3.589 \(\int \frac{\cot ^{\frac{5}{2}}(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx\)

Optimal. Leaf size=325 \[ \frac{(a (A-B)+b (A+B)) \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )}-\frac{(a (A-B)+b (A+B)) \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )}-\frac{2 b^{5/2} (A b-a B) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{\cot (c+d x)}}{\sqrt{b}}\right )}{a^{5/2} d \left (a^2+b^2\right )}+\frac{(b (A-B)-a (A+B)) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d \left (a^2+b^2\right )}-\frac{(b (A-B)-a (A+B)) \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )}+\frac{2 (A b-a B) \sqrt{\cot (c+d x)}}{a^2 d}-\frac{2 A \cot ^{\frac{3}{2}}(c+d x)}{3 a d} \]

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

________________________________________________________________________________________

Rubi [A]  time = 1.12914, antiderivative size = 325, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 14, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.424, Rules used = {3581, 3607, 3647, 3653, 3534, 1168, 1162, 617, 204, 1165, 628, 3634, 63, 205} \[ \frac{(a (A-B)+b (A+B)) \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )}-\frac{(a (A-B)+b (A+B)) \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )}-\frac{2 b^{5/2} (A b-a B) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{\cot (c+d x)}}{\sqrt{b}}\right )}{a^{5/2} d \left (a^2+b^2\right )}+\frac{(b (A-B)-a (A+B)) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d \left (a^2+b^2\right )}-\frac{(b (A-B)-a (A+B)) \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )}+\frac{2 (A b-a B) \sqrt{\cot (c+d x)}}{a^2 d}-\frac{2 A \cot ^{\frac{3}{2}}(c+d x)}{3 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Cot[c + d*x]^(5/2)*(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*b^(5/2)*(A*b - a*B)*ArcTan[(Sqrt[a]
*Sqrt[Cot[c + d*x]])/Sqrt[b]])/(a^(5/2)*(a^2 + b^2)*d) + (2*(A*b - a*B)*Sqrt[Cot[c + d*x]])/(a^2*d) - (2*A*Cot
[c + d*x]^(3/2))/(3*a*d) + ((a*(A - B) + b*(A + B))*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2*Sqr
t[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 3607

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*B*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(d*
f*(m + n)), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[a^2*A*d*(m +
n) - b*B*(b*c*(m - 1) + a*d*(n + 1)) + d*(m + n)*(2*a*A*b + B*(a^2 - b^2))*Tan[e + f*x] - (b*B*(b*c - a*d)*(m
- 1) - b*(A*b + a*B)*d*(m + n))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*
c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&
!(IGtQ[n, 1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3647

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*
tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^m*(c + d
*Tan[e + f*x])^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, 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[m, 0] &&  !(IGtQ[n, 0] && ( !Intege
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

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]

