\(\int \frac {(a+a \sec (c+d x))^2}{(e \csc (c+d x))^{5/2}} \, dx\) [292]

Optimal result

Integrand size = 25, antiderivative size = 236 \[ \int \frac {(a+a \sec (c+d x))^2}{(e \csc (c+d x))^{5/2}} \, dx=-\frac {2 a^2 \arctan \left (\sqrt {\sin (c+d x)}\right )}{d e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}+\frac {2 a^2 \text {arctanh}\left (\sqrt {\sin (c+d x)}\right )}{d e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}-\frac {9 a^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right )}{5 d e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}-\frac {4 a^2 \sin (c+d x)}{3 d e^2 \sqrt {e \csc (c+d x)}}-\frac {2 a^2 \cos (c+d x) \sin (c+d x)}{5 d e^2 \sqrt {e \csc (c+d x)}}+\frac {a^2 \tan (c+d x)}{d e^2 \sqrt {e \csc (c+d x)}} \]

[Out]

-4/3*a^2*sin(d*x+c)/d/e^2/(e*csc(d*x+c))^(1/2)-2/5*a^2*cos(d*x+c)*sin(d*x+c)/d/e^2/(e*csc(d*x+c))^(1/2)-2*a^2*
arctan(sin(d*x+c)^(1/2))/d/e^2/(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^2/(e*
csc(d*x+c))^(1/2)/sin(d*x+c)^(1/2)+9/5*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)*Ellip
ticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))/d/e^2/(e*csc(d*x+c))^(1/2)/sin(d*x+c)^(1/2)+a^2*tan(d*x+c)/d/e^2/(e*cs
c(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3963, 3957, 2952, 2715, 2719, 2644, 327, 335, 304, 209, 212, 2646} \[ \int \frac {(a+a \sec (c+d x))^2}{(e \csc (c+d x))^{5/2}} \, dx=-\frac {2 a^2 \arctan \left (\sqrt {\sin (c+d x)}\right )}{d e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}+\frac {2 a^2 \text {arctanh}\left (\sqrt {\sin (c+d x)}\right )}{d e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}-\frac {4 a^2 \sin (c+d x)}{3 d e^2 \sqrt {e \csc (c+d x)}}+\frac {a^2 \tan (c+d x)}{d e^2 \sqrt {e \csc (c+d x)}}-\frac {2 a^2 \sin (c+d x) \cos (c+d x)}{5 d e^2 \sqrt {e \csc (c+d x)}}-\frac {9 a^2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{5 d e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}} \]

[In]

Int[(a + a*Sec[c + d*x])^2/(e*Csc[c + d*x])^(5/2),x]

[Out]

(-2*a^2*ArcTan[Sqrt[Sin[c + d*x]]])/(d*e^2*Sqrt[e*Csc[c + d*x]]*Sqrt[Sin[c + d*x]]) + (2*a^2*ArcTanh[Sqrt[Sin[
c + d*x]]])/(d*e^2*Sqrt[e*Csc[c + d*x]]*Sqrt[Sin[c + d*x]]) - (9*a^2*EllipticE[(c - Pi/2 + d*x)/2, 2])/(5*d*e^
2*Sqrt[e*Csc[c + d*x]]*Sqrt[Sin[c + d*x]]) - (4*a^2*Sin[c + d*x])/(3*d*e^2*Sqrt[e*Csc[c + d*x]]) - (2*a^2*Cos[
c + d*x]*Sin[c + d*x])/(5*d*e^2*Sqrt[e*Csc[c + d*x]]) + (a^2*Tan[c + d*x])/(d*e^2*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 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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 2646

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-a)*(a*Sin[e
+ f*x])^(m - 1)*((b*Cos[e + f*x])^(n + 1)/(b*f*(n + 1))), x] + Dist[a^2*((m - 1)/(b^2*(n + 1))), Int[(a*Sin[e
+ f*x])^(m - 2)*(b*Cos[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[m, 1] && LtQ[n, -1] && (Int
egersQ[2*m, 2*n] || EqQ[m + n, 0])

