\(\int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\) [587]

Optimal result

Integrand size = 33, antiderivative size = 380 \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}+\frac {2 b^2 (5 A b+9 a B)}{15 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 b \left (15 a A b+14 a^2 B-5 b^2 B\right )}{5 d \sqrt {\cot (c+d x)}}+\frac {2 b B (b+a \cot (c+d x))^2}{5 d \cot ^{\frac {5}{2}}(c+d x)}-\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}+\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d} \]

[Out]

2/15*b^2*(5*A*b+9*B*a)/d/cot(d*x+c)^(3/2)+2/5*b*B*(b+a*cot(d*x+c))^2/d/cot(d*x+c)^(5/2)-1/2*(3*a^2*b*(A-B)-b^3
*(A-B)+a^3*(A+B)-3*a*b^2*(A+B))*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))/d*2^(1/2)-1/2*(3*a^2*b*(A-B)-b^3*(A-B)+a^3
*(A+B)-3*a*b^2*(A+B))*arctan(1+2^(1/2)*cot(d*x+c)^(1/2))/d*2^(1/2)-1/4*(a^3*(A-B)-3*a*b^2*(A-B)-3*a^2*b*(A+B)+
b^3*(A+B))*ln(1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^(1/2))/d*2^(1/2)+1/4*(a^3*(A-B)-3*a*b^2*(A-B)-3*a^2*b*(A+B)+b^3*
(A+B))*ln(1+cot(d*x+c)+2^(1/2)*cot(d*x+c)^(1/2))/d*2^(1/2)+2/5*b*(15*A*a*b+14*B*a^2-5*B*b^2)/d/cot(d*x+c)^(1/2
)

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 380, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3662, 3686, 3716, 3709, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {2 b \left (14 a^2 B+15 a A b-5 b^2 B\right )}{5 d \sqrt {\cot (c+d x)}}+\frac {\left (a^3 (A+B)+3 a^2 b (A-B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (a^3 (A+B)+3 a^2 b (A-B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d}-\frac {\left (a^3 (A-B)-3 a^2 b (A+B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {\left (a^3 (A-B)-3 a^2 b (A+B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {2 b^2 (9 a B+5 A b)}{15 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 b B (a \cot (c+d x)+b)^2}{5 d \cot ^{\frac {5}{2}}(c+d x)} \]

[In]

Int[Sqrt[Cot[c + d*x]]*(a + b*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

((3*a^2*b*(A - B) - b^3*(A - B) + a^3*(A + B) - 3*a*b^2*(A + B))*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt
[2]*d) - ((3*a^2*b*(A - B) - b^3*(A - B) + a^3*(A + B) - 3*a*b^2*(A + B))*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]
]])/(Sqrt[2]*d) + (2*b^2*(5*A*b + 9*a*B))/(15*d*Cot[c + d*x]^(3/2)) + (2*b*(15*a*A*b + 14*a^2*B - 5*b^2*B))/(5
*d*Sqrt[Cot[c + d*x]]) + (2*b*B*(b + a*Cot[c + d*x])^2)/(5*d*Cot[c + d*x]^(5/2)) - ((a^3*(A - B) - 3*a*b^2*(A
- B) - 3*a^2*b*(A + B) + b^3*(A + B))*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2*Sqrt[2]*d) + ((a^
3*(A - B) - 3*a*b^2*(A - B) - 3*a^2*b*(A + B) + b^3*(A + B))*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]
])/(2*Sqrt[2]*d)

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 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 1182

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 3615

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 3662

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 3686

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

Rule 3709

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 + b^2)
)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[b*B + a*(A - C) - (A*b - a*B - b*C)*Tan[e +
f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && LtQ[m, -1] && NeQ[a^2
+ b^2, 0]

