3.597 \(\int \frac {A+B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=390 \[ -\frac {(A b-a B) \sqrt {\cot (c+d x)}}{d \left (a^2+b^2\right ) (a \cot (c+d x)+b)}-\frac {\left (-\left (a^2 (A+B)\right )+2 a b (A-B)+b^2 (A+B)\right ) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^2}+\frac {\left (-\left (a^2 (A+B)\right )+2 a b (A-B)+b^2 (A+B)\right ) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^2}+\frac {\left (a^2 (A-B)+2 a b (A+B)-b^2 (A-B)\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d \left (a^2+b^2\right )^2}-\frac {\left (a^2 (A-B)+2 a b (A+B)-b^2 (A-B)\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d \left (a^2+b^2\right )^2}+\frac {\left (a^3 (-B)+3 a^2 A b+3 a b^2 B-A b^3\right ) \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} d \left (a^2+b^2\right )^2} \]

[Out]

-1/2*(a^2*(A-B)-b^2*(A-B)+2*a*b*(A+B))*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))/(a^2+b^2)^2/d*2^(1/2)-1/2*(a^2*(A-B
)-b^2*(A-B)+2*a*b*(A+B))*arctan(1+2^(1/2)*cot(d*x+c)^(1/2))/(a^2+b^2)^2/d*2^(1/2)-1/4*(2*a*b*(A-B)-a^2*(A+B)+b
^2*(A+B))*ln(1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^(1/2))/(a^2+b^2)^2/d*2^(1/2)+1/4*(2*a*b*(A-B)-a^2*(A+B)+b^2*(A+B)
)*ln(1+cot(d*x+c)+2^(1/2)*cot(d*x+c)^(1/2))/(a^2+b^2)^2/d*2^(1/2)+(3*A*a^2*b-A*b^3-B*a^3+3*B*a*b^2)*arctan(a^(
1/2)*cot(d*x+c)^(1/2)/b^(1/2))/(a^2+b^2)^2/d/a^(1/2)/b^(1/2)-(A*b-B*a)*cot(d*x+c)^(1/2)/(a^2+b^2)/d/(b+a*cot(d
*x+c))

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Rubi [A]  time = 0.92, antiderivative size = 390, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.394, Rules used = {3581, 3608, 3653, 3534, 1168, 1162, 617, 204, 1165, 628, 3634, 63, 205} \[ -\frac {(A b-a B) \sqrt {\cot (c+d x)}}{d \left (a^2+b^2\right ) (a \cot (c+d x)+b)}-\frac {\left (a^2 (-(A+B))+2 a b (A-B)+b^2 (A+B)\right ) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^2}+\frac {\left (a^2 (-(A+B))+2 a b (A-B)+b^2 (A+B)\right ) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^2}+\frac {\left (a^2 (A-B)+2 a b (A+B)-b^2 (A-B)\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d \left (a^2+b^2\right )^2}-\frac {\left (a^2 (A-B)+2 a b (A+B)-b^2 (A-B)\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d \left (a^2+b^2\right )^2}+\frac {\left (3 a^2 A b+a^3 (-B)+3 a b^2 B-A b^3\right ) \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} d \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[c + d*x])/(Sqrt[Cot[c + d*x]]*(a + b*Tan[c + d*x])^2),x]

[Out]

((a^2*(A - B) - b^2*(A - B) + 2*a*b*(A + B))*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^2*d)
- ((a^2*(A - B) - b^2*(A - B) + 2*a*b*(A + B))*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^2
*d) + ((3*a^2*A*b - A*b^3 - a^3*B + 3*a*b^2*B)*ArcTan[(Sqrt[a]*Sqrt[Cot[c + d*x]])/Sqrt[b]])/(Sqrt[a]*Sqrt[b]*
(a^2 + b^2)^2*d) - ((A*b - a*B)*Sqrt[Cot[c + d*x]])/((a^2 + b^2)*d*(b + a*Cot[c + d*x])) - ((2*a*b*(A - B) - a
^2*(A + B) + b^2*(A + B))*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)^2*d) + ((
2*a*b*(A - B) - a^2*(A + B) + b^2*(A + B))*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2*Sqrt[2]*(a^2
+ b^2)^2*d)

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 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 1168

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[-(a*c)]

Rule 3534

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 3581

Int[(cot[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.)
+ (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Cot[e + f*x])^(p - m - n)*(b + a*Cot[e + f*x])^m*(d
+ c*Cot[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] &&  !IntegerQ[p] && IntegerQ[m] && IntegerQ
[n]

