3.591 \(\int \frac{\sqrt{\cot (c+d x)} (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx\)
Optimal. Leaf size=278 \[ -\frac{(a (A-B)+b (A+B)) \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )}+\frac{(a (A-B)+b (A+B)) \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )}-\frac{(b (A-B)-a (A+B)) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d \left (a^2+b^2\right )}+\frac{(b (A-B)-a (A+B)) \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )}-\frac{2 \sqrt{b} (A b-a B) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{\cot (c+d x)}}{\sqrt{b}}\right )}{\sqrt{a} d \left (a^2+b^2\right )} \]
[Out]
-(((b*(A - B) - a*(A + B))*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)*d)) + ((b*(A - B) - a*
(A + B))*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)*d) - (2*Sqrt[b]*(A*b - a*B)*ArcTan[(Sqrt
[a]*Sqrt[Cot[c + d*x]])/Sqrt[b]])/(Sqrt[a]*(a^2 + b^2)*d) - ((a*(A - B) + b*(A + B))*Log[1 - Sqrt[2]*Sqrt[Cot[
c + d*x]] + Cot[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)*d) + ((a*(A - B) + b*(A + B))*Log[1 + Sqrt[2]*Sqrt[Cot[c + d
*x]] + Cot[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)*d)
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Rubi [A] time = 0.459772, antiderivative size = 278, normalized size of antiderivative =
1., number of steps used = 15, number of rules used = 12, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) =
0.364, Rules used = {3581, 3612, 3534, 1168, 1162, 617, 204, 1165, 628, 3634, 63, 205}
\[ -\frac{(a (A-B)+b (A+B)) \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )}+\frac{(a (A-B)+b (A+B)) \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )}-\frac{(b (A-B)-a (A+B)) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d \left (a^2+b^2\right )}+\frac{(b (A-B)-a (A+B)) \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )}-\frac{2 \sqrt{b} (A b-a B) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{\cot (c+d x)}}{\sqrt{b}}\right )}{\sqrt{a} d \left (a^2+b^2\right )} \]
Antiderivative was successfully verified.
[In]
Int[(Sqrt[Cot[c + d*x]]*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x]),x]
[Out]
-(((b*(A - B) - a*(A + B))*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)*d)) + ((b*(A - B) - a*
(A + B))*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)*d) - (2*Sqrt[b]*(A*b - a*B)*ArcTan[(Sqrt
[a]*Sqrt[Cot[c + d*x]])/Sqrt[b]])/(Sqrt[a]*(a^2 + b^2)*d) - ((a*(A - B) + b*(A + B))*Log[1 - Sqrt[2]*Sqrt[Cot[
c + d*x]] + Cot[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)*d) + ((a*(A - B) + b*(A + B))*Log[1 + Sqrt[2]*Sqrt[Cot[c + d
*x]] + Cot[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)*d)
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 3612
Int[(((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]])/((a_.) + (b_.)*tan[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^2), Int[Simp[A*(a*c + b*d) + B*(b*c - a*d) - (A*(b*c - a*d)
- B*(a*c + b*d))*Tan[e + f*x], x]/Sqrt[c + d*Tan[e + f*x]], x], x] - Dist[((b*c - a*d)*(B*a - A*b))/(a^2 + b^2
), Int[(1 + Tan[e + f*x]^2)/((a + b*Tan[e + f*x])*Sqrt[c + d*Tan[e + f*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]
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 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 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 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 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 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 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 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 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 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]
Rubi steps
