Integrand size = 33, antiderivative size = 169 \[ \int \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=-\frac {\sqrt {a} (3 A b+2 a B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {(a-i b)^{3/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}-\frac {(a+i b)^{3/2} (i A-B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}-\frac {a A \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d} \]
(a-I*b)^(3/2)*(I*A+B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/d-(a+I *b)^(3/2)*(I*A-B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/d-(3*A*b+2 *B*a)*arctanh((a+b*tan(d*x+c))^(1/2)/a^(1/2))*a^(1/2)/d-a*A*cot(d*x+c)*(a+ b*tan(d*x+c))^(1/2)/d
Time = 0.64 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.67 \[ \int \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {-\sqrt {a} (3 A b+2 a B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )+(a-i b)^{3/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )-i a A \sqrt {a+i b} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )+A \sqrt {a+i b} b \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )+a \sqrt {a+i b} B \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )+i \sqrt {a+i b} b B \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )-a A \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d} \]
(-(Sqrt[a]*(3*A*b + 2*a*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a]]) + (a - I*b)^(3/2)*(I*A + B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]] - I*a*A*Sqrt[a + I*b]*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]] + A*Sq rt[a + I*b]*b*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]] + a*Sqrt[a + I*b]*B*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]] + I*Sqrt[a + I*b]* b*B*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]] - a*A*Cot[c + d*x]*Sqr t[a + b*Tan[c + d*x]])/d
Time = 1.40 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.93, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.485, Rules used = {3042, 4088, 27, 3042, 4136, 27, 3042, 4022, 3042, 4020, 25, 73, 221, 4117, 73, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan (c+d x)^2}dx\) |
| \(\Big \downarrow \) 4088 |
| \(\displaystyle \int \frac {\cot (c+d x) \left (-b (a A-2 b B) \tan ^2(c+d x)-2 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (3 A b+2 a B)\right )}{2 \sqrt {a+b \tan (c+d x)}}dx-\frac {a A \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \int \frac {\cot (c+d x) \left (-b (a A-2 b B) \tan ^2(c+d x)-2 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (3 A b+2 a B)\right )}{\sqrt {a+b \tan (c+d x)}}dx-\frac {a A \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {-b (a A-2 b B) \tan (c+d x)^2-2 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (3 A b+2 a B)}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx-\frac {a A \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}\) |
| \(\Big \downarrow \) 4136 |
| \(\displaystyle \frac {1}{2} \left (\int -\frac {2 \left (A a^2-2 b B a-A b^2+\left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)\right )}{\sqrt {a+b \tan (c+d x)}}dx+a (2 a B+3 A b) \int \frac {\cot (c+d x) \left (\tan ^2(c+d x)+1\right )}{\sqrt {a+b \tan (c+d x)}}dx\right )-\frac {a A \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (a (2 a B+3 A b) \int \frac {\cot (c+d x) \left (\tan ^2(c+d x)+1\right )}{\sqrt {a+b \tan (c+d x)}}dx-2 \int \frac {A a^2-2 b B a-A b^2+\left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx\right )-\frac {a A \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (a (2 a B+3 A b) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx-2 \int \frac {A a^2-2 b B a-A b^2+\left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx\right )-\frac {a A \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}\) |
| \(\Big \downarrow \) 4022 |
| \(\displaystyle -\frac {a A \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}+\frac {1}{2} \left (a (2 a B+3 A b) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx-2 \left (\frac {1}{2} (a-i b)^2 (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (a+i b)^2 (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx\right )\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {a A \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}+\frac {1}{2} \left (a (2 a B+3 A b) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx-2 \left (\frac {1}{2} (a-i b)^2 (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (a+i b)^2 (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx\right )\right )\) |
| \(\Big \downarrow \) 4020 |
