3.4.67 \(\int \frac {1}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^2} \, dx\) [367]

Optimal. Leaf size=281 \[ \frac {3 \text {ArcTan}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 \sqrt {d} f}+\frac {\text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 \sqrt {d} f}-\frac {\text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 \sqrt {d} f}-\frac {\log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{4 \sqrt {2} a^2 \sqrt {d} f}+\frac {\log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{4 \sqrt {2} a^2 \sqrt {d} f}+\frac {\sqrt {d \tan (e+f x)}}{2 d f \left (a^2+a^2 \tan (e+f x)\right )} \]

[Out]

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

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Rubi [A]
time = 0.35, antiderivative size = 281, 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 = {3650, 3734, 12, 16, 3557, 335, 303, 1176, 631, 210, 1179, 642, 3715, 65, 211} \begin {gather*} \frac {3 \text {ArcTan}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 \sqrt {d} f}+\frac {\text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 \sqrt {d} f}-\frac {\text {ArcTan}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{2 \sqrt {2} a^2 \sqrt {d} f}+\frac {\sqrt {d \tan (e+f x)}}{2 d f \left (a^2 \tan (e+f x)+a^2\right )}-\frac {\log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{4 \sqrt {2} a^2 \sqrt {d} f}+\frac {\log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{4 \sqrt {2} a^2 \sqrt {d} f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[d*Tan[e + f*x]]*(a + a*Tan[e + f*x])^2),x]

[Out]

(3*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/(2*a^2*Sqrt[d]*f) + ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]
]/(2*Sqrt[2]*a^2*Sqrt[d]*f) - ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]/(2*Sqrt[2]*a^2*Sqrt[d]*f) - L
og[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]]/(4*Sqrt[2]*a^2*Sqrt[d]*f) + Log[Sqrt[d] + Sq
rt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]]/(4*Sqrt[2]*a^2*Sqrt[d]*f) + Sqrt[d*Tan[e + f*x]]/(2*d*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 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 65

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 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 211

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

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 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 3650

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

Rule 3715

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 3734

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 {1}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^2} \, dx &=\frac {\sqrt {d \tan (e+f x)}}{2 d f \left (a^2+a^2 \tan (e+f x)\right )}+\frac {\int \frac {\frac {3 a^2 d}{2}-a^2 d \tan (e+f x)+\frac {1}{2} a^2 d \tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{2 a^3 d}\\ &=\frac {\sqrt {d \tan (e+f x)}}{2 d f \left (a^2+a^2 \tan (e+f x)\right )}+\frac {3 \int \frac {1+\tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{4 a}+\frac {\int -\frac {2 a^3 d \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{4 a^5 d}\\ &=\frac {\sqrt {d \tan (e+f x)}}{2 d f \left (a^2+a^2 \tan (e+f x)\right )}-\frac {\int \frac {\tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{2 a^2}+\frac {3 \text {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 d f \left (a^2+a^2 \tan (e+f x)\right )}-\frac {\int \sqrt {d \tan (e+f x)} \, dx}{2 a^2 d}+\frac {3 \text {Subst}\left (\int \frac {1}{a+\frac {a x^2}{d}} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{2 a d f}\\ &=\frac {3 \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 \sqrt {d} f}+\frac {\sqrt {d \tan (e+f x)}}{2 d f \left (a^2+a^2 \tan (e+f x)\right )}-\frac {\text {Subst}\left (\int \frac {\sqrt {x}}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{2 a^2 f}\\ &=\frac {3 \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 \sqrt {d} f}+\frac {\sqrt {d \tan (e+f x)}}{2 d f \left (a^2+a^2 \tan (e+f x)\right )}-\frac {\text {Subst}\left (\int \frac {x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^2 f}\\ &=\frac {3 \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 \sqrt {d} f}+\frac {\sqrt {d \tan (e+f x)}}{2 d f \left (a^2+a^2 \tan (e+f x)\right )}+\frac {\text {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 {\text {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 {3 \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 \sqrt {d} f}+\frac {\sqrt {d \tan (e+f x)}}{2 d f \left (a^2+a^2 \tan (e+f x)\right )}-\frac {\text {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 {\text {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 {\text {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 \sqrt {d} f}-\frac {\text {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 \sqrt {d} f}\\ &=\frac {3 \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 \sqrt {d} f}-\frac {\log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{4 \sqrt {2} a^2 \sqrt {d} f}+\frac {\log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{4 \sqrt {2} a^2 \sqrt {d} f}+\frac {\sqrt {d \tan (e+f x)}}{2 d f \left (a^2+a^2 \tan (e+f x)\right )}-\frac {\text {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 \sqrt {d} f}+\frac {\text {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 \sqrt {d} f}\\ &=\frac {3 \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 \sqrt {d} f}+\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 \sqrt {d} f}-\frac {\tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 \sqrt {d} f}-\frac {\log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{4 \sqrt {2} a^2 \sqrt {d} f}+\frac {\log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{4 \sqrt {2} a^2 \sqrt {d} f}+\frac {\sqrt {d \tan (e+f x)}}{2 d f \left (a^2+a^2 \tan (e+f x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.74, size = 337, normalized size = 1.20 \begin {gather*} \frac {\left (12 \text {ArcTan}\left (\sqrt {\tan (e+f x)}\right ) \cos (e+f x)-\sqrt {2} \cos (e+f x) \log \left (-1+\sqrt {2} \sqrt {\tan (e+f x)}-\tan (e+f x)\right )+\sqrt {2} \cos (e+f x) \log \left (1+\sqrt {2} \sqrt {\tan (e+f x)}+\tan (e+f x)\right )+12 \text {ArcTan}\left (\sqrt {\tan (e+f x)}\right ) \sin (e+f x)-\sqrt {2} \log \left (-1+\sqrt {2} \sqrt {\tan (e+f x)}-\tan (e+f x)\right ) \sin (e+f x)+\sqrt {2} \log \left (1+\sqrt {2} \sqrt {\tan (e+f x)}+\tan (e+f x)\right ) \sin (e+f x)+2 \sqrt {2} \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right ) (\cos (e+f x)+\sin (e+f x))-2 \sqrt {2} \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (e+f x)}\right ) (\cos (e+f x)+\sin (e+f x))+4 \cos (e+f x) \sqrt {\tan (e+f x)}\right ) \sqrt {\tan (e+f x)}}{8 a^2 f (\cos (e+f x)+\sin (e+f x)) \sqrt {d \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[d*Tan[e + f*x]]*(a + a*Tan[e + f*x])^2),x]

