3.635 \(\int \frac{(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\cot ^{\frac{3}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=457 \[ \frac{\left (-5 a^2 B+40 a A b-48 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{96 b d \sqrt{\cot (c+d x)}}+\frac{\left (40 a^2 A b-5 a^3 B-112 a b^2 B-64 A b^3\right ) \sqrt{a+b \tan (c+d x)}}{64 b d \sqrt{\cot (c+d x)}}+\frac{\left (40 a^3 A b-240 a^2 b^2 B-5 a^4 B-320 a A b^3+128 b^4 B\right ) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{64 b^{3/2} d}-\frac{(-b+i a)^{5/2} (-B+i A) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{(8 A b-a B) (a+b \tan (c+d x))^{5/2}}{24 b d \sqrt{\cot (c+d x)}}-\frac{(b+i a)^{5/2} (B+i A) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{B (a+b \tan (c+d x))^{7/2}}{4 b d \sqrt{\cot (c+d x)}} \]

[Out]

-(((I*a - b)^(5/2)*(I*A - B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[Cot[c +
d*x]]*Sqrt[Tan[c + d*x]])/d) + ((40*a^3*A*b - 320*a*A*b^3 - 5*a^4*B - 240*a^2*b^2*B + 128*b^4*B)*ArcTanh[(Sqrt
[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]])/(64*b^(3/2)*d) - ((I*
a + b)^(5/2)*(I*A + B)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]
*Sqrt[Tan[c + d*x]])/d + ((40*a^2*A*b - 64*A*b^3 - 5*a^3*B - 112*a*b^2*B)*Sqrt[a + b*Tan[c + d*x]])/(64*b*d*Sq
rt[Cot[c + d*x]]) + ((40*a*A*b - 5*a^2*B - 48*b^2*B)*(a + b*Tan[c + d*x])^(3/2))/(96*b*d*Sqrt[Cot[c + d*x]]) +
((8*A*b - a*B)*(a + b*Tan[c + d*x])^(5/2))/(24*b*d*Sqrt[Cot[c + d*x]]) + (B*(a + b*Tan[c + d*x])^(7/2))/(4*b*
d*Sqrt[Cot[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 3.01098, antiderivative size = 457, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 11, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.314, Rules used = {4241, 3607, 3647, 3655, 6725, 63, 217, 206, 93, 205, 208} \[ \frac{\left (-5 a^2 B+40 a A b-48 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{96 b d \sqrt{\cot (c+d x)}}+\frac{\left (40 a^2 A b-5 a^3 B-112 a b^2 B-64 A b^3\right ) \sqrt{a+b \tan (c+d x)}}{64 b d \sqrt{\cot (c+d x)}}+\frac{\left (40 a^3 A b-240 a^2 b^2 B-5 a^4 B-320 a A b^3+128 b^4 B\right ) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{64 b^{3/2} d}-\frac{(-b+i a)^{5/2} (-B+i A) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{(8 A b-a B) (a+b \tan (c+d x))^{5/2}}{24 b d \sqrt{\cot (c+d x)}}-\frac{(b+i a)^{5/2} (B+i A) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{B (a+b \tan (c+d x))^{7/2}}{4 b d \sqrt{\cot (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/Cot[c + d*x]^(3/2),x]

[Out]

-(((I*a - b)^(5/2)*(I*A - B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[Cot[c +
d*x]]*Sqrt[Tan[c + d*x]])/d) + ((40*a^3*A*b - 320*a*A*b^3 - 5*a^4*B - 240*a^2*b^2*B + 128*b^4*B)*ArcTanh[(Sqrt
[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]])/(64*b^(3/2)*d) - ((I*
a + b)^(5/2)*(I*A + B)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]
*Sqrt[Tan[c + d*x]])/d + ((40*a^2*A*b - 64*A*b^3 - 5*a^3*B - 112*a*b^2*B)*Sqrt[a + b*Tan[c + d*x]])/(64*b*d*Sq
rt[Cot[c + d*x]]) + ((40*a*A*b - 5*a^2*B - 48*b^2*B)*(a + b*Tan[c + d*x])^(3/2))/(96*b*d*Sqrt[Cot[c + d*x]]) +
((8*A*b - a*B)*(a + b*Tan[c + d*x])^(5/2))/(24*b*d*Sqrt[Cot[c + d*x]]) + (B*(a + b*Tan[c + d*x])^(7/2))/(4*b*
d*Sqrt[Cot[c + d*x]])