| \(\displaystyle -\frac {a A \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}+\frac {1}{2} \left (a (2 a B+3 A b) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx-2 \left (\frac {i (a-i b)^2 (A-i B) \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {i (a+i b)^2 (A+i B) \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}\right )\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {a A \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}+\frac {1}{2} \left (a (2 a B+3 A b) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx-2 \left (\frac {i (a+i b)^2 (A+i B) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}-\frac {i (a-i b)^2 (A-i B) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}\right )\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {a A \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}+\frac {1}{2} \left (a (2 a B+3 A b) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx-2 \left (\frac {(a-i b)^2 (A-i B) \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {(a+i b)^2 (A+i B) \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}\right )\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {a A \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}+\frac {1}{2} \left (a (2 a B+3 A b) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx-2 \left (\frac {(a-i b)^{3/2} (A-i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d}+\frac {(a+i b)^{3/2} (A+i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d}\right )\right )\) |
| \(\Big \downarrow \) 4117 |
| \(\displaystyle -\frac {a A \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}+\frac {1}{2} \left (\frac {a (2 a B+3 A b) \int \frac {\cot (c+d x)}{\sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{d}-2 \left (\frac {(a-i b)^{3/2} (A-i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d}+\frac {(a+i b)^{3/2} (A+i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d}\right )\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {a A \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}+\frac {1}{2} \left (\frac {2 a (2 a B+3 A b) \int \frac {1}{\frac {a+b \tan (c+d x)}{b}-\frac {a}{b}}d\sqrt {a+b \tan (c+d x)}}{b d}-2 \left (\frac {(a-i b)^{3/2} (A-i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d}+\frac {(a+i b)^{3/2} (A+i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d}\right )\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {a A \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}+\frac {1}{2} \left (-\frac {2 \sqrt {a} (2 a B+3 A b) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )}{d}-2 \left (\frac {(a-i b)^{3/2} (A-i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d}+\frac {(a+i b)^{3/2} (A+i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d}\right )\right )\) |
(-2*(((a - I*b)^(3/2)*(A - I*B)*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/d + (( a + I*b)^(3/2)*(A + I*B)*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/d) - (2*Sqrt[ a]*(3*A*b + 2*a*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a]])/d)/2 - (a*A* Cot[c + d*x]*Sqrt[a + b*Tan[c + d*x]])/d
3.4.29.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f) Subst[Int[(a + (b/d)*x)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2 Int[(a + b*Tan[e + f*x])^m*( 1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2 Int[(a + b*Tan[e + f*x])^m *(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && !IntegerQ[m]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(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] - Simp[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] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[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]
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] :> Simp[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] + Simp[ (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]
Leaf count of result is larger than twice the leaf count of optimal. \(1688\) vs. \(2(143)=286\).
Time = 0.24 (sec) , antiderivative size = 1689, normalized size of antiderivative = 9.99
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(1689\) |
| default | \(\text {Expression too large to display}\) | \(1689\) |
-1/4/d/b*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x +c)-a-(a^2+b^2)^(1/2))*A*(a^2+b^2)^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a-1 /d/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*( a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*B*(a^2+b^2)^(1/2)*a+ 1/4/d/b*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^( 1/2)+(a^2+b^2)^(1/2))*A*(a^2+b^2)^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a+1/ d/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b ^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*B*(a^2+b^2)^(1/2)*a+1 /d*b^2/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*( a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*B-1/d/(2*(a^2+b^ 2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c ))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*B*b^2-1/4/d*b*ln((a+b*tan(d*x+c)) ^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*A*(2* (a^2+b^2)^(1/2)+2*a)^(1/2)+1/4/d*b*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2 )*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*A*(2*(a^2+b^2)^(1/2)+2*a) ^(1/2)-1/4/d/b*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2) +2*a)^(1/2)+(a^2+b^2)^(1/2))*A*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^2-1/4/d*ln( b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b ^2)^(1/2))*B*(a^2+b^2)^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+1/2/d*ln(b*tan( d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)...
