\(\int \cot ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx\) [580]

Optimal result

Integrand size = 33, antiderivative size = 276 \[ \int \cot ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}+\frac {2 b^2 B}{d \sqrt {\cot (c+d x)}}-\frac {2 a^2 A \sqrt {\cot (c+d x)}}{d}-\frac {\left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}+\frac {\left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d} \]

[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))/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))/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))/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))/d*2^(1/2)+2*b^2*B/d/cot(d*x+c)^(1/2)-2*a^2*A*cot(d*x+c)^(1/2)/d

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {3662, 3685, 3711, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \cot ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d}-\frac {\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {2 a^2 A \sqrt {\cot (c+d x)}}{d}+\frac {2 b^2 B}{d \sqrt {\cot (c+d x)}} \]

[In]

Int[Cot[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^2*(A + B*Tan[c + d*x]),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]*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]*d) + (2*b^2*B)/(d*Sqrt[C
ot[c + d*x]]) - (2*a^2*A*Sqrt[Cot[c + d*x]])/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]*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]*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 3685

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

Rule 3711

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

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

Mathematica [A] (verified)

Time = 0.90 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.80 \[ \int \cot ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {\sqrt {\cot (c+d x)} \left (2 \sqrt {2} \left (a^2 (A-B)+b^2 (-A+B)-2 a b (A+B)\right ) \left (\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-\arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )\right )-\sqrt {2} \left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \left (\log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-\log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )\right )-\frac {8 a^2 A}{\sqrt {\tan (c+d x)}}+8 b^2 B \sqrt {\tan (c+d x)}\right ) \sqrt {\tan (c+d x)}}{4 d} \]

[In]

Integrate[Cot[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^2*(A + B*Tan[c + d*x]),x]

[Out]

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

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.87

method	result	size
derivativedivides	\(-\frac {2 A \,a^{2} \sqrt {\cot \left (d x +c \right )}+\frac {\left (-A \,a^{2}+A \,b^{2}+2 B a b \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}{1+\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{4}+\frac {\left (2 a b A +B \,a^{2}-B \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}{1+\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{4}-\frac {2 B \,b^{2}}{\sqrt {\cot \left (d x +c \right )}}}{d}\)	\(239\)
default	\(-\frac {2 A \,a^{2} \sqrt {\cot \left (d x +c \right )}+\frac {\left (-A \,a^{2}+A \,b^{2}+2 B a b \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}{1+\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{4}+\frac {\left (2 a b A +B \,a^{2}-B \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}{1+\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{4}-\frac {2 B \,b^{2}}{\sqrt {\cot \left (d x +c \right )}}}{d}\)	\(239\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4248 vs. \(2 (244) = 488\).

