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

Optimal. Leaf size=383 \[ \frac {\left (a^2 (-B)+6 a A b-8 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{8 b d \sqrt {\cot (c+d x)}}+\frac {\left (a^3 (-B)+6 a^2 A b-24 a b^2 B-16 A b^3\right ) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{8 b^{3/2} d}+\frac {(-b+i a)^{3/2} (A+i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(6 A b-a B) (a+b \tan (c+d x))^{3/2}}{12 b d \sqrt {\cot (c+d x)}}+\frac {(b+i a)^{3/2} (A-i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {B (a+b \tan (c+d x))^{5/2}}{3 b d \sqrt {\cot (c+d x)}} \]

[Out]

(I*a-b)^(3/2)*(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+1/8*(6*A*a^2*b-16*A*b^3-B*a^3-24*B*a*b^2)*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*c
ot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/b^(3/2)/d+(I*a+b)^(3/2)*(A-I*B)*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*t
an(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/d+1/8*(6*A*a*b-B*a^2-8*B*b^2)*(a+b*tan(d*x+c))^(1/2)/b/d/c
ot(d*x+c)^(1/2)+1/12*(6*A*b-B*a)*(a+b*tan(d*x+c))^(3/2)/b/d/cot(d*x+c)^(1/2)+1/3*B*(a+b*tan(d*x+c))^(5/2)/b/d/
cot(d*x+c)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 2.53, antiderivative size = 383, normalized size of antiderivative = 1.00, number of steps used = 16, 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} \[ \frac {\left (a^2 (-B)+6 a A b-8 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{8 b d \sqrt {\cot (c+d x)}}+\frac {\left (6 a^2 A b+a^3 (-B)-24 a b^2 B-16 A b^3\right ) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{8 b^{3/2} d}+\frac {(-b+i a)^{3/2} (A+i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(6 A b-a B) (a+b \tan (c+d x))^{3/2}}{12 b d \sqrt {\cot (c+d x)}}+\frac {(b+i a)^{3/2} (A-i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {B (a+b \tan (c+d x))^{5/2}}{3 b d \sqrt {\cot (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I*a - b)^(3/2)*(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 + ((6*a^2*A*b - 16*A*b^3 - a^3*B - 24*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*b^(3/2)*d) + ((I*a + b)^(3/2)*(A - I*B)*Ar
cTanh[(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 +
((6*a*A*b - a^2*B - 8*b^2*B)*Sqrt[a + b*Tan[c + d*x]])/(8*b*d*Sqrt[Cot[c + d*x]]) + ((6*A*b - a*B)*(a + b*Tan[
c + d*x])^(3/2))/(12*b*d*Sqrt[Cot[c + d*x]]) + (B*(a + b*Tan[c + d*x])^(5/2))/(3*b*d*Sqrt[Cot[c + d*x]])

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 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 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 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]

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 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 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 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]

