3.123 \(\int \frac {\sqrt {e \tan (c+d x)}}{a+a \sec (c+d x)} \, dx\)
Optimal. Leaf size=315 \[ -\frac {\sqrt {e} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}+\frac {\sqrt {e} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} a d}+\frac {\sqrt {e} \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d}-\frac {\sqrt {e} \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d}+\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{a d e}+\frac {2 e (1-\sec (c+d x))}{a d \sqrt {e \tan (c+d x)}}-\frac {2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{a d \sqrt {\sin (2 c+2 d x)}} \]
[Out]
-1/2*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))*e^(1/2)/a/d*2^(1/2)+1/2*arctan(1+2^(1/2)*(e*tan(d*x+c))^(1
/2)/e^(1/2))*e^(1/2)/a/d*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))*e^(1/2)/a/d*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))*e^(1/2)/a/d*2^(1/2)+2*e*(1-sec(d*x+c))/
a/d/(e*tan(d*x+c))^(1/2)+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)/a/d/sin(2*d*x+2*c)^(1/2)+2*cos(d*x+c)*(e*tan(d*x+c))^(3/2)/a/d/e
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Rubi [A] time = 0.38, antiderivative size = 315, normalized size of antiderivative = 1.00,
number of steps used = 18, number of rules used = 15, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used
= {3888, 3882, 3884, 3476, 329, 297, 1162, 617, 204, 1165, 628, 2613, 2615, 2572, 2639}
\[ -\frac {\sqrt {e} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}+\frac {\sqrt {e} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} a d}+\frac {\sqrt {e} \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d}-\frac {\sqrt {e} \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d}+\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{a d e}+\frac {2 e (1-\sec (c+d x))}{a d \sqrt {e \tan (c+d x)}}-\frac {2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{a d \sqrt {\sin (2 c+2 d x)}} \]
Antiderivative was successfully verified.
[In]
Int[Sqrt[e*Tan[c + d*x]]/(a + a*Sec[c + d*x]),x]
[Out]
-((Sqrt[e]*ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*a*d)) + (Sqrt[e]*ArcTan[1 + (Sqrt[2]*S
qrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*a*d) + (Sqrt[e]*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] - Sqrt[2]*Sqrt[e*Ta
n[c + d*x]]])/(2*Sqrt[2]*a*d) - (Sqrt[e]*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(
2*Sqrt[2]*a*d) + (2*e*(1 - Sec[c + d*x]))/(a*d*Sqrt[e*Tan[c + d*x]]) - (2*Cos[c + d*x]*EllipticE[c - Pi/4 + d*
x, 2]*Sqrt[e*Tan[c + d*x]])/(a*d*Sqrt[Sin[2*c + 2*d*x]]) + (2*Cos[c + d*x]*(e*Tan[c + d*x])^(3/2))/(a*d*e)
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 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 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 3882
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 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 3888
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*} \int \frac {\sqrt {e \tan (c+d x)}}{a+a \sec (c+d x)} \, dx &=\frac {e^2 \int \frac {-a+a \sec (c+d x)}{(e \tan (c+d x))^{3/2}} \, dx}{a^2}\\ &=\frac {2 e (1-\sec (c+d x))}{a d \sqrt {e \tan (c+d x)}}+\frac {2 \int \left (\frac {a}{2}+\frac {1}{2} a \sec (c+d x)\right ) \sqrt {e \tan (c+d x)} \, dx}{a^2}\\ &=\frac {2 e (1-\sec (c+d x))}{a d \sqrt {e \tan (c+d x)}}+\frac {\int \sqrt {e \tan (c+d x)} \, dx}{a}+\frac {\int \sec (c+d x) \sqrt {e \tan (c+d x)} \, dx}{a}\\ &=\frac {2 e (1-\sec (c+d x))}{a d \sqrt {e \tan (c+d