Integrand size = 35, antiderivative size = 282 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {(i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{5/2} d}+\frac {2 B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^{5/2} d}-\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{5/2} d}+\frac {2 a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}} \]
(I*A-B)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/(I*a -b)^(5/2)/d+2*B*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/b ^(5/2)/d-(I*A+B)*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^( 1/2))/(I*a+b)^(5/2)/d+2*a*(2*A*b^3-a*(a^2+3*b^2)*B)*tan(d*x+c)^(1/2)/b^2/( a^2+b^2)^2/d/(a+b*tan(d*x+c))^(1/2)+2/3*a*(A*b-B*a)*tan(d*x+c)^(3/2)/b/(a^ 2+b^2)/d/(a+b*tan(d*x+c))^(3/2)
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(596\) vs. \(2(282)=564\).
Time = 6.46 (sec) , antiderivative size = 596, normalized size of antiderivative = 2.11 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {(A-i B) \left (\frac {3 \sqrt [4]{-1} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(-a+i b)^{5/2}}+\frac {\tan ^{\frac {3}{2}}(c+d x)}{(a-i b) (a+b \tan (c+d x))^{3/2}}-\frac {3 i \sqrt {\tan (c+d x)}}{(a-i b)^2 \sqrt {a+b \tan (c+d x)}}\right )}{3 d}-\frac {(A+i B) \left (\frac {3 \sqrt [4]{-1} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(a+i b)^{5/2}}-\frac {\tan ^{\frac {3}{2}}(c+d x)}{(a+i b) (a+b \tan (c+d x))^{3/2}}-\frac {3 i \sqrt {\tan (c+d x)}}{(a+i b)^2 \sqrt {a+b \tan (c+d x)}}\right )}{3 d}+\frac {(i A-B) \sqrt {a+b \tan (c+d x)} \left (\frac {b^2 \tan ^2(c+d x)}{(a+b \tan (c+d x))^2}+\frac {3 b \tan (c+d x)}{a+b \tan (c+d x)}-\frac {3 \sqrt {b} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {\tan (c+d x)}}{\sqrt {a} \sqrt {1+\frac {b \tan (c+d x)}{a}}}\right )}{3 b^3 d \sqrt {\tan (c+d x)}}-\frac {(i A+B) \sqrt {a+b \tan (c+d x)} \left (\frac {b^2 \tan ^2(c+d x)}{(a+b \tan (c+d x))^2}+\frac {3 b \tan (c+d x)}{a+b \tan (c+d x)}-\frac {3 \sqrt {b} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {\tan (c+d x)}}{\sqrt {a} \sqrt {1+\frac {b \tan (c+d x)}{a}}}\right )}{3 b^3 d \sqrt {\tan (c+d x)}} \]
((A - I*B)*((3*(-1)^(1/4)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d *x]])/Sqrt[a + b*Tan[c + d*x]]])/(-a + I*b)^(5/2) + Tan[c + d*x]^(3/2)/((a - I*b)*(a + b*Tan[c + d*x])^(3/2)) - ((3*I)*Sqrt[Tan[c + d*x]])/((a - I*b )^2*Sqrt[a + b*Tan[c + d*x]])))/(3*d) - ((A + I*B)*((3*(-1)^(1/4)*ArcTan[( (-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(a + I*b)^(5/2) - Tan[c + d*x]^(3/2)/((a + I*b)*(a + b*Tan[c + d*x])^(3/2)) - ((3*I)*Sqrt[Tan[c + d*x]])/((a + I*b)^2*Sqrt[a + b*Tan[c + d*x]])))/(3*d ) + ((I*A - B)*Sqrt[a + b*Tan[c + d*x]]*((b^2*Tan[c + d*x]^2)/(a + b*Tan[c + d*x])^2 + (3*b*Tan[c + d*x])/(a + b*Tan[c + d*x]) - (3*Sqrt[b]*ArcSinh[ (Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[Tan[c + d*x]])/(Sqrt[a]*Sqrt[1 + (b*Tan[c + d*x])/a])))/(3*b^3*d*Sqrt[Tan[c + d*x]]) - ((I*A + B)*Sqrt[a + b*Tan[c + d*x]]*((b^2*Tan[c + d*x]^2)/(a + b*Tan[c + d*x])^2 + (3*b*Tan[ c + d*x])/(a + b*Tan[c + d*x]) - (3*Sqrt[b]*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[Tan[c + d*x]])/(Sqrt[a]*Sqrt[1 + (b*Tan[c + d*x])/a]) ))/(3*b^3*d*Sqrt[Tan[c + d*x]])
Time = 1.73 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.22, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.314, Rules used = {3042, 4088, 27, 3042, 4128, 27, 3042, 4138, 2035, 2257, 2009}
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.