Rule 4241

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ
[u, x]

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 3655

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

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*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]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\cot ^{\frac{3}{2}}(c+d x)} \, dx &=\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\\ &=\frac{B (a+b \tan (c+d x))^{7/2}}{4 b d \sqrt{\cot (c+d x)}}+\frac{\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{(a+b \tan (c+d x))^{5/2} \left (-\frac{a B}{2}-4 b B \tan (c+d x)+\frac{1}{2} (8 A b-a B) \tan ^2(c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx}{4 b}\\ &=\frac{(8 A b-a B) (a+b \tan (c+d x))^{5/2}}{24 b d \sqrt{\cot (c+d x)}}+\frac{B (a+b \tan (c+d x))^{7/2}}{4 b d \sqrt{\cot (c+d x)}}+\frac{\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{(a+b \tan (c+d x))^{3/2} \left (-\frac{1}{4} a (8 A b+5 a B)-12 b (A b+a B) \tan (c+d x)+\frac{1}{4} \left (40 a A b-5 a^2 B-48 b^2 B\right ) \tan ^2(c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx}{12 b}\\ &=\frac{\left (40 a A b-5 a^2 B-48 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{96 b d \sqrt{\cot (c+d x)}}+\frac{(8 A b-a B) (a+b \tan (c+d x))^{5/2}}{24 b d \sqrt{\cot (c+d x)}}+\frac{B (a+b \tan (c+d x))^{7/2}}{4 b d \sqrt{\cot (c+d x)}}+\frac{\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\sqrt{a+b \tan (c+d x)} \left (-\frac{3}{8} a \left (24 a A b+5 a^2 B-16 b^2 B\right )-24 b \left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)+\frac{3}{8} \left (40 a^2 A b-64 A b^3-5 a^3 B-112 a b^2 B\right ) \tan ^2(c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx}{24 b}\\ &=\frac{\left (40 a^2 A b-64 A b^3-5 a^3 B-112 a b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{64 b d \sqrt{\cot (c+d x)}}+\frac{\left (40 a A b-5 a^2 B-48 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{96 b d \sqrt{\cot (c+d x)}}+\frac{(8 A b-a B) (a+b \tan (c+d x))^{5/2}}{24 b d \sqrt{\cot (c+d x)}}+\frac{B (a+b \tan (c+d x))^{7/2}}{4 b d \sqrt{\cot (c+d x)}}+\frac{\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{-\frac{3}{16} a \left (88 a^2 A b-64 A b^3+5 a^3 B-144 a b^2 B\right )-24 b \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)+\frac{3}{16} \left (40 a^3 A b-320 a A b^3-5 a^4 B-240 a^2 b^2 B+128 b^4 B\right ) \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{24 b}\\ &=\frac{\left (40 a^2 A b-64 A b^3-5 a^3 B-112 a b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{64 b d \sqrt{\cot (c+d x)}}+\frac{\left (40 a A b-5 a^2 B-48 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{96 b d \sqrt{\cot (c+d x)}}+\frac{(8 A b-a B) (a+b \tan (c+d x))^{5/2}}{24 b d \sqrt{\cot (c+d x)}}+\frac{B (a+b \tan (c+d x))^{7/2}}{4 b d \sqrt{\cot (c+d x)}}+\frac{\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{-\frac{3}{16} a \left (88 a^2 A b-64 A b^3+5 a^3 B-144 a b^2 B\right )-24 b \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) x+\frac{3}{16} \left (40 a^3 A b-320 a A b^3-5 a^4 B-240 a^2 b^2 B+128 b^4 B\right ) x^2}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{24 b d}\\ &=\frac{\left (40 a^2 A b-64 A b^3-5 a^3 B-112 a b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{64 b d \sqrt{\cot (c+d x)}}+\frac{\left (40 a A b-5 a^2 B-48 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{96 b d \sqrt{\cot (c+d x)}}+\frac{(8 A b-a B) (a+b \tan (c+d x))^{5/2}}{24 b d \sqrt{\cot (c+d x)}}+\frac{B (a+b \tan (c+d x))^{7/2}}{4 b d \sqrt{\cot (c+d x)}}+\frac{\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \left (\frac{3 \left (40 a^3 A b-320 a A b^3-5 a^4 B-240 a^2 b^2 B+128 b^4 B\right )}{16 \sqrt{x} \sqrt{a+b x}}-\frac{24 \left (b \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )+b \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) x\right )}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{24 b d}\\ &=\frac{\left (40 a^2 A b-64 A b^3-5 a^3 B-112 a b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{64 b d \sqrt{\cot (c+d x)}}+\frac{\left (40 a A b-5 a^2 B-48 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{96 b d \sqrt{\cot (c+d x)}}+\frac{(8 A b-a B) (a+b \tan (c+d x))^{5/2}}{24 b d \sqrt{\cot (c+d x)}}+\frac{B (a+b \tan (c+d x))^{7/2}}{4 b d \sqrt{\cot (c+d x)}}-\frac{\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{b \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )+b \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) x}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{b d}+\frac{\left (\left (40 a^3 A b-320 a A b^3-5 a^4 B-240 a^2 b^2 B+128 b^4 B\right ) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{128 b d}\\ &=\frac{\left (40 a^2 A b-64 A b^3-5 a^3 B-112 a