3.366 \(\int \frac {\sqrt {d \tan (e+f x)}}{(a+a \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=278 \[ -\frac {\sqrt {d} \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 f}-\frac {\sqrt {d} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 f}+\frac {\sqrt {d} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{2 \sqrt {2} a^2 f}-\frac {\sqrt {d \tan (e+f x)}}{2 f \left (a^2 \tan (e+f x)+a^2\right )}-\frac {\sqrt {d} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{4 \sqrt {2} a^2 f}+\frac {\sqrt {d} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{4 \sqrt {2} a^2 f} \]

[Out]

-1/2*arctan((d*tan(f*x+e))^(1/2)/d^(1/2))*d^(1/2)/a^2/f-1/4*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))*d^(
1/2)/a^2/f*2^(1/2)+1/4*arctan(1+2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))*d^(1/2)/a^2/f*2^(1/2)-1/8*ln(d^(1/2)-2^(
1/2)*(d*tan(f*x+e))^(1/2)+d^(1/2)*tan(f*x+e))*d^(1/2)/a^2/f*2^(1/2)+1/8*ln(d^(1/2)+2^(1/2)*(d*tan(f*x+e))^(1/2
)+d^(1/2)*tan(f*x+e))*d^(1/2)/a^2/f*2^(1/2)-1/2*(d*tan(f*x+e))^(1/2)/f/(a^2+a^2*tan(f*x+e))

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Rubi [A]  time = 0.48, antiderivative size = 278, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 14, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.560, Rules used = {3568, 3653, 12, 3476, 329, 211, 1165, 628, 1162, 617, 204, 3634, 63, 205} \[ -\frac {\sqrt {d} \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 f}-\frac {\sqrt {d} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 f}+\frac {\sqrt {d} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{2 \sqrt {2} a^2 f}-\frac {\sqrt {d \tan (e+f x)}}{2 f \left (a^2 \tan (e+f x)+a^2\right )}-\frac {\sqrt {d} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{4 \sqrt {2} a^2 f}+\frac {\sqrt {d} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{4 \sqrt {2} a^2 f} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[d*Tan[e + f*x]]/(a + a*Tan[e + f*x])^2,x]

[Out]

-(Sqrt[d]*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/(2*a^2*f) - (Sqrt[d]*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])
/Sqrt[d]])/(2*Sqrt[2]*a^2*f) + (Sqrt[d]*ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(2*Sqrt[2]*a^2*f)
- (Sqrt[d]*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(4*Sqrt[2]*a^2*f) + (Sqrt[d]*Lo
g[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(4*Sqrt[2]*a^2*f) - Sqrt[d*Tan[e + f*x]]/(2*
f*(a^2 + a^2*Tan[e + f*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 205

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

Rule 211

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 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 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 3568

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n)/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(a^2
+ b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1)*Simp[a*c*(m + 1) - b*d*n - (b*c - a*d)*
(m + 1)*Tan[e + f*x] - b*d*(m + n + 1)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] && GtQ[n, 0] && IntegerQ[2*m]

Rule 3634

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]

Rule 3653

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^2), Int[(c + d*Tan[e + f*
x])^n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 +
b^2), Int[((c + d*Tan[e + f*x])^n*(1 + Tan[e + f*x]^2))/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e,
f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !GtQ[n, 0] &&  !LeQ[n, -
1]