x)}}+\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{a d e}-\frac {2 \int \cos (c+d x) \sqrt {e \tan (c+d x)} \, dx}{a}+\frac {e \operatorname {Subst}\left (\int \frac {\sqrt {x}}{e^2+x^2} \, dx,x,e \tan (c+d x)\right )}{a d}\\ &=\frac {2 e (1-\sec (c+d x))}{a d \sqrt {e \tan (c+d x)}}+\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{a d e}+\frac {(2 e) \operatorname {Subst}\left (\int \frac {x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a d}-\frac {\left (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}{a \sqrt {\sin (c+d x)}}\\ &=\frac {2 e (1-\sec (c+d x))}{a d \sqrt {e \tan (c+d x)}}+\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{a d e}-\frac {e \operatorname {Subst}\left (\int \frac {e-x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a d}+\frac {e \operatorname {Subst}\left (\int \frac {e+x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a d}-\frac {\left (2 \cos (c+d x) \sqrt {e \tan (c+d x)}\right ) \int \sqrt {\sin (2 c+2 d x)} \, dx}{a \sqrt {\sin (2 c+2 d x)}}\\ &=\frac {2 e (1-\sec (c+d x))}{a d \sqrt {e \tan (c+d x)}}-\frac {2 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{a d \sqrt {\sin (2 c+2 d x)}}+\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{a d e}+\frac {\sqrt {e} \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} a d}+\frac {\sqrt {e} \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} a d}+\frac {e \operatorname {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}+\frac {e \operatorname {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}\\ &=\frac {\sqrt {e} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}-\frac {\sqrt {e} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}+\frac {2 e (1-\sec (c+d x))}{a d \sqrt {e \tan (c+d x)}}-\frac {2 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{a d \sqrt {\sin (2 c+2 d x)}}+\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{a d e}+\frac {\sqrt {e} \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} a d}-\frac {\sqrt {e} \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} a d}\\ &=-\frac {\sqrt {e} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}+\frac {\sqrt {e} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}+\frac {\sqrt {e} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}-\frac {\sqrt {e} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}+\frac {2 e (1-\sec (c+d x))}{a d \sqrt {e \tan (c+d x)}}-\frac {2 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{a d \sqrt {\sin (2 c+2 d x)}}+\frac {2 \cos (c+d x) (e \tan (c+d x))^{3/2}}{a d e}\\ \end {align*}
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Mathematica [C] time = 8.09, size = 2715, normalized size = 8.62 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Sqrt[e*Tan[c + d*x]]/(a + a*Sec[c + d*x]),x]
[Out]
(Cos[c/2 + (d*x)/2]^2*Sec[c + d*x]*((-2*Cos[c/2]*Cos[d*x]*Sec[2*c]*(4*Sin[c/2] + Sin[(3*c)/2] + Sin[(5*c)/2]))
/(d*(1 + 2*Cos[c])) - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*Sin[(d*x)/2])/d - ((-2 - 5*Cos[c] - 6*Cos[2*c] + Cos[3*c]
)*Sec[2*c]*Sin[d*x])/(d*(1 + 2*Cos[c])) - (4*Tan[c/2])/d)*Sqrt[e*Tan[c + d*x]])/(a + a*Sec[c + d*x]) + ((E^((2
*I)*c)*Sqrt[-1 + E^((4*I)*(c + d*x))]*ArcTan[Sqrt[-1 + E^((4*I)*(c + d*x))]] - 2*Sqrt[-1 + E^((2*I)*(c + d*x))
]*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]])*Cos[c/2 +
(d*x)/2]^2*Sec[2*c]*Sec[c + d*x]*Sqrt[e*Tan[c + d*x]])/(2*d*E^(I*c)*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1