| \(\displaystyle \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^{5/2} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4088 |
| \(\displaystyle \frac {2 \int -\frac {3 \sqrt {\tan (c+d x)} \left (-\left (\left (a^2+b^2\right ) B \tan ^2(c+d x)\right )-b (A b-a B) \tan (c+d x)+a (A b-a B)\right )}{2 (a+b \tan (c+d x))^{3/2}}dx}{3 b \left (a^2+b^2\right )}+\frac {2 a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {\int \frac {\sqrt {\tan (c+d x)} \left (-\left (\left (a^2+b^2\right ) B \tan ^2(c+d x)\right )-b (A b-a B) \tan (c+d x)+a (A b-a B)\right )}{(a+b \tan (c+d x))^{3/2}}dx}{b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {\int \frac {\sqrt {\tan (c+d x)} \left (-\left (\left (a^2+b^2\right ) B \tan (c+d x)^2\right )-b (A b-a B) \tan (c+d x)+a (A b-a B)\right )}{(a+b \tan (c+d x))^{3/2}}dx}{b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 4128 |
| \(\displaystyle \frac {2 a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {\frac {2 \int \frac {\left (A a^2+2 b B a-A b^2\right ) \tan (c+d x) b^2-\left (a^2+b^2\right )^2 B \tan ^2(c+d x)+a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right )}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{b \left (a^2+b^2\right )}-\frac {2 a \left (2 A b^3-a B \left (a^2+3 b^2\right )\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {\frac {\int \frac {\left (A a^2+2 b B a-A b^2\right ) \tan (c+d x) b^2-\left (a^2+b^2\right )^2 B \tan ^2(c+d x)+a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{b \left (a^2+b^2\right )}-\frac {2 a \left (2 A b^3-a B \left (a^2+3 b^2\right )\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {\frac {\int \frac {\left (A a^2+2 b B a-A b^2\right ) \tan (c+d x) b^2-\left (a^2+b^2\right )^2 B \tan (c+d x)^2+a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{b \left (a^2+b^2\right )}-\frac {2 a \left (2 A b^3-a B \left (a^2+3 b^2\right )\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 4138 |
| \(\displaystyle \frac {2 a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {\frac {\int \frac {\left (A a^2+2 b B a-A b^2\right ) \tan (c+d x) b^2-\left (a^2+b^2\right )^2 B \tan ^2(c+d x)+a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{b d \left (a^2+b^2\right )}-\frac {2 a \left (2 A b^3-a B \left (a^2+3 b^2\right )\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle \frac {2 a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {\frac {2 \int \frac {\left (A a^2+2 b B a-A b^2\right ) \tan (c+d x) b^2-\left (a^2+b^2\right )^2 B \tan ^2(c+d x)+a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right )}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right )}-\frac {2 a \left (2 A b^3-a B \left (a^2+3 b^2\right )\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 2257 |
| \(\displaystyle \frac {2 a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {\frac {2 \int \left (\frac {\left (-B a^2+2 A b a+b^2 B\right ) b^2+\left (A a^2+2 b B a-A b^2\right ) \tan (c+d x) b^2}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}-\frac {\left (a^2+b^2\right )^2 B}{\sqrt {a+b \tan (c+d x)}}\right )d\sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right )}-\frac {2 a \left (2 A b^3-a B \left (a^2+3 b^2\right )\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {-\frac {2 a \left (2 A b^3-a B \left (a^2+3 b^2\right )\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {2 \left (-\frac {B \left (a^2+b^2\right )^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b}}+\frac {b^2 (a-i b)^2 (-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{2 \sqrt {-b+i a}}-\frac {b^2 (a+i b)^2 (B+i A) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{2 \sqrt {b+i a}}\right )}{b d \left (a^2+b^2\right )}}{b \left (a^2+b^2\right )}\) |
(2*a*(A*b - a*B)*Tan[c + d*x]^(3/2))/(3*b*(a^2 + b^2)*d*(a + b*Tan[c + d*x ])^(3/2)) - ((2*(((a - I*b)^2*b^2*(I*A - B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan [c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(2*Sqrt[I*a - b]) - ((a^2 + b^2)^2* B*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[b] - ((a + I*b)^2*b^2*(I*A + B)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sq rt[a + b*Tan[c + d*x]]])/(2*Sqrt[I*a + b])))/(b*(a^2 + b^2)*d) - (2*a*(2*A *b^3 - a*(a^2 + 3*b^2)*B)*Sqrt[Tan[c + d*x]])/(b*(a^2 + b^2)*d*Sqrt[a + b* Tan[c + d*x]]))/(b*(a^2 + b^2))
3.5.63.3.1 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[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol ] :> Int[ExpandIntegrand[Px*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a , c, d, e, q}, x] && PolyQ[Px, x] && IntegerQ[p]
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_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*d^2 + c*(c*C - B*d))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Sim p[1/(d*(n + 1)*(c^2 + d^2)) Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*d*(b*d*m - a*c*(n + 1)) + (c*C - B*d)*(b*c*m + a*d* (n + 1)) - d*(n + 1)*((A - C)*(b*c - a*d) + B*(a*c + b*d))*Tan[e + f*x] - b *(d*(B*c - A*d)*(m + n + 1) - C*(c^2*m - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ [a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
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] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, S imp[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]
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 2.18 (sec) , antiderivative size = 2978162, normalized size of antiderivative = 10560.86
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 27883 vs. \(2 (236) = 472\).
Time = 22.13 (sec) , antiderivative size = 55768, normalized size of antiderivative = 197.76 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}} \,d x \]