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

Optimal. Leaf size=155 \[ \frac{4 e^2 \sqrt{\sin (c+d x)} \text{EllipticF}\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right ),2\right ) \sqrt{e \csc (c+d x)}}{21 a d}-\frac{2 e^2 \csc ^3(c+d x) \sqrt{e \csc (c+d x)}}{7 a d}+\frac{2 e^2 \cot (c+d x) \csc ^2(c+d x) \sqrt{e \csc (c+d x)}}{7 a d}-\frac{4 e^2 \cot (c+d x) \sqrt{e \csc (c+d x)}}{21 a d} \]

[Out]

(-4*e^2*Cot[c + d*x]*Sqrt[e*Csc[c + d*x]])/(21*a*d) + (2*e^2*Cot[c + d*x]*Csc[c + d*x]^2*Sqrt[e*Csc[c + d*x]])
/(7*a*d) - (2*e^2*Csc[c + d*x]^3*Sqrt[e*Csc[c + d*x]])/(7*a*d) + (4*e^2*Sqrt[e*Csc[c + d*x]]*EllipticF[(c - Pi
/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(21*a*d)

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Rubi [A]  time = 0.223577, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {3878, 3872, 2839, 2564, 30, 2567, 2636, 2641} \[ -\frac{2 e^2 \csc ^3(c+d x) \sqrt{e \csc (c+d x)}}{7 a d}+\frac{2 e^2 \cot (c+d x) \csc ^2(c+d x) \sqrt{e \csc (c+d x)}}{7 a d}-\frac{4 e^2 \cot (c+d x) \sqrt{e \csc (c+d x)}}{21 a d}+\frac{4 e^2 \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \csc (c+d x)}}{21 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*e^2*Cot[c + d*x]*Sqrt[e*Csc[c + d*x]])/(21*a*d) + (2*e^2*Cot[c + d*x]*Csc[c + d*x]^2*Sqrt[e*Csc[c + d*x]])
/(7*a*d) - (2*e^2*Csc[c + d*x]^3*Sqrt[e*Csc[c + d*x]])/(7*a*d) + (4*e^2*Sqrt[e*Csc[c + d*x]]*EllipticF[(c - Pi
/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(21*a*d)

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 2567

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 2636

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

Rubi steps

\begin{align*} \int \frac{(e \csc (c+d x))^{5/2}}{a+a \sec (c+d x)} \, dx &=\left (e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \int \frac{1}{(a+a \sec (c+d x)) \sin ^{\frac{5}{2}}(c+d x)} \, dx\\ &=-\left (\left (e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \int \frac{\cos (c+d x)}{(-a-a \cos (c+d x)) \sin ^{\frac{5}{2}}(c+d x)} \, dx\right )\\ &=\frac{\left (e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \int \frac{\cos (c+d x)}{\sin ^{\frac{9}{2}}(c+d x)} \, dx}{a}-\frac{\left (e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \int \frac{\cos ^2(c+d x)}{\sin ^{\frac{9}{2}}(c+d x)} \, dx}{a}\\ &=\frac{2 e^2 \cot (c+d x) \csc ^2(c+d x) \sqrt{e \csc (c+d x)}}{7 a d}+\frac{\left (2 e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sin ^{\frac{5}{2}}(c+d x)} \, dx}{7 a}+\frac{\left (e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{x^{9/2}} \, dx,x,\sin (c+d x)\right )}{a d}\\ &=-\frac{4 e^2 \cot (c+d x) \sqrt{e \csc (c+d x)}}{21 a d}+\frac{2 e^2 \cot (c+d x) \csc ^2(c+d x) \sqrt{e \csc (c+d x)}}{7 a d}-\frac{2 e^2 \csc ^3(c+d x) \sqrt{e \csc (c+d x)}}{7 a d}+\frac{\left (2 e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\sin (c+d x)}} \, dx}{21 a}\\ &=-\frac{4 e^2 \cot (c+d x) \sqrt{e \csc (c+d x)}}{21 a d}+\frac{2 e^2 \cot (c+d x) \csc ^2(c+d x) \sqrt{e \csc (c+d x)}}{7 a d}-\frac{2 e^2 \csc ^3(c+d x) \sqrt{e \csc (c+d x)}}{7 a d}+\frac{4 e^2 \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)}}{21 a d}\\ \end{align*}