Time = 2.28 (sec) , antiderivative size = 4248, normalized size of antiderivative = 15.39 \[ \int \cot ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/2*(d*sqrt((2*A*B*a^4 - 12*A*B*a^2*b^2 + 2*A*B*b^4 + 4*(A^2 - B^2)*a^3*b - 4*(A^2 - B^2)*a*b^3 + d^2*sqrt(-((
A^4 - 2*A^2*B^2 + B^4)*a^8 - 16*(A^3*B - A*B^3)*a^7*b - 4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a^6*b^2 + 112*(A^3*B -
A*B^3)*a^5*b^3 + 2*(19*A^4 - 102*A^2*B^2 + 19*B^4)*a^4*b^4 - 112*(A^3*B - A*B^3)*a^3*b^5 - 4*(3*A^4 - 22*A^2*B
^2 + 3*B^4)*a^2*b^6 + 16*(A^3*B - A*B^3)*a*b^7 + (A^4 - 2*A^2*B^2 + B^4)*b^8)/d^4))/d^2)*log(((A*a^2 - 2*B*a*b
- A*b^2)*d^3*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^8 - 16*(A^3*B - A*B^3)*a^7*b - 4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a
^6*b^2 + 112*(A^3*B - A*B^3)*a^5*b^3 + 2*(19*A^4 - 102*A^2*B^2 + 19*B^4)*a^4*b^4 - 112*(A^3*B - A*B^3)*a^3*b^5
- 4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a^2*b^6 + 16*(A^3*B - A*B^3)*a*b^7 + (A^4 - 2*A^2*B^2 + B^4)*b^8)/d^4) - ((A
^2*B - B^3)*a^6 + 2*(A^3 - 5*A*B^2)*a^5*b - (23*A^2*B - 7*B^3)*a^4*b^2 - 4*(3*A^3 - 7*A*B^2)*a^3*b^3 + (23*A^2
*B - 7*B^3)*a^2*b^4 + 2*(A^3 - 5*A*B^2)*a*b^5 - (A^2*B - B^3)*b^6)*d)*sqrt((2*A*B*a^4 - 12*A*B*a^2*b^2 + 2*A*B
*b^4 + 4*(A^2 - B^2)*a^3*b - 4*(A^2 - B^2)*a*b^3 + d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^8 - 16*(A^3*B - A*B^3)
*a^7*b - 4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a^6*b^2 + 112*(A^3*B - A*B^3)*a^5*b^3 + 2*(19*A^4 - 102*A^2*B^2 + 19*B
^4)*a^4*b^4 - 112*(A^3*B - A*B^3)*a^3*b^5 - 4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a^2*b^6 + 16*(A^3*B - A*B^3)*a*b^7
+ (A^4 - 2*A^2*B^2 + B^4)*b^8)/d^4))/d^2) - ((A^4 - B^4)*a^8 - 8*(A^3*B + A*B^3)*a^7*b - 4*(A^4 - B^4)*a^6*b^2
- 8*(A^3*B + A*B^3)*a^5*b^3 - 10*(A^4 - B^4)*a^4*b^4 + 8*(A^3*B + A*B^3)*a^3*b^5 - 4*(A^4 - B^4)*a^2*b^6 + 8*
(A^3*B + A*B^3)*a*b^7 + (A^4 - B^4)*b^8)*sqrt(tan(d*x + c))) - d*sqrt((2*A*B*a^4 - 12*A*B*a^2*b^2 + 2*A*B*b^4
+ 4*(A^2 - B^2)*a^3*b - 4*(A^2 - B^2)*a*b^3 + d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^8 - 16*(A^3*B - A*B^3)*a^7*
b - 4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a^6*b^2 + 112*(A^3*B - A*B^3)*a^5*b^3 + 2*(19*A^4 - 102*A^2*B^2 + 19*B^4)*a
^4*b^4 - 112*(A^3*B - A*B^3)*a^3*b^5 - 4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a^2*b^6 + 16*(A^3*B - A*B^3)*a*b^7 + (A^
4 - 2*A^2*B^2 + B^4)*b^8)/d^4))/d^2)*log(-((A*a^2 - 2*B*a*b - A*b^2)*d^3*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^8 -
16*(A^3*B - A*B^3)*a^7*b - 4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a^6*b^2 + 112*(A^3*B - A*B^3)*a^5*b^3 + 2*(19*A^4 -
102*A^2*B^2 + 19*B^4)*a^4*b^4 - 112*(A^3*B - A*B^3)*a^3*b^5 - 4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a^2*b^6 + 16*(A^3
*B - A*B^3)*a*b^7 + (A^4 - 2*A^2*B^2 + B^4)*b^8)/d^4) - ((A^2*B - B^3)*a^6 + 2*(A^3 - 5*A*B^2)*a^5*b - (23*A^2
*B - 7*B^3)*a^4*b^2 - 4*(3*A^3 - 7*A*B^2)*a^3*b^3 + (23*A^2*B - 7*B^3)*a^2*b^4 + 2*(A^3 - 5*A*B^2)*a*b^5 - (A^
2*B - B^3)*b^6)*d)*sqrt((2*A*B*a^4 - 12*A*B*a^2*b^2 + 2*A*B*b^4 + 4*(A^2 - B^2)*a^3*b - 4*(A^2 - B^2)*a*b^3 +
d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^8 - 16*(A^3*B - A*B^3)*a^7*b - 4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a^6*b^2 + 1
12*(A^3*B - A*B^3)*a^5*b^3 + 2*(19*A^4 - 102*A^2*B^2 + 19*B^4)*a^4*b^4 - 112*(A^3*B - A*B^3)*a^3*b^5 - 4*(3*A^
4 - 22*A^2*B^2 + 3*B^4)*a^2*b^6 + 16*(A^3*B - A*B^3)*a*b^7 + (A^4 - 2*A^2*B^2 + B^4)*b^8)/d^4))/d^2) - ((A^4 -
B^4)*a^8 - 8*(A^3*B + A*B^3)*a^7*b - 4*(A^4 - B^4)*a^6*b^2 - 8*(A^3*B + A*B^3)*a^5*b^3 - 10*(A^4 - B^4)*a^4*b
^4 + 8*(A^3*B + A*B^3)*a^3*b^5 - 4*(A^4 - B^4)*a^2*b^6 + 8*(A^3*B + A*B^3)*a*b^7 + (A^4 - B^4)*b^8)*sqrt(tan(d
