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

Optimal. Leaf size=145 \[ -\frac{2 e \csc ^2(c+d x) \sqrt{e \csc (c+d x)}}{5 a d}-\frac{4 e \cos (c+d x) \sqrt{e \csc (c+d x)}}{5 a d}+\frac{2 e \cot (c+d x) \csc (c+d x) \sqrt{e \csc (c+d x)}}{5 a d}-\frac{4 e \sqrt{\sin (c+d x)} E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \csc (c+d x)}}{5 a d} \]

[Out]

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

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Rubi [A]  time = 0.229357, antiderivative size = 145, 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, 2639} \[ -\frac{2 e \csc ^2(c+d x) \sqrt{e \csc (c+d x)}}{5 a d}-\frac{4 e \cos (c+d x) \sqrt{e \csc (c+d x)}}{5 a d}+\frac{2 e \cot (c+d x) \csc (c+d x) \sqrt{e \csc (c+d x)}}{5 a d}-\frac{4 e \sqrt{\sin (c+d x)} E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \csc (c+d x)}}{5 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*e*Cos[c + d*x]*Sqrt[e*Csc[c + d*x]])/(5*a*d) + (2*e*Cot[c + d*x]*Csc[c + d*x]*Sqrt[e*Csc[c + d*x]])/(5*a*d
) - (2*e*Csc[c + d*x]^2*Sqrt[e*Csc[c + d*x]])/(5*a*d) - (4*e*Sqrt[e*Csc[c + d*x]]*EllipticE[(c - Pi/2 + d*x)/2
, 2]*Sqrt[Sin[c + d*x]])/(5*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 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{(e \csc (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx &=\left (e \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \int \frac{1}{(a+a \sec (c+d x)) \sin ^{\frac{3}{2}}(c+d x)} \, dx\\ &=-\left (\left (e \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{3}{2}}(c+d x)} \, dx\right )\\ &=\frac{\left (e \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \int \frac{\cos (c+d x)}{\sin ^{\frac{7}{2}}(c+d x)} \, dx}{a}-\frac{\left (e \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \int \frac{\cos ^2(c+d x)}{\sin ^{\frac{7}{2}}(c+d x)} \, dx}{a}\\ &=\frac{2 e \cot (c+d x) \csc (c+d x) \sqrt{e \csc (c+d x)}}{5 a d}+\frac{\left (2 e \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sin ^{\frac{3}{2}}(c+d x)} \, dx}{5 a}+\frac{\left (e \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{x^{7/2}} \, dx,x,\sin (c+d x)\right )}{a d}\\ &=-\frac{4 e \cos (c+d x) \sqrt{e \csc (c+d x)}}{5 a d}+\frac{2 e \cot (c+d x) \csc (c+d x) \sqrt{e \csc (c+d x)}}{5 a d}-\frac{2 e \csc ^2(c+d x) \sqrt{e \csc (c+d x)}}{5 a d}-\frac{\left (2 e \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}\right ) \int \sqrt{\sin (c+d x)} \, dx}{5 a}\\ &=-\frac{4 e \cos (c+d x) \sqrt{e \csc (c+d x)}}{5 a d}+\frac{2 e \cot (c+d x) \csc (c+d x) \sqrt{e \csc (c+d x)}}{5 a d}-\frac{2 e \csc ^2(c+d x) \sqrt{e \csc (c+d x)}}{5 a d}-\frac{4 e \sqrt{e \csc (c+d x)} E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{5 a d}\\ \end{align*}

Mathematica [C]  time = 1.33974, size = 230, normalized size = 1.59 \[ \frac{\cos ^2\left (\frac{1}{2} (c+d x)\right ) (e \csc (c+d x))^{3/2} \left (-\frac{6 \tan (c+d x) \left (\sec ^2\left (\frac{1}{2} (c+d x)\right )+4 \sec (c) \cos (d x)\right )}{d}+\frac{8 \sqrt{2} e^{i (c-d x)} \sqrt{\frac{i e^{i (c+d x)}}{-1+e^{2 i (c+d x)}}} \sec (c+d x) \left (\left (1+e^{2 i c}\right ) e^{2 i d x} \sqrt{1-e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},e^{2 i (c+d x)}\right )-3 e^{2 i (c+d x)}+3\right )}{\left (1+e^{2 i c}\right ) d \csc ^{\frac{3}{2}}(c+d x)}\right )}{15 a (\sec (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^2*(e*Csc[c + d*x])^(3/2)*((8*Sqrt[2]*E^(I*(c - d*x))*Sqrt[(I*E^(I*(c + d*x)))/(-1 + E^((2*I)
*(c + d*x)))]*(3 - 3*E^((2*I)*(c + d*x)) + E^((2*I)*d*x)*(1 + E^((2*I)*c))*Sqrt[1 - E^((2*I)*(c + d*x))]*Hyper
geometric2F1[1/2, 3/4, 7/4, E^((2*I)*(c + d*x))])*Sec[c + d*x])/(d*(1 + E^((2*I)*c))*Csc[c + d*x]^(3/2)) - (6*
(4*Cos[d*x]*Sec[c] + Sec[(c + d*x)/2]^2)*Tan[c + d*x])/d))/(15*a*(1 + Sec[c + d*x]))

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Maple [C]  time = 0.21, size = 781, normalized size = 5.4 \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))^(3/2)/(a+a*sec(d*x+c)),x)

[Out]

-1/5/a/d*2^(1/2)*(-1+cos(d*x+c))*(4*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)*EllipticE(((I*cos(d*x+c)+sin(d*x+c)-I)
/sin(d*x+c))^(1/2),1/2*2^(1/2))-2*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)/s
in(d*x+c))^(1/2),1/2*2^(1/2))+8*(-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))-4*(-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))+4*((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)*EllipticE(((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2),1/2*2^
(1/2))-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)*((-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))-2*cos(d*
x+c)*2^(1/2)-3*2^(1/2))*(e/sin(d*x+c))^(3/2)/sin(d*x+c)

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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))^(3/2)/(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 \csc \left (d x + c\right )}{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))^(3/2)/(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

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

[Out]

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