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

Optimal. Leaf size=310 \[ \frac{a e^{5/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} d}-\frac{a e^{5/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}+1\right )}{\sqrt{2} d}-\frac{a e^{5/2} \log \left (\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} d}+\frac{a e^{5/2} \log \left (\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} d}+\frac{6 a e^2 \cos (c+d x) E\left (\left .c+d x-\frac{\pi }{4}\right |2\right ) \sqrt{e \tan (c+d x)}}{5 d \sqrt{\sin (2 c+2 d x)}}-\frac{6 a e \cos (c+d x) (e \tan (c+d x))^{3/2}}{5 d}+\frac{2 e (3 a \sec (c+d x)+5 a) (e \tan (c+d x))^{3/2}}{15 d} \]

[Out]

(a*e^(5/2)*ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d) - (a*e^(5/2)*ArcTan[1 + (Sqrt[2]*Sq
rt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d) - (a*e^(5/2)*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] - Sqrt[2]*Sqrt[e*Tan
[c + d*x]]])/(2*Sqrt[2]*d) + (a*e^(5/2)*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(2
*Sqrt[2]*d) + (6*a*e^2*Cos[c + d*x]*EllipticE[c - Pi/4 + d*x, 2]*Sqrt[e*Tan[c + d*x]])/(5*d*Sqrt[Sin[2*c + 2*d
*x]]) - (6*a*e*Cos[c + d*x]*(e*Tan[c + d*x])^(3/2))/(5*d) + (2*e*(5*a + 3*a*Sec[c + d*x])*(e*Tan[c + d*x])^(3/
2))/(15*d)

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Rubi [A]  time = 0.320431, antiderivative size = 310, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 14, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.609, Rules used = {3881, 3884, 3476, 329, 297, 1162, 617, 204, 1165, 628, 2613, 2615, 2572, 2639} \[ \frac{a e^{5/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} d}-\frac{a e^{5/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}+1\right )}{\sqrt{2} d}-\frac{a e^{5/2} \log \left (\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} d}+\frac{a e^{5/2} \log \left (\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} d}+\frac{6 a e^2 \cos (c+d x) E\left (\left .c+d x-\frac{\pi }{4}\right |2\right ) \sqrt{e \tan (c+d x)}}{5 d \sqrt{\sin (2 c+2 d x)}}-\frac{6 a e \cos (c+d x) (e \tan (c+d x))^{3/2}}{5 d}+\frac{2 e (3 a \sec (c+d x)+5 a) (e \tan (c+d x))^{3/2}}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*e^(5/2)*ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d) - (a*e^(5/2)*ArcTan[1 + (Sqrt[2]*Sq
rt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d) - (a*e^(5/2)*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] - Sqrt[2]*Sqrt[e*Tan
[c + d*x]]])/(2*Sqrt[2]*d) + (a*e^(5/2)*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(2
*Sqrt[2]*d) + (6*a*e^2*Cos[c + d*x]*EllipticE[c - Pi/4 + d*x, 2]*Sqrt[e*Tan[c + d*x]])/(5*d*Sqrt[Sin[2*c + 2*d
*x]]) - (6*a*e*Cos[c + d*x]*(e*Tan[c + d*x])^(3/2))/(5*d) + (2*e*(5*a + 3*a*Sec[c + d*x])*(e*Tan[c + d*x])^(3/
2))/(15*d)

Rule 3881

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[(e*(e*Cot[
c + d*x])^(m - 1)*(a*m + b*(m - 1)*Csc[c + d*x]))/(d*m*(m - 1)), x] - Dist[e^2/m, Int[(e*Cot[c + d*x])^(m - 2)
*(a*m + b*(m - 1)*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[m, 1]

Rule 3884

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

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 297

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

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 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 2613

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a^2*(a*Sec[
e + f*x])^(m - 2)*(b*Tan[e + f*x])^(n + 1))/(b*f*(m + n - 1)), x] + Dist[(a^2*(m - 2))/(m + n - 1), Int[(a*Sec
[e + f*x])^(m - 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (GtQ[m, 1] || (EqQ[m, 1] && EqQ[
n, 1/2])) && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]

Rule 2615

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

Rule 2572

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

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]

