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

Optimal. Leaf size=316 \[ -\frac{2 a^2 \sqrt{\sin (2 c+2 d x)} \sec (c+d x) \text{EllipticF}\left (c+d x-\frac{\pi }{4},2\right )}{3 d e^2 \sqrt{e \tan (c+d x)}}+\frac{a^2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} d e^{5/2}}-\frac{a^2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}+1\right )}{\sqrt{2} d e^{5/2}}+\frac{a^2 \log \left (\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} d e^{5/2}}-\frac{a^2 \log \left (\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} d e^{5/2}}-\frac{4 a^2}{3 d e (e \tan (c+d x))^{3/2}}-\frac{4 a^2 \sec (c+d x)}{3 d e (e \tan (c+d x))^{3/2}} \]

[Out]

(a^2*ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d*e^(5/2)) - (a^2*ArcTan[1 + (Sqrt[2]*Sqrt[e
*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d*e^(5/2)) + (a^2*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] - Sqrt[2]*Sqrt[e*Tan[c
+ d*x]]])/(2*Sqrt[2]*d*e^(5/2)) - (a^2*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(2
*Sqrt[2]*d*e^(5/2)) - (4*a^2)/(3*d*e*(e*Tan[c + d*x])^(3/2)) - (4*a^2*Sec[c + d*x])/(3*d*e*(e*Tan[c + d*x])^(3
/2)) - (2*a^2*EllipticF[c - Pi/4 + d*x, 2]*Sec[c + d*x]*Sqrt[Sin[2*c + 2*d*x]])/(3*d*e^2*Sqrt[e*Tan[c + d*x]])

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Rubi [A]  time = 0.386216, antiderivative size = 316, normalized size of antiderivative = 1., number of steps used = 20, number of rules used = 16, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.64, Rules used = {3886, 3474, 3476, 329, 211, 1165, 628, 1162, 617, 204, 2609, 2614, 2573, 2641, 2607, 32} \[ \frac{a^2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} d e^{5/2}}-\frac{a^2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}+1\right )}{\sqrt{2} d e^{5/2}}+\frac{a^2 \log \left (\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} d e^{5/2}}-\frac{a^2 \log \left (\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} d e^{5/2}}-\frac{2 a^2 \sqrt{\sin (2 c+2 d x)} \sec (c+d x) F\left (\left .c+d x-\frac{\pi }{4}\right |2\right )}{3 d e^2 \sqrt{e \tan (c+d x)}}-\frac{4 a^2}{3 d e (e \tan (c+d x))^{3/2}}-\frac{4 a^2 \sec (c+d x)}{3 d e (e \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d*e^(5/2)) - (a^2*ArcTan[1 + (Sqrt[2]*Sqrt[e
*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d*e^(5/2)) + (a^2*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] - Sqrt[2]*Sqrt[e*Tan[c
+ d*x]]])/(2*Sqrt[2]*d*e^(5/2)) - (a^2*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(2
*Sqrt[2]*d*e^(5/2)) - (4*a^2)/(3*d*e*(e*Tan[c + d*x])^(3/2)) - (4*a^2*Sec[c + d*x])/(3*d*e*(e*Tan[c + d*x])^(3
/2)) - (2*a^2*EllipticF[c - Pi/4 + d*x, 2]*Sec[c + d*x]*Sqrt[Sin[2*c + 2*d*x]])/(3*d*e^2*Sqrt[e*Tan[c + d*x]])

Rule 3886

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[ExpandI
ntegrand[(e*Cot[c + d*x])^m, (a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0]

Rule 3474

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Tan[c + d*x])^(n + 1)/(b*d*(n + 1)), x] - Dist[
1/b^2, Int[(b*Tan[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1]

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[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 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 2609

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*Sec[e +
f*x])^m*(b*Tan[e + f*x])^(n + 1))/(b*f*(n + 1)), x] - Dist[(m + n + 1)/(b^2*(n + 1)), Int[(a*Sec[e + f*x])^m*(
b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, -1] && IntegersQ[2*m, 2*n]