Mathematica [A]  time = 0.998386, size = 131, normalized size = 0.85 \[ -\frac{\sin ^{\frac{5}{2}}(c+d x) \csc ^2\left (\frac{1}{2} (c+d x)\right ) \sec ^4\left (\frac{1}{2} (c+d x)\right ) (e \csc (c+d x))^{5/2} \left ((\cos (c+d x)-2 \cos (2 (c+d x))-\cos (3 (c+d x))+2) \text{EllipticF}\left (\frac{1}{4} (-2 c-2 d x+\pi ),2\right )+2 \sqrt{\sin (c+d x)} (2 \cos (c+d x)+\cos (2 (c+d x))+4)\right )}{168 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Csc[(c + d*x)/2]^2*(e*Csc[c + d*x])^(5/2)*Sec[(c + d*x)/2]^4*((2 + Cos[c + d*x] - 2*Cos[2*(c + d*x)] - Cos[3
*(c + d*x)])*EllipticF[(-2*c + Pi - 2*d*x)/4, 2] + 2*(4 + 2*Cos[c + d*x] + Cos[2*(c + d*x)])*Sqrt[Sin[c + d*x]
])*Sin[c + d*x]^(5/2))/(168*a*d)

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Maple [C]  time = 0.224, size = 465, normalized size = 3. \begin{align*} -{\frac{\sqrt{2} \left ( -1+\cos \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2}}{21\,da \left ( \sin \left ( dx+c \right ) \right ) ^{5}} \left ( 2\,i\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{{\frac{-i \left ( -1+\cos \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }}}\sqrt{{\frac{i\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) -i}{\sin \left ( dx+c \right ) }}}\sqrt{{\frac{-i\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) +i}{\sin \left ( dx+c \right ) }}}{\it EllipticF} \left ( \sqrt{{\frac{i\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) -i}{\sin \left ( dx+c \right ) }}},{\frac{\sqrt{2}}{2}} \right ) +4\,i\sqrt{{\frac{-i\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) +i}{\sin \left ( dx+c \right ) }}}\sqrt{{\frac{i\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) -i}{\sin \left ( dx+c \right ) }}}\sqrt{{\frac{-i \left ( -1+\cos \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }}}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ){\it EllipticF} \left ( \sqrt{{\frac{i\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) -i}{\sin \left ( dx+c \right ) }}},{\frac{\sqrt{2}}{2}} \right ) +2\,i{\it EllipticF} \left ( \sqrt{{\frac{i\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) -i}{\sin \left ( dx+c \right ) }}},{\frac{\sqrt{2}}{2}} \right ) \sqrt{{\frac{-i\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) +i}{\sin \left ( dx+c \right ) }}}\sqrt{{\frac{i\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) -i}{\sin \left ( dx+c \right ) }}}\sqrt{{\frac{-i \left ( -1+\cos \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }}}\sin \left ( dx+c \right ) -2\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{2}-2\,\cos \left ( dx+c \right ) \sqrt{2}-3\,\sqrt{2} \right ) \left ({\frac{e}{\sin \left ( dx+c \right ) }} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/21/a/d*2^(1/2)*(-1+cos(d*x+c))^3*(2*I*sin(d*x+c)*cos(d*x+c)^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)*((-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))+4*I*((-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)*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)*cos(d*x+c)*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*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)*sin(d*x+c)-2*cos(d*x+c)^2*2^(1/2)-2*cos(d*x+c)*2^(1/2)-3*2^(1/2))*
(cos(d*x+c)+1)^2*(e/sin(d*x+c))^(5/2)/sin(d*x+c)^5

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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((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 )} e^{2} \csc \left (d x + c\right )^{2}}{a \sec \left (d x + c\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(e*csc(d*x + c))*e^2*csc(d*x + c)^2/(a*sec(d*x + c) + a), 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((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{\left (e \csc \left (d x + c\right )\right )^{\frac{5}{2}}}{a \sec \left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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