\begin{align*} \int \frac{\sqrt{\cot (c+d x)} (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx &=\int \frac{\sqrt{\cot (c+d x)} (B+A \cot (c+d x))}{b+a \cot (c+d x)} \, dx\\ &=\frac{\int \frac{-A b+a B+(a A+b B) \cot (c+d x)}{\sqrt{\cot (c+d x)}} \, dx}{a^2+b^2}+\frac{(b (A b-a B)) \int \frac{1+\cot ^2(c+d x)}{\sqrt{\cot (c+d x)} (b+a \cot (c+d x))} \, dx}{a^2+b^2}\\ &=\frac{2 \operatorname{Subst}\left (\int \frac{A b-a B+(-a A-b B) x^2}{1+x^4} \, dx,x,\sqrt{\cot (c+d x)}\right )}{\left (a^2+b^2\right ) d}+\frac{(b (A b-a B)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-x} (b-a x)} \, dx,x,-\cot (c+d x)\right )}{\left (a^2+b^2\right ) d}\\ &=-\frac{(2 b (A b-a B)) \operatorname{Subst}\left (\int \frac{1}{b+a x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{\left (a^2+b^2\right ) d}+\frac{(b (A-B)-a (A+B)) \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 ) d}+\frac{(a (A-B)+b (A+B)) \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 ) d}\\ &=-\frac{2 \sqrt{b} (A b-a B) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{\cot (c+d x)}}{\sqrt{b}}\right )}{\sqrt{a} \left (a^2+b^2\right ) d}+\frac{(b (A-B)-a (A+B)) \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 ) d}+\frac{(b (A-B)-a (A+B)) \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 ) d}-\frac{(a (A-B)+b (A+B)) \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 ) d}-\frac{(a (A-B)+b (A+B)) \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 ) d}\\ &=-\frac{2 \sqrt{b} (A b-a B) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{\cot (c+d x)}}{\sqrt{b}}\right )}{\sqrt{a} \left (a^2+b^2\right ) d}-\frac{(a (A-B)+b (A+B)) \log \left (1-\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right ) d}+\frac{(a (A-B)+b (A+B)) \log \left (1+\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right ) d}+\frac{(b (A-B)-a (A+B)) \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 ) d}-\frac{(b (A-B)-a (A+B)) \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 ) d}\\ &=-\frac{(b (A-B)-a (A+B)) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right ) d}+\frac{(b (A-B)-a (A+B)) \tan ^{-1}\left (1+\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right ) d}-\frac{2 \sqrt{b} (A b-a B) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{\cot (c+d x)}}{\sqrt{b}}\right )}{\sqrt{a} \left (a^2+b^2\right ) d}-\frac{(a (A-B)+b (A+B)) \log \left (1-\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right ) d}+\frac{(a (A-B)+b (A+B)) \log \left (1+\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right ) d}\\ \end{align*}
Mathematica [A] time = 0.376939, size = 215, normalized size = 0.77 \[ \frac{\sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \left (-2 \sqrt{2} (a (A+B)+b (B-A)) \left (\tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )-\tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )\right )+\frac{8 \sqrt{b} (A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{\sqrt{a}}-\sqrt{2} (a (A-B)+b (A+B)) \left (\log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )-\log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )\right )\right )}{4 d \left (a^2+b^2\right )} \]
Antiderivative was successfully verified.
[In]
Integrate[(Sqrt[Cot[c + d*x]]*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x]),x]
[Out]
(Sqrt[Cot[c + d*x]]*(-2*Sqrt[2]*(b*(-A + B) + a*(A + B))*(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] - ArcTan[1 +
Sqrt[2]*Sqrt[Tan[c + d*x]]]) + (8*Sqrt[b]*(A*b - a*B)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/Sqrt[a] -
Sqrt[2]*(a*(A - B) + b*(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]]))*Sqrt[Tan[c + d*x]])/(4*(a^2 + b^2)*d)
________________________________________________________________________________________
Maple [C] time = 0.438, size = 4107, normalized size = 14.8 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c)),x)
[Out]
1/2/d*2^(1/2)/a/(a^2+b^2)^(3/2)/(a+b+(a^2+b^2)^(1/2))/(-b+(a^2+b^2)^(1/2)-a)*(cos(d*x+c)/sin(d*x+c))^(1/2)*(co
s(d*x+c)-1)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d
*x+c)-1)/sin(d*x+c))^(1/2)*(2*a^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),-a/(-b+(a^2+b^2)^(1
/2)-a),1/2*2^(1/2))*B*b^3+2*A*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*(a^2+b^2)^(
3/2)*a^2+2*A*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*(a^2+b^2)^(3/2)*b^2-2*A*Elli
pticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*(a^2+b^2)^(1/2)*a^4-2*A*EllipticF((-(cos(d*x+