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[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*} \text {integral}& = \frac {\int (a+a \sec (c+d x))^2 \sin ^{\frac {5}{2}}(c+d x) \, dx}{e^2 \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) \sin ^{\frac {5}{2}}(c+d x) \, dx}{e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}} \\ & = \frac {\int \left (a^2 \sin ^{\frac {5}{2}}(c+d x)+2 a^2 \sec (c+d x) \sin ^{\frac {5}{2}}(c+d x)+a^2 \sec ^2(c+d x) \sin ^{\frac {5}{2}}(c+d x)\right ) \, dx}{e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}} \\ & = \frac {a^2 \int \sin ^{\frac {5}{2}}(c+d x) \, dx}{e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}+\frac {a^2 \int \sec ^2(c+d x) \sin ^{\frac {5}{2}}(c+d x) \, dx}{e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}+\frac {\left (2 a^2\right ) \int \sec (c+d x) \sin ^{\frac {5}{2}}(c+d x) \, dx}{e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}} \\ & = -\frac {2 a^2 \cos (c+d x) \sin (c+d x)}{5 d e^2 \sqrt {e \csc (c+d x)}}+\frac {a^2 \tan (c+d x)}{d e^2 \sqrt {e \csc (c+d x)}}+\frac {\left (3 a^2\right ) \int \sqrt {\sin (c+d x)} \, dx}{5 e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}-\frac {\left (3 a^2\right ) \int \sqrt {\sin (c+d x)} \, dx}{2 e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}+\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {x^{5/2}}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}} \\ & = -\frac {9 a^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right )}{5 d e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}-\frac {4 a^2 \sin (c+d x)}{3 d e^2 \sqrt {e \csc (c+d x)}}-\frac {2 a^2 \cos (c+d x) \sin (c+d x)}{5 d e^2 \sqrt {e \csc (c+d x)}}+\frac {a^2 \tan (c+d x)}{d e^2 \sqrt {e \csc (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 e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}} \\ & = -\frac {9 a^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right )}{5 d e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}-\frac {4 a^2 \sin (c+d x)}{3 d e^2 \sqrt {e \csc (c+d x)}}-\frac {2 a^2 \cos (c+d x) \sin (c+d x)}{5 d e^2 \sqrt {e \csc (c+d x)}}+\frac {a^2 \tan (c+d x)}{d e^2 \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 e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}} \\ & = -\frac {9 a^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right )}{5 d e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}-\frac {4 a^2 \sin (c+d x)}{3 d e^2 \sqrt {e \csc (c+d x)}}-\frac {2 a^2 \cos (c+d x) \sin (c+d x)}{5 d e^2 \sqrt {e \csc (c+d x)}}+\frac {a^2 \tan (c+d x)}{d e^2 \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 e^2 \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 e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}} \\ & = -\frac {2 a^2 \arctan \left (\sqrt {\sin (c+d x)}\right )}{d e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}+\frac {2 a^2 \text {arctanh}\left (\sqrt {\sin (c+d x)}\right )}{d e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}-\frac {9 a^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right )}{5 d e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}-\frac {4 a^2 \sin (c+d x)}{3 d e^2 \sqrt {e \csc (c+d x)}}-\frac {2 a^2 \cos (c+d x) \sin (c+d x)}{5 d e^2 \sqrt {e \csc (c+d x)}}+\frac {a^2 \tan (c+d x)}{d e^2 \sqrt {e \csc (c+d x)}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 15.03 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.64 \[ \int \frac {(a+a \sec (c+d x))^2}{(e \csc (c+d x))^{5/2}} \, dx=\frac {2 a^2 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^4\left (\frac {1}{2} \csc ^{-1}(\csc (c+d x))\right ) \left (-10 \sqrt {\cos ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},1,\frac {1}{4},\csc ^2(c+d x)\right )+3 \sqrt {-\cot ^2(c+d x)} \left (-10 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {3}{2},\frac {3}{4},\csc ^2(c+d x)\right )+\operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {3}{2},-\frac {1}{4},\csc ^2(c+d x)\right ) \sin ^2(c+d x)\right )\right ) \tan (c+d x)}{15 d e^2 \sqrt {e \csc (c+d x)}} \]

[In]

Integrate[(a + a*Sec[c + d*x])^2/(e*Csc[c + d*x])^(5/2),x]

[Out]

(2*a^2*Cos[(c + d*x)/2]^4*Sec[ArcCsc[Csc[c + d*x]]/2]^4*(-10*Sqrt[Cos[c + d*x]^2]*Hypergeometric2F1[-3/4, 1, 1
/4, Csc[c + d*x]^2] + 3*Sqrt[-Cot[c + d*x]^2]*(-10*Hypergeometric2F1[-1/4, 3/2, 3/4, Csc[c + d*x]^2] + Hyperge
ometric2F1[-5/4, 3/2, -1/4, Csc[c + d*x]^2]*Sin[c + d*x]^2))*Tan[c + d*x])/(15*d*e^2*Sqrt[e*Csc[c + d*x]])