[Out]

((12*ArcTan[Sqrt[Tan[e + f*x]]]*Cos[e + f*x] - Sqrt[2]*Cos[e + f*x]*Log[-1 + Sqrt[2]*Sqrt[Tan[e + f*x]] - Tan[
e + f*x]] + Sqrt[2]*Cos[e + f*x]*Log[1 + Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]] + 12*ArcTan[Sqrt[Tan[e + f
*x]]]*Sin[e + f*x] - Sqrt[2]*Log[-1 + Sqrt[2]*Sqrt[Tan[e + f*x]] - Tan[e + f*x]]*Sin[e + f*x] + Sqrt[2]*Log[1
+ Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]]*Sin[e + f*x] + 2*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]]]*(
Cos[e + f*x] + Sin[e + f*x]) - 2*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[e + f*x]]]*(Cos[e + f*x] + Sin[e + f*x])
+ 4*Cos[e + f*x]*Sqrt[Tan[e + f*x]])*Sqrt[Tan[e + f*x]])/(8*a^2*f*(Cos[e + f*x] + Sin[e + f*x])*Sqrt[d*Tan[e +
f*x]])

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Maple [A]
time = 0.20, size = 197, normalized size = 0.70

method	result	size
derivativedivides	\(\frac {2 d^{3} \left (\frac {\frac {\sqrt {d \tan \left (f x +e \right )}}{2 d \tan \left (f x +e \right )+2 d}+\frac {3 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{2 \sqrt {d}}}{2 d^{3}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{16 d^{3} \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f \,a^{2}}\)	\(197\)
default	\(\frac {2 d^{3} \left (\frac {\frac {\sqrt {d \tan \left (f x +e \right )}}{2 d \tan \left (f x +e \right )+2 d}+\frac {3 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{2 \sqrt {d}}}{2 d^{3}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{16 d^{3} \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f \,a^{2}}\)	\(197\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/f/a^2*d^3*(1/2/d^3*(1/2*(d*tan(f*x+e))^(1/2)/(d*tan(f*x+e)+d)+3/2/d^(1/2)*arctan((d*tan(f*x+e))^(1/2)/d^(1/2
)))-1/16/d^3/(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*ta
n(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))+2*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1
/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 = 0.53, size = 226, normalized size = 0.80 \begin {gather*} \frac {\frac {4 \, \sqrt {d \tan \left (f x + e\right )} d}{a^{2} d \tan \left (f x + e\right ) + a^{2} d} - \frac {d {\left (\frac {2 \, \sqrt {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 {d}} + \frac {2 \, \sqrt {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 {d}} - \frac {\sqrt {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 {d}} + \frac {\sqrt {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 {d}}\right )}}{a^{2}} + \frac {12 \, \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a^{2}}}{8 \, d f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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/(a^2*d*tan(f*x + e) + a^2*d) - d*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d)
+ 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x
+ e)))/sqrt(d))/sqrt(d) - sqrt(2)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d) + sq
rt(2)*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d))/a^2 + 12*sqrt(d)*arctan(sqrt(d*t