Rubi steps

\begin {align*} \int \frac {\sqrt {d \tan (e+f x)}}{(a+a \tan (e+f x))^2} \, dx &=-\frac {\sqrt {d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )}-\frac {\int \frac {-\frac {a d}{2}-a d \tan (e+f x)+\frac {1}{2} a d \tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{2 a^2}\\ &=-\frac {\sqrt {d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )}-\frac {\int -\frac {2 a^2 d}{\sqrt {d \tan (e+f x)}} \, dx}{4 a^4}-\frac {d \int \frac {1+\tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{4 a}\\ &=-\frac {\sqrt {d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )}+\frac {d \int \frac {1}{\sqrt {d \tan (e+f x)}} \, dx}{2 a^2}-\frac {d \operatorname {Subst}\left (\int \frac {1}{\sqrt {d x} (a+a x)} \, dx,x,\tan (e+f x)\right )}{4 a f}\\ &=-\frac {\sqrt {d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {1}{a+\frac {a x^2}{d}} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{2 a f}+\frac {d^2 \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (d^2+x^2\right )} \, dx,x,d \tan (e+f x)\right )}{2 a^2 f}\\ &=-\frac {\sqrt {d} \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 f}-\frac {\sqrt {d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )}+\frac {d^2 \operatorname {Subst}\left (\int \frac {1}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^2 f}\\ &=-\frac {\sqrt {d} \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 f}-\frac {\sqrt {d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )}+\frac {d \operatorname {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{2 a^2 f}+\frac {d \operatorname {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{2 a^2 f}\\ &=-\frac {\sqrt {d} \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 f}-\frac {\sqrt {d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )}-\frac {\sqrt {d} \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}+2 x}{-d-\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{4 \sqrt {2} a^2 f}-\frac {\sqrt {d} \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}-2 x}{-d+\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{4 \sqrt {2} a^2 f}+\frac {d \operatorname {Subst}\left (\int \frac {1}{d-\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{4 a^2 f}+\frac {d \operatorname {Subst}\left (\int \frac {1}{d+\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{4 a^2 f}\\ &=-\frac {\sqrt {d} \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 f}-\frac {\sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{4 \sqrt {2} a^2 f}+\frac {\sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{4 \sqrt {2} a^2 f}-\frac {\sqrt {d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )}+\frac {\sqrt {d} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 f}-\frac {\sqrt {d} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 f}\\ &=-\frac {\sqrt {d} \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 f}-\frac {\sqrt {d} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 f}+\frac {\sqrt {d} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 f}-\frac {\sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{4 \sqrt {2} a^2 f}+\frac {\sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{4 \sqrt {2} a^2 f}-\frac {\sqrt {d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )}\\ \end {align*}

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Mathematica [A]  time = 1.38, size = 192, normalized size = 0.69 \[ \frac {\sqrt {d \tan (e+f x)} \left (-\frac {2 \sqrt {2} \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right )-2 \sqrt {2} \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (e+f x)}+1\right )+4 \tan ^{-1}\left (\sqrt {\tan (e+f x)}\right )+\sqrt {2} \log \left (-\tan (e+f x)+\sqrt {2} \sqrt {\tan (e+f x)}-1\right )-\sqrt {2} \log \left (\tan (e+f x)+\sqrt {2} \sqrt {\tan (e+f x)}+1\right )}{2 \sqrt {\tan (e+f x)}}-\frac {2 \cos (e+f x)}{\sin (e+f x)+\cos (e+f x)}\right )}{4 a^2 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[d*Tan[e + f*x]]/(a + a*Tan[e + f*x])^2,x]

[Out]

(((-2*Cos[e + f*x])/(Cos[e + f*x] + Sin[e + f*x]) - (2*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]]] - 2*Sqrt
[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[e + f*x]]] + 4*ArcTan[Sqrt[Tan[e + f*x]]] + Sqrt[2]*Log[-1 + Sqrt[2]*Sqrt[Tan[
e + f*x]] - Tan[e + f*x]] - Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]])/(2*Sqrt[Tan[e + f*x]])
)*Sqrt[d*Tan[e + f*x]])/(4*a^2*f)