b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{64 b d \sqrt{\cot (c+d x)}}+\frac{\left (40 a A b-5 a^2 B-48 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{96 b d \sqrt{\cot (c+d x)}}+\frac{(8 A b-a B) (a+b \tan (c+d x))^{5/2}}{24 b d \sqrt{\cot (c+d x)}}+\frac{B (a+b \tan (c+d x))^{7/2}}{4 b d \sqrt{\cot (c+d x)}}-\frac{\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \left (\frac{-b \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )+i b \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )}{2 (i-x) \sqrt{x} \sqrt{a+b x}}+\frac{b \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )+i b \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )}{2 \sqrt{x} (i+x) \sqrt{a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{b d}+\frac{\left (\left (40 a^3 A b-320 a A b^3-5 a^4 B-240 a^2 b^2 B+128 b^4 B\right ) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{64 b d}\\ &=\frac{\left (40 a^2 A b-64 A b^3-5 a^3 B-112 a b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{64 b d \sqrt{\cot (c+d x)}}+\frac{\left (40 a A b-5 a^2 B-48 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{96 b d \sqrt{\cot (c+d x)}}+\frac{(8 A b-a B) (a+b \tan (c+d x))^{5/2}}{24 b d \sqrt{\cot (c+d x)}}+\frac{B (a+b \tan (c+d x))^{7/2}}{4 b d \sqrt{\cot (c+d x)}}+\frac{\left ((i a+b)^3 (A-i B) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} (i+x) \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{\left (\left (40 a^3 A b-320 a A b^3-5 a^4 B-240 a^2 b^2 B+128 b^4 B\right ) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{64 b d}-\frac{\left (\left (-b \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )+i b \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )\right ) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{(i-x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 b d}\\ &=\frac{\left (40 a^3 A b-320 a A b^3-5 a^4 B-240 a^2 b^2 B+128 b^4 B\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right ) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}}{64 b^{3/2} d}+\frac{\left (40 a^2 A b-64 A b^3-5 a^3 B-112 a b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{64 b d \sqrt{\cot (c+d x)}}+\frac{\left (40 a A b-5 a^2 B-48 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{96 b d \sqrt{\cot (c+d x)}}+\frac{(8 A b-a B) (a+b \tan (c+d x))^{5/2}}{24 b d \sqrt{\cot (c+d x)}}+\frac{B (a+b \tan (c+d x))^{7/2}}{4 b d \sqrt{\cot (c+d x)}}+\frac{\left ((i a+b)^3 (A-i B) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{i-(-a+i b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{\left (\left (-b \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )+i b \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )\right ) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{i-(a+i b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{b d}\\ &=-\frac{(i a-b)^{5/2} (i A-B) \tan ^{-1}\left (\frac{\sqrt{i a-b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right ) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}}{d}+\frac{\left (40 a^3 A b-320 a A b^3-5 a^4 B-240 a^2 b^2 B+128 b^4 B\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right ) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}}{64 b^{3/2} d}-\frac{(i a+b)^{5/2} (i A+B) \tanh ^{-1}\left (\frac{\sqrt{i a+b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right ) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}}{d}+\frac{\left (40 a^2 A b-64 A b^3-5 a^3 B-112 a b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{64 b d \sqrt{\cot (c+d x)}}+\frac{\left (40 a A b-5 a^2 B-48 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{96 b d \sqrt{\cot (c+d x)}}+\frac{(8 A b-a B) (a+b \tan (c+d x))^{5/2}}{24 b d \sqrt{\cot (c+d x)}}+\frac{B (a+b \tan (c+d x))^{7/2}}{4 b d \sqrt{\cot (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 5.32911, size = 431, normalized size = 0.94 \[ \frac{\sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \left (-2 \left (5 a^2 B-40 a A b+48 b^2 B\right ) \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}-3 \left (-40 a^2 A b+5 a^3 B+112 a b^2 B+64 A b^3\right ) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}-\frac{3 \sqrt{a} \left (-40 a^3 A b+240 a^2 b^2 B+5 a^4 B+320 a A b^3-128 b^4 B\right ) \sqrt{\frac{b \tan (c+d x)}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{\sqrt{b} \sqrt{a+b \tan (c+d x)}}+8 (8 A b-a B) \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{5/2}-192 (-1)^{3/4} b (-a-i b)^{5/2} (A+i B) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )+192 \sqrt [4]{-1} b (a-i b)^{5/2} (B+i A) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )+48 B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{7/2}\right )}{192 b d} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/Cot[c + d*x]^(3/2),x]