Rule 3716

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {(b+a \cot (c+d x))^3 (B+A \cot (c+d x))}{\cot ^{\frac {7}{2}}(c+d x)} \, dx \\ & = \frac {2 b B (b+a \cot (c+d x))^2}{5 d \cot ^{\frac {5}{2}}(c+d x)}-\frac {2}{5} \int \frac {(b+a \cot (c+d x)) \left (-\frac {1}{2} b (5 A b+9 a B)-\frac {5}{2} \left (2 a A b+a^2 B-b^2 B\right ) \cot (c+d x)-\frac {1}{2} a (5 a A-b B) \cot ^2(c+d x)\right )}{\cot ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 b^2 (5 A b+9 a B)}{15 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 b B (b+a \cot (c+d x))^2}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {\frac {1}{2} b \left (15 a A b+14 a^2 B-5 b^2 B\right )+\frac {5}{2} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \cot (c+d x)+\frac {1}{2} a^2 (5 a A-b B) \cot ^2(c+d x)}{\cot ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 b^2 (5 A b+9 a B)}{15 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 b \left (15 a A b+14 a^2 B-5 b^2 B\right )}{5 d \sqrt {\cot (c+d x)}}+\frac {2 b B (b+a \cot (c+d x))^2}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {\frac {5}{2} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )+\frac {5}{2} \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx \\ & = \frac {2 b^2 (5 A b+9 a B)}{15 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 b \left (15 a A b+14 a^2 B-5 b^2 B\right )}{5 d \sqrt {\cot (c+d x)}}+\frac {2 b B (b+a \cot (c+d x))^2}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {4 \text {Subst}\left (\int \frac {-\frac {5}{2} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )-\frac {5}{2} \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{5 d} \\ & = \frac {2 b^2 (5 A b+9 a B)}{15 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 b \left (15 a A b+14 a^2 B-5 b^2 B\right )}{5 d \sqrt {\cot (c+d x)}}+\frac {2 b B (b+a \cot (c+d x))^2}{5 d \cot ^{\frac {5}{2}}(c+d x)}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d}+\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d} \\ & = \frac {2 b^2 (5 A b+9 a B)}{15 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 b \left (15 a A b+14 a^2 B-5 b^2 B\right )}{5 d \sqrt {\cot (c+d x)}}+\frac {2 b B (b+a \cot (c+d x))^2}{5 d \cot ^{\frac {5}{2}}(c+d x)}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 d}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 d}-\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \text {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} d}-\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \text {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} d} \\ & = \frac {2 b^2 (5 A b+9 a B)}{15 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 b \left (15 a A b+14 a^2 B-5 b^2 B\right )}{5 d \sqrt {\cot (c+d x)}}+\frac {2 b B (b+a \cot (c+d x))^2}{5 d \cot ^{\frac {5}{2}}(c+d x)}-\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}+\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d} \\ & = \frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}+\frac {2 b^2 (5 A b+9 a B)}{15 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 b \left (15 a A b+14 a^2 B-5 b^2 B\right )}{5 d \sqrt {\cot (c+d x)}}+\frac {2 b B (b+a \cot (c+d x))^2}{5 d \cot ^{\frac {5}{2}}(c+d x)}-\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}+\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.32 (sec) , antiderivative size = 287, normalized size of antiderivative = 0.76 \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (-\frac {\left (3 a^2 b (A-B)+b^3 (-A+B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \left (\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-\arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )\right )}{2 \sqrt {2}}-\frac {\left (a^3 (A-B)+3 a b^2 (-A+B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \left (\log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-\log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )\right )}{4 \sqrt {2}}+b \left (3 a A b+3 a^2 B-b^2 B\right ) \sqrt {\tan (c+d x)}+\frac {1}{3} b^2 (A b+3 a B) \tan ^{\frac {3}{2}}(c+d x)+\frac {1}{5} b^3 B \tan ^{\frac {5}{2}}(c+d x)\right )}{d} \]

[In]

Integrate[Sqrt[Cot[c + d*x]]*(a + b*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

(2*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(-1/2*((3*a^2*b*(A - B) + b^3*(-A + B) + a^3*(A + B) - 3*a*b^2*(A + B
))*(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] - ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]))/Sqrt[2] - ((a^3*(A - B)
+ 3*a*b^2*(-A + B) - 3*a^2*b*(A + B) + b^3*(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*Sqrt[2]) + b*(3*a*A*b + 3*a^2*B - b^2*B)*Sqrt[Tan[c + d*x]
] + (b^2*(A*b + 3*a*B)*Tan[c + d*x]^(3/2))/3 + (b^3*B*Tan[c + d*x]^(5/2))/5))/d

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(955\) vs. \(2(340)=680\).

Time = 0.45 (sec) , antiderivative size = 956, normalized size of antiderivative = 2.52

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(956\)
default	\(\text {Expression too large to display}\)	\(956\)

[In]

int(cot(d*x+c)^(1/2)*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/60/d*(1/tan(d*x+c))^(1/2)*tan(d*x+c)^(1/2)*(15*A*ln(-(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(2^(1/2)*tan(d*
x+c)^(1/2)-tan(d*x+c)-1))*2^(1/2)*a^3+30*A*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*a^3-30*A*arctan(1+2^(1/2
)*tan(d*x+c)^(1/2))*2^(1/2)*b^3+30*A*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*a^3-30*A*arctan(-1+2^(1/2)*ta
n(d*x+c)^(1/2))*2^(1/2)*b^3-15*A*ln(-(2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c)-1)/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d
*x+c)))*2^(1/2)*b^3+15*B*ln(-(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c)-1))*
2^(1/2)*b^3+30*B*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*a^3+30*B*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2
)*b^3+30*B*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*a^3+30*B*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*b^
3+15*B*ln(-(2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c)-1)/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*2^(1/2)*a^3+120*B*
a*b^2*tan(d*x+c)^(3/2)+360*A*a*b^2*tan(d*x+c)^(1/2)+360*B*a^2*b*tan(d*x+c)^(1/2)+24*B*b^3*tan(d*x+c)^(5/2)+40*
A*b^3*tan(d*x+c)^(3/2)-120*B*b^3*tan(d*x+c)^(1/2)-45*A*ln(-(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(2^(1/2)*ta
n(d*x+c)^(1/2)-tan(d*x+c)-1))*2^(1/2)*a*b^2+90*A*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*a^2*b-90*A*arctan(
1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*a*b^2+90*A*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*a^2*b-90*A*arctan(-
1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*a*b^2+45*A*ln(-(2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c)-1)/(1+2^(1/2)*tan(d*x+
c)^(1/2)+tan(d*x+c)))*2^(1/2)*a^2*b-45*B*ln(-(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(2^(1/2)*tan(d*x+c)^(1/2)
-tan(d*x+c)-1))*2^(1/2)*a^2*b-90*B*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*a^2*b-90*B*arctan(1+2^(1/2)*tan(
d*x+c)^(1/2))*2^(1/2)*a*b^2-90*B*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*a^2*b-90*B*arctan(-1+2^(1/2)*tan(
d*x+c)^(1/2))*2^(1/2)*a*b^2-45*B*ln(-(2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c)-1)/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d
*x+c)))*2^(1/2)*a*b^2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 6147 vs. \(2 (340) = 680\).