+ E^((2*I)*(c + d*x)))]*(1 + E^((2*I)*(c + d*x)))*(1 + 2*Cos[c])*(a + a*Sec[c + d*x])*Sqrt[Tan[c + d*x]]) - (
(-(E^((4*I)*c)*Sqrt[-1 + E^((4*I)*(c + d*x))]*ArcTan[Sqrt[-1 + E^((4*I)*(c + d*x))]]) + 2*Sqrt[-1 + E^((2*I)*(
c + d*x))]*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]])*
Cos[c/2 + (d*x)/2]^2*Sec[2*c]*Sec[c + d*x]*Sqrt[e*Tan[c + d*x]])/(2*d*E^((2*I)*c)*Sqrt[((-I)*(-1 + E^((2*I)*(c
+ d*x))))/(1 + E^((2*I)*(c + d*x)))]*(1 + E^((2*I)*(c + d*x)))*(1 + 2*Cos[c])*(a + a*Sec[c + d*x])*Sqrt[Tan[c
+ d*x]]) - ((-(E^((6*I)*c)*Sqrt[-1 + E^((4*I)*(c + d*x))]*ArcTan[Sqrt[-1 + E^((4*I)*(c + d*x))]]) + 2*Sqrt[-1
+ E^((2*I)*(c + d*x))]*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c
+ d*x)))]])*Cos[c/2 + (d*x)/2]^2*Sec[2*c]*Sec[c + d*x]*Sqrt[e*Tan[c + d*x]])/(2*d*E^((3*I)*c)*Sqrt[((-I)*(-1
+ E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*(1 + E^((2*I)*(c + d*x)))*(1 + 2*Cos[c])*(a + a*Sec[c + d*x
])*Sqrt[Tan[c + d*x]]) + ((Sqrt[-1 + E^((4*I)*(c + d*x))]*ArcTan[Sqrt[-1 + E^((4*I)*(c + d*x))]] - 2*E^((2*I)*
c)*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E
^((2*I)*(c + d*x)))]])*Cos[c/2 + (d*x)/2]^2*Sec[2*c]*Sec[c + d*x]*Sqrt[e*Tan[c + d*x]])/(2*d*E^(I*c)*Sqrt[((-I
)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*(1 + E^((2*I)*(c + d*x)))*(1 + 2*Cos[c])*(a + a*Sec[c
+ d*x])*Sqrt[Tan[c + d*x]]) + ((Sqrt[-1 + E^((4*I)*(c + d*x))]*ArcTan[Sqrt[-1 + E^((4*I)*(c + d*x))]] - 2*E^(
(4*I)*c)*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/
(1 + E^((2*I)*(c + d*x)))]])*Cos[c/2 + (d*x)/2]^2*Sec[2*c]*Sec[c + d*x]*Sqrt[e*Tan[c + d*x]])/(2*d*E^((2*I)*c)
*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*(1 + E^((2*I)*(c + d*x)))*(1 + 2*Cos[c])*(a
+ a*Sec[c + d*x])*Sqrt[Tan[c + d*x]]) + ((Sqrt[-1 + E^((4*I)*(c + d*x))]*ArcTan[Sqrt[-1 + E^((4*I)*(c + d*x))
]] - 2*E^((6*I)*c)*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[Sqrt[(-1 + E^((2*I)*(c
+ d*x)))/(1 + E^((2*I)*(c + d*x)))]])*Cos[c/2 + (d*x)/2]^2*Sec[2*c]*Sec[c + d*x]*Sqrt[e*Tan[c + d*x]])/(2*d*E
^((3*I)*c)*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*(1 + E^((2*I)*(c + d*x)))*(1 + 2*
Cos[c])*(a + a*Sec[c + d*x])*Sqrt[Tan[c + d*x]]) + (Cos[c/2 + (d*x)/2]^2*(3 - 3*E^((4*I)*(c + d*x)) + E^((4*I)
*(c + d*x))*(1 + E^((2*I)*c))*Sqrt[1 - E^((4*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, E^((4*I)*(c + d*x
))])*Sec[2*c]*Sec[c + d*x]*Sqrt[e*Tan[c + d*x]])/(3*d*E^(I*(2*c + d*x))*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))
/(1 + E^((2*I)*(c + d*x)))]*(1 + E^((2*I)*(c + d*x)))*(1 + 2*Cos[c])*(a + a*Sec[c + d*x])*Sqrt[Tan[c + d*x]])
+ (Cos[c/2 + (d*x)/2]^2*(3 - 3*E^((4*I)*(c + d*x)) + E^((2*I)*(c + 2*d*x))*(1 + E^((2*I)*c))*Sqrt[1 - E^((4*I)
*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, E^((4*I)*(c + d*x))])*Sec[2*c]*Sec[c + d*x]*Sqrt[e*Tan[c + d*x]]
)/(3*d*E^(I*d*x)*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*(1 + E^((2*I)*(c + d*x)))*(
1 + 2*Cos[c])*(a + a*Sec[c + d*x])*Sqrt[Tan[c + d*x]]) + (5*E^(I*(c - d*x))*Cos[c/2 + (d*x)/2]^2*(3 - 3*E^((4*
I)*(c + d*x)) + E^((4*I)*d*x)*(1 + E^((4*I)*c))*Sqrt[1 - E^((4*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4,
E^((4*I)*(c + d*x))])*Sec[2*c]*Sec[c + d*x]*Sqrt[e*Tan[c + d*x]])/(6*d*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))
/(1 + E^((2*I)*(c + d*x)))]*(1 + E^((2*I)*(c + d*x)))*(1 + 2*Cos[c])*(a + a*Sec[c + d*x])*Sqrt[Tan[c + d*x]])
- (Cos[c/2 + (d*x)/2]^2*(3 - 3*E^((4*I)*(c + d*x)) + E^((4*I)*(c + d*x))*(1 + E^((4*I)*c))*Sqrt[1 - E^((4*I)*(
