\left (\left .\frac {1}{4} (-2 c-2 d x+\pi )\right |2\right )-2 \sqrt {\sin (c+d x)} (\cos (c+d x)-3)}{3 a d \sin ^{\frac {3}{2}}(c+d x) (e \csc (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((e*Csc[c + d*x])^(3/2)*(a + a*Sec[c + d*x])),x]

[Out]

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

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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e \csc \left (d x + c\right )}}{a e^{2} \csc \left (d x + c\right )^{2} \sec \left (d x + c\right ) + a e^{2} \csc \left (d x + c\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*csc(d*x+c))^(3/2)/(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

integral(sqrt(e*csc(d*x + c))/(a*e^2*csc(d*x + c)^2*sec(d*x + c) + a*e^2*csc(d*x + c)^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (e \csc \left (d x + c\right )\right )^{\frac {3}{2}} {\left (a \sec \left (d x + c\right ) + a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*csc(d*x+c))^(3/2)/(a+a*sec(d*x+c)),x, algorithm="giac")

[Out]

integrate(1/((e*csc(d*x + c))^(3/2)*(a*sec(d*x + c) + a)), x)

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maple [C]  time = 1.15, size = 195, normalized size = 1.84 \[ \frac {\left (2 i \EllipticF \left (\sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right ) \sin \left (d x +c \right ) \sqrt {\frac {-i \cos \left (d x +c \right )+\sin \left (d x +c \right )+i}{\sin \left (d x +c \right )}}\, \sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}}-\left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}+4 \cos \left (d x +c \right ) \sqrt {2}-3 \sqrt {2}\right ) \sqrt {2}}{3 a d \left (-1+\cos \left (d x +c \right )\right ) \left (\frac {e}{\sin \left (d x +c \right )}\right )^{\frac {3}{2}} \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*csc(d*x+c))^(3/2)/(a+a*sec(d*x+c)),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (e \csc \left (d x + c\right )\right )^{\frac {3}{2}} {\left (a \sec \left (d x + c\right ) + a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*csc(d*x+c))^(3/2)/(a+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

integrate(1/((e*csc(d*x + c))^(3/2)*(a*sec(d*x + c) + a)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\cos \left (c+d\,x\right )}{a\,{\left (\frac {e}{\sin \left (c+d\,x\right )}\right )}^{3/2}\,\left (\cos \left (c+d\,x\right )+1\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + a/cos(c + d*x))*(e/sin(c + d*x))^(3/2)),x)

[Out]

int(cos(c + d*x)/(a*(e/sin(c + d*x))^(3/2)*(cos(c + d*x) + 1)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {1}{\left (e \csc {\left (c + d x \right )}\right )^{\frac {3}{2}} \sec {\left (c + d x \right )} + \left (e \csc {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*csc(d*x+c))**(3/2)/(a+a*sec(d*x+c)),x)

[Out]

Integral(1/((e*csc(c + d*x))**(3/2)*sec(c + d*x) + (e*csc(c + d*x))**(3/2)), x)/a

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