3.3.90 \(\int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \csc (c+d x)}} \, dx\) [290]

Optimal. Leaf size=153 \[ -\frac {2 a^2 \text {ArcTan}\left (\sqrt {\sin (c+d x)}\right )}{d \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}+\frac {2 a^2 \tanh ^{-1}\left (\sqrt {\sin (c+d x)}\right )}{d \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}+\frac {a^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right )}{d \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}+\frac {a^2 \tan (c+d x)}{d \sqrt {e \csc (c+d x)}} \]

[Out]

-2*a^2*arctan(sin(d*x+c)^(1/2))/d/(e*csc(d*x+c))^(1/2)/sin(d*x+c)^(1/2)+2*a^2*arctanh(sin(d*x+c)^(1/2))/d/(e*c
sc(d*x+c))^(1/2)/sin(d*x+c)^(1/2)-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)*EllipticE(
cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))/d/(e*csc(d*x+c))^(1/2)/sin(d*x+c)^(1/2)+a^2*tan(d*x+c)/d/(e*csc(d*x+c))^(1/
2)

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Rubi [A]
time = 0.19, antiderivative size = 153, 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, 2719, 2644, 335, 304, 209, 212, 2651} \begin {gather*} -\frac {2 a^2 \text {ArcTan}\left (\sqrt {\sin (c+d x)}\right )}{d \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}+\frac {a^2 \tan (c+d x)}{d \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)}} \end {gather*}

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 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 304

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 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 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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 \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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 27.70, size = 288, normalized size = 1.88 \begin {gather*} \frac {\left (1+\cos \left (2 \left (\frac {c}{2}+\frac {d x}{2}\right )\right )\right )^2 \cos (c+d x) \left (-1+\csc ^2(c+d x)\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (-\frac {-2+\csc ^2(c+d x)}{d \csc ^{\frac {3}{2}}(c+d x) \sqrt {1-\sin ^2(c+d x)}}-\frac {-6 \text {ArcTan}\left (\sqrt {\csc (c+d x)}\right )+3 \log \left (1-\sqrt {\csc (c+d x)}\right )-3 \log \left (1+\sqrt {\csc (c+d x)}\right )+\frac {\csc ^{\frac {5}{2}}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};\csc ^2(c+d x)\right ) \sqrt {1-\sin ^2(c+d x)}}{\sqrt {1-\csc ^2(c+d x)}}}{3 d}\right )}{4 \left (1+\cos \left (2 \left (\frac {c}{2}+\frac {1}{2} \left (-c+\csc ^{-1}(\csc (c+d x))\right )\right )\right )\right )^2 \csc ^{\frac {3}{2}}(c+d x) \sqrt {e \csc (c+d x)} \sqrt {1-\sin ^2(c+d x)}} \end {gather*}

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
*(-((-2 + Csc[c + d*x]^2)/(d*Csc[c + d*x]^(3/2)*Sqrt[1 - Sin[c + d*x]^2])) - (-6*ArcTan[Sqrt[Csc[c + d*x]]] +
3*Log[1 - Sqrt[Csc[c + d*x]]] - 3*Log[1 + Sqrt[Csc[c + d*x]]] + (Csc[c + d*x]^(5/2)*Hypergeometric2F1[1/2, 3/4
, 7/4, Csc[c + d*x]^2]*Sqrt[1 - Sin[c + d*x]^2])/Sqrt[1 - Csc[c + d*x]^2])/(3*d)))/(4*(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] Result contains complex when optimal does not.
time = 0.23, size = 1605, normalized size = 10.49

method	result	size
default	\(\text {Expression too large to display}\)	\(1605\)

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*(2*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)*
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))*cos(d*x+c)^2*(-I*(-1+cos(d*x+
c))/sin(d*x+c))^(1/2)-2*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)*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))*cos(d*x+c)^2*(-I*(
-1+cos(d*x+c))/sin(d*x+c))^(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)*(-(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*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-s
in(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)
)*cos(d*x+c)*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)+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))*cos(d*x+c)^2*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)+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))*cos(d*x+c)^2*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)-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))*cos(d*x+c)^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)*(-(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)^2*(-I*(-1+cos(d*x+c))/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)*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*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)*Elli
pticPi(((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)*(-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))+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)*Ell
ipticF(((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)^2*2^(1/2)+2^(1/2)*cos(d*x+c)+2
^(1/2))/sin(d*x+c)/cos(d*x+c)/(e/sin(d*x+c))^(1/2)*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 [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.04, size = 272, normalized size = 1.78 \begin {gather*} \frac {{\left (\sqrt {2 i} a^{2} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + \sqrt {-2 i} a^{2} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, a^{2} \arctan \left (\frac {76 \, \cos \left (d x + c\right )^{2} + \frac {425 \, {\left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right ) - 1\right )}}{\sqrt {\sin \left (d x + c\right )}} - 152 \, \sin \left (d x + c\right ) - 152}{2 \, {\left (16 \, \cos \left (d x + c\right )^{2} + 393 \, \sin \left (d x + c\right ) - 32\right )}}\right ) \cos \left (d x + c\right ) + a^{2} \cos \left (d x + c\right ) \log \left (\frac {\cos \left (d x + c\right )^{2} + \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - \sin \left (d x + c\right ) - 1\right )}}{\sqrt {\sin \left (d x + c\right )}} - 6 \, \sin \left (d x + c\right ) - 2}{\cos \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) - 2}\right ) - \frac {2 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )}}{\sqrt {\sin \left (d x + c\right )}}\right )} e^{\left (-\frac {1}{2}\right )}}{2 \, d \cos \left (d x + c\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, algorithm="fricas")

[Out]

1/2*(sqrt(2*I)*a^2*cos(d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c))
) + sqrt(-2*I)*a^2*cos(d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c))
) + 2*a^2*arctan(1/2*(76*cos(d*x + c)^2 + 425*(cos(d*x + c)^2 + sin(d*x + c) - 1)/sqrt(sin(d*x + c)) - 152*sin
(d*x + c) - 152)/(16*cos(d*x + c)^2 + 393*sin(d*x + c) - 32))*cos(d*x + c) + a^2*cos(d*x + c)*log((cos(d*x + c
)^2 + 4*(cos(d*x + c)^2 - sin(d*x + c) - 1)/sqrt(sin(d*x + c)) - 6*sin(d*x + c) - 2)/(cos(d*x + c)^2 + 2*sin(d
*x + c) - 2)) - 2*(a^2*cos(d*x + c)^2 - a^2)/sqrt(sin(d*x + c)))*e^(-1/2)/(d*cos(d*x + c))

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\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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