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

Optimal. Leaf size=268 \[ \frac{4 e^2 \sqrt{\sin (c+d x)} \text{EllipticF}\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right ),2\right ) \sqrt{e \csc (c+d x)}}{231 a^2 d}+\frac{4 e^2 \csc ^5(c+d x) \sqrt{e \csc (c+d x)}}{11 a^2 d}-\frac{4 e^2 \csc ^3(c+d x) \sqrt{e \csc (c+d x)}}{7 a^2 d}-\frac{2 e^2 \cot ^3(c+d x) \csc ^2(c+d x) \sqrt{e \csc (c+d x)}}{11 a^2 d}-\frac{2 e^2 \cot (c+d x) \csc ^4(c+d x) \sqrt{e \csc (c+d x)}}{11 a^2 d}+\frac{16 e^2 \cot (c+d x) \csc ^2(c+d x) \sqrt{e \csc (c+d x)}}{77 a^2 d}-\frac{4 e^2 \cot (c+d x) \sqrt{e \csc (c+d x)}}{231 a^2 d} \]

[Out]

(-4*e^2*Cot[c + d*x]*Sqrt[e*Csc[c + d*x]])/(231*a^2*d) + (16*e^2*Cot[c + d*x]*Csc[c + d*x]^2*Sqrt[e*Csc[c + d*
x]])/(77*a^2*d) - (2*e^2*Cot[c + d*x]^3*Csc[c + d*x]^2*Sqrt[e*Csc[c + d*x]])/(11*a^2*d) - (4*e^2*Csc[c + d*x]^
3*Sqrt[e*Csc[c + d*x]])/(7*a^2*d) - (2*e^2*Cot[c + d*x]*Csc[c + d*x]^4*Sqrt[e*Csc[c + d*x]])/(11*a^2*d) + (4*e
^2*Csc[c + d*x]^5*Sqrt[e*Csc[c + d*x]])/(11*a^2*d) + (4*e^2*Sqrt[e*Csc[c + d*x]]*EllipticF[(c - Pi/2 + d*x)/2,
2]*Sqrt[Sin[c + d*x]])/(231*a^2*d)

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Rubi [A]  time = 0.504997, antiderivative size = 268, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, Rules used = {3878, 3872, 2875, 2873, 2567, 2636, 2641, 2564, 14} \[ \frac{4 e^2 \csc ^5(c+d x) \sqrt{e \csc (c+d x)}}{11 a^2 d}-\frac{4 e^2 \csc ^3(c+d x) \sqrt{e \csc (c+d x)}}{7 a^2 d}-\frac{2 e^2 \cot ^3(c+d x) \csc ^2(c+d x) \sqrt{e \csc (c+d x)}}{11 a^2 d}-\frac{2 e^2 \cot (c+d x) \csc ^4(c+d x) \sqrt{e \csc (c+d x)}}{11 a^2 d}+\frac{16 e^2 \cot (c+d x) \csc ^2(c+d x) \sqrt{e \csc (c+d x)}}{77 a^2 d}-\frac{4 e^2 \cot (c+d x) \sqrt{e \csc (c+d x)}}{231 a^2 d}+\frac{4 e^2 \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \csc (c+d x)}}{231 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*e^2*Cot[c + d*x]*Sqrt[e*Csc[c + d*x]])/(231*a^2*d) + (16*e^2*Cot[c + d*x]*Csc[c + d*x]^2*Sqrt[e*Csc[c + d*
x]])/(77*a^2*d) - (2*e^2*Cot[c + d*x]^3*Csc[c + d*x]^2*Sqrt[e*Csc[c + d*x]])/(11*a^2*d) - (4*e^2*Csc[c + d*x]^
3*Sqrt[e*Csc[c + d*x]])/(7*a^2*d) - (2*e^2*Cot[c + d*x]*Csc[c + d*x]^4*Sqrt[e*Csc[c + d*x]])/(11*a^2*d) + (4*e
^2*Csc[c + d*x]^5*Sqrt[e*Csc[c + d*x]])/(11*a^2*d) + (4*e^2*Sqrt[e*Csc[c + d*x]]*EllipticF[(c - Pi/2 + d*x)/2,
2]*Sqrt[Sin[c + d*x]])/(231*a^2*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 2875

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Dist[(a/g)^(2*m), Int[((g*Cos[e + f*x])^(2*m + p)*(d*Sin[e + f*x])^n)/(a - b*Sin[e +
f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 0]

Rule 2873

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x]
/; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]

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]