Rubi steps

\begin{align*} \int (a+a \sec (c+d x)) (e \tan (c+d x))^{5/2} \, dx &=\frac{2 e (5 a+3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d}-\frac{1}{5} \left (2 e^2\right ) \int \left (\frac{5 a}{2}+\frac{3}{2} a \sec (c+d x)\right ) \sqrt{e \tan (c+d x)} \, dx\\ &=\frac{2 e (5 a+3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d}-\frac{1}{5} \left (3 a e^2\right ) \int \sec (c+d x) \sqrt{e \tan (c+d x)} \, dx-\left (a e^2\right ) \int \sqrt{e \tan (c+d x)} \, dx\\ &=-\frac{6 a e \cos (c+d x) (e \tan (c+d x))^{3/2}}{5 d}+\frac{2 e (5 a+3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d}+\frac{1}{5} \left (6 a e^2\right ) \int \cos (c+d x) \sqrt{e \tan (c+d x)} \, dx-\frac{\left (a e^3\right ) \operatorname{Subst}\left (\int \frac{\sqrt{x}}{e^2+x^2} \, dx,x,e \tan (c+d x)\right )}{d}\\ &=-\frac{6 a e \cos (c+d x) (e \tan (c+d x))^{3/2}}{5 d}+\frac{2 e (5 a+3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d}-\frac{\left (2 a e^3\right ) \operatorname{Subst}\left (\int \frac{x^2}{e^2+x^4} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{d}+\frac{\left (6 a e^2 \sqrt{\cos (c+d x)} \sqrt{e \tan (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \sqrt{\sin (c+d x)} \, dx}{5 \sqrt{\sin (c+d x)}}\\ &=-\frac{6 a e \cos (c+d x) (e \tan (c+d x))^{3/2}}{5 d}+\frac{2 e (5 a+3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d}+\frac{\left (a e^3\right ) \operatorname{Subst}\left (\int \frac{e-x^2}{e^2+x^4} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{d}-\frac{\left (a e^3\right ) \operatorname{Subst}\left (\int \frac{e+x^2}{e^2+x^4} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{d}+\frac{\left (6 a e^2 \cos (c+d x) \sqrt{e \tan (c+d x)}\right ) \int \sqrt{\sin (2 c+2 d x)} \, dx}{5 \sqrt{\sin (2 c+2 d x)}}\\ &=\frac{6 a e^2 \cos (c+d x) E\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sqrt{e \tan (c+d x)}}{5 d \sqrt{\sin (2 c+2 d x)}}-\frac{6 a e \cos (c+d x) (e \tan (c+d x))^{3/2}}{5 d}+\frac{2 e (5 a+3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d}-\frac{\left (a e^{5/2}\right ) \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}-\frac{\left (a e^{5/2}\right ) \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}-\frac{\left (a e^3\right ) \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}-\frac{\left (a e^3\right ) \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}\\ &=-\frac{a e^{5/2} \log \left (\sqrt{e}+\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} d}+\frac{a e^{5/2} \log \left (\sqrt{e}+\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} d}+\frac{6 a e^2 \cos (c+d x) E\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sqrt{e \tan (c+d x)}}{5 d \sqrt{\sin (2 c+2 d x)}}-\frac{6 a e \cos (c+d x) (e \tan (c+d x))^{3/2}}{5 d}+\frac{2 e (5 a+3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d}-\frac{\left (a e^{5/2}\right ) \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}+\frac{\left (a e^{5/2}\right ) \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}\\ &=\frac{a e^{5/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} d}-\frac{a e^{5/2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} d}-\frac{a e^{5/2} \log \left (\sqrt{e}+\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} d}+\frac{a e^{5/2} \log \left (\sqrt{e}+\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} d}+\frac{6 a e^2 \cos (c+d x) E\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sqrt{e \tan (c+d x)}}{5 d \sqrt{\sin (2 c+2 d x)}}-\frac{6 a e \cos (c+d x) (e \tan (c+d x))^{3/2}}{5 d}+\frac{2 e (5 a+3 a \sec (c+d x)) (e \tan (c+d x))^{3/2}}{15 d}\\ \end{align*}

Mathematica [C]  time = 2.23151, size = 186, normalized size = 0.6 \[ \frac{a (\cos (c+d x)+1) \csc (c+d x) \sec ^2\left (\frac{1}{2} (c+d x)\right ) (e \tan (c+d x))^{5/2} \left (\frac{24 \text{Hypergeometric2F1}\left (\frac{3}{4},\frac{3}{2},\frac{7}{4},-\tan ^2(c+d x)\right )}{\sqrt{\sec ^2(c+d x)}}-36 \cos ^2(c+d x)+20 \cos (c+d x)+15 \sqrt{\sin (2 (c+d x))} \cot ^2(c+d x) \sin ^{-1}(\cos (c+d x)-\sin (c+d x))+15 \sqrt{\sin (2 (c+d x))} \cot ^2(c+d x) \log \left (\sin (c+d x)+\sqrt{\sin (2 (c+d x))}+\cos (c+d x)\right )+12\right )}{60 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(1 + Cos[c + d*x])*Csc[c + d*x]*Sec[(c + d*x)/2]^2*(12 + 20*Cos[c + d*x] - 36*Cos[c + d*x]^2 + (24*Hypergeo
metric2F1[3/4, 3/2, 7/4, -Tan[c + d*x]^2])/Sqrt[Sec[c + d*x]^2] + 15*ArcSin[Cos[c + d*x] - Sin[c + d*x]]*Cot[c
+ d*x]^2*Sqrt[Sin[2*(c + d*x)]] + 15*Cot[c + d*x]^2*Log[Cos[c + d*x] + Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]]
*Sqrt[Sin[2*(c + d*x)]])*(e*Tan[c + d*x])^(5/2))/(60*d)

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Maple [C]  time = 0.296, size = 1495, normalized size = 4.8 \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))*(e*tan(d*x+c))^(5/2),x)

[Out]

1/30*a/d*2^(1/2)*(-1+cos(d*x+c))^2*(15*I*((-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))*cos(d*x+c)^3-15*I*((-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))*cos(d*x+c)^3+15*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/si
n(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))*cos(d*x+c)^3+15*((-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))*cos(d*x+c)^3-36*((-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)*EllipticE(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c)
)^(1/2),1/2*2^(1/2))*cos(d*x+c)^3+18*((-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))*cos(d*x+c)^3+15*I*((-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))*cos(d*x+c)^2-15*I*((-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))*cos(d*x+c)^2+15*((-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))*cos(d*x+c)^2+15*((-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))*cos(d*x+c)^2-36*((-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)*EllipticE(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*
2^(1/2))*cos(d*x+c)^2+18*((-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))*c
os(d*x+c)^2+8*cos(d*x+c)^3*2^(1/2)-24*cos(d*x+c)^2*2^(1/2)+10*cos(d*x+c)*2^(1/2)+6*2^(1/2))*(cos(d*x+c)+1)^2*(
e*sin(d*x+c)/cos(d*x+c))^(5/2)/sin(d*x+c)^7

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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))*(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))*(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))*(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{\left (a \sec \left (d x + c\right ) + a\right )} \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))*(e*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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