Maple [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 13.64 (sec) , antiderivative size = 1069, normalized size of antiderivative = 4.53

method	result	size
parts	\(\text {Expression too large to display}\)	\(1069\)
default	\(\text {Expression too large to display}\)	\(1333\)

[In]

int((a+a*sec(d*x+c))^2/(e*csc(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/5*a^2/d*2^(1/2)*(-6*(-I*(I-cot(d*x+c)+csc(d*x+c)))^(1/2)*(-I*(I+cot(d*x+c)-csc(d*x+c)))^(1/2)*(-I*(cot(d*x+c
)-csc(d*x+c)))^(1/2)*EllipticE((-I*(I-cot(d*x+c)+csc(d*x+c)))^(1/2),1/2*2^(1/2))*cos(d*x+c)+3*(-I*(I-cot(d*x+c
)+csc(d*x+c)))^(1/2)*(-I*(I+cot(d*x+c)-csc(d*x+c)))^(1/2)*(-I*(cot(d*x+c)-csc(d*x+c)))^(1/2)*EllipticF((-I*(I-
cot(d*x+c)+csc(d*x+c)))^(1/2),1/2*2^(1/2))*cos(d*x+c)+cos(d*x+c)^3*2^(1/2)-6*(-I*(I-cot(d*x+c)+csc(d*x+c)))^(1
/2)*(-I*(I+cot(d*x+c)-csc(d*x+c)))^(1/2)*(-I*(cot(d*x+c)-csc(d*x+c)))^(1/2)*EllipticE((-I*(I-cot(d*x+c)+csc(d*
x+c)))^(1/2),1/2*2^(1/2))+3*(-I*(I-cot(d*x+c)+csc(d*x+c)))^(1/2)*(-I*(I+cot(d*x+c)-csc(d*x+c)))^(1/2)*(-I*(cot
(d*x+c)-csc(d*x+c)))^(1/2)*EllipticF((-I*(I-cot(d*x+c)+csc(d*x+c)))^(1/2),1/2*2^(1/2))-4*2^(1/2)*cos(d*x+c)+3*
2^(1/2))/(e*csc(d*x+c))^(1/2)/e^2*csc(d*x+c)+1/2*a^2/d*2^(1/2)/(e*csc(d*x+c))^(1/2)/e^2*(6*(-I*(I-cot(d*x+c)+c
sc(d*x+c)))^(1/2)*(-I*(I+cot(d*x+c)-csc(d*x+c)))^(1/2)*(-I*(cot(d*x+c)-csc(d*x+c)))^(1/2)*EllipticE((-I*(I-cot
(d*x+c)+csc(d*x+c)))^(1/2),1/2*2^(1/2))*cot(d*x+c)-3*(-I*(I-cot(d*x+c)+csc(d*x+c)))^(1/2)*(-I*(I+cot(d*x+c)-cs
c(d*x+c)))^(1/2)*(-I*(cot(d*x+c)-csc(d*x+c)))^(1/2)*EllipticF((-I*(I-cot(d*x+c)+csc(d*x+c)))^(1/2),1/2*2^(1/2)
)*cot(d*x+c)+6*(-I*(I-cot(d*x+c)+csc(d*x+c)))^(1/2)*(-I*(I+cot(d*x+c)-csc(d*x+c)))^(1/2)*(-I*(cot(d*x+c)-csc(d
*x+c)))^(1/2)*EllipticE((-I*(I-cot(d*x+c)+csc(d*x+c)))^(1/2),1/2*2^(1/2))*csc(d*x+c)-3*(-I*(I-cot(d*x+c)+csc(d
*x+c)))^(1/2)*(-I*(I+cot(d*x+c)-csc(d*x+c)))^(1/2)*(-I*(cot(d*x+c)-csc(d*x+c)))^(1/2)*EllipticF((-I*(I-cot(d*x
+c)+csc(d*x+c)))^(1/2),1/2*2^(1/2))*csc(d*x+c)+2*2^(1/2)*cot(d*x+c)-3*2^(1/2)*csc(d*x+c)+2^(1/2)*sec(d*x+c)*cs
c(d*x+c))-2/3*a^2/d*(3*arctan((sin(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(cot(d*x+c)+csc(d*x+c)))*(sin(d*x+c)/(cos(d*
x+c)+1)^2)^(1/2)+3*(sin(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*arctanh((sin(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(cot(d*x+c)
+csc(d*x+c)))+2*cos(d*x+c)-2)/(cos(d*x+c)-1)/e^2/(e*csc(d*x+c))^(1/2)*sin(d*x+c)