Rule 3608

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

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 {A+B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))^2} \, dx &=\int \frac {\sqrt {\cot (c+d x)} (B+A \cot (c+d x))}{(b+a \cot (c+d x))^2} \, dx\\ &=-\frac {(A b-a B) \sqrt {\cot (c+d x)}}{\left (a^2+b^2\right ) d (b+a \cot (c+d x))}+\frac {\int \frac {-\frac {1}{2} a (A b-a B)+a (a A+b B) \cot (c+d x)+\frac {1}{2} a (A b-a B) \cot ^2(c+d x)}{\sqrt {\cot (c+d x)} (b+a \cot (c+d x))} \, dx}{a \left (a^2+b^2\right )}\\ &=-\frac {(A b-a B) \sqrt {\cot (c+d x)}}{\left (a^2+b^2\right ) d (b+a \cot (c+d x))}+\frac {\int \frac {a \left (a^2 A-A b^2+2 a b B\right )+a \left (2 a A b-a^2 B+b^2 B\right ) \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx}{a \left (a^2+b^2\right )^2}-\frac {\left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right ) \int \frac {1+\cot ^2(c+d x)}{\sqrt {\cot (c+d x)} (b+a \cot (c+d x))} \, dx}{2 \left (a^2+b^2\right )^2}\\ &=-\frac {(A b-a B) \sqrt {\cot (c+d x)}}{\left (a^2+b^2\right ) d (b+a \cot (c+d x))}+\frac {2 \operatorname {Subst}\left (\int \frac {-a \left (a^2 A-A b^2+2 a b B\right )-a \left (2 a A b-a^2 B+b^2 B\right ) x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{a \left (a^2+b^2\right )^2 d}-\frac {\left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-x} (b-a x)} \, dx,x,-\cot (c+d x)\right )}{2 \left (a^2+b^2\right )^2 d}\\ &=-\frac {(A b-a B) \sqrt {\cot (c+d x)}}{\left (a^2+b^2\right ) d (b+a \cot (c+d x))}+\frac {\left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right ) \operatorname {Subst}\left (\int \frac {1}{b+a x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{\left (a^2+b^2\right )^2 d}-\frac {\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{\left (a^2+b^2\right )^2 d}+\frac {\left (2 a b (A-B)-a^2 (A+B)+b^2 (A+B)\right ) \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{\left (a^2+b^2\right )^2 d}\\ &=\frac {\left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right ) \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \left (a^2+b^2\right )^2 d}-\frac {(A b-a B) \sqrt {\cot (c+d x)}}{\left (a^2+b^2\right ) d (b+a \cot (c+d x))}-\frac {\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 \left (a^2+b^2\right )^2 d}-\frac {\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 \left (a^2+b^2\right )^2 d}-\frac {\left (2 a b (A-B)-a^2 (A+B)+b^2 (A+B)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d}-\frac {\left (2 a b (A-B)-a^2 (A+B)+b^2 (A+B)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d}\\ &=\frac {\left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right ) \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \left (a^2+b^2\right )^2 d}-\frac {(A b-a B) \sqrt {\cot (c+d x)}}{\left (a^2+b^2\right ) d (b+a \cot (c+d x))}-\frac {\left (2 a b (A-B)-a^2 (A+B)+b^2 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d}+\frac {\left (2 a b (A-B)-a^2 (A+B)+b^2 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d}-\frac {\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}+\frac {\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}\\ &=\frac {\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}-\frac {\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}+\frac {\left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right ) \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \left (a^2+b^2\right )^2 d}-\frac {(A b-a B) \sqrt {\cot (c+d x)}}{\left (a^2+b^2\right ) d (b+a \cot (c+d x))}-\frac {\left (2 a b (A-B)-a^2 (A+B)+b^2 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d}+\frac {\left (2 a b (A-B)-a^2 (A+B)+b^2 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d}\\ \end {align*}

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Mathematica [A]  time = 3.00, size = 336, normalized size = 0.86 \[ -\frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {4 \left (a^2+b^2\right ) (a B-A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b}}+2 \sqrt {2} \left (a^2 (A-B)+2 a b (A+B)+b^2 (B-A)\right ) \left (\tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-\tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )\right )+\frac {8 \sqrt {b} \left (a^2 A+2 a b B-A b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {4 \left (a^2+b^2\right ) (a B-A b) \sqrt {\tan (c+d x)}}{a+b \tan (c+d x)}+\sqrt {2} \left (-\left (a^2 (A+B)\right )+2 a b (A-B)+b^2 (A+B)\right ) \left (\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )-\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )\right )\right )}{4 d \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Tan[c + d*x])/(Sqrt[Cot[c + d*x]]*(a + b*Tan[c + d*x])^2),x]

[Out]