c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*(a^2+b^2)^(1/2)*b^4+I*A*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))
/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(a^2+b^2)^(1/2)*a^2*b^2-A*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin
(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(a^2+b^2)^(3/2)*a^2+A*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))
^(1/2),1/2+1/2*I,1/2*2^(1/2))*(a^2+b^2)^(1/2)*a^4-A*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1
/2-1/2*I,1/2*2^(1/2))*(a^2+b^2)^(3/2)*a^2+A*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I
,1/2*2^(1/2))*(a^2+b^2)^(1/2)*a^4+2*a^3*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),a/(a+b+(a^2+b
^2)^(1/2)),1/2*2^(1/2))*A*b^2+2*a*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),a/(a+b+(a^2+b^2)^(1
/2)),1/2*2^(1/2))*b^4*A-2*a^3*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),-a/(-b+(a^2+b^2)^(1/2)-
a),1/2*2^(1/2))*A*b^2-2*a*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),-a/(-b+(a^2+b^2)^(1/2)-a),1
/2*2^(1/2))*b^4*A+B*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(a^2+b^2)^
(3/2)*a^2-B*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(a^2+b^2)^(1/2)*a^
4+B*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*(a^2+b^2)^(3/2)*a^2-B*Elli
pticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*(a^2+b^2)^(1/2)*a^4-2*a^4*Elliptic
Pi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),a/(a+b+(a^2+b^2)^(1/2)),1/2*2^(1/2))*B*b-2*a^2*EllipticPi((-(
cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),a/(a+b+(a^2+b^2)^(1/2)),1/2*2^(1/2))*B*b^3+2*a^4*EllipticPi((-(cos(
d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),-a/(-b+(a^2+b^2)^(1/2)-a),1/2*2^(1/2))*B*b-2*a^2*EllipticPi((-(cos(d*x+
c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),-a/(-b+(a^2+b^2)^(1/2)-a),1/2*2^(1/2))*B*b^2*(a^2+b^2)^(1/2)-I*A*EllipticPi
((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(a^2+b^2)^(1/2)*a^4-I*A*EllipticPi((-(co
s(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*(a^2+b^2)^(3/2)*a^2-I*B*EllipticPi((-(cos(d*x+
c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(a^2+b^2)^(1/2)*a^4-I*B*EllipticPi((-(cos(d*x+c)-1-s
in(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*(a^2+b^2)^(3/2)*a^2+I*A*EllipticPi((-(cos(d*x+c)-1-sin(d*x
+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(a^2+b^2)^(1/2)*a*b^3+I*A*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))
/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(a^2+b^2)^(3/2)*a^2+I*A*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d
*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*(a^2+b^2)^(1/2)*a^4+3*A*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))
^(1/2),1/2+1/2*I,1/2*2^(1/2))*(a^2+b^2)^(1/2)*a^2*b^2+A*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/
2),1/2+1/2*I,1/2*2^(1/2))*(a^2+b^2)^(1/2)*a*b^3-A*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2
-1/2*I,1/2*2^(1/2))*(a^2+b^2)^(3/2)*a*b+3*A*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I
,1/2*2^(1/2))*(a^2+b^2)^(1/2)*a^3*b+3*A*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2
*2^(1/2))*(a^2+b^2)^(1/2)*a^2*b^2+A*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(
1/2))*(a^2+b^2)^(1/2)*a*b^3-2*a^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),a/(a+b+(a^2+b^2)^(1
/2)),1/2*2^(1/2))*A*b^2*(a^2+b^2)^(1/2)+2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),a/(a+b+(a^2
+b^2)^(1/2)),1/2*2^(1/2))*b^3*A*(a^2+b^2)^(1/2)*a-2*a^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/
2),a/(a+b+(a^2+b^2)^(1/2)),1/2*2^(1/2))*B*b^2*(a^2+b^2)^(1/2)+2*a^3*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin
(d*x+c))^(1/2),-a/(-b+(a^2+b^2)^(1/2)-a),1/2*2^(1/2))*B*b*(a^2+b^2)^(1/2)+I*B*EllipticPi((-(cos(d*x+c)-1-sin(d
*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(a^2+b^2)^(3/2)*a^2+I*B*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))
/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*(a^2+b^2)^(1/2)*a^4+3*I*B*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin