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fricas [B]  time = 0.98, size = 1662, normalized size = 5.98 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*((2*cos(f*x + e)*sin(f*x + e) + 1)*sqrt(-d)*log(-(6*d^2*cos(f*x + e)*sin(f*x + e) - d^2 - 4*(d*cos(f*x +
e)^2 - d*cos(f*x + e)*sin(f*x + e))*sqrt(-d)*sqrt(d*sin(f*x + e)/cos(f*x + e)))/(2*cos(f*x + e)*sin(f*x + e) +
1)) - 4*(2*sqrt(2)*a^2*f*cos(f*x + e)*sin(f*x + e) + sqrt(2)*a^2*f)*(d^2/(a^8*f^4))^(1/4)*arctan(-(sqrt(2)*a^
6*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(d^2/(a^8*f^4))^(3/4) - sqrt(2)*a^6*f^3*sqrt((a^4*f^2*sqrt(d^2/(a^8*f^
4))*cos(f*x + e) + sqrt(2)*a^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(d^2/(a^8*f^4))^(1/4)*cos(f*x + e) + d*sin(
f*x + e))/cos(f*x + e))*(d^2/(a^8*f^4))^(3/4) + d^2)/d^2) - 4*(2*sqrt(2)*a^2*f*cos(f*x + e)*sin(f*x + e) + sqr
t(2)*a^2*f)*(d^2/(a^8*f^4))^(1/4)*arctan(-(sqrt(2)*a^6*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(d^2/(a^8*f^4))^(
3/4) - sqrt(2)*a^6*f^3*sqrt((a^4*f^2*sqrt(d^2/(a^8*f^4))*cos(f*x + e) - sqrt(2)*a^2*f*sqrt(d*sin(f*x + e)/cos(
f*x + e))*(d^2/(a^8*f^4))^(1/4)*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))*(d^2/(a^8*f^4))^(3/4) - d^2)/d^2)
+ (2*sqrt(2)*a^2*f*cos(f*x + e)*sin(f*x + e) + sqrt(2)*a^2*f)*(d^2/(a^8*f^4))^(1/4)*log((a^4*f^2*sqrt(d^2/(a^
8*f^4))*cos(f*x + e) + sqrt(2)*a^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(d^2/(a^8*f^4))^(1/4)*cos(f*x + e) + d*
sin(f*x + e))/cos(f*x + e)) - (2*sqrt(2)*a^2*f*cos(f*x + e)*sin(f*x + e) + sqrt(2)*a^2*f)*(d^2/(a^8*f^4))^(1/4
)*log((a^4*f^2*sqrt(d^2/(a^8*f^4))*cos(f*x + e) - sqrt(2)*a^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(d^2/(a^8*f^
4))^(1/4)*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e)) - 4*(cos(f*x + e)^2 + cos(f*x + e)*sin(f*x + e))*sqrt(d
*sin(f*x + e)/cos(f*x + e)))/(2*a^2*f*cos(f*x + e)*sin(f*x + e) + a^2*f), -1/8*(4*(2*cos(f*x + e)*sin(f*x + e)
+ 1)*sqrt(d)*arctan(sqrt(d*sin(f*x + e)/cos(f*x + e))/sqrt(d)) + 4*(2*sqrt(2)*a^2*f*cos(f*x + e)*sin(f*x + e)
+ sqrt(2)*a^2*f)*(d^2/(a^8*f^4))^(1/4)*arctan(-(sqrt(2)*a^6*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(d^2/(a^8*f
^4))^(3/4) - sqrt(2)*a^6*f^3*sqrt((a^4*f^2*sqrt(d^2/(a^8*f^4))*cos(f*x + e) + sqrt(2)*a^2*f*sqrt(d*sin(f*x + e
)/cos(f*x + e))*(d^2/(a^8*f^4))^(1/4)*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))*(d^2/(a^8*f^4))^(3/4) + d^2
)/d^2) + 4*(2*sqrt(2)*a^2*f*cos(f*x + e)*sin(f*x + e) + sqrt(2)*a^2*f)*(d^2/(a^8*f^4))^(1/4)*arctan(-(sqrt(2)*
a^6*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(d^2/(a^8*f^4))^(3/4) - sqrt(2)*a^6*f^3*sqrt((a^4*f^2*sqrt(d^2/(a^8*
f^4))*cos(f*x + e) - sqrt(2)*a^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(d^2/(a^8*f^4))^(1/4)*cos(f*x + e) + d*si
n(f*x + e))/cos(f*x + e))*(d^2/(a^8*f^4))^(3/4) - d^2)/d^2) - (2*sqrt(2)*a^2*f*cos(f*x + e)*sin(f*x + e) + sqr
t(2)*a^2*f)*(d^2/(a^8*f^4))^(1/4)*log((a^4*f^2*sqrt(d^2/(a^8*f^4))*cos(f*x + e) + sqrt(2)*a^2*f*sqrt(d*sin(f*x
+ e)/cos(f*x + e))*(d^2/(a^8*f^4))^(1/4)*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e)) + (2*sqrt(2)*a^2*f*cos(
f*x + e)*sin(f*x + e) + sqrt(2)*a^2*f)*(d^2/(a^8*f^4))^(1/4)*log((a^4*f^2*sqrt(d^2/(a^8*f^4))*cos(f*x + e) - s
qrt(2)*a^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(d^2/(a^8*f^4))^(1/4)*cos(f*x + e) + d*sin(f*x + e))/cos(f*x +
e)) + 4*(cos(f*x + e)^2 + cos(f*x + e)*sin(f*x + e))*sqrt(d*sin(f*x + e)/cos(f*x + e)))/(2*a^2*f*cos(f*x + e)*
sin(f*x + e) + a^2*f)]