*x + c))) - d*sqrt((2*A*B*a^4 - 12*A*B*a^2*b^2 + 2*A*B*b^4 + 4*(A^2 - B^2)*a^3*b - 4*(A^2 - B^2)*a*b^3 - d^2*s
qrt(-((A^4 - 2*A^2*B^2 + B^4)*a^8 - 16*(A^3*B - A*B^3)*a^7*b - 4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a^6*b^2 + 112*(A
^3*B - A*B^3)*a^5*b^3 + 2*(19*A^4 - 102*A^2*B^2 + 19*B^4)*a^4*b^4 - 112*(A^3*B - A*B^3)*a^3*b^5 - 4*(3*A^4 - 2
2*A^2*B^2 + 3*B^4)*a^2*b^6 + 16*(A^3*B - A*B^3)*a*b^7 + (A^4 - 2*A^2*B^2 + B^4)*b^8)/d^4))/d^2)*log(((A*a^2 -
2*B*a*b - A*b^2)*d^3*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^8 - 16*(A^3*B - A*B^3)*a^7*b - 4*(3*A^4 - 22*A^2*B^2 + 3
*B^4)*a^6*b^2 + 112*(A^3*B - A*B^3)*a^5*b^3 + 2*(19*A^4 - 102*A^2*B^2 + 19*B^4)*a^4*b^4 - 112*(A^3*B - A*B^3)*
a^3*b^5 - 4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a^2*b^6 + 16*(A^3*B - A*B^3)*a*b^7 + (A^4 - 2*A^2*B^2 + B^4)*b^8)/d^4
) + ((A^2*B - B^3)*a^6 + 2*(A^3 - 5*A*B^2)*a^5*b - (23*A^2*B - 7*B^3)*a^4*b^2 - 4*(3*A^3 - 7*A*B^2)*a^3*b^3 +
(23*A^2*B - 7*B^3)*a^2*b^4 + 2*(A^3 - 5*A*B^2)*a*b^5 - (A^2*B - B^3)*b^6)*d)*sqrt((2*A*B*a^4 - 12*A*B*a^2*b^2
+ 2*A*B*b^4 + 4*(A^2 - B^2)*a^3*b - 4*(A^2 - B^2)*a*b^3 - d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^8 - 16*(A^3*B -
A*B^3)*a^7*b - 4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a^6*b^2 + 112*(A^3*B - A*B^3)*a^5*b^3 + 2*(19*A^4 - 102*A^2*B^2
+ 19*B^4)*a^4*b^4 - 112*(A^3*B - A*B^3)*a^3*b^5 - 4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a^2*b^6 + 16*(A^3*B - A*B^3)
*a*b^7 + (A^4 - 2*A^2*B^2 + B^4)*b^8)/d^4))/d^2) - ((A^4 - B^4)*a^8 - 8*(A^3*B + A*B^3)*a^7*b - 4*(A^4 - B^4)*
a^6*b^2 - 8*(A^3*B + A*B^3)*a^5*b^3 - 10*(A^4 - B^4)*a^4*b^4 + 8*(A^3*B + A*B^3)*a^3*b^5 - 4*(A^4 - B^4)*a^2*b
^6 + 8*(A^3*B + A*B^3)*a*b^7 + (A^4 - B^4)*b^8)*sqrt(tan(d*x + c))) + d*sqrt((2*A*B*a^4 - 12*A*B*a^2*b^2 + 2*A
*B*b^4 + 4*(A^2 - B^2)*a^3*b - 4*(A^2 - B^2)*a*b^3 - d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^8 - 16*(A^3*B - A*B^
3)*a^7*b - 4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a^6*b^2 + 112*(A^3*B - A*B^3)*a^5*b^3 + 2*(19*A^4 - 102*A^2*B^2 + 19
*B^4)*a^4*b^4 - 112*(A^3*B - A*B^3)*a^3*b^5 - 4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a^2*b^6 + 16*(A^3*B - A*B^3)*a*b^
7 + (A^4 - 2*A^2*B^2 + B^4)*b^8)/d^4))/d^2)*log(-((A*a^2 - 2*B*a*b - A*b^2)*d^3*sqrt(-((A^4 - 2*A^2*B^2 + B^4)
*a^8 - 16*(A^3*B - A*B^3)*a^7*b - 4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a^6*b^2 + 112*(A^3*B - A*B^3)*a^5*b^3 + 2*(19
*A^4 - 102*A^2*B^2 + 19*B^4)*a^4*b^4 - 112*(A^3*B - A*B^3)*a^3*b^5 - 4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a^2*b^6 +
16*(A^3*B - A*B^3)*a*b^7 + (A^4 - 2*A^2*B^2 + B^4)*b^8)/d^4) + ((A^2*B - B^3)*a^6 + 2*(A^3 - 5*A*B^2)*a^5*b -
(23*A^2*B - 7*B^3)*a^4*b^2 - 4*(3*A^3 - 7*A*B^2)*a^3*b^3 + (23*A^2*B - 7*B^3)*a^2*b^4 + 2*(A^3 - 5*A*B^2)*a*b^
5 - (A^2*B - B^3)*b^6)*d)*sqrt((2*A*B*a^4 - 12*A*B*a^2*b^2 + 2*A*B*b^4 + 4*(A^2 - B^2)*a^3*b - 4*(A^2 - B^2)*a
*b^3 - d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^8 - 16*(A^3*B - A*B^3)*a^7*b - 4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a^6*
b^2 + 112*(A^3*B - A*B^3)*a^5*b^3 + 2*(19*A^4 - 102*A^2*B^2 + 19*B^4)*a^4*b^4 - 112*(A^3*B - A*B^3)*a^3*b^5 -
4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a^2*b^6 + 16*(A^3*B - A*B^3)*a*b^7 + (A^4 - 2*A^2*B^2 + B^4)*b^8)/d^4))/d^2) -
((A^4 - B^4)*a^8 - 8*(A^3*B + A*B^3)*a^7*b - 4*(A^4 - B^4)*a^6*b^2 - 8*(A^3*B + A*B^3)*a^5*b^3 - 10*(A^4 - B^4
)*a^4*b^4 + 8*(A^3*B + A*B^3)*a^3*b^5 - 4*(A^4 - B^4)*a^2*b^6 + 8*(A^3*B + A*B^3)*a*b^7 + (A^4 - B^4)*b^8)*sqr
t(tan(d*x + c))) + 4*(B*b^2*tan(d*x + c) - A*a^2)/sqrt(tan(d*x + c)))/d

