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

Optimal. Leaf size=366 \[ \frac {a^2 e^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d}-\frac {a^2 e^{5/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} d}-\frac {a^2 e^{5/2} \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d}+\frac {a^2 e^{5/2} \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d}+\frac {12 a^2 e^2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{5 d \sqrt {\sin (2 c+2 d x)}}+\frac {2 a^2 (e \tan (c+d x))^{7/2}}{7 d e}+\frac {2 a^2 e (e \tan (c+d x))^{3/2}}{3 d}-\frac {12 a^2 e \cos (c+d x) (e \tan (c+d x))^{3/2}}{5 d}+\frac {4 a^2 e \sec (c+d x) (e \tan (c+d x))^{3/2}}{5 d} \]

[Out]

1/2*a^2*e^(5/2)*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/d*2^(1/2)-1/2*a^2*e^(5/2)*arctan(1+2^(1/2)*(e*t
an(d*x+c))^(1/2)/e^(1/2))/d*2^(1/2)-1/4*a^2*e^(5/2)*ln(e^(1/2)-2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c)
)/d*2^(1/2)+1/4*a^2*e^(5/2)*ln(e^(1/2)+2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c))/d*2^(1/2)-12/5*a^2*e^2
*cos(d*x+c)*(sin(c+1/4*Pi+d*x)^2)^(1/2)/sin(c+1/4*Pi+d*x)*EllipticE(cos(c+1/4*Pi+d*x),2^(1/2))*(e*tan(d*x+c))^
(1/2)/d/sin(2*d*x+2*c)^(1/2)+2/3*a^2*e*(e*tan(d*x+c))^(3/2)/d-12/5*a^2*e*cos(d*x+c)*(e*tan(d*x+c))^(3/2)/d+4/5
*a^2*e*sec(d*x+c)*(e*tan(d*x+c))^(3/2)/d+2/7*a^2*(e*tan(d*x+c))^(7/2)/d/e

________________________________________________________________________________________

Rubi [A]  time = 0.43, antiderivative size = 366, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 17, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.680, Rules used = {3886, 3473, 3476, 329, 297, 1162, 617, 204, 1165, 628, 2611, 2613, 2615, 2572, 2639, 2607, 32} \[ \frac {a^2 e^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d}-\frac {a^2 e^{5/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} d}-\frac {a^2 e^{5/2} \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d}+\frac {a^2 e^{5/2} \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d}+\frac {12 a^2 e^2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{5 d \sqrt {\sin (2 c+2 d x)}}+\frac {2 a^2 (e \tan (c+d x))^{7/2}}{7 d e}+\frac {2 a^2 e (e \tan (c+d x))^{3/2}}{3 d}-\frac {12 a^2 e \cos (c+d x) (e \tan (c+d x))^{3/2}}{5 d}+\frac {4 a^2 e \sec (c+d x) (e \tan (c+d x))^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*e^(5/2)*ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d) - (a^2*e^(5/2)*ArcTan[1 + (Sqrt[2
]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d) - (a^2*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^2*e^(5/2)*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*
x]]])/(2*Sqrt[2]*d) + (12*a^2*e^2*Cos[c + d*x]*EllipticE[c - Pi/4 + d*x, 2]*Sqrt[e*Tan[c + d*x]])/(5*d*Sqrt[Si
n[2*c + 2*d*x]]) + (2*a^2*e*(e*Tan[c + d*x])^(3/2))/(3*d) - (12*a^2*e*Cos[c + d*x]*(e*Tan[c + d*x])^(3/2))/(5*
d) + (4*a^2*e*Sec[c + d*x]*(e*Tan[c + d*x])^(3/2))/(5*d) + (2*a^2*(e*Tan[c + d*x])^(7/2))/(7*d*e)

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]

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 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 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 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 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 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 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 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 2611

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

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 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 3473

