3.3.89 \(\int \sqrt {e \csc (c+d x)} (a+a \sec (c+d x))^2 \, dx\) [289]

Optimal. Leaf size=154 \[ \frac {2 a^2 \text {ArcTan}\left (\sqrt {\sin (c+d x)}\right ) \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}{d}+\frac {2 a^2 \tanh ^{-1}\left (\sqrt {\sin (c+d x)}\right ) \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}{d}+\frac {3 a^2 \sqrt {e \csc (c+d x)} F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{d}+\frac {a^2 \sqrt {e \csc (c+d x)} \tan (c+d x)}{d} \]

[Out]

2*a^2*arctan(sin(d*x+c)^(1/2))*(e*csc(d*x+c))^(1/2)*sin(d*x+c)^(1/2)/d+2*a^2*arctanh(sin(d*x+c)^(1/2))*(e*csc(
d*x+c))^(1/2)*sin(d*x+c)^(1/2)/d-3*a^2*(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)*EllipticF
(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)/d+a^2*(e*csc(d*x+c))^(1/2)*tan(d*x+c
)/d

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Rubi [A]
time = 0.18, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3963, 3957, 2952, 2720, 2644, 335, 218, 212, 209, 2651} \begin {gather*} \frac {2 a^2 \sqrt {\sin (c+d x)} \text {ArcTan}\left (\sqrt {\sin (c+d x)}\right ) \sqrt {e \csc (c+d x)}}{d}+\frac {a^2 \tan (c+d x) \sqrt {e \csc (c+d x)}}{d}+\frac {2 a^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \tanh ^{-1}\left (\sqrt {\sin (c+d x)}\right )}{d}+\frac {3 a^2 \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \csc (c+d x)}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2644

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 2651

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b*Sin[e +
f*x])^(n + 1))*((a*Cos[e + f*x])^(m + 1)/(a*b*f*(m + 1))), x] + Dist[(m + n + 2)/(a^2*(m + 1)), Int[(b*Sin[e +
f*x])^n*(a*Cos[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2952

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 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[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 3963

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 \sqrt {e \csc (c+d x)} (a+a \sec (c+d x))^2 \, dx &=\left (\sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {(a+a \sec (c+d x))^2}{\sqrt {\sin (c+d x)}} \, dx\\ &=\left (\sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {(-a-a \cos (c+d x))^2 \sec ^2(c+d x)}{\sqrt {\sin (c+d x)}} \, dx\\ &=\left (\sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \left (\frac {a^2}{\sqrt {\sin (c+d x)}}+\frac {2 a^2 \sec (c+d x)}{\sqrt {\sin (c+d x)}}+\frac {a^2 \sec ^2(c+d x)}{\sqrt {\sin (c+d x)}}\right ) \, dx\\ &=\left (a^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx+\left (a^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {\sec ^2(c+d x)}{\sqrt {\sin (c+d x)}} \, dx+\left (2 a^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {\sec (c+d x)}{\sqrt {\sin (c+d x)}} \, dx\\ &=\frac {2 a^2 \sqrt {e \csc (c+d x)} F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{d}+\frac {a^2 \sqrt {e \csc (c+d x)} \tan (c+d x)}{d}+\frac {1}{2} \left (a^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx+\frac {\left (2 a^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (1-x^2\right )} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {3 a^2 \sqrt {e \csc (c+d x)} F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{d}+\frac {a^2 \sqrt {e \csc (c+d x)} \tan (c+d x)}{d}+\frac {\left (4 a^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,\sqrt {\sin (c+d x)}\right )}{d}\\ &=\frac {3 a^2 \sqrt {e \csc (c+d x)} F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{d}+\frac {a^2 \sqrt {e \csc (c+d x)} \tan (c+d x)}{d}+\frac {\left (2 a^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {\sin (c+d x)}\right )}{d}+\frac {\left (2 a^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {\sin (c+d x)}\right )}{d}\\ &=\frac {2 a^2 \tan ^{-1}\left (\sqrt {\sin (c+d x)}\right ) \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}{d}+\frac {2 a^2 \tanh ^{-1}\left (\sqrt {\sin (c+d x)}\right ) \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}{d}+\frac {3 a^2 \sqrt {e \csc (c+d x)} F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{d}+\frac {a^2 \sqrt {e \csc (c+d x)} \tan (c+d x)}{d}\\ \end {align*}