Sympy [F]

\[ \int \cot ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\int \left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{2} \cot ^{\frac {3}{2}}{\left (c + d x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 244, normalized size of antiderivative = 0.88 \[ \int \cot ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {8 \, B b^{2} \sqrt {\tan \left (d x + c\right )} + 2 \, \sqrt {2} {\left ({\left (A - B\right )} a^{2} - 2 \, {\left (A + B\right )} a b - {\left (A - B\right )} b^{2}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 2 \, \sqrt {2} {\left ({\left (A - B\right )} a^{2} - 2 \, {\left (A + B\right )} a b - {\left (A - B\right )} b^{2}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + \sqrt {2} {\left ({\left (A + B\right )} a^{2} + 2 \, {\left (A - B\right )} a b - {\left (A + B\right )} b^{2}\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - \sqrt {2} {\left ({\left (A + B\right )} a^{2} + 2 \, {\left (A - B\right )} a b - {\left (A + B\right )} b^{2}\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - \frac {8 \, A a^{2}}{\sqrt {\tan \left (d x + c\right )}}}{4 \, d} \]

[In]

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

[Out]

1/4*(8*B*b^2*sqrt(tan(d*x + c)) + 2*sqrt(2)*((A - B)*a^2 - 2*(A + B)*a*b - (A - B)*b^2)*arctan(1/2*sqrt(2)*(sq
rt(2) + 2/sqrt(tan(d*x + c)))) + 2*sqrt(2)*((A - B)*a^2 - 2*(A + B)*a*b - (A - B)*b^2)*arctan(-1/2*sqrt(2)*(sq
rt(2) - 2/sqrt(tan(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) - 8*A*a^2/sqrt(tan(d*x + c)))/d

Giac [F]

\[ \int \cot ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{2} \cot \left (d x + c\right )^{\frac {3}{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \cot ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^{3/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^2 \,d x \]

[In]

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

[Out]

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