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.46 (sec) , antiderivative size = 776, normalized size of antiderivative = 3.29 \[ \int \frac {(a+a \sec (c+d x))^2}{(e \csc (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

[In]

integrate((a+a*sec(d*x+c))^2/(e*csc(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[-1/60*(30*a^2*sqrt(-e)*arctan(-1/4*(cos(d*x + c)^2 - 6*sin(d*x + c) - 2)*sqrt(-e)*sqrt(e/sin(d*x + c))/(e*sin
(d*x + c) + e))*cos(d*x + c) + 15*a^2*sqrt(-e)*cos(d*x + c)*log((e*cos(d*x + c)^4 - 72*e*cos(d*x + c)^2 + 8*(c
os(d*x + c)^4 - 9*cos(d*x + c)^2 + (7*cos(d*x + c)^2 - 8)*sin(d*x + c) + 8)*sqrt(-e)*sqrt(e/sin(d*x + c)) + 28
*(e*cos(d*x + c)^2 - 2*e)*sin(d*x + c) + 72*e)/(cos(d*x + c)^4 - 8*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 - 2)*sin
(d*x + c) + 8)) + 54*a^2*sqrt(2*I*e)*cos(d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c)
+ I*sin(d*x + c))) + 54*a^2*sqrt(-2*I*e)*cos(d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x
+ c) - I*sin(d*x + c))) - 4*(6*a^2*cos(d*x + c)^4 + 20*a^2*cos(d*x + c)^3 - 21*a^2*cos(d*x + c)^2 - 20*a^2*co
s(d*x + c) + 15*a^2)*sqrt(e/sin(d*x + c)))/(d*e^3*cos(d*x + c)), 1/60*(30*a^2*sqrt(e)*arctan(1/4*(cos(d*x + c)
^2 + 6*sin(d*x + c) - 2)*sqrt(e)*sqrt(e/sin(d*x + c))/(e*sin(d*x + c) - e))*cos(d*x + c) + 15*a^2*sqrt(e)*cos(
d*x + c)*log((e*cos(d*x + c)^4 - 72*e*cos(d*x + c)^2 + 8*(cos(d*x + c)^4 - 9*cos(d*x + c)^2 - (7*cos(d*x + c)^
2 - 8)*sin(d*x + c) + 8)*sqrt(e)*sqrt(e/sin(d*x + c)) - 28*(e*cos(d*x + c)^2 - 2*e)*sin(d*x + c) + 72*e)/(cos(
d*x + c)^4 - 8*cos(d*x + c)^2 + 4*(cos(d*x + c)^2 - 2)*sin(d*x + c) + 8)) - 54*a^2*sqrt(2*I*e)*cos(d*x + c)*we
ierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c))) - 54*a^2*sqrt(-2*I*e)*cos(d*x +
c)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c))) + 4*(6*a^2*cos(d*x + c)^4 +
20*a^2*cos(d*x + c)^3 - 21*a^2*cos(d*x + c)^2 - 20*a^2*cos(d*x + c) + 15*a^2)*sqrt(e/sin(d*x + c)))/(d*e^3*co
s(d*x + c))]

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^2}{(e \csc (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sec(d*x+c))**2/(e*csc(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(a+a \sec (c+d x))^2}{(e \csc (c+d x))^{5/2}} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\left (e \csc \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))^2/(e*csc(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^2/(e*csc(d*x + c))^(5/2), x)

Giac [F]

\[ \int \frac {(a+a \sec (c+d x))^2}{(e \csc (c+d x))^{5/2}} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\left (e \csc \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))^2/(e*csc(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^2/(e*csc(d*x + c))^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^2}{(e \csc (c+d x))^{5/2}} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2}{{\left (\frac {e}{\sin \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

[In]

int((a + a/cos(c + d*x))^2/(e/sin(c + d*x))^(5/2),x)

[Out]

int((a + a/cos(c + d*x))^2/(e/sin(c + d*x))^(5/2), x)