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Mathematica [A]
time = 23.10, size = 216, normalized size = 1.40 \begin {gather*} \frac {16 a^2 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \csc ^{\frac {5}{2}}(c+d x) \sqrt {e \csc (c+d x)} \left (3 \cot ^2(c+d x) F\left (\left .\text {ArcSin}\left (\sqrt {\csc (c+d x)}\right )\right |-1\right )+\sqrt {-\cot ^2(c+d x)} \sqrt {\csc (c+d x)} \left (1-2 \text {ArcTan}\left (\sqrt {\csc (c+d x)}\right ) \sqrt {\cos ^2(c+d x)} \sqrt {\csc (c+d x)}+\sqrt {\cos ^2(c+d x)} \sqrt {\csc (c+d x)} \left (-\log \left (1-\sqrt {\csc (c+d x)}\right )+\log \left (1+\sqrt {\csc (c+d x)}\right )\right )\right )\right ) \sec (c+d x) \sin ^4\left (\frac {1}{2} \csc ^{-1}(\csc (c+d x))\right )}{d \sqrt {-\cot ^2(c+d x)}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(16*a^2*Cos[(c + d*x)/2]^4*Csc[c + d*x]^(5/2)*Sqrt[e*Csc[c + d*x]]*(3*Cot[c + d*x]^2*EllipticF[ArcSin[Sqrt[Csc
[c + d*x]]], -1] + Sqrt[-Cot[c + d*x]^2]*Sqrt[Csc[c + d*x]]*(1 - 2*ArcTan[Sqrt[Csc[c + d*x]]]*Sqrt[Cos[c + d*x
]^2]*Sqrt[Csc[c + d*x]] + Sqrt[Cos[c + d*x]^2]*Sqrt[Csc[c + d*x]]*(-Log[1 - Sqrt[Csc[c + d*x]]] + Log[1 + Sqrt
[Csc[c + d*x]]])))*Sec[c + d*x]*Sin[ArcCsc[Csc[c + d*x]]/2]^4)/(d*Sqrt[-Cot[c + d*x]^2])

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Maple [C] Result contains complex when optimal does not.
time = 0.24, size = 744, normalized size = 4.83

method	result	size
default	\(\frac {a^{2} \left (-1+\cos \left (d x +c \right )\right ) \left (i \cos \left (d x +c \right ) \sin \left (d x +c \right ) \EllipticF \left (\sqrt {\frac {i \cos \left (d x +c \right )+\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {i \cos \left (d x +c \right )+\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \cos \left (d x +c \right )-\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}-2 i \cos \left (d x +c \right ) \sin \left (d x +c \right ) \EllipticPi \left (\sqrt {\frac {i \cos \left (d x +c \right )+\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {i \cos \left (d x +c \right )+\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \cos \left (d x +c \right )-\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}-2 i \cos \left (d x +c \right ) \sin \left (d x +c \right ) \EllipticPi \left (\sqrt {\frac {i \cos \left (d x +c \right )+\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {i \cos \left (d x +c \right )+\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \cos \left (d x +c \right )-\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}-2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \EllipticPi \left (\sqrt {\frac {i \cos \left (d x +c \right )+\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {i \cos \left (d x +c \right )+\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \cos \left (d x +c \right )-\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}+2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \EllipticPi \left (\sqrt {\frac {i \cos \left (d x +c \right )+\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {i \cos \left (d x +c \right )+\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \cos \left (d x +c \right )-\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}+\sqrt {2}\, \cos \left (d x +c \right )-\sqrt {2}\right ) \left (1+\cos \left (d x +c \right )\right )^{2} \sqrt {\frac {e}{\sin \left (d x +c \right )}}\, \sqrt {2}}{2 d \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}}\)	\(744\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2/d*(-1+cos(d*x+c))*(I*cos(d*x+c)*sin(d*x+c)*((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*I*cos(d*x+c)*sin(d*x+c)*((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)*EllipticPi(((I*cos(d*x+c)+s
in(d*x+c)-I)/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))-2*I*cos(d*x+c)*sin(d*x+c)*((I*cos(d*x+c)+sin(d*x+c)-I)/s
in(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)*Ellipti
cPi(((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))-2*cos(d*x+c)*sin(d*x+c)*EllipticPi((
(I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(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)*(-(I*cos(d*x+c)-sin(d*x+c)-I)/sin(d*x+c))^(1/2)+2*cos(d*x+c)*sin(d
*x+c)*EllipticPi(((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(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)*(-(I*cos(d*x+c)-sin(d*x+c)-I)/sin(d*x+c))^(1/2)+
2^(1/2)*cos(d*x+c)-2^(1/2))*(1+cos(d*x+c))^2*(e/sin(d*x+c))^(1/2)/cos(d*x+c)/sin(d*x+c)^3*2^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{2} \left (\int \sqrt {e \csc {\left (c + d x \right )}}\, dx + \int 2 \sqrt {e \csc {\left (c + d x \right )}} \sec {\left (c + d x \right )}\, dx + \int \sqrt {e \csc {\left (c + d x \right )}} \sec ^{2}{\left (c + d x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(sqrt(e*csc(c + d*x)), x) + Integral(2*sqrt(e*csc(c + d*x))*sec(c + d*x), x) + Integral(sqrt(e*c
sc(c + d*x))*sec(c + d*x)**2, x))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2\,\sqrt {\frac {e}{\sin \left (c+d\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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