(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*(a^2+b^2)^(1/2)*a^2*b^2+I*A*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d
*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*(a^2+b^2)^(1/2)*a^3*b+I*B*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c
))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(a^2+b^2)^(3/2)*a*b+I*B*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/
2),1/2-1/2*I,1/2*2^(1/2))*(a^2+b^2)^(1/2)*a*b^3-I*A*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1
/2+1/2*I,1/2*2^(1/2))*(a^2+b^2)^(3/2)*a*b-I*A*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2
*I,1/2*2^(1/2))*(a^2+b^2)^(1/2)*a^3*b-I*A*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1
/2*2^(1/2))*(a^2+b^2)^(1/2)*a^2*b^2-I*A*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2
*2^(1/2))*(a^2+b^2)^(1/2)*a*b^3-3*I*B*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2
^(1/2))*(a^2+b^2)^(1/2)*a^3*b-3*I*B*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(
1/2))*(a^2+b^2)^(1/2)*a^2*b^2-I*B*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/
2))*(a^2+b^2)^(1/2)*a*b^3-I*B*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*
(a^2+b^2)^(3/2)*a*b+3*I*B*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*(a^2
+b^2)^(1/2)*a^3*b+I*A*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*(a^2+b^2
)^(3/2)*a*b-4*A*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*(a^2+b^2)^(1/2)*a^3*b-4*A
*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*(a^2+b^2)^(1/2)*a^2*b^2-4*A*EllipticF((-
(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*(a^2+b^2)^(1/2)*a*b^3-A*EllipticPi((-(cos(d*x+c)-1-si
n(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(a^2+b^2)^(3/2)*a*b+3*A*EllipticPi((-(cos(d*x+c)-1-sin(d*x+
c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(a^2+b^2)^(1/2)*a^3*b-2*a^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c)
)/sin(d*x+c))^(1/2),-a/(-b+(a^2+b^2)^(1/2)-a),1/2*2^(1/2))*A*b^2*(a^2+b^2)^(1/2)+2*EllipticPi((-(cos(d*x+c)-1-
sin(d*x+c))/sin(d*x+c))^(1/2),-a/(-b+(a^2+b^2)^(1/2)-a),1/2*2^(1/2))*b^3*A*(a^2+b^2)^(1/2)*a-B*EllipticPi((-(c
os(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(a^2+b^2)^(3/2)*a*b-B*EllipticPi((-(cos(d*x+c
)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(a^2+b^2)^(1/2)*a^3*b+B*EllipticPi((-(cos(d*x+c)-1-si
n(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(a^2+b^2)^(1/2)*a^2*b^2+B*EllipticPi((-(cos(d*x+c)-1-sin(d*
x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(a^2+b^2)^(1/2)*a*b^3-B*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/
sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*(a^2+b^2)^(3/2)*a*b-B*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+
c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*(a^2+b^2)^(1/2)*a^3*b+B*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1
/2),1/2-1/2*I,1/2*2^(1/2))*(a^2+b^2)^(1/2)*a^2*b^2+B*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),
1/2-1/2*I,1/2*2^(1/2))*(a^2+b^2)^(1/2)*a*b^3+2*a^3*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),a/
(a+b+(a^2+b^2)^(1/2)),1/2*2^(1/2))*B*b*(a^2+b^2)^(1/2))/sin(d*x+c)^2/cos(d*x+c)*(cos(d*x+c)+1)^2
________________________________________________________________________________________
Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c)),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c)),x, algorithm="fricas")
[Out]
Timed out
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \tan{\left (c + d x \right )}\right ) \sqrt{\cot{\left (c + d x \right )}}}{a + b \tan{\left (c + d x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)**(1/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c)),x)
[Out]
Integral((A + B*tan(c + d*x))*sqrt(cot(c + d*x))/(a + b*tan(c + d*x)), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \tan \left (d x + c\right ) + A\right )} \sqrt{\cot \left (d x + c\right )}}{b \tan \left (d x + c\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c)),x, algorithm="giac")
[Out]
integrate((B*tan(d*x + c) + A)*sqrt(cot(d*x + c))/(b*tan(d*x + c) + a), x)