\(\int (a+a \sec (c+d x))^2 (e \tan (c+d x))^{3/2} \, dx\) [112]

Optimal result

Integrand size = 25, antiderivative size = 335 \[ \int (a+a \sec (c+d x))^2 (e \tan (c+d x))^{3/2} \, dx=\frac {a^2 e^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d}-\frac {a^2 e^{3/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d}+\frac {a^2 e^{3/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} d}-\frac {a^2 e^{3/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} d}-\frac {2 a^2 e^2 \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{3 d \sqrt {e \tan (c+d x)}}+\frac {2 a^2 e \sqrt {e \tan (c+d x)}}{d}+\frac {4 a^2 e \sec (c+d x) \sqrt {e \tan (c+d x)}}{3 d}+\frac {2 a^2 (e \tan (c+d x))^{5/2}}{5 d e} \]

[Out]

1/2*a^2*e^(3/2)*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/d*2^(1/2)-1/2*a^2*e^(3/2)*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^(3/2)*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^(3/2)*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)+2/3*a^2*e^2*
(sin(c+1/4*Pi+d*x)^2)^(1/2)/sin(c+1/4*Pi+d*x)*EllipticF(cos(c+1/4*Pi+d*x),2^(1/2))*sec(d*x+c)*sin(2*d*x+2*c)^(
1/2)/d/(e*tan(d*x+c))^(1/2)+2*a^2*e*(e*tan(d*x+c))^(1/2)/d+4/3*a^2*e*sec(d*x+c)*(e*tan(d*x+c))^(1/2)/d+2/5*a^2
*(e*tan(d*x+c))^(5/2)/d/e

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.640, Rules used = {3971, 3554, 3557, 335, 217, 1179, 642, 1176, 631, 210, 2691, 2694, 2653, 2720, 2687, 32} \[ \int (a+a \sec (c+d x))^2 (e \tan (c+d x))^{3/2} \, dx=\frac {a^2 e^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d}-\frac {a^2 e^{3/2} \arctan \left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} d}+\frac {a^2 e^{3/2} \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d}-\frac {a^2 e^{3/2} \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d}-\frac {2 a^2 e^2 \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{3 d \sqrt {e \tan (c+d x)}}+\frac {2 a^2 (e \tan (c+d x))^{5/2}}{5 d e}+\frac {2 a^2 e \sqrt {e \tan (c+d x)}}{d}+\frac {4 a^2 e \sec (c+d x) \sqrt {e \tan (c+d x)}}{3 d} \]

[In]

Int[(a + a*Sec[c + d*x])^2*(e*Tan[c + d*x])^(3/2),x]

[Out]

(a^2*e^(3/2)*ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d) - (a^2*e^(3/2)*ArcTan[1 + (Sqrt[2
]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d) + (a^2*e^(3/2)*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] - Sqrt[2]*Sqrt
[e*Tan[c + d*x]]])/(2*Sqrt[2]*d) - (a^2*e^(3/2)*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*
x]]])/(2*Sqrt[2]*d) - (2*a^2*e^2*EllipticF[c - Pi/4 + d*x, 2]*Sec[c + d*x]*Sqrt[Sin[2*c + 2*d*x]])/(3*d*Sqrt[e
*Tan[c + d*x]]) + (2*a^2*e*Sqrt[e*Tan[c + d*x]])/d + (4*a^2*e*Sec[c + d*x]*Sqrt[e*Tan[c + d*x]])/(3*d) + (2*a^
2*(e*Tan[c + d*x])^(5/2))/(5*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 210

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

Rule 217

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 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 631

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 642

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 1176

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 1179

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 2653

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 2687

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 2691

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 2694

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 2720

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

Rule 3554

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 3557

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 3971

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

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 26.79 (sec) , antiderivative size = 257, normalized size of antiderivative = 0.77 \[ \int (a+a \sec (c+d x))^2 (e \tan (c+d x))^{3/2} \, dx=\frac {a^2 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^4\left (\frac {1}{2} \arctan (\tan (c+d x))\right ) (e \tan (c+d x))^{3/2} \left (30 \sqrt {2} \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-30 \sqrt {2} \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )+15 \sqrt {2} \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-15 \sqrt {2} \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )+120 \sqrt {\tan (c+d x)}-80 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\tan ^2(c+d x)\right ) \sqrt {\tan (c+d x)}+80 \sqrt {\sec ^2(c+d x)} \sqrt {\tan (c+d x)}+24 \tan ^{\frac {5}{2}}(c+d x)\right )}{60 d \tan ^{\frac {3}{2}}(c+d x)} \]

[In]

Integrate[(a + a*Sec[c + d*x])^2*(e*Tan[c + d*x])^(3/2),x]

[Out]

