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

Optimal. Leaf size=213 \[ -\frac{4 \text{EllipticF}\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right ),2\right )}{a^2 d e \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)}}+\frac{4 \csc ^2(c+d x)}{3 a^2 d e \sqrt{e \csc (c+d x)}}+\frac{4}{a^2 d e \sqrt{e \csc (c+d x)}}-\frac{4 \cos (c+d x)}{3 a^2 d e \sqrt{e \csc (c+d x)}}-\frac{2 \cot (c+d x) \csc (c+d x)}{3 a^2 d e \sqrt{e \csc (c+d x)}}-\frac{2 \cos (c+d x) \cot ^2(c+d x)}{3 a^2 d e \sqrt{e \csc (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.475118, antiderivative size = 213, 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, 2641, 2564, 14, 2569} \[ \frac{4 \csc ^2(c+d x)}{3 a^2 d e \sqrt{e \csc (c+d x)}}+\frac{4}{a^2 d e \sqrt{e \csc (c+d x)}}-\frac{4 \cos (c+d x)}{3 a^2 d e \sqrt{e \csc (c+d x)}}-\frac{2 \cot (c+d x) \csc (c+d x)}{3 a^2 d e \sqrt{e \csc (c+d x)}}-\frac{2 \cos (c+d x) \cot ^2(c+d x)}{3 a^2 d e \sqrt{e \csc (c+d x)}}-\frac{4 F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{a^2 d e \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.536797, size = 101, normalized size = 0.47 \[ \frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) \left (12 (\cos (c+d x)+1) \text{EllipticF}\left (\frac{1}{4} (-2 c-2 d x+\pi ),2\right )+\sqrt{\sin (c+d x)} (10 \cos (c+d x)-\cos (2 (c+d x))+15)\right )}{6 a^2 d \sin ^{\frac{3}{2}}(c+d x) (e \csc (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[(c + d*x)/2]^2*(12*(1 + Cos[c + d*x])*EllipticF[(-2*c + Pi - 2*d*x)/4, 2] + (15 + 10*Cos[c + d*x] - Cos[2
*(c + d*x)])*Sqrt[Sin[c + d*x]]))/(6*a^2*d*(e*Csc[c + d*x])^(3/2)*Sin[c + d*x]^(3/2))

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Maple [C]  time = 0.221, size = 327, normalized size = 1.5 \begin{align*} -{\frac{\sqrt{2}}{3\,d{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}} \left ( 6\,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 ) +6\,i\sin \left ( dx+c \right ) \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 ) - \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sqrt{2}+6\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{2}+3\,\cos \left ( dx+c \right ) \sqrt{2}-8\,\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))^2,x)

[Out]

-1/3/a^2/d*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)/si
n(d*x+c))^(1/2),1/2*2^(1/2))+6*I*sin(d*x+c)*(-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))-cos(d*x+c)^3*2^(1/2)+6*cos(d*x+c)^2*2^(1/2)+3*cos(d*x+c)*2^(1/2)-8*2^(1/2))/(e/sin(
d*x+c))^(3/2)/sin(d*x+c)^3

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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 )}^{2}}\,{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))^2,x, algorithm="maxima")

[Out]

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

[Out]

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

[Out]

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