\(\int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx\) [634]

Optimal result

Integrand size = 35, antiderivative size = 376 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=-\frac {(i a-b)^{5/2} (A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {\left (30 a^2 A b-16 A b^3+5 a^3 B-40 a b^2 B\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{8 \sqrt {b} d}+\frac {(i a+b)^{5/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {\left (14 a A b+5 a^2 B-8 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{8 d \sqrt {\cot (c+d x)}}+\frac {b B (a+b \tan (c+d x))^{3/2}}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {(2 A b+3 a B) (a+b \tan (c+d x))^{3/2}}{4 d \sqrt {\cot (c+d x)}} \]

[Out]

-(I*a-b)^(5/2)*(A+I*B)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+
c)^(1/2)/d+(I*a+b)^(5/2)*(A-I*B)*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/
2)*tan(d*x+c)^(1/2)/d+1/8*(30*A*a^2*b-16*A*b^3+5*B*a^3-40*B*a*b^2)*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d
*x+c))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/d/b^(1/2)+1/8*(14*A*a*b+5*B*a^2-8*B*b^2)*(a+b*tan(d*x+c))^(1/2
)/d/cot(d*x+c)^(1/2)+1/3*b*B*(a+b*tan(d*x+c))^(3/2)/d/cot(d*x+c)^(3/2)+1/4*(2*A*b+3*B*a)*(a+b*tan(d*x+c))^(3/2
)/d/cot(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 3.60 (sec) , antiderivative size = 376, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.314, Rules used = {4326, 3688, 3728, 3736, 6857, 65, 223, 212, 95, 211, 214} \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\frac {\left (5 a^2 B+14 a A b-8 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{8 d \sqrt {\cot (c+d x)}}+\frac {\left (5 a^3 B+30 a^2 A b-40 a b^2 B-16 A b^3\right ) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{8 \sqrt {b} d}-\frac {(-b+i a)^{5/2} (A+i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(b+i a)^{5/2} (A-i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(3 a B+2 A b) (a+b \tan (c+d x))^{3/2}}{4 d \sqrt {\cot (c+d x)}}+\frac {b B (a+b \tan (c+d x))^{3/2}}{3 d \cot ^{\frac {3}{2}}(c+d x)} \]

[In]

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

[Out]

-(((I*a - b)^(5/2)*(A + I*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) + ((30*a^2*A*b - 16*A*b^3 + 5*a^3*B - 40*a*b^2*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]])/(8*Sqrt[b]*d) + ((I*a + b)^(5/2)*(A - I
*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 + ((14*a*A*b + 5*a^2*B - 8*b^2*B)*Sqrt[a + b*Tan[c + d*x]])/(8*d*Sqrt[Cot[c + d*x]]) + (b*B*(a + b*Tan[c +
d*x])^(3/2))/(3*d*Cot[c + d*x]^(3/2)) + ((2*A*b + 3*a*B)*(a + b*Tan[c + d*x])^(3/2))/(4*d*Sqrt[Cot[c + d*x]])

