3.290 \(\int \frac{(a+a \sec (c+d x))^2}{\sqrt{e \csc (c+d x)}} \, dx\)
Optimal. Leaf size=153 \[ \frac{a^2 \tan (c+d x)}{d \sqrt{e \csc (c+d x)}}-\frac{2 a^2 \tan ^{-1}\left (\sqrt{\sin (c+d x)}\right )}{d \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)}}+\frac{2 a^2 \tanh ^{-1}\left (\sqrt{\sin (c+d x)}\right )}{d \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)}}+\frac{a^2 E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{d \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)}} \]
[Out]
(-2*a^2*ArcTan[Sqrt[Sin[c + d*x]]])/(d*Sqrt[e*Csc[c + d*x]]*Sqrt[Sin[c + d*x]]) + (2*a^2*ArcTanh[Sqrt[Sin[c +
d*x]]])/(d*Sqrt[e*Csc[c + d*x]]*Sqrt[Sin[c + d*x]]) + (a^2*EllipticE[(c - Pi/2 + d*x)/2, 2])/(d*Sqrt[e*Csc[c +
d*x]]*Sqrt[Sin[c + d*x]]) + (a^2*Tan[c + d*x])/(d*Sqrt[e*Csc[c + d*x]])
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Rubi [A] time = 0.271704, antiderivative size = 153, normalized size of antiderivative = 1.,
number of steps used = 12, number of rules used = 10, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used =
{3878, 3872, 2873, 2639, 2564, 329, 298, 203, 206, 2571} \[ \frac{a^2 \tan (c+d x)}{d \sqrt{e \csc (c+d x)}}-\frac{2 a^2 \tan ^{-1}\left (\sqrt{\sin (c+d x)}\right )}{d \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)}}+\frac{2 a^2 \tanh ^{-1}\left (\sqrt{\sin (c+d x)}\right )}{d \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)}}+\frac{a^2 E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{d \sqrt{\sin (c+d x)} \sqrt{e \csc (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Sec[c + d*x])^2/Sqrt[e*Csc[c + d*x]],x]
[Out]
(-2*a^2*ArcTan[Sqrt[Sin[c + d*x]]])/(d*Sqrt[e*Csc[c + d*x]]*Sqrt[Sin[c + d*x]]) + (2*a^2*ArcTanh[Sqrt[Sin[c +
d*x]]])/(d*Sqrt[e*Csc[c + d*x]]*Sqrt[Sin[c + d*x]]) + (a^2*EllipticE[(c - Pi/2 + d*x)/2, 2])/(d*Sqrt[e*Csc[c +
d*x]]*Sqrt[Sin[c + d*x]]) + (a^2*Tan[c + d*x])/(d*Sqrt[e*Csc[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 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 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 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 329
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 298
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
& !GtQ[a/b, 0]
Rule 203
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 2571
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]
Rubi steps
\begin{align*} \int \frac{(a+a \sec (c+d x))^2}{\sqrt{e \csc (c+d x)}} \, dx &=\frac{\int (a+a \sec (c+d x))^2 \sqrt{\sin (c+d x)} \, dx}{\sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}\\ &=\frac{\int (-a-a \cos (c+d x))^2 \sec ^2(c+d x) \sqrt{\sin (c+d x)} \, dx}{\sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}\\ &=\frac{\int \left (a^2 \sqrt{\sin (c+d x)}+2 a^2 \sec (c+d x) \sqrt{\sin (c+d x)}+a^2 \sec ^2(c+d x) \sqrt{\sin (c+d x)}\right ) \, dx}{\sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}\\ &=\frac{a^2 \int \sqrt{\sin (c+d x)} \, dx}{\sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}+\frac{a^2 \int \sec ^2(c+d x) \sqrt{\sin (c+d x)} \, dx}{\sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}+\frac{\left (2 a^2\right ) \int \sec (c+d x) \sqrt{\sin (c+d x)} \, dx}{\sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}\\ &=\frac{2 a^2 E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right )}{d \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}+\frac{a^2 \tan (c+d x)}{d \sqrt{e \csc (c+d x)}}-\frac{a^2 \int \sqrt{\sin (c+d x)} \, dx}{2 \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}+\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt{x}}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}\\ &=\frac{a^2 E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right )}{d \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}+\frac{a^2 \tan (c+d x)}{d \sqrt{e \csc (c+d x)}}+\frac{\left (4 a^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{1-x^4} \, dx,x,\sqrt{\sin (c+d x)}\right )}{d \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}\\ &=\frac{a^2 E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right )}{d \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}+\frac{a^2 \tan (c+d x)}{d \sqrt{e \csc (c+d x)}}+\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt{\sin (c+d x)}\right )}{d \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}-\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt{\sin (c+d x)}\right )}{d \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}\\ &=-\frac{2 a^2 \tan ^{-1}\left (\sqrt{\sin (c+d x)}\right )}{d \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}+\frac{2 a^2 \tanh ^{-1}\left (\sqrt{\sin (c+d x)}\right )}{d \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}+\frac{a^2 E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right )}{d \sqrt{e \csc (c+d x)} \sqrt{\sin (c+d x)}}+\frac{a^2 \tan (c+d x)}{d \sqrt{e \csc (c+d x)}}\\ \end{align*}