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

Rubi steps

\begin {align*} \int (a+a \sec (c+d x))^2 (e \tan (c+d x))^{5/2} \, dx &=\int \left (a^2 (e \tan (c+d x))^{5/2}+2 a^2 \sec (c+d x) (e \tan (c+d x))^{5/2}+a^2 \sec ^2(c+d x) (e \tan (c+d x))^{5/2}\right ) \, dx\\ &=a^2 \int (e \tan (c+d x))^{5/2} \, dx+a^2 \int \sec ^2(c+d x) (e \tan (c+d x))^{5/2} \, dx+\left (2 a^2\right ) \int \sec (c+d x) (e \tan (c+d x))^{5/2} \, dx\\ &=\frac {2 a^2 e (e \tan (c+d x))^{3/2}}{3 d}+\frac {4 a^2 e \sec (c+d x) (e \tan (c+d x))^{3/2}}{5 d}+\frac {a^2 \operatorname {Subst}\left (\int (e x)^{5/2} \, dx,x,\tan (c+d x)\right )}{d}-\left (a^2 e^2\right ) \int \sqrt {e \tan (c+d x)} \, dx-\frac {1}{5} \left (6 a^2 e^2\right ) \int \sec (c+d x) \sqrt {e \tan (c+d x)} \, dx\\ &=\frac {2 a^2 e (e \tan (c+d x))^{3/2}}{3 d}-\frac {12 a^2 e \cos (c+d x) (e \tan (c+d x))^{3/2}}{5 d}+\frac {4 a^2 e \sec (c+d x) (e \tan (c+d x))^{3/2}}{5 d}+\frac {2 a^2 (e \tan (c+d x))^{7/2}}{7 d e}+\frac {1}{5} \left (12 a^2 e^2\right ) \int \cos (c+d x) \sqrt {e \tan (c+d x)} \, dx-\frac {\left (a^2 e^3\right ) \operatorname {Subst}\left (\int \frac {\sqrt {x}}{e^2+x^2} \, dx,x,e \tan (c+d x)\right )}{d}\\ &=\frac {2 a^2 e (e \tan (c+d x))^{3/2}}{3 d}-\frac {12 a^2 e \cos (c+d x) (e \tan (c+d x))^{3/2}}{5 d}+\frac {4 a^2 e \sec (c+d x) (e \tan (c+d x))^{3/2}}{5 d}+\frac {2 a^2 (e \tan (c+d x))^{7/2}}{7 d e}-\frac {\left (2 a^2 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 (12 a^2 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 {2 a^2 e (e \tan (c+d x))^{3/2}}{3 d}-\frac {12 a^2 e \cos (c+d x) (e \tan (c+d x))^{3/2}}{5 d}+\frac {4 a^2 e \sec (c+d x) (e \tan (c+d x))^{3/2}}{5 d}+\frac {2 a^2 (e \tan (c+d x))^{7/2}}{7 d e}+\frac {\left (a^2 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^2 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 (12 a^2 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 {12 a^2 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 {2 a^2 e (e \tan (c+d x))^{3/2}}{3 d}-\frac {12 a^2 e \cos (c+d x) (e \tan (c+d x))^{3/2}}{5 d}+\frac {4 a^2 e \sec (c+d x) (e \tan (c+d x))^{3/2}}{5 d}+\frac {2 a^2 (e \tan (c+d x))^{7/2}}{7 d e}-\frac {\left (a^2 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^2 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^2 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^2 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^2 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^2 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 {12 a^2 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 {2 a^2 e (e \tan (c+d x))^{3/2}}{3 d}-\frac {12 a^2 e \cos (c+d x) (e \tan (c+d x))^{3/2}}{5 d}+\frac {4 a^2 e \sec (c+d x) (e \tan (c+d x))^{3/2}}{5 d}+\frac {2 a^2 (e \tan (c+d x))^{7/2}}{7 d e}-\frac {\left (a^2 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^2 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^2 e^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d}-\frac {a^2 e^{5/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d}-\frac {a^2 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^2 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 {12 a^2 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 {2 a^2 e (e \tan (c+d x))^{3/2}}{3 d}-\frac {12 a^2 e \cos (c+d x) (e \tan (c+d x))^{3/2}}{5 d}+\frac {4 a^2 e \sec (c+d x) (e \tan (c+d x))^{3/2}}{5 d}+\frac {2 a^2 (e \tan (c+d x))^{7/2}}{7 d e}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 6.21, size = 117, normalized size = 0.32 \[ \frac {2 a^2 e \cos ^4\left (\frac {1}{2} (c+d x)\right ) (e \tan (c+d x))^{3/2} \sec ^4\left (\frac {1}{2} \tan ^{-1}(\tan (c+d x))\right ) \left (-42 \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\tan ^2(c+d x)\right )-35 \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\tan ^2(c+d x)\right )+15 \tan ^2(c+d x)+42 \sqrt {\sec ^2(c+d x)}+35\right )}{105 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*a^2*e*Cos[(c + d*x)/2]^4*Sec[ArcTan[Tan[c + d*x]]/2]^4*(e*Tan[c + d*x])^(3/2)*(35 - 42*Hypergeometric2F1[1/
2, 3/4, 7/4, -Tan[c + d*x]^2] - 35*Hypergeometric2F1[3/4, 1, 7/4, -Tan[c + d*x]^2] + 42*Sqrt[Sec[c + d*x]^2] +
15*Tan[c + d*x]^2))/(105*d)

