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

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

[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 = 0.32, 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 = {3878, 3872, 2873, 2635, 2639, 2564, 321, 329, 298, 203, 206, 2566} \[ -\frac {4 a^2 \sin (c+d x)}{3 d e^2 \sqrt {e \csc (c+d x)}}+\frac {a^2 \tan (c+d x)}{d e^2 \sqrt {e \csc (c+d x)}}-\frac {2 a^2 \sin (c+d x) \cos (c+d x)}{5 d e^2 \sqrt {e \csc (c+d x)}}-\frac {2 a^2 \tan ^{-1}\left (\sqrt {\sin (c+d x)}\right )}{d e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}+\frac {2 a^2 \tanh ^{-1}\left (\sqrt {\sin (c+d x)}\right )}{d e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}-\frac {9 a^2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{5 d e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}} \]

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 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 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 321

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 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 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 2566

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] && (Integ
ersQ[2*m, 2*n] || EqQ[m + n, 0])

Rule 2635

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] && Integer
Q[2*n]

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 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 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 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]

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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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]  time = 11.31, size = 152, normalized size = 0.64 \[ \frac {2 a^2 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \tan (c+d x) \sec ^4\left (\frac {1}{2} \csc ^{-1}(\csc (c+d x))\right ) \left (3 \sqrt {-\cot ^2(c+d x)} \left (\sin ^2(c+d x) \, _2F_1\left (-\frac {5}{4},\frac {3}{2};-\frac {1}{4};\csc ^2(c+d x)\right )-10 \, _2F_1\left (-\frac {1}{4},\frac {3}{2};\frac {3}{4};\csc ^2(c+d x)\right )\right )-10 \sqrt {\cos ^2(c+d x)} \, _2F_1\left (-\frac {3}{4},1;\frac {1}{4};\csc ^2(c+d x)\right )\right )}{15 d e^2 \sqrt {e \csc (c+d x)}} \]

Warning: Unable to verify antiderivative.

[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]])

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fricas [F]  time = 1.05, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{2} \sec \left (d x + c\right )^{2} + 2 \, a^{2} \sec \left (d x + c\right ) + a^{2}\right )} \sqrt {e \csc \left (d x + c\right )}}{e^{3} \csc \left (d x + c\right )^{3}}, x\right ) \]

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]

integral((a^2*sec(d*x + c)^2 + 2*a^2*sec(d*x + c) + a^2)*sqrt(e*csc(d*x + c))/(e^3*csc(d*x + c)^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\left (e \csc \left (d x + c\right )\right )^{\frac {5}{2}}}\,{d x} \]

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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maple [C]  time = 1.32, size = 1636, normalized size = 6.93 \[ \text {result too large to display} \]

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)

[Out]

-1/30*a^2/d*(-30*I*cos(d*x+c)*((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin(d*x+c))/sin
(d*x+c))^(1/2)*EllipticPi(((I*cos(d*x+c)-I+sin(d*x+c))/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)-30*I*cos(d*x+c)^2*((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin
(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((I*cos(d*x+c)-I+sin(d*x+c))/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)+30*I*cos(d*x+c)^2*((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(
d*x+c)-I-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((I*cos(d*x+c)-I+sin(d*x+c))/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)+30*I*cos(d*x+c)*((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2)
*(-(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((I*cos(d*x+c)-I+sin(d*x+c))/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)+27*cos(d*x+c)^2*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2
)*((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticF(((I
*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))-54*cos(d*x+c)^2*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)
*((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticE(((I*
cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))-30*cos(d*x+c)^2*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*
((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((I*
cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))-30*cos(d*x+c)^2*(-I*(-1+cos(d*x+c))/sin(d*x+
c))^(1/2)*((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c))^(1/2)*Ellip
ticPi(((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))-6*cos(d*x+c)^4*2^(1/2)+27*cos(d*x+
c)*((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticF(((
I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)-54*cos(d*x+c)*
((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticE(((I*c
os(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)-30*cos(d*x+c)*((I
*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((I*cos
(d*x+c)-I+sin(d*x+c))/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)-30*cos(d*
x+c)*((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi
(((I*cos(d*x+c)-I+sin(d*x+c))/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)-2
0*2^(1/2)*cos(d*x+c)^3-6*cos(d*x+c)^2*2^(1/2)+47*cos(d*x+c)*2^(1/2)-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(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2}{{\left (\frac {e}{\sin \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

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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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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