\(\int \frac {1}{(a+a \sec (c+d x)) (e \tan (c+d x))^{3/2}} \, dx\) [125]

Optimal result

Integrand size = 25, antiderivative size = 359 \[ \int \frac {1}{(a+a \sec (c+d x)) (e \tan (c+d x))^{3/2}} \, dx=\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d e^{3/2}}-\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d e^{3/2}}-\frac {\log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d e^{3/2}}+\frac {\log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d e^{3/2}}+\frac {2 e (1-\sec (c+d x))}{5 a d (e \tan (c+d x))^{5/2}}-\frac {2 (5-3 \sec (c+d x))}{5 a d e \sqrt {e \tan (c+d x)}}+\frac {6 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{5 a d e^2 \sqrt {\sin (2 c+2 d x)}}-\frac {6 \cos (c+d x) (e \tan (c+d x))^{3/2}}{5 a d e^3} \]

[Out]

1/2*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/a/d/e^(3/2)*2^(1/2)-1/2*arctan(1+2^(1/2)*(e*tan(d*x+c))^(1/
2)/e^(1/2))/a/d/e^(3/2)*2^(1/2)-1/4*ln(e^(1/2)-2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c))/a/d/e^(3/2)*2^
(1/2)+1/4*ln(e^(1/2)+2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c))/a/d/e^(3/2)*2^(1/2)-2/5*(5-3*sec(d*x+c))
/a/d/e/(e*tan(d*x+c))^(1/2)-6/5*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*P
i+d*x),2^(1/2))*(e*tan(d*x+c))^(1/2)/a/d/e^2/sin(2*d*x+2*c)^(1/2)+2/5*e*(1-sec(d*x+c))/a/d/(e*tan(d*x+c))^(5/2
)-6/5*cos(d*x+c)*(e*tan(d*x+c))^(3/2)/a/d/e^3

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 359, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3973, 3967, 3969, 3557, 335, 303, 1176, 631, 210, 1179, 642, 2693, 2695, 2652, 2719} \[ \int \frac {1}{(a+a \sec (c+d x)) (e \tan (c+d x))^{3/2}} \, dx=\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d e^{3/2}}-\frac {\arctan \left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} a d e^{3/2}}-\frac {\log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d e^{3/2}}+\frac {\log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d e^{3/2}}-\frac {6 \cos (c+d x) (e \tan (c+d x))^{3/2}}{5 a d e^3}+\frac {6 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{5 a d e^2 \sqrt {\sin (2 c+2 d x)}}-\frac {2 (5-3 \sec (c+d x))}{5 a d e \sqrt {e \tan (c+d x)}}+\frac {2 e (1-\sec (c+d x))}{5 a d (e \tan (c+d x))^{5/2}} \]

[In]

Int[1/((a + a*Sec[c + d*x])*(e*Tan[c + d*x])^(3/2)),x]

[Out]

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

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 303

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 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 2652

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 2693

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 2695

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 2719

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

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 3967

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

Rule 3969

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

Mathematica [C] (warning: unable to verify)

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

Time = 13.99 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.50 \[ \int \frac {1}{(a+a \sec (c+d x)) (e \tan (c+d x))^{3/2}} \, dx=-\frac {4 \csc (c+d x) \left (15 \cot ^2(c+d x)-3 \cot ^4(c+d x)+3 \cot ^4(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},-\frac {1}{2},-\frac {1}{4},-\tan ^2(c+d x)\right )-15 \cot ^2(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{4},\frac {3}{4},-\tan ^2(c+d x)\right )-5 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\tan ^2(c+d x)\right )+5 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},-\tan ^2(c+d x)\right )\right ) \left (1+\sqrt {\sec ^2(c+d x)}\right ) \sin ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {e \tan (c+d x)}}{15 a d e^2} \]

[In]

Integrate[1/((a + a*Sec[c + d*x])*(e*Tan[c + d*x])^(3/2)),x]

[Out]

(-4*Csc[c + d*x]*(15*Cot[c + d*x]^2 - 3*Cot[c + d*x]^4 + 3*Cot[c + d*x]^4*Hypergeometric2F1[-5/4, -1/2, -1/4,
-Tan[c + d*x]^2] - 15*Cot[c + d*x]^2*Hypergeometric2F1[-1/2, -1/4, 3/4, -Tan[c + d*x]^2] - 5*Hypergeometric2F1
[1/2, 3/4, 7/4, -Tan[c + d*x]^2] + 5*Hypergeometric2F1[3/4, 1, 7/4, -Tan[c + d*x]^2])*(1 + Sqrt[Sec[c + d*x]^2
])*Sin[(c + d*x)/2]^2*Sqrt[e*Tan[c + d*x]])/(15*a*d*e^2)

Maple [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 2.99 (sec) , antiderivative size = 1116, normalized size of antiderivative = 3.11

method	result	size
default	\(\text {Expression too large to display}\)	\(1116\)