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{\cot ^{\frac{5}{2}}(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx &=\int \frac{\cot ^{\frac{5}{2}}(c+d x) (B+A \cot (c+d x))}{b+a \cot (c+d x)} \, dx\\ &=-\frac{2 A \cot ^{\frac{3}{2}}(c+d x)}{3 a d}-\frac{2 \int \frac{\sqrt{\cot (c+d x)} \left (\frac{3 A b}{2}+\frac{3}{2} a A \cot (c+d x)+\frac{3}{2} (A b-a B) \cot ^2(c+d x)\right )}{b+a \cot (c+d x)} \, dx}{3 a}\\ &=\frac{2 (A b-a B) \sqrt{\cot (c+d x)}}{a^2 d}-\frac{2 A \cot ^{\frac{3}{2}}(c+d x)}{3 a d}+\frac{4 \int \frac{\frac{3}{4} b (A b-a B)-\frac{3}{4} a^2 B \cot (c+d x)-\frac{3}{4} \left (a^2 A-A b^2+a b B\right ) \cot ^2(c+d x)}{\sqrt{\cot (c+d x)} (b+a \cot (c+d x))} \, dx}{3 a^2}\\ &=\frac{2 (A b-a B) \sqrt{\cot (c+d x)}}{a^2 d}-\frac{2 A \cot ^{\frac{3}{2}}(c+d x)}{3 a d}+\frac{4 \int \frac{\frac{3}{4} a^2 (A b-a B)-\frac{3}{4} a^2 (a A+b B) \cot (c+d x)}{\sqrt{\cot (c+d x)}} \, dx}{3 a^2 \left (a^2+b^2\right )}+\frac{\left (b^3 (A b-a B)\right ) \int \frac{1+\cot ^2(c+d x)}{\sqrt{\cot (c+d x)} (b+a \cot (c+d x))} \, dx}{a^2 \left (a^2+b^2\right )}\\ &=\frac{2 (A b-a B) \sqrt{\cot (c+d x)}}{a^2 d}-\frac{2 A \cot ^{\frac{3}{2}}(c+d x)}{3 a d}+\frac{8 \operatorname{Subst}\left (\int \frac{-\frac{3}{4} a^2 (A b-a B)+\frac{3}{4} a^2 (a A+b B) x^2}{1+x^4} \, dx,x,\sqrt{\cot (c+d x)}\right )}{3 a^2 \left (a^2+b^2\right ) d}+\frac{\left (b^3 (A b-a B)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-x} (b-a x)} \, dx,x,-\cot (c+d x)\right )}{a^2 \left (a^2+b^2\right ) d}\\ &=\frac{2 (A b-a B) \sqrt{\cot (c+d x)}}{a^2 d}-\frac{2 A \cot ^{\frac{3}{2}}(c+d x)}{3 a d}-\frac{\left (2 b^3 (A b-a B)\right ) \operatorname{Subst}\left (\int \frac{1}{b+a x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{a^2 \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 b^{5/2} (A b-a B) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{\cot (c+d x)}}{\sqrt{b}}\right )}{a^{5/2} \left (a^2+b^2\right ) d}+\frac{2 (A b-a B) \sqrt{\cot (c+d x)}}{a^2 d}-\frac{2 A \cot ^{\frac{3}{2}}(c+d x)}{3 a 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 b^{5/2} (A b-a B) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{\cot (c+d x)}}{\sqrt{b}}\right )}{a^{5/2} \left (a^2+b^2\right ) d}+\frac{2 (A b-a B) \sqrt{\cot (c+d x)}}{a^2 d}-\frac{2 A \cot ^{\frac{3}{2}}(c+d x)}{3 a 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 b^{5/2} (A b-a B) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{\cot (c+d x)}}{\sqrt{b}}\right )}{a^{5/2} \left (a^2+b^2\right ) d}+\frac{2 (A b-a B) \sqrt{\cot (c+d x)}}{a^2 d}-\frac{2 A \cot ^{\frac{3}{2}}(c+d x)}{3 a 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 = 1.64012, size = 272, normalized size = 0.84 \[ -\frac{\sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \left (\frac{24 b^{5/2} (a B-A b) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{a^{5/2} \left (a^2+b^2\right )}-\frac{6 \sqrt{2} (a (A+B)+b (B-A)) \left (\tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )-\tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )\right )}{a^2+b^2}-\frac{3 \sqrt{2} (a (A-B)+b (A+B)) \left (\log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )-\log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )\right )}{a^2+b^2}+\frac{24 (a B-A b)}{a^2 \sqrt{\tan (c+d x)}}+\frac{8 A}{a \tan ^{\frac{3}{2}}(c+d x)}\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cot[c + d*x]^(5/2)*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x]),x]

[Out]

-(Sqrt[Cot[c + d*x]]*((-6*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]]]))/(a^2 + b^2) + (24*b^(5/2)*(-(A*b) + a*B)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/S
qrt[a]])/(a^(5/2)*(a^2 + b^2)) - (3*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]]))/(a^2 + b^2) + (8*A)/(a*Tan[c + d*x]^(3/2)) +
(24*(-(A*b) + a*B))/(a^2*Sqrt[Tan[c + d*x]]))*Sqrt[Tan[c + d*x]])/(12*d)

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Maple [C]  time = 1.073, size = 22300, normalized size = 68.6 \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)^(5/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c)),x)

[Out]

result too large to display

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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)^(5/2)*(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)^(5/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-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)**(5/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c)),x)

[Out]

Timed out

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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 )} \cot \left (d x + c\right )^{\frac{5}{2}}}{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)^(5/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*cot(d*x + c)^(5/2)/(b*tan(d*x + c) + a), x)