3.297 \(\int \frac{1}{(e \csc (c+d x))^{3/2} (a+a \sec (c+d x))} \, dx\)

Optimal. Leaf size=106 \[ -\frac{4 \text{EllipticF}\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right ),2\right )}{3 a d e \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)}}+\frac{2}{a d e \sqrt{e \csc (c+d x)}}-\frac{2 \cos (c+d x)}{3 a d e \sqrt{e \csc (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.227105, antiderivative size = 106, 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, 2641} \[ \frac{2}{a d e \sqrt{e \csc (c+d x)}}-\frac{2 \cos (c+d x)}{3 a d e \sqrt{e \csc (c+d x)}}-\frac{4 F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{3 a d e \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/((e*Csc[c + d*x])^(3/2)*(a + a*Sec[c + d*x])),x]

[Out]

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

Mathematica [A]  time = 0.374567, size = 70, normalized size = 0.66 \[ \frac{4 \text{EllipticF}\left (\frac{1}{4} (-2 c-2 d x+\pi ),2\right )-2 \sqrt{\sin (c+d x)} (\cos (c+d x)-3)}{3 a d \sin ^{\frac{3}{2}}(c+d x) (e \csc (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((e*Csc[c + d*x])^(3/2)*(a + a*Sec[c + d*x])),x]

[Out]

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

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Maple [C]  time = 0.196, size = 195, normalized size = 1.8 \begin{align*}{\frac{\sqrt{2}}{3\,da \left ( -1+\cos \left ( dx+c \right ) \right ) \sin \left ( dx+c \right ) } \left ( 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 ) - \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{2}+4\,\cos \left ( dx+c \right ) \sqrt{2}-3\,\sqrt{2} \right ) \left ({\frac{e}{\sin \left ( dx+c \right ) }} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*csc(d*x+c))^(3/2)/(a+a*sec(d*x+c)),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (e \csc \left (d x + c\right )\right )^{\frac{3}{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))^(3/2)/(a+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

integrate(1/((e*csc(d*x + c))^(3/2)*(a*sec(d*x + c) + a)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*csc(d*x+c))^(3/2)/(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

integral(sqrt(e*csc(d*x + c))/(a*e^2*csc(d*x + c)^2*sec(d*x + c) + a*e^2*csc(d*x + c)^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(1/(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{1}{\left (e \csc \left (d x + c\right )\right )^{\frac{3}{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))^(3/2)/(a+a*sec(d*x+c)),x, algorithm="giac")

[Out]

integrate(1/((e*csc(d*x + c))^(3/2)*(a*sec(d*x + c) + a)), x)