-1/4*(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(2*Sqrt[2]*(a^2*(A - B) + b^2*(-A + B) + 2*a*b*(A + B))*(ArcTan[1
- Sqrt[2]*Sqrt[Tan[c + d*x]]] - ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]) - (4*(a^2 + b^2)*(-(A*b) + a*B)*ArcTan
[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(Sqrt[a]*Sqrt[b]) + (8*Sqrt[b]*(a^2*A - A*b^2 + 2*a*b*B)*ArcTan[(Sqrt[
b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/Sqrt[a] + Sqrt[2]*(2*a*b*(A - B) - a^2*(A + B) + b^2*(A + B))*(Log[1 - Sqrt[2
]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] - Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]) - (4*(a^2 + b^2)*(-
(A*b) + a*B)*Sqrt[Tan[c + d*x]])/(a + b*Tan[c + d*x])))/((a^2 + b^2)^2*d)

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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((A+B*tan(d*x+c))/cot(d*x+c)^(1/2)/(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \tan \left (d x + c\right ) + A}{{\left (b \tan \left (d x + c\right ) + a\right )}^{2} \sqrt {\cot \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/cot(d*x+c)^(1/2)/(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)/((b*tan(d*x + c) + a)^2*sqrt(cot(d*x + c))), x)

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maple [C]  time = 4.26, size = 40753, normalized size = 104.49 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tan(d*x+c))/cot(d*x+c)^(1/2)/(a+b*tan(d*x+c))^2,x)

[Out]

result too large to display

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maxima [A]  time = 0.84, size = 347, normalized size = 0.89 \[ -\frac {\frac {4 \, {\left (B a^{3} - 3 \, A a^{2} b - 3 \, B a b^{2} + A b^{3}\right )} \arctan \left (\frac {a}{\sqrt {a b} \sqrt {\tan \left (d x + c\right )}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a b}} + \frac {2 \, \sqrt {2} {\left ({\left (A - B\right )} a^{2} + 2 \, {\left (A + B\right )} a b - {\left (A - B\right )} b^{2}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 2 \, \sqrt {2} {\left ({\left (A - B\right )} a^{2} + 2 \, {\left (A + B\right )} a b - {\left (A - B\right )} b^{2}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + \sqrt {2} {\left ({\left (A + B\right )} a^{2} - 2 \, {\left (A - B\right )} a b - {\left (A + B\right )} b^{2}\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - \sqrt {2} {\left ({\left (A + B\right )} a^{2} - 2 \, {\left (A - B\right )} a b - {\left (A + B\right )} b^{2}\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {4 \, {\left (B a - A b\right )}}{{\left (a^{2} b + b^{3} + \frac {a^{3} + a b^{2}}{\tan \left (d x + c\right )}\right )} \sqrt {\tan \left (d x + c\right )}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/cot(d*x+c)^(1/2)/(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/4*(4*(B*a^3 - 3*A*a^2*b - 3*B*a*b^2 + A*b^3)*arctan(a/(sqrt(a*b)*sqrt(tan(d*x + c))))/((a^4 + 2*a^2*b^2 + b
^4)*sqrt(a*b)) + (2*sqrt(2)*((A - B)*a^2 + 2*(A + B)*a*b - (A - B)*b^2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(t
an(d*x + c)))) + 2*sqrt(2)*((A - B)*a^2 + 2*(A + B)*a*b - (A - B)*b^2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2/sqrt(t
an(d*x + c)))) + sqrt(2)*((A + B)*a^2 - 2*(A - B)*a*b - (A + B)*b^2)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*
x + c) + 1) - sqrt(2)*((A + B)*a^2 - 2*(A - B)*a*b - (A + B)*b^2)*log(-sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x
+ c) + 1))/(a^4 + 2*a^2*b^2 + b^4) - 4*(B*a - A*b)/((a^2*b + b^3 + (a^3 + a*b^2)/tan(d*x + c))*sqrt(tan(d*x +
c))))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+B\,\mathrm {tan}\left (c+d\,x\right )}{\sqrt {\mathrm {cot}\left (c+d\,x\right )}\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*tan(c + d*x))/(cot(c + d*x)^(1/2)*(a + b*tan(c + d*x))^2),x)

[Out]

int((A + B*tan(c + d*x))/(cot(c + d*x)^(1/2)*(a + b*tan(c + d*x))^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B \tan {\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{2} \sqrt {\cot {\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/cot(d*x+c)**(1/2)/(a+b*tan(d*x+c))**2,x)

[Out]

Integral((A + B*tan(c + d*x))/((a + b*tan(c + d*x))**2*sqrt(cot(c + d*x))), x)

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