Mathematica [C] time = 8.42174, size = 287, normalized size = 1.88 \[ -\frac{\left (\cos \left (2 \left (\frac{c}{2}+\frac{d x}{2}\right )\right )+1\right )^2 \cos (c+d x) \left (\csc ^2(c+d x)-1\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^2 \left (\frac{\sqrt{1-\sin ^2(c+d x)} \sqrt{\csc (c+d x)} \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{3}{2},\frac{3}{4},\csc ^2(c+d x)\right )}{\sqrt{1-\csc ^2(c+d x)}}-\frac{2 \sqrt{\csc (c+d x)} \sqrt{1-\csc ^2(c+d x)} \text{Hypergeometric2F1}\left (\frac{3}{4},\frac{3}{2},\frac{7}{4},\csc ^2(c+d x)\right )}{3 \sqrt{1-\sin ^2(c+d x)}}-\tan ^{-1}\left (\sqrt{\csc (c+d x)}\right )-\tanh ^{-1}\left (\sqrt{\csc (c+d x)}\right )\right )}{2 d \sqrt{1-\sin ^2(c+d x)} \csc ^{\frac{3}{2}}(c+d x) \sqrt{e \csc (c+d x)} \left (\cos \left (2 \left (\frac{1}{2} \left (\csc ^{-1}(\csc (c+d x))-c\right )+\frac{c}{2}\right )\right )+1\right )^2} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(a + a*Sec[c + d*x])^2/Sqrt[e*Csc[c + d*x]],x]
[Out]
-((1 + Cos[2*(c/2 + (d*x)/2)])^2*Cos[c + d*x]*(-1 + Csc[c + d*x]^2)*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^
2*(-ArcTan[Sqrt[Csc[c + d*x]]] - ArcTanh[Sqrt[Csc[c + d*x]]] - (2*Sqrt[Csc[c + d*x]]*Sqrt[1 - Csc[c + d*x]^2]*
Hypergeometric2F1[3/4, 3/2, 7/4, Csc[c + d*x]^2])/(3*Sqrt[1 - Sin[c + d*x]^2]) + (Sqrt[Csc[c + d*x]]*Hypergeom
etric2F1[-1/4, 3/2, 3/4, Csc[c + d*x]^2]*Sqrt[1 - Sin[c + d*x]^2])/Sqrt[1 - Csc[c + d*x]^2]))/(2*d*(1 + Cos[2*
(c/2 + (-c + ArcCsc[Csc[c + d*x]])/2)])^2*Csc[c + d*x]^(3/2)*Sqrt[e*Csc[c + d*x]]*Sqrt[1 - Sin[c + d*x]^2])
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Maple [C] time = 0.231, size = 1610, normalized size = 10.5 \begin{align*} \text{result too large to display} \end{align*}
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)
[Out]
-1/2*a^2/d*2^(1/2)*(-2*I*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^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)*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))+2*I*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^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)*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))-2*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^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)*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*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^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))+2*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^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*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^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)*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))-2*I*(-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)*EllipticPi(((I*co
s(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*I*cos(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)*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)-2*cos(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)*E
llipticPi(((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))-cos(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))+2*cos(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)*
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)*(-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)+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)+2*cos(d*x+c)^2*2^(1/2)-cos(d*x+c)
*2^(1/2)-2^(1/2))/cos(d*x+c)/(e/sin(d*x+c))^(1/2)/sin(d*x+c)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\sqrt{e \csc \left (d x + c\right )}}\,{d x} \end{align*}
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]
integrate((a*sec(d*x + c) + a)^2/sqrt(e*csc(d*x + c)), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (a^{2} \sec \left (d x + c\right )^{2} + 2 \, a^{2} \sec \left (d x + c\right ) + a^{2}\right )} \sqrt{e \csc \left (d x + c\right )}}{e \csc \left (d x + c\right )}, x\right ) \end{align*}
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]
integral((a^2*sec(d*x + c)^2 + 2*a^2*sec(d*x + c) + a^2)*sqrt(e*csc(d*x + c))/(e*csc(d*x + c)), x)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int \frac{1}{\sqrt{e \csc{\left (c + d x \right )}}}\, dx + \int \frac{2 \sec{\left (c + d x \right )}}{\sqrt{e \csc{\left (c + d x \right )}}}\, dx + \int \frac{\sec ^{2}{\left (c + d x \right )}}{\sqrt{e \csc{\left (c + d x \right )}}}\, dx\right ) \end{align*}
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(1/sqrt(e*csc(c + d*x)), x) + Integral(2*sec(c + d*x)/sqrt(e*csc(c + d*x)), x) + Integral(sec(c
+ d*x)**2/sqrt(e*csc(c + d*x)), x))
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\sqrt{e \csc \left (d x + c\right )}}\,{d x} \end{align*}
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((a*sec(d*x + c) + a)^2/sqrt(e*csc(d*x + c)), x)