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

Optimal. Leaf size=250 \[ \frac {4 e \csc ^4(c+d x) \sqrt {e \csc (c+d x)}}{9 a^2 d}-\frac {4 e \csc ^2(c+d x) \sqrt {e \csc (c+d x)}}{5 a^2 d}-\frac {4 e \cos (c+d x) \sqrt {e \csc (c+d x)}}{15 a^2 d}-\frac {2 e \cot ^3(c+d x) \csc (c+d x) \sqrt {e \csc (c+d x)}}{9 a^2 d}-\frac {2 e \cot (c+d x) \csc ^3(c+d x) \sqrt {e \csc (c+d x)}}{9 a^2 d}+\frac {16 e \cot (c+d x) \csc (c+d x) \sqrt {e \csc (c+d x)}}{45 a^2 d}-\frac {4 e \sqrt {\sin (c+d x)} E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \csc (c+d x)}}{15 a^2 d} \]

[Out]

-4/15*e*cos(d*x+c)*(e*csc(d*x+c))^(1/2)/a^2/d+16/45*e*cot(d*x+c)*csc(d*x+c)*(e*csc(d*x+c))^(1/2)/a^2/d-2/9*e*c
ot(d*x+c)^3*csc(d*x+c)*(e*csc(d*x+c))^(1/2)/a^2/d-4/5*e*csc(d*x+c)^2*(e*csc(d*x+c))^(1/2)/a^2/d-2/9*e*cot(d*x+
c)*csc(d*x+c)^3*(e*csc(d*x+c))^(1/2)/a^2/d+4/9*e*csc(d*x+c)^4*(e*csc(d*x+c))^(1/2)/a^2/d+4/15*e*(sin(1/2*c+1/4
*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*(e*csc(d*x+c))^(1
/2)*sin(d*x+c)^(1/2)/a^2/d

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Rubi [A]  time = 0.49, antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3878, 3872, 2875, 2873, 2567, 2636, 2639, 2564, 14} \[ \frac {4 e \csc ^4(c+d x) \sqrt {e \csc (c+d x)}}{9 a^2 d}-\frac {4 e \csc ^2(c+d x) \sqrt {e \csc (c+d x)}}{5 a^2 d}-\frac {4 e \cos (c+d x) \sqrt {e \csc (c+d x)}}{15 a^2 d}-\frac {2 e \cot ^3(c+d x) \csc (c+d x) \sqrt {e \csc (c+d x)}}{9 a^2 d}-\frac {2 e \cot (c+d x) \csc ^3(c+d x) \sqrt {e \csc (c+d x)}}{9 a^2 d}+\frac {16 e \cot (c+d x) \csc (c+d x) \sqrt {e \csc (c+d x)}}{45 a^2 d}-\frac {4 e \sqrt {\sin (c+d x)} E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \csc (c+d x)}}{15 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*e*Cos[c + d*x]*Sqrt[e*Csc[c + d*x]])/(15*a^2*d) + (16*e*Cot[c + d*x]*Csc[c + d*x]*Sqrt[e*Csc[c + d*x]])/(4
5*a^2*d) - (2*e*Cot[c + d*x]^3*Csc[c + d*x]*Sqrt[e*Csc[c + d*x]])/(9*a^2*d) - (4*e*Csc[c + d*x]^2*Sqrt[e*Csc[c
+ d*x]])/(5*a^2*d) - (2*e*Cot[c + d*x]*Csc[c + d*x]^3*Sqrt[e*Csc[c + d*x]])/(9*a^2*d) + (4*e*Csc[c + d*x]^4*S
qrt[e*Csc[c + d*x]])/(9*a^2*d) - (4*e*Sqrt[e*Csc[c + d*x]]*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]]
)/(15*a^2*d)

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 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 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]

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 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 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 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]

