\(\int \frac {(e \tan (c+d x))^{11/2}}{(a+a \sec (c+d x))^2} \, dx\) [128]

Optimal result

Integrand size = 25, antiderivative size = 339 \[ \int \frac {(e \tan (c+d x))^{11/2}}{(a+a \sec (c+d x))^2} \, dx=\frac {e^{11/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}-\frac {e^{11/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}+\frac {e^{11/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a^2 d}-\frac {e^{11/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a^2 d}+\frac {2 e^6 \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{3 a^2 d \sqrt {e \tan (c+d x)}}+\frac {2 e^5 \sqrt {e \tan (c+d x)}}{a^2 d}-\frac {4 e^5 \sec (c+d x) \sqrt {e \tan (c+d x)}}{3 a^2 d}+\frac {2 e^3 (e \tan (c+d x))^{5/2}}{5 a^2 d} \]

[Out]

1/2*e^(11/2)*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/a^2/d*2^(1/2)-1/2*e^(11/2)*arctan(1+2^(1/2)*(e*tan
(d*x+c))^(1/2)/e^(1/2))/a^2/d*2^(1/2)+1/4*e^(11/2)*ln(e^(1/2)-2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c))
/a^2/d*2^(1/2)-1/4*e^(11/2)*ln(e^(1/2)+2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c))/a^2/d*2^(1/2)-2/3*e^6*
(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)/a^2/d/(e*tan(d*x+c))^(1/2)+2*e^5*(e*tan(d*x+c))^(1/2)/a^2/d-4/3*e^5*sec(d*x+c)*(e*tan(d*x+c))^(1/2)/a^2/d
+2/5*e^3*(e*tan(d*x+c))^(5/2)/a^2/d

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.680, Rules used = {3973, 3971, 3554, 3557, 335, 217, 1179, 642, 1176, 631, 210, 2691, 2694, 2653, 2720, 2687, 32} \[ \int \frac {(e \tan (c+d x))^{11/2}}{(a+a \sec (c+d x))^2} \, dx=\frac {e^{11/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}-\frac {e^{11/2} \arctan \left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} a^2 d}+\frac {e^{11/2} \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a^2 d}-\frac {e^{11/2} \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a^2 d}+\frac {2 e^6 \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{3 a^2 d \sqrt {e \tan (c+d x)}}+\frac {2 e^5 \sqrt {e \tan (c+d x)}}{a^2 d}-\frac {4 e^5 \sec (c+d x) \sqrt {e \tan (c+d x)}}{3 a^2 d}+\frac {2 e^3 (e \tan (c+d x))^{5/2}}{5 a^2 d} \]

[In]

Int[(e*Tan[c + d*x])^(11/2)/(a + a*Sec[c + d*x])^2,x]

[Out]

(e^(11/2)*ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*a^2*d) - (e^(11/2)*ArcTan[1 + (Sqrt[2]*
Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*a^2*d) + (e^(11/2)*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] - Sqrt[2]*Sqrt[
e*Tan[c + d*x]]])/(2*Sqrt[2]*a^2*d) - (e^(11/2)*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*
x]]])/(2*Sqrt[2]*a^2*d) + (2*e^6*EllipticF[c - Pi/4 + d*x, 2]*Sec[c + d*x]*Sqrt[Sin[2*c + 2*d*x]])/(3*a^2*d*Sq
rt[e*Tan[c + d*x]]) + (2*e^5*Sqrt[e*Tan[c + d*x]])/(a^2*d) - (4*e^5*Sec[c + d*x]*Sqrt[e*Tan[c + d*x]])/(3*a^2*
d) + (2*e^3*(e*Tan[c + d*x])^(5/2))/(5*a^2*d)

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]