[Out]

(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(-192*(-1)^(3/4)*(-a - I*b)^(5/2)*b*(A + I*B)*ArcTanh[((-1)^(1/4)*Sqrt[
-a - I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + 192*(-1)^(1/4)*(a - I*b)^(5/2)*b*(I*A + B)*ArcTanh[(
(-1)^(1/4)*Sqrt[a - I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] - 3*(-40*a^2*A*b + 64*A*b^3 + 5*a^3*B +
112*a*b^2*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]] - 2*(-40*a*A*b + 5*a^2*B + 48*b^2*B)*Sqrt[Tan[c + d*
x]]*(a + b*Tan[c + d*x])^(3/2) + 8*(8*A*b - a*B)*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(5/2) + 48*B*Sqrt[Tan
[c + d*x]]*(a + b*Tan[c + d*x])^(7/2) - (3*Sqrt[a]*(-40*a^3*A*b + 320*a*A*b^3 + 5*a^4*B + 240*a^2*b^2*B - 128*
b^4*B)*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[1 + (b*Tan[c + d*x])/a])/(Sqrt[b]*Sqrt[a + b*Tan[c +
d*x]])))/(192*b*d)

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Maple [C]  time = 3.753, size = 39339, normalized size = 86.1 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/cot(d*x+c)^(3/2),x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \tan \left (d x + c\right ) + A\right )}{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}{\cot \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/cot(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

integrate((B*tan(d*x + c) + A)*(b*tan(d*x + c) + a)^(5/2)/cot(d*x + c)^(3/2), x)

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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((a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/cot(d*x+c)^(3/2),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((a+b*tan(d*x+c))**(5/2)*(A+B*tan(d*x+c))/cot(d*x+c)**(3/2),x)

[Out]

Timed out

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Giac [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((a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/cot(d*x+c)^(3/2),x, algorithm="giac")

[Out]

Timed out