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giac [A]  time = 1.20, size = 260, normalized size = 0.94 \[ \frac {\frac {2 \, \sqrt {2} d \sqrt {{\left | d \right |}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{a^{2} f} + \frac {2 \, \sqrt {2} d \sqrt {{\left | d \right |}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{a^{2} f} + \frac {\sqrt {2} d \sqrt {{\left | d \right |}} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{a^{2} f} - \frac {\sqrt {2} d \sqrt {{\left | d \right |}} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{a^{2} f} - \frac {4 \, d^{\frac {3}{2}} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a^{2} f} - \frac {4 \, \sqrt {d \tan \left (f x + e\right )} d^{2}}{{\left (d \tan \left (f x + e\right ) + d\right )} a^{2} f}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*sqrt(2)*d*sqrt(abs(d))*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) + 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d)))
/(a^2*f) + 2*sqrt(2)*d*sqrt(abs(d))*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) - 2*sqrt(d*tan(f*x + e)))/sqrt(a
bs(d)))/(a^2*f) + sqrt(2)*d*sqrt(abs(d))*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d)) + abs(
d))/(a^2*f) - sqrt(2)*d*sqrt(abs(d))*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d)) + abs(d))/
(a^2*f) - 4*d^(3/2)*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/(a^2*f) - 4*sqrt(d*tan(f*x + e))*d^2/((d*tan(f*x + e)
+ d)*a^2*f))/d

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maple [A]  time = 0.29, size = 223, normalized size = 0.80 \[ -\frac {d \sqrt {d \tan \left (f x +e \right )}}{2 f \,a^{2} \left (d \tan \left (f x +e \right )+d \right )}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right ) \sqrt {d}}{2 a^{2} f}+\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{8 f \,a^{2}}+\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{4 f \,a^{2}}-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{4 f \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(f*x+e))^(1/2)/(a+a*tan(f*x+e))^2,x)

[Out]

-1/2/f/a^2*d*(d*tan(f*x+e))^(1/2)/(d*tan(f*x+e)+d)-1/2*arctan((d*tan(f*x+e))^(1/2)/d^(1/2))*d^(1/2)/a^2/f+1/8/
f/a^2*(d^2)^(1/4)*2^(1/2)*ln((d*tan(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*tan(f*x+e)
-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))+1/4/f/a^2*(d^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(d^2)^(1/
4)*(d*tan(f*x+e))^(1/2)+1)-1/4/f/a^2*(d^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)

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maxima [A]  time = 0.62, size = 218, normalized size = 0.78 \[ -\frac {\frac {4 \, \sqrt {d \tan \left (f x + e\right )} d^{2}}{a^{2} d \tan \left (f x + e\right ) + a^{2} d} + \frac {4 \, d^{\frac {3}{2}} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a^{2}} - \frac {2 \, \sqrt {2} d^{\frac {3}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right ) + 2 \, \sqrt {2} d^{\frac {3}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right ) + \sqrt {2} d^{\frac {3}{2}} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right ) - \sqrt {2} d^{\frac {3}{2}} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{a^{2}}}{8 \, d f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(4*sqrt(d*tan(f*x + e))*d^2/(a^2*d*tan(f*x + e) + a^2*d) + 4*d^(3/2)*arctan(sqrt(d*tan(f*x + e))/sqrt(d))
/a^2 - (2*sqrt(2)*d^(3/2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e)))/sqrt(d)) + 2*sqrt(2)*d
^(3/2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d)) + sqrt(2)*d^(3/2)*log(d*tan(f*x
+ e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d) - sqrt(2)*d^(3/2)*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*
x + e))*sqrt(d) + d))/a^2)/(d*f)