[In]

int(1/(a+a*sec(d*x+c))/(e*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/10/a/d*2^(1/2)*(5*I*(csc(d*x+c)-cot(d*x+c)+1)^(1/2)*(2-2*csc(d*x+c)+2*cot(d*x+c))^(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))*((1-cos(d*x+c))*((1-cos(d*x+c))^2
*csc(d*x+c)^2-1)*csc(d*x+c))^(1/2)-5*I*(csc(d*x+c)-cot(d*x+c)+1)^(1/2)*(2-2*csc(d*x+c)+2*cot(d*x+c))^(1/2)*(co
t(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))*((1-cos(d*x+c))*(
(1-cos(d*x+c))^2*csc(d*x+c)^2-1)*csc(d*x+c))^(1/2)+12*(csc(d*x+c)-cot(d*x+c)+1)^(1/2)*(2-2*csc(d*x+c)+2*cot(d*
x+c))^(1/2)*(cot(d*x+c)-csc(d*x+c))^(1/2)*EllipticE((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2*2^(1/2))*((1-cos(d*x+c
))*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)*csc(d*x+c))^(1/2)-6*(csc(d*x+c)-cot(d*x+c)+1)^(1/2)*(2-2*csc(d*x+c)+2*cot
(d*x+c))^(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))*((1-cos(d*
x+c))*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)*csc(d*x+c))^(1/2)-5*(csc(d*x+c)-cot(d*x+c)+1)^(1/2)*(2-2*csc(d*x+c)+2*
cot(d*x+c))^(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))*((1-cos(d*x+c))*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)*csc(d*x+c))^(1/2)-5*(csc(d*x+c)-cot(d*x+c)+1)^(1/2)*(2-2
*csc(d*x+c)+2*cot(d*x+c))^(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))*((1-cos(d*x+c))*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)*csc(d*x+c))^(1/2)+((1-cos(d*x+c))*((1-cos(
d*x+c))^2*csc(d*x+c)^2-1)*csc(d*x+c))^(1/2)*(1-cos(d*x+c))^4*csc(d*x+c)^4-((1-cos(d*x+c))*((1-cos(d*x+c))^2*cs
c(d*x+c)^2-1)*csc(d*x+c))^(1/2)*(1-cos(d*x+c))^2*csc(d*x+c)^2-5*(1-cos(d*x+c))^2*((1-cos(d*x+c))^3*csc(d*x+c)^
3+cot(d*x+c)-csc(d*x+c))^(1/2)*csc(d*x+c)^2+5*((1-cos(d*x+c))^3*csc(d*x+c)^3+cot(d*x+c)-csc(d*x+c))^(1/2))*(1-
cos(d*x+c))/((1-cos(d*x+c))^3*csc(d*x+c)^3+cot(d*x+c)-csc(d*x+c))^(1/2)/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^2/(-
e/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)*(-cot(d*x+c)+csc(d*x+c)))^(3/2)*csc(d*x+c)

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{(a+a \sec (c+d x)) (e \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(1/(a+a*sec(d*x+c))/(e*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

Timed out

Sympy [F]

\[ \int \frac {1}{(a+a \sec (c+d x)) (e \tan (c+d x))^{3/2}} \, dx=\frac {\int \frac {1}{\left (e \tan {\left (c + d x \right )}\right )^{\frac {3}{2}} \sec {\left (c + d x \right )} + \left (e \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx}{a} \]

[In]

integrate(1/(a+a*sec(d*x+c))/(e*tan(d*x+c))**(3/2),x)

[Out]

Integral(1/((e*tan(c + d*x))**(3/2)*sec(c + d*x) + (e*tan(c + d*x))**(3/2)), x)/a

Maxima [F]

\[ \int \frac {1}{(a+a \sec (c+d x)) (e \tan (c+d x))^{3/2}} \, dx=\int { \frac {1}{{\left (a \sec \left (d x + c\right ) + a\right )} \left (e \tan \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate(1/(a+a*sec(d*x+c))/(e*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((a*sec(d*x + c) + a)*(e*tan(d*x + c))^(3/2)), x)

Giac [F]

\[ \int \frac {1}{(a+a \sec (c+d x)) (e \tan (c+d x))^{3/2}} \, dx=\int { \frac {1}{{\left (a \sec \left (d x + c\right ) + a\right )} \left (e \tan \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate(1/(a+a*sec(d*x+c))/(e*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(1/((a*sec(d*x + c) + a)*(e*tan(d*x + c))^(3/2)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+a \sec (c+d x)) (e \tan (c+d x))^{3/2}} \, dx=\int \frac {\cos \left (c+d\,x\right )}{a\,{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}\,\left (\cos \left (c+d\,x\right )+1\right )} \,d x \]

[In]

int(1/((e*tan(c + d*x))^(3/2)*(a + a/cos(c + d*x))),x)

[Out]

int(cos(c + d*x)/(a*(e*tan(c + d*x))^(3/2)*(cos(c + d*x) + 1)), x)