Rule 3973

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[a^(2*n
)/e^(2*n), Int[(e*Cot[c + d*x])^(m + 2*n)/(-a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && E
qQ[a^2 - b^2, 0] && ILtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {e^4 \int (-a+a \sec (c+d x))^2 (e \tan (c+d x))^{3/2} \, dx}{a^4} \\ & = \frac {e^4 \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^4} \\ & = \frac {e^4 \int (e \tan (c+d x))^{3/2} \, dx}{a^2}+\frac {e^4 \int \sec ^2(c+d x) (e \tan (c+d x))^{3/2} \, dx}{a^2}-\frac {\left (2 e^4\right ) \int \sec (c+d x) (e \tan (c+d x))^{3/2} \, dx}{a^2} \\ & = \frac {2 e^5 \sqrt {e \tan (c+d x)}}{a^2 d}-\frac {4 e^5 \sec (c+d x) \sqrt {e \tan (c+d x)}}{3 a^2 d}+\frac {e^4 \text {Subst}\left (\int (e x)^{3/2} \, dx,x,\tan (c+d x)\right )}{a^2 d}+\frac {\left (2 e^6\right ) \int \frac {\sec (c+d x)}{\sqrt {e \tan (c+d x)}} \, dx}{3 a^2}-\frac {e^6 \int \frac {1}{\sqrt {e \tan (c+d x)}} \, dx}{a^2} \\ & = \frac {2 e^5 \sqrt {e \tan (c+d x)}}{a^2 d}-\frac {4 e^5 \sec (c+d x) \sqrt {e \tan (c+d x)}}{3 a^2 d}+\frac {2 e^3 (e \tan (c+d x))^{5/2}}{5 a^2 d}-\frac {e^7 \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (e^2+x^2\right )} \, dx,x,e \tan (c+d x)\right )}{a^2 d}+\frac {\left (2 e^6 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\sin (c+d x)}} \, dx}{3 a^2 \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}} \\ & = \frac {2 e^5 \sqrt {e \tan (c+d x)}}{a^2 d}-\frac {4 e^5 \sec (c+d x) \sqrt {e \tan (c+d x)}}{3 a^2 d}+\frac {2 e^3 (e \tan (c+d x))^{5/2}}{5 a^2 d}-\frac {\left (2 e^7\right ) \text {Subst}\left (\int \frac {1}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a^2 d}+\frac {\left (2 e^6 \sec (c+d x) \sqrt {\sin (2 c+2 d x)}\right ) \int \frac {1}{\sqrt {\sin (2 c+2 d x)}} \, dx}{3 a^2 \sqrt {e \tan (c+d x)}} \\ & = \frac {2 e^6 \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{3 a^2 d \sqrt {e \tan (c+d x)}}+\frac {2 e^5 \sqrt {e \tan (c+d x)}}{a^2 d}-\frac {4 e^5 \sec (c+d x) \sqrt {e \tan (c+d x)}}{3 a^2 d}+\frac {2 e^3 (e \tan (c+d x))^{5/2}}{5 a^2 d}-\frac {e^6 \text {Subst}\left (\int \frac {e-x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a^2 d}-\frac {e^6 \text {Subst}\left (\int \frac {e+x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a^2 d} \\ & = \frac {2 e^6 \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{3 a^2 d \sqrt {e \tan (c+d x)}}+\frac {2 e^5 \sqrt {e \tan (c+d x)}}{a^2 d}-\frac {4 e^5 \sec (c+d x) \sqrt {e \tan (c+d x)}}{3 a^2 d}+\frac {2 e^3 (e \tan (c+d x))^{5/2}}{5 a^2 d}+\frac {e^{11/2} \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} a^2 d}+\frac {e^{11/2} \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} a^2 d}-\frac {e^6 \text {Subst}\left (\int \frac {1}{e-\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 a^2 d}-\frac {e^6 \text {Subst}\left (\int \frac {1}{e+\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 a^2 d} \\ & = \frac {e^{11/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a^2 d}-\frac {e^{11/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a^2 d}+\frac {2 e^6 \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{3 a^2 d \sqrt {e \tan (c+d x)}}+\frac {2 e^5 \sqrt {e \tan (c+d x)}}{a^2 d}-\frac {4 e^5 \sec (c+d x) \sqrt {e \tan (c+d x)}}{3 a^2 d}+\frac {2 e^3 (e \tan (c+d x))^{5/2}}{5 a^2 d}-\frac {e^{11/2} \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} a^2 d}+\frac {e^{11/2} \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} a^2 d} \\ & = \frac {e^{11/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}-\frac {e^{11/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}+\frac {e^{11/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a^2 d}-\frac {e^{11/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a^2 d}+\frac {2 e^6 \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{3 a^2 d \sqrt {e \tan (c+d x)}}+\frac {2 e^5 \sqrt {e \tan (c+d x)}}{a^2 d}-\frac {4 e^5 \sec (c+d x) \sqrt {e \tan (c+d x)}}{3 a^2 d}+\frac {2 e^3 (e \tan (c+d x))^{5/2}}{5 a^2 d} \\ \end{align*}

Mathematica [F]

\[ \int \frac {(e \tan (c+d x))^{11/2}}{(a+a \sec (c+d x))^2} \, dx=\int \frac {(e \tan (c+d x))^{11/2}}{(a+a \sec (c+d x))^2} \, dx \]

[In]

Integrate[(e*Tan[c + d*x])^(11/2)/(a + a*Sec[c + d*x])^2,x]

