\(\int \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\) [621]

Optimal result

Integrand size = 35, antiderivative size = 299 \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {(a+i b)^2 (i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {i a-b} d}+\frac {(i a+b)^{3/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {2 \left (15 a^2 A-3 A b^2-20 a b B\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{15 a d}-\frac {2 (6 A b+5 a B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 d}-\frac {2 a A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d} \]

[Out]

(I*a+b)^(3/2)*(I*A+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+(a+I*b)^2*(I*A-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)*ta
n(d*x+c)^(1/2)/d/(I*a-b)^(1/2)-2/15*(6*A*b+5*B*a)*cot(d*x+c)^(3/2)*(a+b*tan(d*x+c))^(1/2)/d-2/5*a*A*cot(d*x+c)
^(5/2)*(a+b*tan(d*x+c))^(1/2)/d+2/15*(15*A*a^2-3*A*b^2-20*B*a*b)*cot(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2)/a/d

Rubi [A] (verified)

Time = 1.45 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {4326, 3686, 3730, 3697, 3696, 95, 209, 212} \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {2 \left (15 a^2 A-20 a b B-3 A b^2\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{15 a d}+\frac {(a+i b)^2 (-B+i A) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}+\frac {(b+i a)^{3/2} (B+i A) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 (5 a B+6 A b) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 d}-\frac {2 a A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d} \]

[In]

Int[Cot[c + d*x]^(7/2)*(a + b*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]),x]

[Out]

((a + I*b)^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]])/(Sqrt[I*a - b]*d) + ((I*a + b)^(3/2)*(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 + (2*(15*a^2*A - 3*A*b^2 - 20*a*b*B)*Sqrt[C
ot[c + d*x]]*Sqrt[a + b*Tan[c + d*x]])/(15*a*d) - (2*(6*A*b + 5*a*B)*Cot[c + d*x]^(3/2)*Sqrt[a + b*Tan[c + d*x
]])/(15*d) - (2*a*A*Cot[c + d*x]^(5/2)*Sqrt[a + b*Tan[c + d*x]])/(5*d)

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 209

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

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 3686

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*c - a*d)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e
+ f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Dist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m -
2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a*A*d*(b*d*(m - 1) - a*c*(n + 1)) + (b*B*c - (A*b + a*B)*d)*(b*c*(m - 1)
+ a*d*(n + 1)) - d*((a*A - b*B)*(b*c - a*d) + (A*b + a*B)*(a*c + b*d))*(n + 1)*Tan[e + f*x] - b*(d*(A*b*c + a
*B*c - a*A*d)*(m + n) - b*B*(c^2*(m - 1) - d^2*(n + 1)))*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] && GtQ[m, 1] && LtQ[n, -1] && (Inte
gerQ[m] || IntegersQ[2*m, 2*n])