Rule 65

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 95

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 211

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

Rule 212

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

Rule 214

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

Rule 223

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 3688

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 3728

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 3736

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 4326

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 6857

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]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx \\ & = \frac {b B (a+b \tan (c+d x))^{3/2}}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\frac {3}{2} a (2 a A-b B)+3 \left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)+\frac {3}{2} b (2 A b+3 a B) \tan ^2(c+d x)\right ) \, dx \\ & = \frac {b B (a+b \tan (c+d x))^{3/2}}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {(2 A b+3 a B) (a+b \tan (c+d x))^{3/2}}{4 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}{4} a b (2 A b+3 a B)+6 b \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)+\frac {3}{4} b \left (14 a A b+5 a^2 B-8 b^2 B\right ) \tan ^2(c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{6 b} \\ & = \frac {\left (14 a A b+5 a^2 B-8 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{8 d \sqrt {\cot (c+d x)}}+\frac {b B (a+b \tan (c+d x))^{3/2}}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {(2 A b+3 a B) (a+b \tan (c+d x))^{3/2}}{4 d \sqrt {\cot (c+d x)}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {-\frac {3}{8} a b \left (18 a A b+11 a^2 B-8 b^2 B\right )+6 b \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \tan (c+d x)+\frac {3}{8} b \left (30 a^2 A b-16 A b^3+5 a^3 B-40 a b^2 B\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{6 b} \\ & = \frac {\left (14 a A b+5 a^2 B-8 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{8 d \sqrt {\cot (c+d x)}}+\frac {b B (a+b \tan (c+d x))^{3/2}}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {(2 A b+3 a B) (a+b \tan (c+d x))^{3/2}}{4 d \sqrt {\cot (c+d x)}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {-\frac {3}{8} a b \left (18 a A b+11 a^2 B-8 b^2 B\right )+6 b \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) x+\frac {3}{8} b \left (30 a^2 A b-16 A b^3+5 a^3 B-40 a b^2 B\right ) x^2}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{6 b d} \\ & = \frac {\left (14 a A b+5 a^2 B-8 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{8 d \sqrt {\cot (c+d x)}}+\frac {b B (a+b \tan (c+d x))^{3/2}}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {(2 A b+3 a B) (a+b \tan (c+d x))^{3/2}}{4 d \sqrt {\cot (c+d x)}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \left (\frac {3 b \left (30 a^2 A b-16 A b^3+5 a^3 B-40 a b^2 B\right )}{8 \sqrt {x} \sqrt {a+b x}}-\frac {6 \left (b \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )-b \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) x\right )}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{6 b d} \\ & = \frac {\left (14 a A b+5 a^2 B-8 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{8 d \sqrt {\cot (c+d x)}}+\frac {b B (a+b \tan (c+d x))^{3/2}}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {(2 A b+3 a B) (a+b \tan (c+d x))^{3/2}}{4 d \sqrt {\cot (c+d x)}}-\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {b \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )-b \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 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 (30 a^2 A b-16 A b^3+5 a^3 B-40 a b^2 B\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{16 d} \\ & = \frac {\left (14 a A b+5 a^2 B-8 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{8 d \sqrt {\cot (c+d x)}}+\frac {b B (a+b \tan (c+d x))^{3/2}}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {(2 A b+3 a B) (a+b \tan (c+d x))^{3/2}}{4 d \sqrt {\cot (c+d x)}}-\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \left (\frac {i b \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )+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 {i b \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )-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 (30 a^2 A b-16 A b^3+5 a^3 B-40 a b^2 B\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{8 d} \\ & = \frac {\left (14 a A b+5 a^2 B-8 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{8 d \sqrt {\cot (c+d x)}}+\frac {b B (a+b \tan (c+d x))^{3/2}}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {(2 A b+3 a B) (a+b \tan (c+d x))^{3/2}}{4 d \sqrt {\cot (c+d x)}}+\frac {\left ((a-i b)^3 (A-i B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} (i+x) \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}-\frac {\left ((a+i b)^3 (A+i B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{(i-x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {\left (\left (30 a^2 A b-16 A b^3+5 a^3 B-40 a b^2 B\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {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 )}{8 d} \\ & = \frac {\left (30 a^2 A b-16 A b^3+5 a^3 B-40 a b^2 B\right ) \text {arctanh}\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)}}{8 \sqrt {b} d}+\frac {\left (14 a A b+5 a^2 B-8 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{8 d \sqrt {\cot (c+d x)}}+\frac {b B (a+b \tan (c+d x))^{3/2}}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {(2 A b+3 a B) (a+b \tan (c+d x))^{3/2}}{4 d \sqrt {\cot (c+d x)}}+\frac {\left ((a-i b)^3 (A-i B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {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 ((a+i b)^3 (A+i B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {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 {(i a-b)^{5/2} (A+i B) \arctan \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 (30 a^2 A b-16 A b^3+5 a^3 B-40 a b^2 B\right ) \text {arctanh}\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)}}{8 \sqrt {b} d}+\frac {(i a+b)^{5/2} (A-i B) \text {arctanh}\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 (14 a A b+5 a^2 B-8 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{8 d \sqrt {\cot (c+d x)}}+\frac {b B (a+b \tan (c+d x))^{3/2}}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {(2 A b+3 a B) (a+b \tan (c+d x))^{3/2}}{4 d \sqrt {\cot (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.84 (sec) , antiderivative size = 365, normalized size of antiderivative = 0.97 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (24 \sqrt [4]{-1} (-a+i b)^{5/2} (A-i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+24 \sqrt [4]{-1} (a+i b)^{5/2} (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+3 \left (14 a A b+5 a^2 B-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}+6 (2 A b+3 a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}+8 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}+\frac {3 \sqrt {a} \left (30 a^2 A b-16 A b^3+5 a^3 B-40 a b^2 B\right ) \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {b} \sqrt {a+b \tan (c+d x)}}\right )}{24 d} \]

[In]

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

[Out]

(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(24*(-1)^(1/4)*(-a + I*b)^(5/2)*(A - I*B)*ArcTan[((-1)^(1/4)*Sqrt[-a +
I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + 24*(-1)^(1/4)*(a + I*b)^(5/2)*(A + I*B)*ArcTan[((-1)^(1/4
)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + 3*(14*a*A*b + 5*a^2*B - 8*b^2*B)*Sqrt[Tan[c +
d*x]]*Sqrt[a + b*Tan[c + d*x]] + 6*(2*A*b + 3*a*B)*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(3/2) + 8*b*B*Tan[c
+ d*x]^(3/2)*(a + b*Tan[c + d*x])^(3/2) + (3*Sqrt[a]*(30*a^2*A*b - 16*A*b^3 + 5*a^3*B - 40*a*b^2*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]])))/(24*d)

Maple [B] (warning: unable to verify)

result has leaf size over 500,000. Avoiding possible recursion issues.

Time = 1.42 (sec) , antiderivative size = 2657129, normalized size of antiderivative = 7066.83

\[\text {output too large to display}\]

[In]

int((a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/cot(d*x+c)^(1/2),x)

[Out]

result too large to display

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 17750 vs. \(2 (308) = 616\).

Time = 7.92 (sec) , antiderivative size = 35533, normalized size of antiderivative = 94.50 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\text {Too large to display} \]

[In]

integrate((a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/cot(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

Too large to include

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\text {Timed out} \]

[In]

integrate((a+b*tan(d*x+c))**(5/2)*(A+B*tan(d*x+c))/cot(d*x+c)**(1/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\sqrt {\cot \left (d x + c\right )}} \,d x } \]

[In]

integrate((a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/cot(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F(-1)]

Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\text {Timed out} \]

[In]

integrate((a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/cot(d*x+c)^(1/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{\sqrt {\mathrm {cot}\left (c+d\,x\right )}} \,d x \]

[In]

int(((A + B*tan(c + d*x))*(a + b*tan(c + d*x))^(5/2))/cot(c + d*x)^(1/2),x)

[Out]

int(((A + B*tan(c + d*x))*(a + b*tan(c + d*x))^(5/2))/cot(c + d*x)^(1/2), x)