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mupad [B]  time = 4.58, size = 367, normalized size = 1.32 \[ -\frac {\mathrm {atan}\left (\frac {4\,d^{12}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {d^2}{a^8\,f^4}\right )}^{1/4}}{\frac {4\,d^{13}}{a^2\,f}-4\,a^2\,d^{12}\,f\,\sqrt {-\frac {d^2}{a^8\,f^4}}}+\frac {4\,d^{11}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {d^2}{a^8\,f^4}\right )}^{3/4}}{\frac {4\,d^{13}}{a^6\,f^3}-\frac {4\,d^{12}\,\sqrt {-\frac {d^2}{a^8\,f^4}}}{a^2\,f}}\right )\,{\left (-\frac {d^2}{a^8\,f^4}\right )}^{1/4}}{2}-\mathrm {atan}\left (\frac {d^{12}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {d^2}{256\,a^8\,f^4}\right )}^{1/4}\,16{}\mathrm {i}}{\frac {4\,d^{13}}{a^2\,f}+64\,a^2\,d^{12}\,f\,\sqrt {-\frac {d^2}{256\,a^8\,f^4}}}-\frac {d^{11}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {d^2}{256\,a^8\,f^4}\right )}^{3/4}\,256{}\mathrm {i}}{\frac {4\,d^{13}}{a^6\,f^3}+\frac {64\,d^{12}\,\sqrt {-\frac {d^2}{256\,a^8\,f^4}}}{a^2\,f}}\right )\,{\left (-\frac {d^2}{256\,a^8\,f^4}\right )}^{1/4}\,2{}\mathrm {i}-\frac {d\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\left (a^2\,d\,f+a^2\,d\,f\,\mathrm {tan}\left (e+f\,x\right )\right )}+\frac {\sqrt {-d}\,\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{\sqrt {-d}}\right )\,1{}\mathrm {i}}{2\,a^2\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(e + f*x))^(1/2)/(a + a*tan(e + f*x))^2,x)

[Out]

((-d)^(1/2)*atan(((d*tan(e + f*x))^(1/2)*1i)/(-d)^(1/2))*1i)/(2*a^2*f) - atan((d^12*(d*tan(e + f*x))^(1/2)*(-d
^2/(256*a^8*f^4))^(1/4)*16i)/((4*d^13)/(a^2*f) + 64*a^2*d^12*f*(-d^2/(256*a^8*f^4))^(1/2)) - (d^11*(d*tan(e +
f*x))^(1/2)*(-d^2/(256*a^8*f^4))^(3/4)*256i)/((4*d^13)/(a^6*f^3) + (64*d^12*(-d^2/(256*a^8*f^4))^(1/2))/(a^2*f
)))*(-d^2/(256*a^8*f^4))^(1/4)*2i - (d*(d*tan(e + f*x))^(1/2))/(2*(a^2*d*f + a^2*d*f*tan(e + f*x))) - (atan((4
*d^12*(d*tan(e + f*x))^(1/2)*(-d^2/(a^8*f^4))^(1/4))/((4*d^13)/(a^2*f) - 4*a^2*d^12*f*(-d^2/(a^8*f^4))^(1/2))
+ (4*d^11*(d*tan(e + f*x))^(1/2)*(-d^2/(a^8*f^4))^(3/4))/((4*d^13)/(a^6*f^3) - (4*d^12*(-d^2/(a^8*f^4))^(1/2))
/(a^2*f)))*(-d^2/(a^8*f^4))^(1/4))/2

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sqrt {d \tan {\left (e + f x \right )}}}{\tan ^{2}{\left (e + f x \right )} + 2 \tan {\left (e + f x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))**(1/2)/(a+a*tan(f*x+e))**2,x)

[Out]

Integral(sqrt(d*tan(e + f*x))/(tan(e + f*x)**2 + 2*tan(e + f*x) + 1), x)/a**2

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