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

Optimal. Leaf size=236 \[ -\frac {2 a^2 \text {ArcTan}\left (\sqrt {\sin (c+d x)}\right )}{d e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}+\frac {2 a^2 \tanh ^{-1}\left (\sqrt {\sin (c+d x)}\right )}{d e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}-\frac {9 a^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right )}{5 d e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}-\frac {4 a^2 \sin (c+d x)}{3 d e^2 \sqrt {e \csc (c+d x)}}-\frac {2 a^2 \cos (c+d x) \sin (c+d x)}{5 d e^2 \sqrt {e \csc (c+d x)}}+\frac {a^2 \tan (c+d x)}{d e^2 \sqrt {e \csc (c+d x)}} \]

[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)

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Rubi [A]
time = 0.22, antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 12, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3963, 3957, 2952, 2715, 2719, 2644, 327, 335, 304, 209, 212, 2646} \begin {gather*} -\frac {2 a^2 \text {ArcTan}\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 {2 a^2 \tanh ^{-1}\left (\sqrt {\sin (c+d x)}\right )}{d e^2 \sqrt {\sin (c+d x)} \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)}} \end {gather*}

Antiderivative was successfully verified.

[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*} \int \frac {(a+a \sec (c+d x))^2}{(e \csc (c+d x))^{5/2}} \, dx &=\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 \tan ^{-1}\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 \tanh ^{-1}\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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 31.63, size = 322, normalized size = 1.36 \begin {gather*} \frac {\left (1+\cos \left (2 \left (\frac {c}{2}+\frac {d x}{2}\right )\right )\right )^2 \cos (c+d x) \sqrt {\csc (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 {-10 \text {ArcTan}\left (\sqrt {\csc (c+d x)}\right )+5 \log \left (1-\sqrt {\csc (c+d x)}\right )-5 \log \left (1+\sqrt {\csc (c+d x)}\right )-\frac {3 \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)}}}{5 d}-\frac {-6-27 \csc ^4(c+d x)+\csc ^2(c+d x) \left (18+20 \sqrt {1-\sin ^2(c+d x)}\right )}{15 d \csc ^{\frac {7}{2}}(c+d x) \sqrt {1-\sin ^2(c+d x)}}\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 (e \csc (c+d x))^{5/2} \sqrt {1-\sin ^2(c+d x)}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((1 + Cos[2*(c/2 + (d*x)/2)])^2*Cos[c + d*x]*Sqrt[Csc[c + d*x]]*(-1 + Csc[c + d*x]^2)*Sec[c/2 + (d*x)/2]^4*(a
+ a*Sec[c + d*x])^2*(-1/5*(-10*ArcTan[Sqrt[Csc[c + d*x]]] + 5*Log[1 - Sqrt[Csc[c + d*x]]] - 5*Log[1 + Sqrt[Csc
[c + d*x]]] - (3*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])/d - (-6 - 27*Csc[c + d*x]^4 + Csc[c + d*x]^2*(18 + 20*Sqrt[1 - Sin[c + d*x]^2]))/(1
5*d*Csc[c + d*x]^(7/2)*Sqrt[1 - Sin[c + d*x]^2])))/(4*(1 + Cos[2*(c/2 + (-c + ArcCsc[Csc[c + d*x]])/2)])^2*(e*
Csc[c + d*x])^(5/2)*Sqrt[1 - Sin[c + d*x]^2])

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Maple [C] Result contains complex when optimal does not.
time = 0.20, size = 1636, normalized size = 6.93

method	result	size
default	\(\text {Expression too large to display}\)	\(1636\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*a^2/d*(30*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))+30*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)+30*((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)+30*((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)+54*((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2)*(-(I*c
os(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)-27*((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2)*(-(I*co
s(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)-30*I*((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2)*(-(I*c
os(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)*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)-30*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)+6*cos(d*x+c)^4*2^(1/2)+30*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))+30*co
s(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
))+54*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))-
27*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))+20*
cos(d*x+c)^3*2^(1/2)+6*cos(d*x+c)^2*2^(1/2)-47*2^(1/2)*cos(d*x+c)+15*2^(1/2))/cos(d*x+c)/(e/sin(d*x+c))^(5/2)/
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))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.21, size = 313, normalized size = 1.33 \begin {gather*} -\frac {{\left (27 \, \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 ) + 27 \, \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 ) - 30 \, 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 ) - 15 \, 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 (6 \, a^{2} \cos \left (d x + c\right )^{4} + 20 \, a^{2} \cos \left (d x + c\right )^{3} - 21 \, a^{2} \cos \left (d x + c\right )^{2} - 20 \, a^{2} \cos \left (d x + c\right ) + 15 \, a^{2}\right )}}{\sqrt {\sin \left (d x + c\right )}}\right )} e^{\left (-\frac {5}{2}\right )}}{30 \, 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))^(5/2),x, algorithm="fricas")

[Out]

-1/30*(27*sqrt(2*I)*a^2*cos(d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x
+ c))) + 27*sqrt(-2*I)*a^2*cos(d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d
*x + c))) - 30*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) - 15*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*(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*c
os(d*x + c) + 15*a^2)/sqrt(sin(d*x + c)))*e^(-5/2)/(d*cos(d*x + c))

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Sympy [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))**(5/2),x)

[Out]

Timed out

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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))^(5/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \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 \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))^(5/2),x)

[Out]

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

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