(a^2*Cos[(c + d*x)/2]^4*Sec[ArcTan[Tan[c + d*x]]/2]^4*(e*Tan[c + d*x])^(3/2)*(30*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sq
rt[Tan[c + d*x]]] - 30*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]] + 15*Sqrt[2]*Log[1 - Sqrt[2]*Sqrt[Tan[c
+ d*x]] + Tan[c + d*x]] - 15*Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] + 120*Sqrt[Tan[c + d*x
]] - 80*Hypergeometric2F1[1/4, 1/2, 5/4, -Tan[c + d*x]^2]*Sqrt[Tan[c + d*x]] + 80*Sqrt[Sec[c + d*x]^2]*Sqrt[Ta
n[c + d*x]] + 24*Tan[c + d*x]^(5/2)))/(60*d*Tan[c + d*x]^(3/2))

Maple [A] (verified)

Time = 6.59 (sec) , antiderivative size = 397, normalized size of antiderivative = 1.19

method	result	size
parts	\(\frac {2 a^{2} e \left (\sqrt {e \tan \left (d x +c \right )}-\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {e \tan \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \tan \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \tan \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \tan \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8}\right )}{d}+\frac {2 a^{2} \left (e \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5 d e}-\frac {2 a^{2} \sqrt {2}\, \left (-\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \cos \left (d x +c \right )^{2}-\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \cos \left (d x +c \right )+\sqrt {2}\, \sin \left (d x +c \right )\right ) e \tan \left (d x +c \right ) \sqrt {e \tan \left (d x +c \right )}}{3 d \left (\cos \left (d x +c \right )^{2}-1\right )}\)	\(397\)
default	\(\text {Expression too large to display}\)	\(931\)

[In]

int((a+a*sec(d*x+c))^2*(e*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

2*a^2/d*e*((e*tan(d*x+c))^(1/2)-1/8*(e^2)^(1/4)*2^(1/2)*(ln((e*tan(d*x+c)+(e^2)^(1/4)*(e*tan(d*x+c))^(1/2)*2^(
1/2)+(e^2)^(1/2))/(e*tan(d*x+c)-(e^2)^(1/4)*(e*tan(d*x+c))^(1/2)*2^(1/2)+(e^2)^(1/2)))+2*arctan(2^(1/2)/(e^2)^
(1/4)*(e*tan(d*x+c))^(1/2)+1)-2*arctan(-2^(1/2)/(e^2)^(1/4)*(e*tan(d*x+c))^(1/2)+1)))+2/5*a^2*(e*tan(d*x+c))^(
5/2)/d/e-2/3*a^2/d*2^(1/2)*(-(csc(d*x+c)-cot(d*x+c)+1)^(1/2)*(cot(d*x+c)-csc(d*x+c)+1)^(1/2)*(cot(d*x+c)-csc(d
*x+c))^(1/2)*EllipticF((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2*2^(1/2))*cos(d*x+c)^2-(csc(d*x+c)-cot(d*x+c)+1)^(1/
2)*(cot(d*x+c)-csc(d*x+c)+1)^(1/2)*(cot(d*x+c)-csc(d*x+c))^(1/2)*EllipticF((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2
*2^(1/2))*cos(d*x+c)+2^(1/2)*sin(d*x+c))*e*tan(d*x+c)*(e*tan(d*x+c))^(1/2)/(cos(d*x+c)^2-1)

Fricas [F(-1)]

Timed out. \[ \int (a+a \sec (c+d x))^2 (e \tan (c+d x))^{3/2} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sec(d*x+c))^2*(e*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

Timed out

Sympy [F]

\[ \int (a+a \sec (c+d x))^2 (e \tan (c+d x))^{3/2} \, dx=a^{2} \left (\int \left (e \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}\, dx + \int 2 \left (e \tan {\left (c + d x \right )}\right )^{\frac {3}{2}} \sec {\left (c + d x \right )}\, dx + \int \left (e \tan {\left (c + d x \right )}\right )^{\frac {3}{2}} \sec ^{2}{\left (c + d x \right )}\, dx\right ) \]

[In]

integrate((a+a*sec(d*x+c))**2*(e*tan(d*x+c))**(3/2),x)

[Out]

a**2*(Integral((e*tan(c + d*x))**(3/2), x) + Integral(2*(e*tan(c + d*x))**(3/2)*sec(c + d*x), x) + Integral((e
*tan(c + d*x))**(3/2)*sec(c + d*x)**2, x))

Maxima [F(-2)]

Exception generated. \[ \int (a+a \sec (c+d x))^2 (e \tan (c+d x))^{3/2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((a+a*sec(d*x+c))^2*(e*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [F]

\[ \int (a+a \sec (c+d x))^2 (e \tan (c+d x))^{3/2} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \left (e \tan \left (d x + c\right )\right )^{\frac {3}{2}} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))^2*(e*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^2*(e*tan(d*x + c))^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int (a+a \sec (c+d x))^2 (e \tan (c+d x))^{3/2} \, dx=\int {\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2 \,d x \]

[In]

int((e*tan(c + d*x))^(3/2)*(a + a/cos(c + d*x))^2,x)

[Out]

int((e*tan(c + d*x))^(3/2)*(a + a/cos(c + d*x))^2, x)