c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, E^((4*I)*(c + d*x))])*Sec[2*c]*Sec[c + d*x]*Sqrt[e*Tan[c + d*x]])/
(6*d*E^(I*(3*c + d*x))*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*(1 + E^((2*I)*(c + d*
x)))*(1 + 2*Cos[c])*(a + a*Sec[c + d*x])*Sqrt[Tan[c + d*x]]) + (2*Cos[c/2 + (d*x)/2]^2*(-3*E^((2*I)*c)*(-1 + E
^((4*I)*(c + d*x))) + E^((4*I)*d*x)*(1 + E^((6*I)*c))*Sqrt[1 - E^((4*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4
, 7/4, E^((4*I)*(c + d*x))])*Sec[2*c]*Sec[c + d*x]*Sqrt[e*Tan[c + d*x]])/(3*d*E^(I*d*x)*Sqrt[((-I)*(-1 + E^((2
*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*(1 + E^((2*I)*(c + d*x)))*(1 + 2*Cos[c])*(a + a*Sec[c + d*x])*Sqrt
[Tan[c + d*x]])
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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((e*tan(d*x+c))^(1/2)/(a+a*sec(d*x+c)),x, algorithm="fricas")
[Out]
Timed out
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {e \tan \left (d x + c\right )}}{a \sec \left (d x + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*tan(d*x+c))^(1/2)/(a+a*sec(d*x+c)),x, algorithm="giac")
[Out]
integrate(sqrt(e*tan(d*x + c))/(a*sec(d*x + c) + a), x)
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maple [C] time = 1.77, size = 352, normalized size = 1.12 \[ -\frac {\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {e \sin \left (d x +c \right )}{\cos \left (d x +c \right )}}\, \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right ) \left (i \EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-i \EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+4 \EllipticE \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right )-2 \EllipticF \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right )-\EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-\EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )\right ) \sqrt {2}}{2 a d \sin \left (d x +c \right )^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((e*tan(d*x+c))^(1/2)/(a+a*sec(d*x+c)),x)
[Out]
-1/2/a/d*((-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)*(e*sin(d*x+c)/cos(d*x+c))^(1/2)*(1+cos(d*x+c))^2*(-1+cos(d*x+c))*(I*EllipticPi(((1-co
s(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))-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))+4*EllipticE(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))-2*Ellip
ticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(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))-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))
)/sin(d*x+c)^3*2^(1/2)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {e \tan \left (d x + c\right )}}{a \sec \left (d x + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*tan(d*x+c))^(1/2)/(a+a*sec(d*x+c)),x, algorithm="maxima")
[Out]
integrate(sqrt(e*tan(d*x + c))/(a*sec(d*x + c) + a), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\cos \left (c+d\,x\right )\,\sqrt {e\,\mathrm {tan}\left (c+d\,x\right )}}{a\,\left (\cos \left (c+d\,x\right )+1\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((e*tan(c + d*x))^(1/2)/(a + a/cos(c + d*x)),x)
[Out]
int((cos(c + d*x)*(e*tan(c + d*x))^(1/2))/(a*(cos(c + d*x) + 1)), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sqrt {e \tan {\left (c + d x \right )}}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*tan(d*x+c))**(1/2)/(a+a*sec(d*x+c)),x)
[Out]
Integral(sqrt(e*tan(c + d*x))/(sec(c + d*x) + 1), x)/a
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