an(f*x + e))/sqrt(d))/a^2)/(d*f)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 863 vs. \(2 (226) = 452\).
time = 2.63, size = 1808, normalized size = 6.43 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(3*(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*d*f*cos(f*x + e)*sin(f*x + e) + sqrt(2)*a^2*d*f)*(1/(a^8*d^2*f^4))^(1/4)*arctan(-sq
rt(2)*a^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^8*d^2*f^4))^(1/4) + sqrt(2)*a^2*f*sqrt((sqrt(2)*a^6*d^2*f^
3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^8*d^2*f^4))^(3/4)*cos(f*x + e) + a^4*d^2*f^2*sqrt(1/(a^8*d^2*f^4))*c
os(f*x + e) + d*sin(f*x + e))/cos(f*x + e))*(1/(a^8*d^2*f^4))^(1/4) - 1) - 4*(2*sqrt(2)*a^2*d*f*cos(f*x + e)*s
in(f*x + e) + sqrt(2)*a^2*d*f)*(1/(a^8*d^2*f^4))^(1/4)*arctan(-sqrt(2)*a^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))
*(1/(a^8*d^2*f^4))^(1/4) + sqrt(2)*a^2*f*sqrt(-(sqrt(2)*a^6*d^2*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^8*
d^2*f^4))^(3/4)*cos(f*x + e) - a^4*d^2*f^2*sqrt(1/(a^8*d^2*f^4))*cos(f*x + e) - d*sin(f*x + e))/cos(f*x + e))*
(1/(a^8*d^2*f^4))^(1/4) + 1) - (2*sqrt(2)*a^2*d*f*cos(f*x + e)*sin(f*x + e) + sqrt(2)*a^2*d*f)*(1/(a^8*d^2*f^4
))^(1/4)*log((sqrt(2)*a^6*d^2*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^8*d^2*f^4))^(3/4)*cos(f*x + e) + a^4
*d^2*f^2*sqrt(1/(a^8*d^2*f^4))*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e)) + (2*sqrt(2)*a^2*d*f*cos(f*x + e)*
sin(f*x + e) + sqrt(2)*a^2*d*f)*(1/(a^8*d^2*f^4))^(1/4)*log(-(sqrt(2)*a^6*d^2*f^3*sqrt(d*sin(f*x + e)/cos(f*x
+ e))*(1/(a^8*d^2*f^4))^(3/4)*cos(f*x + e) - a^4*d^2*f^2*sqrt(1/(a^8*d^2*f^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*d*f*c
os(f*x + e)*sin(f*x + e) + a^2*d*f), 1/8*(12*(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*d*f*cos(f*x + e)*sin(f*x + e) + sqrt(2)*a^2*d*f)*(1/(a^8*d^2*f^4
))^(1/4)*arctan(-sqrt(2)*a^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^8*d^2*f^4))^(1/4) + sqrt(2)*a^2*f*sqrt(
(sqrt(2)*a^6*d^2*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^8*d^2*f^4))^(3/4)*cos(f*x + e) + a^4*d^2*f^2*sqrt
(1/(a^8*d^2*f^4))*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))*(1/(a^8*d^2*f^4))^(1/4) - 1) + 4*(2*sqrt(2)*a^2
*d*f*cos(f*x + e)*sin(f*x + e) + sqrt(2)*a^2*d*f)*(1/(a^8*d^2*f^4))^(1/4)*arctan(-sqrt(2)*a^2*f*sqrt(d*sin(f*x
+ e)/cos(f*x + e))*(1/(a^8*d^2*f^4))^(1/4) + sqrt(2)*a^2*f*sqrt(-(sqrt(2)*a^6*d^2*f^3*sqrt(d*sin(f*x + e)/cos
(f*x + e))*(1/(a^8*d^2*f^4))^(3/4)*cos(f*x + e) - a^4*d^2*f^2*sqrt(1/(a^8*d^2*f^4))*cos(f*x + e) - d*sin(f*x +
e))/cos(f*x + e))*(1/(a^8*d^2*f^4))^(1/4) + 1) + (2*sqrt(2)*a^2*d*f*cos(f*x + e)*sin(f*x + e) + sqrt(2)*a^2*d
*f)*(1/(a^8*d^2*f^4))^(1/4)*log((sqrt(2)*a^6*d^2*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^8*d^2*f^4))^(3/4)
*cos(f*x + e) + a^4*d^2*f^2*sqrt(1/(a^8*d^2*f^4))*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e)) - (2*sqrt(2)*a^
2*d*f*cos(f*x + e)*sin(f*x + e) + sqrt(2)*a^2*d*f)*(1/(a^8*d^2*f^4))^(1/4)*log(-(sqrt(2)*a^6*d^2*f^3*sqrt(d*si