[Out]

Integrate[(e*Tan[c + d*x])^(11/2)/(a + a*Sec[c + d*x])^2, x]

Maple [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 4.51 (sec) , antiderivative size = 1007, normalized size of antiderivative = 2.97

method	result	size
default	\(\text {Expression too large to display}\)	\(1007\)

[In]

int((e*tan(d*x+c))^(11/2)/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

-1/30/a^2/d*2^(1/2)*(15*I*(cot(d*x+c)-csc(d*x+c)+1)^(1/2)*(cot(d*x+c)-csc(d*x+c))^(1/2)*EllipticPi((csc(d*x+c)
-cot(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*(csc(d*x+c)-cot(d*x+c)+1)^(1/2)*cos(d*x+c)^3-15*I*(cot(d*x+c)-csc(
d*x+c)+1)^(1/2)*(cot(d*x+c)-csc(d*x+c))^(1/2)*(csc(d*x+c)-cot(d*x+c)+1)^(1/2)*EllipticPi((csc(d*x+c)-cot(d*x+c
)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(d*x+c)^3+15*I*(cot(d*x+c)-csc(d*x+c)+1)^(1/2)*(cot(d*x+c)-csc(d*x+c))^(1
/2)*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*(csc(d*x+c)-cot(d*x+c)+1)^(1/2)*cos(d*x+
c)^2-15*I*(cot(d*x+c)-csc(d*x+c)+1)^(1/2)*(cot(d*x+c)-csc(d*x+c))^(1/2)*(csc(d*x+c)-cot(d*x+c)+1)^(1/2)*Ellipt
icPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(d*x+c)^2-15*(cot(d*x+c)-csc(d*x+c)+1)^(1/2)*(c
ot(d*x+c)-csc(d*x+c))^(1/2)*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*(csc(d*x+c)-cot(
d*x+c)+1)^(1/2)*cos(d*x+c)^3-15*(cot(d*x+c)-csc(d*x+c)+1)^(1/2)*(cot(d*x+c)-csc(d*x+c))^(1/2)*(csc(d*x+c)-cot(
d*x+c)+1)^(1/2)*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(d*x+c)^3+50*(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)^3-15*(cot(d*x+c)-csc(d*x+c)+1)^(1/2)*(cot(d*x+c)-csc(d*x+c))^(1/2)*(csc(
d*x+c)-cot(d*x+c)+1)^(1/2)*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)^2-15*(
cot(d*x+c)-csc(d*x+c)+1)^(1/2)*(cot(d*x+c)-csc(d*x+c))^(1/2)*(csc(d*x+c)-cot(d*x+c)+1)^(1/2)*EllipticPi((csc(d
*x+c)-cot(d*x+c)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(d*x+c)^2+50*(csc(d*x+c)-cot(d*x+c)+1)^(1/2)*(cot(d*x+c)-c
sc(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+24*cos(d*x+c)^2*2^(1/2)*sin(d*x+c)-20*cos(d*x+c)*sin(d*x+c)*2^(1/2)+6*2^(1/2)*sin(d*x+c))*(e*tan(d*x+c)
)^(1/2)*e^5/(cos(d*x+c)-1)^3/(cos(d*x+c)+1)^3*sin(d*x+c)^3*tan(d*x+c)^2

Fricas [F(-1)]

Timed out. \[ \int \frac {(e \tan (c+d x))^{11/2}}{(a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate((e*tan(d*x+c))^(11/2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

Timed out

Sympy [F(-1)]

Timed out. \[ \int \frac {(e \tan (c+d x))^{11/2}}{(a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate((e*tan(d*x+c))**(11/2)/(a+a*sec(d*x+c))**2,x)

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \frac {(e \tan (c+d x))^{11/2}}{(a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate((e*tan(d*x+c))^(11/2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

Timed out

Giac [F]

\[ \int \frac {(e \tan (c+d x))^{11/2}}{(a+a \sec (c+d x))^2} \, dx=\int { \frac {\left (e \tan \left (d x + c\right )\right )^{\frac {11}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]

[In]

integrate((e*tan(d*x+c))^(11/2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((e*tan(d*x + c))^(11/2)/(a*sec(d*x + c) + a)^2, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \tan (c+d x))^{11/2}}{(a+a \sec (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^{11/2}}{a^2\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \,d x \]

[In]

int((e*tan(c + d*x))^(11/2)/(a + a/cos(c + d*x))^2,x)

[Out]

int((cos(c + d*x)^2*(e*tan(c + d*x))^(11/2))/(a^2*(cos(c + d*x) + 1)^2), x)