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

Optimal. Leaf size=215 \[ \frac{4 \csc (c+d x)}{a^2 d e^2 \sqrt{e \csc (c+d x)}}-\frac{2 \cot (c+d x)}{a^2 d e^2 \sqrt{e \csc (c+d x)}}+\frac{4 \sin (c+d x)}{3 a^2 d e^2 \sqrt{e \csc (c+d x)}}-\frac{2 \cos ^2(c+d x) \cot (c+d x)}{a^2 d e^2 \sqrt{e \csc (c+d x)}}-\frac{12 \sin (c+d x) \cos (c+d x)}{5 a^2 d e^2 \sqrt{e \csc (c+d x)}}-\frac{44 E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{5 a^2 d e^2 \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)}} \]

[Out]

(-2*Cot[c + d*x])/(a^2*d*e^2*Sqrt[e*Csc[c + d*x]]) - (2*Cos[c + d*x]^2*Cot[c + d*x])/(a^2*d*e^2*Sqrt[e*Csc[c +
d*x]]) + (4*Csc[c + d*x])/(a^2*d*e^2*Sqrt[e*Csc[c + d*x]]) - (44*EllipticE[(c - Pi/2 + d*x)/2, 2])/(5*a^2*d*e
^2*Sqrt[e*Csc[c + d*x]]*Sqrt[Sin[c + d*x]]) + (4*Sin[c + d*x])/(3*a^2*d*e^2*Sqrt[e*Csc[c + d*x]]) - (12*Cos[c
+ d*x]*Sin[c + d*x])/(5*a^2*d*e^2*Sqrt[e*Csc[c + d*x]])

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Rubi [A]  time = 0.471265, antiderivative size = 215, normalized size of antiderivative = 1., number of steps used = 13, 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, 2639, 2564, 14, 2569} \[ \frac{4 \csc (c+d x)}{a^2 d e^2 \sqrt{e \csc (c+d x)}}-\frac{2 \cot (c+d x)}{a^2 d e^2 \sqrt{e \csc (c+d x)}}+\frac{4 \sin (c+d x)}{3 a^2 d e^2 \sqrt{e \csc (c+d x)}}-\frac{2 \cos ^2(c+d x) \cot (c+d x)}{a^2 d e^2 \sqrt{e \csc (c+d x)}}-\frac{12 \sin (c+d x) \cos (c+d x)}{5 a^2 d e^2 \sqrt{e \csc (c+d x)}}-\frac{44 E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{5 a^2 d e^2 \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cot[c + d*x])/(a^2*d*e^2*Sqrt[e*Csc[c + d*x]]) - (2*Cos[c + d*x]^2*Cot[c + d*x])/(a^2*d*e^2*Sqrt[e*Csc[c +
d*x]]) + (4*Csc[c + d*x])/(a^2*d*e^2*Sqrt[e*Csc[c + d*x]]) - (44*EllipticE[(c - Pi/2 + d*x)/2, 2])/(5*a^2*d*e
^2*Sqrt[e*Csc[c + d*x]]*Sqrt[Sin[c + d*x]]) + (4*Sin[c + d*x])/(3*a^2*d*e^2*Sqrt[e*Csc[c + d*x]]) - (12*Cos[c
+ d*x]*Sin[c + d*x])/(5*a^2*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 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 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]

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]]

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]