Rule 2614

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[e + f*x]]/(Sqrt[Co
s[e + f*x]]*Sqrt[b*Tan[e + f*x]]), Int[1/(Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]]), x], x] /; FreeQ[{b, e, f}, x
]

Rule 2573

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

\begin{align*} \int \frac{(a+a \sec (c+d x))^2}{(e \tan (c+d x))^{5/2}} \, dx &=\int \left (\frac{a^2}{(e \tan (c+d x))^{5/2}}+\frac{2 a^2 \sec (c+d x)}{(e \tan (c+d x))^{5/2}}+\frac{a^2 \sec ^2(c+d x)}{(e \tan (c+d x))^{5/2}}\right ) \, dx\\ &=a^2 \int \frac{1}{(e \tan (c+d x))^{5/2}} \, dx+a^2 \int \frac{\sec ^2(c+d x)}{(e \tan (c+d x))^{5/2}} \, dx+\left (2 a^2\right ) \int \frac{\sec (c+d x)}{(e \tan (c+d x))^{5/2}} \, dx\\ &=-\frac{2 a^2}{3 d e (e \tan (c+d x))^{3/2}}-\frac{4 a^2 \sec (c+d x)}{3 d e (e \tan (c+d x))^{3/2}}+\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{(e x)^{5/2}} \, dx,x,\tan (c+d x)\right )}{d}-\frac{\left (2 a^2\right ) \int \frac{\sec (c+d x)}{\sqrt{e \tan (c+d x)}} \, dx}{3 e^2}-\frac{a^2 \int \frac{1}{\sqrt{e \tan (c+d x)}} \, dx}{e^2}\\ &=-\frac{4 a^2}{3 d e (e \tan (c+d x))^{3/2}}-\frac{4 a^2 \sec (c+d x)}{3 d e (e \tan (c+d x))^{3/2}}-\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (e^2+x^2\right )} \, dx,x,e \tan (c+d x)\right )}{d e}-\frac{\left (2 a^2 \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{\sin (c+d x)}} \, dx}{3 e^2 \sqrt{\cos (c+d x)} \sqrt{e \tan (c+d x)}}\\ &=-\frac{4 a^2}{3 d e (e \tan (c+d x))^{3/2}}-\frac{4 a^2 \sec (c+d x)}{3 d e (e \tan (c+d x))^{3/2}}-\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{e^2+x^4} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{d e}-\frac{\left (2 a^2 \sec (c+d x) \sqrt{\sin (2 c+2 d x)}\right ) \int \frac{1}{\sqrt{\sin (2 c+2 d x)}} \, dx}{3 e^2 \sqrt{e \tan (c+d x)}}\\ &=-\frac{4 a^2}{3 d e (e \tan (c+d x))^{3/2}}-\frac{4 a^2 \sec (c+d x)}{3 d e (e \tan (c+d x))^{3/2}}-\frac{2 a^2 F\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt{\sin (2 c+2 d x)}}{3 d e^2 \sqrt{e \tan (c+d x)}}-\frac{a^2 \operatorname{Subst}\left (\int \frac{e-x^2}{e^2+x^4} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{d e^2}-\frac{a^2 \operatorname{Subst}\left (\int \frac{e+x^2}{e^2+x^4} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{d e^2}\\ &=-\frac{4 a^2}{3 d e (e \tan (c+d x))^{3/2}}-\frac{4 a^2 \sec (c+d x)}{3 d e (e \tan (c+d x))^{3/2}}-\frac{2 a^2 F\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt{\sin (2 c+2 d x)}}{3 d e^2 \sqrt{e \tan (c+d x)}}+\frac{a^2 \operatorname{Subst}\left (\int \frac{\sqrt{2} \sqrt{e}+2 x}{-e-\sqrt{2} \sqrt{e} x-x^2} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} d e^{5/2}}+\frac{a^2 \operatorname{Subst}\left (\int \frac{\sqrt{2} \sqrt{e}-2 x}{-e+\sqrt{2} \sqrt{e} x-x^2} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} d e^{5/2}}-\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{e-\sqrt{2} \sqrt{e} x+x^2} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{2 d e^2}-\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{e+\sqrt{2} \sqrt{e} x+x^2} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{2 d e^2}\\ &=\frac{a^2 \log \left (\sqrt{e}+\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} d e^{5/2}}-\frac{a^2 \log \left (\sqrt{e}+\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} d e^{5/2}}-\frac{4 a^2}{3 d e (e \tan (c+d x))^{3/2}}-\frac{4 a^2 \sec (c+d x)}{3 d e (e \tan (c+d x))^{3/2}}-\frac{2 a^2 