Time = 6.19 (sec) , antiderivative size = 6147, normalized size of antiderivative = 16.18 \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

[In]

integrate(cot(d*x+c)^(1/2)*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

Too large to include

Sympy [F]

\[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\int \left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{3} \sqrt {\cot {\left (c + d x \right )}}\, dx \]

[In]

integrate(cot(d*x+c)**(1/2)*(a+b*tan(d*x+c))**3*(A+B*tan(d*x+c)),x)

[Out]

Integral((A + B*tan(c + d*x))*(a + b*tan(c + d*x))**3*sqrt(cot(c + d*x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 334, normalized size of antiderivative = 0.88 \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {8 \, {\left (3 \, B b^{3} + \frac {5 \, {\left (3 \, B a b^{2} + A b^{3}\right )}}{\tan \left (d x + c\right )} + \frac {15 \, {\left (3 \, B a^{2} b + 3 \, A a b^{2} - B b^{3}\right )}}{\tan \left (d x + c\right )^{2}}\right )} \tan \left (d x + c\right )^{\frac {5}{2}} - 30 \, \sqrt {2} {\left ({\left (A + B\right )} a^{3} + 3 \, {\left (A - B\right )} a^{2} b - 3 \, {\left (A + B\right )} a b^{2} - {\left (A - B\right )} b^{3}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) - 30 \, \sqrt {2} {\left ({\left (A + B\right )} a^{3} + 3 \, {\left (A - B\right )} a^{2} b - 3 \, {\left (A + B\right )} a b^{2} - {\left (A - B\right )} b^{3}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 15 \, \sqrt {2} {\left ({\left (A - B\right )} a^{3} - 3 \, {\left (A + B\right )} a^{2} b - 3 \, {\left (A - B\right )} a b^{2} + {\left (A + B\right )} b^{3}\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - 15 \, \sqrt {2} {\left ({\left (A - B\right )} a^{3} - 3 \, {\left (A + B\right )} a^{2} b - 3 \, {\left (A - B\right )} a b^{2} + {\left (A + B\right )} b^{3}\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )}{60 \, d} \]

[In]

integrate(cot(d*x+c)^(1/2)*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/60*(8*(3*B*b^3 + 5*(3*B*a*b^2 + A*b^3)/tan(d*x + c) + 15*(3*B*a^2*b + 3*A*a*b^2 - B*b^3)/tan(d*x + c)^2)*tan
(d*x + c)^(5/2) - 30*sqrt(2)*((A + B)*a^3 + 3*(A - B)*a^2*b - 3*(A + B)*a*b^2 - (A - B)*b^3)*arctan(1/2*sqrt(2
)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) - 30*sqrt(2)*((A + B)*a^3 + 3*(A - B)*a^2*b - 3*(A + B)*a*b^2 - (A - B)*b^
3)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2/sqrt(tan(d*x + c)))) + 15*sqrt(2)*((A - B)*a^3 - 3*(A + B)*a^2*b - 3*(A -
B)*a*b^2 + (A + B)*b^3)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) - 15*sqrt(2)*((A - B)*a^3 - 3*(A
+ B)*a^2*b - 3*(A - B)*a*b^2 + (A + B)*b^3)*log(-sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1))/d

Giac [F]

\[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{3} \sqrt {\cot \left (d x + c\right )} \,d x } \]

[In]

integrate(cot(d*x+c)^(1/2)*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(b*tan(d*x + c) + a)^3*sqrt(cot(d*x + c)), x)

Mupad [F(-1)]

Timed out. \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\int \sqrt {\mathrm {cot}\left (c+d\,x\right )}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^3 \,d x \]

[In]

int(cot(c + d*x)^(1/2)*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^3,x)

[Out]

int(cot(c + d*x)^(1/2)*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^3, x)