Rule 3696

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

Rule 3697

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

Rule 3730

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] :> Simp[(A*b^2 - a*(b*B - a*C))*(a + b*Ta
n[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2))), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*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] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 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]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx \\ & = -\frac {2 a A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d}+\frac {1}{5} \left (2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\frac {1}{2} a (6 A b+5 a B)-\frac {5}{2} \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)-\frac {1}{2} b (4 a A-5 b B) \tan ^2(c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx \\ & = -\frac {2 (6 A b+5 a B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 d}-\frac {2 a A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d}-\frac {\left (4 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\frac {1}{4} a \left (15 a^2 A-3 A b^2-20 a b B\right )+\frac {15}{4} a \left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)+\frac {1}{2} a b (6 A b+5 a B) \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx}{15 a} \\ & = \frac {2 \left (15 a^2 A-3 A b^2-20 a b B\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{15 a d}-\frac {2 (6 A b+5 a B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 d}-\frac {2 a A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d}+\frac {\left (8 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {-\frac {15}{8} a^2 \left (2 a A b+a^2 B-b^2 B\right )+\frac {15}{8} a^2 \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{15 a^2} \\ & = \frac {2 \left (15 a^2 A-3 A b^2-20 a b B\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{15 a d}-\frac {2 (6 A b+5 a B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 d}-\frac {2 a A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d}+\frac {1}{2} \left ((a+i b)^2 (i A-B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx-\frac {1}{2} \left ((a-i b)^2 (i A+B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {1+i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx \\ & = \frac {2 \left (15 a^2 A-3 A b^2-20 a b B\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{15 a d}-\frac {2 (6 A b+5 a B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 d}-\frac {2 a A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d}+\frac {\left ((a+i b)^2 (i A-B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{(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 ) \text {Subst}\left (\int \frac {1}{(1-i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d} \\ & = \frac {2 \left (15 a^2 A-3 A b^2-20 a b B\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{15 a d}-\frac {2 (6 A b+5 a B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 d}-\frac {2 a A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d}+\frac {\left ((a+i b)^2 (i A-B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{1-(-i a+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 ) \text {Subst}\left (\int \frac {1}{1-(i a+b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \\ & = \frac {(a+i b)^2 (i A-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)}}{\sqrt {i a-b} d}+\frac {(i a+b)^{3/2} (i A+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 {2 \left (15 a^2 A-3 A b^2-20 a b B\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{15 a d}-\frac {2 (6 A b+5 a B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 d}-\frac {2 a A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.21 (sec) , antiderivative size = 286, normalized size of antiderivative = 0.96 \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=-\frac {\cot ^{\frac {5}{2}}(c+d x) \left (30 \sqrt [4]{-1} a \left ((-a+i b)^{3/2} (A-i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-(a+i b)^{3/2} (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )\right ) \tan ^{\frac {5}{2}}(c+d x)+15 a b B \sqrt {a+b \tan (c+d x)}+3 a (4 a A-5 b B) \sqrt {a+b \tan (c+d x)}+4 a (6 A b+5 a B) \tan (c+d x) \sqrt {a+b \tan (c+d x)}-4 \left (15 a^2 A-3 A b^2-20 a b B\right ) \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}\right )}{30 a d} \]

[In]

Integrate[Cot[c + d*x]^(7/2)*(a + b*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]),x]

[Out]

-1/30*(Cot[c + d*x]^(5/2)*(30*(-1)^(1/4)*a*((-a + I*b)^(3/2)*(A - I*B)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[
Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] - (a + I*b)^(3/2)*(A + I*B)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan
[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])*Tan[c + d*x]^(5/2) + 15*a*b*B*Sqrt[a + b*Tan[c + d*x]] + 3*a*(4*a*A - 5
*b*B)*Sqrt[a + b*Tan[c + d*x]] + 4*a*(6*A*b + 5*a*B)*Tan[c + d*x]*Sqrt[a + b*Tan[c + d*x]] - 4*(15*a^2*A - 3*A
*b^2 - 20*a*b*B)*Tan[c + d*x]^2*Sqrt[a + b*Tan[c + d*x]]))/(a*d)

Maple [B] (warning: unable to verify)

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

Time = 1.49 (sec) , antiderivative size = 2401287, normalized size of antiderivative = 8031.06

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

[In]

int(cot(d*x+c)^(7/2)*(a+b*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),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. 12233 vs. \(2 (243) = 486\).

Time = 2.37 (sec) , antiderivative size = 12233, normalized size of antiderivative = 40.91 \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

[In]

integrate(cot(d*x+c)^(7/2)*(a+b*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

Too large to include

Sympy [F(-1)]

Timed out. \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(cot(d*x+c)**(7/2)*(a+b*tan(d*x+c))**(3/2)*(A+B*tan(d*x+c)),x)

[Out]

Timed out

Maxima [F]

\[ \int \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cot \left (d x + c\right )^{\frac {7}{2}} \,d x } \]

[In]

integrate(cot(d*x+c)^(7/2)*(a+b*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*tan(d*x + c) + A)*(b*tan(d*x + c) + a)^(3/2)*cot(d*x + c)^(7/2), x)

Giac [F(-1)]

Timed out. \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(cot(d*x+c)^(7/2)*(a+b*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^{7/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2} \,d x \]

[In]

int(cot(c + d*x)^(7/2)*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^(3/2),x)

[Out]

int(cot(c + d*x)^(7/2)*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^(3/2), x)