Rubi steps

\begin {align*} \int \frac {(e \csc (c+d x))^{3/2}}{(a+a \sec (c+d x))^2} \, dx &=\left (e \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {1}{(a+a \sec (c+d x))^2 \sin ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\left (e \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 {3}{2}}(c+d x)} \, dx\\ &=\frac {\left (e \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 {11}{2}}(c+d x)} \, dx}{a^4}\\ &=\frac {\left (e \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \left (\frac {a^2 \cos ^2(c+d x)}{\sin ^{\frac {11}{2}}(c+d x)}-\frac {2 a^2 \cos ^3(c+d x)}{\sin ^{\frac {11}{2}}(c+d x)}+\frac {a^2 \cos ^4(c+d x)}{\sin ^{\frac {11}{2}}(c+d x)}\right ) \, dx}{a^4}\\ &=\frac {\left (e \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {\cos ^2(c+d x)}{\sin ^{\frac {11}{2}}(c+d x)} \, dx}{a^2}+\frac {\left (e \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {\cos ^4(c+d x)}{\sin ^{\frac {11}{2}}(c+d x)} \, dx}{a^2}-\frac {\left (2 e \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {\cos ^3(c+d x)}{\sin ^{\frac {11}{2}}(c+d x)} \, dx}{a^2}\\ &=-\frac {2 e \cot ^3(c+d x) \csc (c+d x) \sqrt {e \csc (c+d x)}}{9 a^2 d}-\frac {2 e \cot (c+d x) \csc ^3(c+d x) \sqrt {e \csc (c+d x)}}{9 a^2 d}-\frac {\left (2 e \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sin ^{\frac {7}{2}}(c+d x)} \, dx}{9 a^2}-\frac {\left (2 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}{3 a^2}-\frac {\left (2 e \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1-x^2}{x^{11/2}} \, dx,x,\sin (c+d x)\right )}{a^2 d}\\ &=\frac {16 e \cot (c+d x) \csc (c+d x) \sqrt {e \csc (c+d x)}}{45 a^2 d}-\frac {2 e \cot ^3(c+d x) \csc (c+d x) \sqrt {e \csc (c+d x)}}{9 a^2 d}-\frac {2 e \cot (c+d x) \csc ^3(c+d x) \sqrt {e \csc (c+d x)}}{9 a^2 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}{15 a^2}+\frac {\left (4 e \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sin ^{\frac {3}{2}}(c+d x)} \, dx}{15 a^2}-\frac {\left (2 e \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{x^{11/2}}-\frac {1}{x^{7/2}}\right ) \, dx,x,\sin (c+d x)\right )}{a^2 d}\\ &=-\frac {4 e \cos (c+d x) \sqrt {e \csc (c+d x)}}{15 a^2 d}+\frac {16 e \cot (c+d x) \csc (c+d x) \sqrt {e \csc (c+d x)}}{45 a^2 d}-\frac {2 e \cot ^3(c+d x) \csc (c+d x) \sqrt {e \csc (c+d x)}}{9 a^2 d}-\frac {4 e \csc ^2(c+d x) \sqrt {e \csc (c+d x)}}{5 a^2 d}-\frac {2 e \cot (c+d x) \csc ^3(c+d x) \sqrt {e \csc (c+d x)}}{9 a^2 d}+\frac {4 e \csc ^4(c+d x) \sqrt {e \csc (c+d x)}}{9 a^2 d}+\frac {\left (2 e \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \sqrt {\sin (c+d x)} \, dx}{15 a^2}-\frac {\left (4 e \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \sqrt {\sin (c+d x)} \, dx}{15 a^2}\\ &=-\frac {4 e \cos (c+d x) \sqrt {e \csc (c+d x)}}{15 a^2 d}+\frac {16 e \cot (c+d x) \csc (c+d x) \sqrt {e \csc (c+d x)}}{45 a^2 d}-\frac {2 e \cot ^3(c+d x) \csc (c+d x) \sqrt {e \csc (c+d x)}}{9 a^2 d}-\frac {4 e \csc ^2(c+d x) \sqrt {e \csc (c+d x)}}{5 a^2 d}-\frac {2 e \cot (c+d x) \csc ^3(c+d x) \sqrt {e \csc (c+d x)}}{9 a^2 d}+\frac {4 e \csc ^4(c+d x) \sqrt {e \csc (c+d x)}}{9 a^2 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)}}{15 a^2 d}\\ \end {align*}