Rubi steps

\begin{align*} \int \frac{1}{(e \csc (c+d x))^{5/2} (a+a \sec (c+d x))^2} \, dx &=\frac{\int \frac{\sin ^{\frac{5}{2}}(c+d x)}{(a+a \sec (c+d x))^2} \, dx}{e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}\\ &=\frac{\int \frac{\cos ^2(c+d x) \sin ^{\frac{5}{2}}(c+d x)}{(-a-a \cos (c+d x))^2} \, dx}{e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}\\ &=\frac{\int \frac{\cos ^2(c+d x) (-a+a \cos (c+d x))^2}{\sin ^{\frac{3}{2}}(c+d x)} \, dx}{a^4 e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}\\ &=\frac{\int \left (\frac{a^2 \cos ^2(c+d x)}{\sin ^{\frac{3}{2}}(c+d x)}-\frac{2 a^2 \cos ^3(c+d x)}{\sin ^{\frac{3}{2}}(c+d x)}+\frac{a^2 \cos ^4(c+d x)}{\sin ^{\frac{3}{2}}(c+d x)}\right ) \, dx}{a^4 e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}\\ &=\frac{\int \frac{\cos ^2(c+d x)}{\sin ^{\frac{3}{2}}(c+d x)} \, dx}{a^2 e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}+\frac{\int \frac{\cos ^4(c+d x)}{\sin ^{\frac{3}{2}}(c+d x)} \, dx}{a^2 e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}-\frac{2 \int \frac{\cos ^3(c+d x)}{\sin ^{\frac{3}{2}}(c+d x)} \, dx}{a^2 e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}\\ &=-\frac{2 \cot (c+d x)}{a^2 d e^2 \sqrt{e \csc (c+d x)}}-\frac{2 \cos ^2(c+d x) \cot (c+d x)}{a^2 d e^2 \sqrt{e \csc (c+d x)}}-\frac{2 \int \sqrt{\sin (c+d x)} \, dx}{a^2 e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}-\frac{6 \int \cos ^2(c+d x) \sqrt{\sin (c+d x)} \, dx}{a^2 e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}-\frac{2 \operatorname{Subst}\left (\int \frac{1-x^2}{x^{3/2}} \, dx,x,\sin (c+d x)\right )}{a^2 d e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}\\ &=-\frac{2 \cot (c+d x)}{a^2 d e^2 \sqrt{e \csc (c+d x)}}-\frac{2 \cos ^2(c+d x) \cot (c+d x)}{a^2 d e^2 \sqrt{e \csc (c+d x)}}-\frac{4 E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right )}{a^2 d e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}-\frac{12 \cos (c+d x) \sin (c+d x)}{5 a^2 d e^2 \sqrt{e \csc (c+d x)}}-\frac{12 \int \sqrt{\sin (c+d x)} \, dx}{5 a^2 e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}-\frac{2 \operatorname{Subst}\left (\int \left (\frac{1}{x^{3/2}}-\sqrt{x}\right ) \, dx,x,\sin (c+d x)\right )}{a^2 d e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}\\ &=-\frac{2 \cot (c+d x)}{a^2 d e^2 \sqrt{e \csc (c+d x)}}-\frac{2 \cos ^2(c+d x) \cot (c+d x)}{a^2 d e^2 \sqrt{e \csc (c+d x)}}+\frac{4 \csc (c+d x)}{a^2 d e^2 \sqrt{e \csc (c+d x)}}-\frac{44 E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right )}{5 a^2 d e^2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}+\frac{4 \sin (c+d x)}{3 a^2 d e^2 \sqrt{e \csc (c+d x)}}-\frac{12 \cos (c+d x) \sin (c+d x)}{5 a^2 d e^2 \sqrt{e \csc (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 2.13475, size = 125, normalized size = 0.58 \[ \frac{88 \sqrt{1-e^{2 i (c+d x)}} (\cot (c+d x)+i) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},e^{2 i (c+d x)}\right )-123 \cot (c+d x)+\csc (c+d x) (-264 i \sin (c+d x)-20 \cos (2 (c+d x))+3 \cos (3 (c+d x))+140)}{30 a^2 d e^2 \sqrt{e \csc (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-123*Cot[c + d*x] + 88*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))] + Csc[c + d*x]*(140 - 20*Cos[2*(c + d*x)] + 3*Cos[3*(c + d*x)] - (264*I)*Sin[c + d*x]))/(30*a
^2*d*e^2*Sqrt[e*Csc[c + d*x]])

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Maple [C]  time = 0.239, size = 551, normalized size = 2.6 \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))^2,x)

[Out]

1/15/a^2/d*2^(1/2)*(132*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)*((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))-66*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)*((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))+3*cos(d*x+c)^3*2^(1/2)+132*((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2)*(-I*(-1+cos(d*x+c))/s
in(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))-66*((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))-10*cos(d*x+c)^2*2^(1/2)+33*cos(d*x+c)*2^(1/2)-26*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))^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 )}}{a^{2} e^{3} \csc \left (d x + c\right )^{3} \sec \left (d x + c\right )^{2} + 2 \, a^{2} e^{3} \csc \left (d x + c\right )^{3} \sec \left (d x + c\right ) + a^{2} 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))^2,x, algorithm="fricas")

[Out]

integral(sqrt(e*csc(d*x + c))/(a^2*e^3*csc(d*x + c)^3*sec(d*x + c)^2 + 2*a^2*e^3*csc(d*x + c)^3*sec(d*x + c) +
a^2*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))**2,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 )}^{2}}\,{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))^2,x, algorithm="giac")

[Out]

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