3.298 \(\int \frac{1}{(e \csc (c+d x))^{5/2} (a+a \sec (c+d x))} \, dx\)

Optimal. Leaf size=120 \[ \frac{2 \sin (c+d x)}{3 a d e^2 \sqrt{e \csc (c+d x)}}-\frac{2 \sin (c+d x) \cos (c+d x)}{5 a d e^2 \sqrt{e \csc (c+d x)}}-\frac{4 E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{5 a d e^2 \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)}} \]

[Out]

(-4*EllipticE[(c - Pi/2 + d*x)/2, 2])/(5*a*d*e^2*Sqrt[e*Csc[c + d*x]]*Sqrt[Sin[c + d*x]]) + (2*Sin[c + d*x])/(
3*a*d*e^2*Sqrt[e*Csc[c + d*x]]) - (2*Cos[c + d*x]*Sin[c + d*x])/(5*a*d*e^2*Sqrt[e*Csc[c + d*x]])

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Rubi [A]  time = 0.222214, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {3878, 3872, 2839, 2564, 30, 2569, 2639} \[ \frac{2 \sin (c+d x)}{3 a d e^2 \sqrt{e \csc (c+d x)}}-\frac{2 \sin (c+d x) \cos (c+d x)}{5 a d e^2 \sqrt{e \csc (c+d x)}}-\frac{4 E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{5 a d e^2 \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/((e*Csc[c + d*x])^(5/2)*(a + a*Sec[c + d*x])),x]

[Out]

(-4*EllipticE[(c - Pi/2 + d*x)/2, 2])/(5*a*d*e^2*Sqrt[e*Csc[c + d*x]]*Sqrt[Sin[c + d*x]]) + (2*Sin[c + d*x])/(
3*a*d*e^2*Sqrt[e*Csc[c + d*x]]) - (2*Cos[c + d*x]*Sin[c + d*x])/(5*a*d*e^2*Sqrt[e*Csc[c + d*x]])

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]

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 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 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 30

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

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 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]

Rubi steps

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

Mathematica [C]  time = 0.874674, size = 100, normalized size = 0.83 \[ \frac{8 \sqrt{1-e^{2 i (c+d x)}} (\cot (c+d x)+i) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},e^{2 i (c+d x)}\right )+20 \sin (c+d x)-6 (\sin (2 (c+d x))+4 i)}{30 a d e^2 \sqrt{e \csc (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((e*Csc[c + d*x])^(5/2)*(a + a*Sec[c + d*x])),x]

[Out]

(8*Sqrt[1 - E^((2*I)*(c + d*x))]*(I + Cot[c + d*x])*Hypergeometric2F1[1/2, 3/4, 7/4, E^((2*I)*(c + d*x))] + 20
*Sin[c + d*x] - 6*(4*I + Sin[2*(c + d*x)]))/(30*a*d*e^2*Sqrt[e*Csc[c + d*x]])

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Maple [C]  time = 0.227, size = 563, normalized size = 4.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*csc(d*x+c))^(5/2)/(a+a*sec(d*x+c)),x)

[Out]

1/15/a/d*2^(1/2)*(-6*cos(d*x+c)*(-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)*((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))+12*cos(d*x+c)*(-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)*((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2)*EllipticE(((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2),1
/2*2^(1/2))+3*cos(d*x+c)^3*2^(1/2)-6*(-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))*((I*cos(d*x+c)+sin(d*x+c)-I
)/sin(d*x+c))^(1/2)+12*(-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)*E
llipticE(((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)-5*cos(d*x+c)^2*2^(1/2)+3*cos(d*x+c)*2^(1/2)-2^(1/2))/(e/sin(d*x+c))^(5/2)/sin(d*x+c)^3

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*csc(d*x+c))^(5/2)/(a+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{e \csc \left (d x + c\right )}}{a e^{3} \csc \left (d x + c\right )^{3} \sec \left (d x + c\right ) + a e^{3} \csc \left (d x + c\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*csc(d*x+c))^(5/2)/(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

integral(sqrt(e*csc(d*x + c))/(a*e^3*csc(d*x + c)^3*sec(d*x + c) + a*e^3*csc(d*x + c)^3), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*csc(d*x+c))**(5/2)/(a+a*sec(d*x+c)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (e \csc \left (d x + c\right )\right )^{\frac{5}{2}}{\left (a \sec \left (d x + c\right ) + a\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*csc(d*x+c))^(5/2)/(a+a*sec(d*x+c)),x, algorithm="giac")

[Out]

integrate(1/((e*csc(d*x + c))^(5/2)*(a*sec(d*x + c) + a)), x)