Leaf count of result is larger than twice the leaf count of optimal. 3116 vs. \(2 (137) = 274\).
Time = 3.34 (sec) , antiderivative size = 6248, normalized size of antiderivative = 36.97 \[ \int \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]
\[ \int \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\int \left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}} \cot ^{2}{\left (c + d x \right )}\, dx \]
\[ \int \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cot \left (d x + c\right )^{2} \,d x } \]
Timed out. \[ \int \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]
Time = 9.84 (sec) , antiderivative size = 21319, normalized size of antiderivative = 126.15 \[ \int \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]
(a^(1/2)*atan(((a^(1/2)*((16*(a + b*tan(c + d*x))^(1/2)*(2*A^4*b^16 + 2*B^ 4*b^16 + 4*A^2*B^2*b^16 - A^4*a^2*b^14 + 66*A^4*a^4*b^12 - A^4*a^6*b^10 + 2*A^4*a^8*b^8 + 8*B^4*a^2*b^14 + 16*B^4*a^4*b^12 - 16*B^4*a^6*b^10 + 6*B^4 *a^8*b^8 + 25*A^2*B^2*a^2*b^14 - 130*A^2*B^2*a^4*b^12 + 145*A^2*B^2*a^6*b^ 10 + 12*A*B^3*a^3*b^13 - 104*A*B^3*a^5*b^11 + 44*A*B^3*a^7*b^9 - 84*A^3*B* a^3*b^13 + 144*A^3*B*a^5*b^11 - 12*A^3*B*a^7*b^9))/d^4 + (a^(1/2)*(3*A*b + 2*B*a)*((8*(100*A^3*a^2*b^13*d^2 + 44*A^3*a^4*b^11*d^2 - 56*A^3*a^6*b^9*d ^2 - 92*B^3*a^3*b^12*d^2 - 84*B^3*a^5*b^10*d^2 + 12*B^3*a^7*b^8*d^2 + 4*B^ 3*a*b^14*d^2 - 92*A^2*B*a*b^14*d^2 - 216*A*B^2*a^2*b^13*d^2 - 48*A*B^2*a^4 *b^11*d^2 + 168*A*B^2*a^6*b^9*d^2 + 208*A^2*B*a^3*b^12*d^2 + 264*A^2*B*a^5 *b^10*d^2 - 36*A^2*B*a^7*b^8*d^2))/d^5 - (a^(1/2)*(3*A*b + 2*B*a)*((16*(a + b*tan(c + d*x))^(1/2)*(92*A^2*a^3*b^10*d^2 - 20*A^2*a^5*b^8*d^2 - 56*B^2 *a^3*b^10*d^2 + 36*B^2*a^5*b^8*d^2 - 32*A*B*b^13*d^2 + 44*A^2*a*b^12*d^2 - 44*B^2*a*b^12*d^2 + 32*A*B*a^2*b^11*d^2 + 176*A*B*a^4*b^9*d^2))/d^4 + (a^ (1/2)*(3*A*b + 2*B*a)*((8*(80*A*a*b^11*d^4 + 80*A*a^3*b^9*d^4 + 48*B*a^2*b ^10*d^4 + 48*B*a^4*b^8*d^4))/d^5 - (8*a^(1/2)*(3*A*b + 2*B*a)*(32*b^10*d^4 + 48*a^2*b^8*d^4)*(a + b*tan(c + d*x))^(1/2))/d^5))/(2*d)))/(2*d)))/(2*d) )*(3*A*b + 2*B*a)*1i)/(2*d) + (a^(1/2)*((16*(a + b*tan(c + d*x))^(1/2)*(2* A^4*b^16 + 2*B^4*b^16 + 4*A^2*B^2*b^16 - A^4*a^2*b^14 + 66*A^4*a^4*b^12 - A^4*a^6*b^10 + 2*A^4*a^8*b^8 + 8*B^4*a^2*b^14 + 16*B^4*a^4*b^12 - 16*B^...