F\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt{\sin (2 c+2 d x)}}{3 d e^2 \sqrt{e \tan (c+d x)}}-\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} d e^{5/2}}+\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} d e^{5/2}}\\ &=\frac{a^2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} d e^{5/2}}-\frac{a^2 \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} d e^{5/2}}+\frac{a^2 \log \left (\sqrt{e}+\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} d e^{5/2}}-\frac{a^2 \log \left (\sqrt{e}+\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} d e^{5/2}}-\frac{4 a^2}{3 d e (e \tan (c+d x))^{3/2}}-\frac{4 a^2 \sec (c+d x)}{3 d e (e \tan (c+d x))^{3/2}}-\frac{2 a^2 F\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt{\sin (2 c+2 d x)}}{3 d e^2 \sqrt{e \tan (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 4.8991, size = 224, normalized size = 0.71 \[ -\frac{a^2 \cos ^2\left (\frac{1}{2} (c+d x)\right ) \cos (c+d x) \cot \left (\frac{1}{2} (c+d x)\right ) \sec ^4\left (\frac{1}{2} \tan ^{-1}(\tan (c+d x))\right ) \left (16 \text{Hypergeometric2F1}\left (-\frac{3}{4},\frac{1}{2},\frac{1}{4},-\tan ^2(c+d x)\right )+16 \text{Hypergeometric2F1}\left (-\frac{3}{4},1,\frac{1}{4},-\tan ^2(c+d x)\right )+3 \sqrt{2} \tan ^{\frac{3}{2}}(c+d x) \left (2 \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )-2 \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )+\log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )-\log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )\right )\right )}{24 d e^2 \sqrt{e \tan (c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(a^2*Cos[(c + d*x)/2]^2*Cos[c + d*x]*Cot[(c + d*x)/2]*Sec[ArcTan[Tan[c + d*x]]/2]^4*(16*Hypergeometric2F1[-3/
4, 1/2, 1/4, -Tan[c + d*x]^2] + 16*Hypergeometric2F1[-3/4, 1, 1/4, -Tan[c + d*x]^2] + 3*Sqrt[2]*(2*ArcTan[1 -
Sqrt[2]*Sqrt[Tan[c + d*x]]] - 2*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]] + Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] +
Tan[c + d*x]] - Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])*Tan[c + d*x]^(3/2)))/(24*d*e^2*Sqrt[e*Tan[
c + d*x]])

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Maple [C]  time = 0.226, size = 650, normalized size = 2.1 \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*tan(d*x+c))^(5/2),x)

[Out]

1/6*a^2/d*2^(1/2)*(-1+cos(d*x+c))*(3*I*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2
^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*
x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)-3*I*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c
))/sin(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*
x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))+3*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))
/sin(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+
c))^(1/2),1/2-1/2*I,1/2*2^(1/2))+3*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/s
in(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c)
)^(1/2),1/2+1/2*I,1/2*2^(1/2))-2*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin
(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(
1/2),1/2*2^(1/2))+4*cos(d*x+c)*2^(1/2))*(cos(d*x+c)+1)^2/sin(d*x+c)/cos(d*x+c)^3/(e*sin(d*x+c)/cos(d*x+c))^(5/
2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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}}{\left (e \tan \left (d x + c\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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