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Mathematica [C]  time = 1.83, size = 247, normalized size = 0.99 \[ \frac {\cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) (e \csc (c+d x))^{3/2} \left (-\frac {2 \tan (c+d x) \left ((13 \cos (c+d x)+8) \sec ^4\left (\frac {1}{2} (c+d x)\right )+24 \sec (c) \cos (d x)\right )}{d}+\frac {16 \sqrt {2} e^{i (c-d x)} \sqrt {\frac {i e^{i (c+d x)}}{-1+e^{2 i (c+d x)}}} \left (\left (1+e^{2 i c}\right ) e^{2 i d x} \sqrt {1-e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};e^{2 i (c+d x)}\right )-3 e^{2 i (c+d x)}+3\right ) \sec (c+d x)}{\left (1+e^{2 i c}\right ) d \csc ^{\frac {3}{2}}(c+d x)}\right )}{45 a^2 (\sec (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^4*(e*Csc[c + d*x])^(3/2)*Sec[c + d*x]*((16*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))]*Hypergeometric2F1[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)) - (2*(24*Cos[d*x]*Sec[c] + (8 + 13*Cos[c + d*x])*Sec[(c + d*x)/2]^4)*Tan[c + d*x])/d))/(45*a^2*(1 + S
ec[c + d*x])^2)

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e \csc \left (d x + c\right )} e \csc \left (d x + c\right )}{a^{2} \sec \left (d x + c\right )^{2} + 2 \, a^{2} \sec \left (d x + c\right ) + a^{2}}, x\right ) \]

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))^2,x, algorithm="fricas")

[Out]

integral(sqrt(e*csc(d*x + c))*e*csc(d*x + c)/(a^2*sec(d*x + c)^2 + 2*a^2*sec(d*x + c) + a^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \csc \left (d x + c\right )\right )^{\frac {3}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

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))^2,x, algorithm="giac")

[Out]

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

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maple [C]  time = 1.34, size = 1044, normalized size = 4.18 \[ \text {result too large to display} \]

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))^2,x)

[Out]

1/45/a^2/d*(-1+cos(d*x+c))^2*(12*((I*cos(d*x+c)-I+sin(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)-I+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*cos(d*x+c)^3*(-I*(-1+c
os(d*x+c))/sin(d*x+c))^(1/2)-6*((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*((-I*cos(d*x+c)+sin(d*x+c)+I)/si
n(d*x+c))^(1/2)*EllipticF(((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*cos(d*x+c)^3*(-I*(-1+cos
(d*x+c))/sin(d*x+c))^(1/2)+36*((I*cos(d*x+c)-I+sin(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)-I+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*cos(d*x+c)^2*(-I*(-1+cos(
d*x+c))/sin(d*x+c))^(1/2)-18*((I*cos(d*x+c)-I+sin(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)-I+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*cos(d*x+c)^2*(-I*(-1+cos(d
*x+c))/sin(d*x+c))^(1/2)+36*cos(d*x+c)*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((I*cos(d*x+c)-I+sin(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)-I+sin(d*x+c))/sin(d*x+c
))^(1/2),1/2*2^(1/2))-18*cos(d*x+c)*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((I*cos(d*x+c)-I+sin(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)-I+sin(d*x+c))/sin(d*x+c))^
(1/2),1/2*2^(1/2))+12*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((I*cos(d*x+c)-I+sin(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)-I+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1
/2))-6*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*((-I*cos(d*x+c)+si
n(d*x+c)+I)/sin(d*x+c))^(1/2)*EllipticF(((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))-6*cos(d*x+
c)^2*2^(1/2)-25*cos(d*x+c)*2^(1/2)-14*2^(1/2))*(e/sin(d*x+c))^(3/2)/sin(d*x+c)^3*2^(1/2)

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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))^2,x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^2\,{\left (\frac {e}{\sin \left (c+d\,x\right )}\right )}^{3/2}}{a^2\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e/sin(c + d*x))^(3/2)/(a + a/cos(c + d*x))^2,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\left (e \csc {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

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))**2,x)

[Out]

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

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