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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \frac{(e \csc (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, 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))^2 \sin ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\left (e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \int \frac{\cos ^2(c+d x)}{(-a-a \cos (c+d x))^2 \sin ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{\left (e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \int \frac{\cos ^2(c+d x) (-a+a \cos (c+d x))^2}{\sin ^{\frac{13}{2}}(c+d x)} \, dx}{a^4}\\ &=\frac{\left (e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \int \left (\frac{a^2 \cos ^2(c+d x)}{\sin ^{\frac{13}{2}}(c+d x)}-\frac{2 a^2 \cos ^3(c+d x)}{\sin ^{\frac{13}{2}}(c+d x)}+\frac{a^2 \cos ^4(c+d x)}{\sin ^{\frac{13}{2}}(c+d x)}\right ) \, dx}{a^4}\\ &=\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{13}{2}}(c+d x)} \, dx}{a^2}+\frac{\left (e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \int \frac{\cos ^4(c+d x)}{\sin ^{\frac{13}{2}}(c+d x)} \, dx}{a^2}-\frac{\left (2 e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \int \frac{\cos ^3(c+d x)}{\sin ^{\frac{13}{2}}(c+d x)} \, dx}{a^2}\\ &=-\frac{2 e^2 \cot ^3(c+d x) \csc ^2(c+d x) \sqrt{e \csc (c+d x)}}{11 a^2 d}-\frac{2 e^2 \cot (c+d x) \csc ^4(c+d x) \sqrt{e \csc (c+d x)}}{11 a^2 d}-\frac{\left (2 e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sin ^{\frac{9}{2}}(c+d x)} \, dx}{11 a^2}-\frac{\left (6 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}{11 a^2}-\frac{\left (2 e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1-x^2}{x^{13/2}} \, dx,x,\sin (c+d x)\right )}{a^2 d}\\ &=\frac{16 e^2 \cot (c+d x) \csc ^2(c+d x) \sqrt{e \csc (c+d x)}}{77 a^2 d}-\frac{2 e^2 \cot ^3(c+d x) \csc ^2(c+d x) \sqrt{e \csc (c+d x)}}{11 a^2 d}-\frac{2 e^2 \cot (c+d x) \csc ^4(c+d x) \sqrt{e \csc (c+d x)}}{11 a^2 d}-\frac{\left (10 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}{77 a^2}+\frac{\left (12 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}{77 a^2}-\frac{\left (2 e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{x^{13/2}}-\frac{1}{x^{9/2}}\right ) \, dx,x,\sin (c+d x)\right )}{a^2 d}\\ &=-\frac{4 e^2 \cot (c+d x) \sqrt{e \csc (c+d x)}}{231 a^2 d}+\frac{16 e^2 \cot (c+d x) \csc ^2(c+d x) \sqrt{e \csc (c+d x)}}{77 a^2 d}-\frac{2 e^2 \cot ^3(c+d x) \csc ^2(c+d x) \sqrt{e \csc (c+d x)}}{11 a^2 d}-\frac{4 e^2 \csc ^3(c+d x) \sqrt{e \csc (c+d x)}}{7 a^2 d}-\frac{2 e^2 \cot (c+d x) \csc ^4(c+d x) \sqrt{e \csc (c+d x)}}{11 a^2 d}+\frac{4 e^2 \csc ^5(c+d x) \sqrt{e \csc (c+d x)}}{11 a^2 d}-\frac{\left (10 e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\sin (c+d x)}} \, dx}{231 a^2}+\frac{\left (4 e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\sin (c+d x)}} \, dx}{77 a^2}\\ &=-\frac{4 e^2 \cot (c+d x) \sqrt{e \csc (c+d x)}}{231 a^2 d}+\frac{16 e^2 \cot (c+d x) \csc ^2(c+d x) \sqrt{e \csc (c+d x)}}{77 a^2 d}-\frac{2 e^2 \cot ^3(c+d x) \csc ^2(c+d x) \sqrt{e \csc (c+d x)}}{11 a^2 d}-\frac{4 e^2 \csc ^3(c+d x) \sqrt{e \csc (c+d x)}}{7 a^2 d}-\frac{2 e^2 \cot (c+d x) \csc ^4(c+d x) \sqrt{e \csc (c+d x)}}{11 a^2 d}+\frac{4 e^2 \csc ^5(c+d x) \sqrt{e \csc (c+d x)}}{11 a^2 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)}}{231 a^2 d}\\ \end{align*}

Mathematica [A]  time = 1.09839, size = 115, normalized size = 0.43 \[ -\frac{e^3 \csc ^2\left (\frac{1}{2} (c+d x)\right ) \sec ^6\left (\frac{1}{2} (c+d x)\right ) \left (\sin ^{\frac{11}{2}}(c+d x) \csc ^4\left (\frac{1}{2} (c+d x)\right ) \text{EllipticF}\left (\frac{1}{4} (-2 c-2 d x+\pi ),2\right )+97 \cos (c+d x)+4 \cos (2 (c+d x))+\cos (3 (c+d x))+52\right )}{3696 a^2 d \sqrt{e \csc (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(e^3*Csc[(c + d*x)/2]^2*Sec[(c + d*x)/2]^6*(52 + 97*Cos[c + d*x] + 4*Cos[2*(c + d*x)] + Cos[3*(c + d*x)] + Cs
c[(c + d*x)/2]^4*EllipticF[(-2*c + Pi - 2*d*x)/4, 2]*Sin[c + d*x]^(11/2)))/(3696*a^2*d*Sqrt[e*Csc[c + d*x]])

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Maple [C]  time = 0.247, size = 609, normalized size = 2.3 \begin{align*} \text{result too large to display} \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))^2,x)

[Out]

1/231/a^2/d*2^(1/2)*(-1+cos(d*x+c))^4*(2*I*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)*cos(d*x+c)^3*((I*c
os(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))+6*I*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)*cos(d*x
+c)^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))+6*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)^3*2^(1/2)-4*cos(d*x+c)^2*2^(1
/2)-47*cos(d*x+c)*2^(1/2)-24*2^(1/2))*(cos(d*x+c)+1)^2*(e/sin(d*x+c))^(5/2)/sin(d*x+c)^7

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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))^2,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^{2} \sec \left (d x + c\right )^{2} + 2 \, a^{2} \sec \left (d x + c\right ) + a^{2}}, 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))^2,x, algorithm="fricas")

[Out]

integral(sqrt(e*csc(d*x + c))*e^2*csc(d*x + c)^2/(a^2*sec(d*x + c)^2 + 2*a^2*sec(d*x + c) + a^2), 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))**2,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}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}\,{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))^2,x, algorithm="giac")

[Out]

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