________________________________________________________________________________________

fricas [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*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

maple [C]  time = 1.96, size = 1518, normalized size = 4.15 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/210*a^2/d*(-1+cos(d*x+c))^2*(-105*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)*cos(d*x+c)^4+105*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)*cos(d*x+c)^4+105*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)+s
in(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^4+105*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)*cos(d*x+c)^3+105*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)*cos(d*x+c)^4-105*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-co
s(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^3+252*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)
,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)*cos(d*x+c)^4-504*EllipticE(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),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)*cos(d*x+c)^4+105*((-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)*cos(d*x+c)^3*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))+105*((-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)*cos(d*x+c)^3*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))+252*((-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)*cos(d*x+c)^3*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/si
n(d*x+c))^(1/2),1/2*2^(1/2))-504*((-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)*cos(d*x+c)^3*EllipticE(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))
^(1/2),1/2*2^(1/2))+212*cos(d*x+c)^4*2^(1/2)-336*2^(1/2)*cos(d*x+c)^3+10*cos(d*x+c)^2*2^(1/2)+84*cos(d*x+c)*2^
(1/2)+30*2^(1/2))*(1+cos(d*x+c))^2*(e*sin(d*x+c)/cos(d*x+c))^(5/2)/sin(d*x+c)^7/cos(d*x+c)*2^(1/2)

________________________________________________________________________________________

maxima [A]  time = 0.61, size = 200, normalized size = 0.55 \[ \frac {\frac {24 \, \left (e \tan \left (d x + c\right )\right )^{\frac {7}{2}} a^{2}}{e} - \frac {7 \, {\left (3 \, e^{4} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} + 2 \, \sqrt {e \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {e}}\right )}{\sqrt {e}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} - 2 \, \sqrt {e \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {e}}\right )}{\sqrt {e}} - \frac {\sqrt {2} \log \left (e \tan \left (d x + c\right ) + \sqrt {2} \sqrt {e \tan \left (d x + c\right )} \sqrt {e} + e\right )}{\sqrt {e}} + \frac {\sqrt {2} \log \left (e \tan \left (d x + c\right ) - \sqrt {2} \sqrt {e \tan \left (d x + c\right )} \sqrt {e} + e\right )}{\sqrt {e}}\right )} - 8 \, \left (e \tan \left (d x + c\right )\right )^{\frac {3}{2}} e^{2}\right )} a^{2}}{e}}{84 \, d} \]

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]

1/84*(24*(e*tan(d*x + c))^(7/2)*a^2/e - 7*(3*e^4*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(e) + 2*sqrt(e*tan
(d*x + c)))/sqrt(e))/sqrt(e) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(e) - 2*sqrt(e*tan(d*x + c)))/sqrt(e
))/sqrt(e) - sqrt(2)*log(e*tan(d*x + c) + sqrt(2)*sqrt(e*tan(d*x + c))*sqrt(e) + e)/sqrt(e) + sqrt(2)*log(e*ta
n(d*x + c) - sqrt(2)*sqrt(e*tan(d*x + c))*sqrt(e) + e)/sqrt(e)) - 8*(e*tan(d*x + c))^(3/2)*e^2)*a^2/e)/d

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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*tan(d*x+c))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________