n(f*x + e)/cos(f*x + e))*(1/(a^8*d^2*f^4))^(3/4)*cos(f*x + e) - a^4*d^2*f^2*sqrt(1/(a^8*d^2*f^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*d*f*cos(f*x + e)*sin(f*x + e) + a^2*d*f)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.68, size = 261, normalized size = 0.93 \begin {gather*} -\frac {\sqrt {2} {\left | d \right |}^{\frac {3}{2}} \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 )}{4 \, a^{2} d^{2} f} - \frac {\sqrt {2} {\left | d \right |}^{\frac {3}{2}} \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 )}{4 \, a^{2} d^{2} f} + \frac {\sqrt {2} {\left | d \right |}^{\frac {3}{2}} \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 )}{8 \, a^{2} d^{2} f} - \frac {\sqrt {2} {\left | d \right |}^{\frac {3}{2}} \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 )}{8 \, a^{2} d^{2} f} + \frac {3 \, \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{2 \, a^{2} \sqrt {d} f} + \frac {\sqrt {d \tan \left (f x + e\right )}}{2 \, {\left (d \tan \left (f x + e\right ) + d\right )} a^{2} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 4.65, size = 365, normalized size = 1.30 \begin {gather*} \frac {\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 {\mathrm {atan}\left (\frac {4\,d^8\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {1}{a^8\,d^2\,f^4}\right )}^{1/4}}{\frac {4\,d^8}{a^2\,f}+36\,a^2\,d^9\,f\,\sqrt {-\frac {1}{a^8\,d^2\,f^4}}}+\frac {36\,d^9\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {1}{a^8\,d^2\,f^4}\right )}^{3/4}}{\frac {4\,d^8}{a^6\,f^3}+\frac {36\,d^9\,\sqrt {-\frac {1}{a^8\,d^2\,f^4}}}{a^2\,f}}\right )\,{\left (-\frac {1}{a^8\,d^2\,f^4}\right )}^{1/4}}{2}-\mathrm {atan}\left (\frac {d^8\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {1}{256\,a^8\,d^2\,f^4}\right )}^{1/4}\,16{}\mathrm {i}}{\frac {4\,d^8}{a^2\,f}-576\,a^2\,d^9\,f\,\sqrt {-\frac {1}{256\,a^8\,d^2\,f^4}}}-\frac {d^9\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {1}{256\,a^8\,d^2\,f^4}\right )}^{3/4}\,2304{}\mathrm {i}}{\frac {4\,d^8}{a^6\,f^3}-\frac {576\,d^9\,\sqrt {-\frac {1}{256\,a^8\,d^2\,f^4}}}{a^2\,f}}\right )\,{\left (-\frac {1}{256\,a^8\,d^2\,f^4}\right )}^{1/4}\,2{}\mathrm {i}+\frac {\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{\sqrt {-d}}\right )\,3{}\mathrm {i}}{2\,a^2\,\sqrt {-d}\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d*tan(e + f*x))^(1/2)/(2*(a^2*d*f + a^2*d*f*tan(e + f*x))) - (atan((4*d^8*(d*tan(e + f*x))^(1/2)*(-1/(a^8*d^2
*f^4))^(1/4))/((4*d^8)/(a^2*f) + 36*a^2*d^9*f*(-1/(a^8*d^2*f^4))^(1/2)) + (36*d^9*(d*tan(e + f*x))^(1/2)*(-1/(
a^8*d^2*f^4))^(3/4))/((4*d^8)/(a^6*f^3) + (36*d^9*(-1/(a^8*d^2*f^4))^(1/2))/(a^2*f)))*(-1/(a^8*d^2*f^4))^(1/4)
)/2 - atan((d^8*(d*tan(e + f*x))^(1/2)*(-1/(256*a^8*d^2*f^4))^(1/4)*16i)/((4*d^8)/(a^2*f) - 576*a^2*d^9*f*(-1/
(256*a^8*d^2*f^4))^(1/2)) - (d^9*(d*tan(e + f*x))^(1/2)*(-1/(256*a^8*d^2*f^4))^(3/4)*2304i)/((4*d^8)/(a^6*f^3)
- (576*d^9*(-1/(256*a^8*d^2*f^4))^(1/2))/(a^2*f)))*(-1/(256*a^8*d^2*f^4))^(1/4)*2i + (atan(((d*tan(e + f*x))^
(1/2)*1i)/(-d)^(1/2))*3i)/(2*a^2*(-d)^(1/2)*f)

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