Rubi steps

\begin {align*} \int \frac {(a+b \tan (c+d x))^{3/2} (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))^{3/2} (A+B \tan (c+d x)) \, dx\\ &=\frac {B (a+b \tan (c+d x))^{5/2}}{3 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 {a B}{2}-3 b B \tan (c+d x)+\frac {1}{2} (6 A b-a B) \tan ^2(c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{3 b}\\ &=\frac {(6 A b-a B) (a+b \tan (c+d x))^{3/2}}{12 b d \sqrt {\cot (c+d x)}}+\frac {B (a+b \tan (c+d x))^{5/2}}{3 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}{4} a (2 A b+a B)-6 b (A b+a B) \tan (c+d x)+\frac {3}{4} \left (6 a A b-a^2 B-8 b^2 B\right ) \tan ^2(c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{6 b}\\ &=\frac {\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{8 b d \sqrt {\cot (c+d x)}}+\frac {(6 A b-a B) (a+b \tan (c+d x))^{3/2}}{12 b d \sqrt {\cot (c+d x)}}+\frac {B (a+b \tan (c+d x))^{5/2}}{3 b 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 \left (10 a A b+a^2 B-8 b^2 B\right )-6 b \left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)+\frac {3}{8} \left (6 a^2 A b-16 A b^3-a^3 B-24 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 (6 a A b-a^2 B-8 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{8 b d \sqrt {\cot (c+d x)}}+\frac {(6 A b-a B) (a+b \tan (c+d x))^{3/2}}{12 b d \sqrt {\cot (c+d x)}}+\frac {B (a+b \tan (c+d x))^{5/2}}{3 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}{8} a \left (10 a A b+a^2 B-8 b^2 B\right )-6 b \left (2 a A b+a^2 B-b^2 B\right ) x+\frac {3}{8} \left (6 a^2 A b-16 A b^3-a^3 B-24 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 (6 a A b-a^2 B-8 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{8 b d \sqrt {\cot (c+d x)}}+\frac {(6 A b-a B) (a+b \tan (c+d x))^{3/2}}{12 b d \sqrt {\cot (c+d x)}}+\frac {B (a+b \tan (c+d x))^{5/2}}{3 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 (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 B\right )}{8 \sqrt {x} \sqrt {a+b x}}-\frac {6 \left (b \left (a^2 A-A b^2-2 a b B\right )+b \left (2 a A b+a^2 B-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 )}{6 b d}\\ &=\frac {\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{8 b d \sqrt {\cot (c+d x)}}+\frac {(6 A b-a B) (a+b \tan (c+d x))^{3/2}}{12 b d \sqrt {\cot (c+d x)}}+\frac {B (a+b \tan (c+d x))^{5/2}}{3 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^2 A-A b^2-2 a b B\right )+b \left (2 a A b+a^2 B-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 (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 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 )}{16 b d}\\ &=\frac {\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{8 b d \sqrt {\cot (c+d x)}}+\frac {(6 A b-a B) (a+b \tan (c+d x))^{3/2}}{12 b d \sqrt {\cot (c+d x)}}+\frac {B (a+b \tan (c+d x))^{5/2}}{3 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 {i b \left (a^2 A-A b^2-2 a b B\right )-b \left (2 a A b+a^2 B-b^2 B\right )}{2 (i-x) \sqrt {x} \sqrt {a+b x}}+\frac {i b \left (a^2 A-A b^2-2 a b B\right )+b \left (2 a A b+a^2 B-b^2 B\right )}{2 \sqrt {x} (i+x) \sqrt {a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{b d}+\frac {\left (\left (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 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 )}{8 b d}\\ &=\frac {\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{8 b d \sqrt {\cot (c+d x)}}+\frac {(6 A b-a B) (a+b \tan (c+d x))^{3/2}}{12 b d \sqrt {\cot (c+d x)}}+\frac {B (a+b \tan (c+d x))^{5/2}}{3 b d \sqrt {\cot (c+d x)}}-\frac {\left ((a+i b)^2 (i A-B) \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 d}-\frac {\left ((a-i b)^2 (i A+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 (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 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 )}{8 b d}\\ &=\frac {\left (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 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)}}{8 b^{3/2} d}+\frac {\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{8 b d \sqrt {\cot (c+d x)}}+\frac {(6 A b-a B) (a+b \tan (c+d x))^{3/2}}{12 b d \sqrt {\cot (c+d x)}}+\frac {B (a+b \tan (c+d x))^{5/2}}{3 b d \sqrt {\cot (c+d x)}}-\frac {\left ((a+i b)^2 (i A-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 ((a-i b)^2 (i A+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 {(i a-b)^{3/2} (A+i 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 (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 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)}}{8 b^{3/2} d}+\frac {(i a+b)^{3/2} (A-i 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 (6 a A b-a^2 B-8 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{8 b d \sqrt {\cot (c+d x)}}+\frac {(6 A b-a B) (a+b \tan (c+d x))^{3/2}}{12 b d \sqrt {\cot (c+d x)}}+\frac {B (a+b \tan (c+d x))^{5/2}}{3 b d \sqrt {\cot (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 5.67, size = 367, normalized size = 0.96 \[ \frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-3 \left (a^2 B-6 a A b+8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}-\frac {3 \sqrt {a} \left (a^3 B-6 a^2 A b+24 a b^2 B+16 A b^3\right ) \sqrt {\frac {b \tan (c+d x)}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {b} \sqrt {a+b \tan (c+d x)}}+24 \sqrt [4]{-1} b (-a+i b)^{3/2} (B+i A) \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+24 (-1)^{3/4} b (a+i b)^{3/2} (A+i B) \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+2 (6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}+8 B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}\right )}{24 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [C]  time = 5.21, size = 34780, normalized size = 90.81 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(d*x+c))^(3/2)*(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\cot \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}}{{\mathrm {cot}\left (c+d\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\cot ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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