Integrand size = 35, antiderivative size = 247 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {(i a-b)^{5/2} (A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(i a+b)^{5/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a (8 A b+5 a B) \sqrt {a+b \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2 A-23 A b^2-35 a b B\right ) \sqrt {a+b \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)} \]
-(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))/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))/d+2/15*(15*A*a^2-23*A*b^2-35*B*a*b)*(a+b*tan(d*x+c ))^(1/2)/d/tan(d*x+c)^(1/2)-2/15*a*(8*A*b+5*B*a)*(a+b*tan(d*x+c))^(1/2)/d/ tan(d*x+c)^(3/2)-2/5*a*A*(a+b*tan(d*x+c))^(3/2)/d/tan(d*x+c)^(5/2)
Time = 2.29 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.30 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\frac {60 \sqrt [4]{-1} \left ((-a+i b)^{5/2} (A-i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+(a+i b)^{5/2} (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )\right ) \tan ^{\frac {5}{2}}(c+d x)+15 b (-2 A b+a B) \sqrt {a+b \tan (c+d x)}-3 \left (8 a^2 A-10 A b^2-15 a b B\right ) \sqrt {a+b \tan (c+d x)}-4 \left (22 a A b+10 a^2 B-15 b^2 B\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}+8 \left (15 a^2 A-23 A b^2-35 a b B\right ) \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}-60 b B (a+b \tan (c+d x))^{3/2}}{60 d \tan ^{\frac {5}{2}}(c+d x)} \]
(60*(-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]]] + (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]]])*Tan[c + d*x]^(5/2) + 15*b*(-2*A*b + a*B)*Sqrt[a + b*Tan[c + d*x]] - 3*(8*a^2*A - 10*A*b^2 - 15*a*b*B)*Sqrt[a + b*Tan[c + d*x]] - 4*(22*a*A* b + 10*a^2*B - 15*b^2*B)*Tan[c + d*x]*Sqrt[a + b*Tan[c + d*x]] + 8*(15*a^2 *A - 23*A*b^2 - 35*a*b*B)*Tan[c + d*x]^2*Sqrt[a + b*Tan[c + d*x]] - 60*b*B *(a + b*Tan[c + d*x])^(3/2))/(60*d*Tan[c + d*x]^(5/2))
Time = 1.89 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.15, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.457, Rules used = {3042, 4088, 27, 3042, 4128, 27, 3042, 4132, 27, 3042, 4099, 3042, 4098, 104, 216, 219}
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.
| \(\displaystyle \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan (c+d x)^{7/2}}dx\) |
| \(\Big \downarrow \) 4088 |
| \(\displaystyle \frac {2}{5} \int \frac {\sqrt {a+b \tan (c+d x)} \left (-b (2 a A-5 b B) \tan ^2(c+d x)-5 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (8 A b+5 a B)\right )}{2 \tan ^{\frac {5}{2}}(c+d x)}dx-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \frac {\sqrt {a+b \tan (c+d x)} \left (-b (2 a A-5 b B) \tan ^2(c+d x)-5 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (8 A b+5 a B)\right )}{\tan ^{\frac {5}{2}}(c+d x)}dx-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {\sqrt {a+b \tan (c+d x)} \left (-b (2 a A-5 b B) \tan (c+d x)^2-5 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (8 A b+5 a B)\right )}{\tan (c+d x)^{5/2}}dx-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4128 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int -\frac {b \left (10 B a^2+22 A b a-15 b^2 B\right ) \tan ^2(c+d x)+15 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (15 A a^2-35 b B a-23 A b^2\right )}{2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx-\frac {2 a (5 a B+8 A b) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (-\frac {1}{3} \int \frac {b \left (10 B a^2+22 A b a-15 b^2 B\right ) \tan ^2(c+d x)+15 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (15 A a^2-35 b B a-23 A b^2\right )}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx-\frac {2 a (5 a B+8 A b) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (-\frac {1}{3} \int \frac {b \left (10 B a^2+22 A b a-15 b^2 B\right ) \tan (c+d x)^2+15 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (15 A a^2-35 b B a-23 A b^2\right )}{\tan (c+d x)^{3/2} \sqrt {a+b \tan (c+d x)}}dx-\frac {2 a (5 a B+8 A b) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4132 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {2 \int -\frac {15 \left (a \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right )-a \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)\right )}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}+\frac {2 \left (15 a^2 A-35 a b B-23 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 a (5 a B+8 A b) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {2 \left (15 a^2 A-35 a b B-23 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {15 \int \frac {a \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right )-a \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}\right )-\frac {2 a (5 a B+8 A b) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {2 \left (15 a^2 A-35 a b B-23 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {15 \int \frac {a \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right )-a \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}\right )-\frac {2 a (5 a B+8 A b) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4099 |
| \(\displaystyle -\frac {2 a A (a+b \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (-\frac {2 a (5 a B+8 A b) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (\frac {2 \left (15 a^2 A-35 a b B-23 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {15 \left (\frac {1}{2} a (a-i b)^3 (B+i A) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} a (a+i b)^3 (-B+i A) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )}{a}\right )\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 a A (a+b \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (-\frac {2 a (5 a B+8 A b) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (\frac {2 \left (15 a^2 A-35 a b B-23 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {15 \left (\frac {1}{2} a (a-i b)^3 (B+i A) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} a (a+i b)^3 (-B+i A) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )}{a}\right )\right )\) |
| \(\Big \downarrow \) 4098 |
| \(\displaystyle -\frac {2 a A (a+b \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (-\frac {2 a (5 a B+8 A b) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (\frac {2 \left (15 a^2 A-35 a b B-23 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {15 \left (\frac {a (a-i b)^3 (B+i A) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}-\frac {a (a+i b)^3 (-B+i A) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}\right )}{a}\right )\right )\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle -\frac {2 a A (a+b \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (-\frac {2 a (5 a B+8 A b) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (\frac {2 \left (15 a^2 A-35 a b B-23 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {15 \left (\frac {a (a-i b)^3 (B+i A) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}-\frac {a (a+i b)^3 (-B+i A) \int \frac {1}{\frac {(i a-b) \tan (c+d x)}{a+b \tan (c+d x)}+1}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}\right )}{a}\right )\right )\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {2 a A (a+b \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (-\frac {2 a (5 a B+8 A b) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (\frac {2 \left (15 a^2 A-35 a b B-23 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {15 \left (\frac {a (a-i b)^3 (B+i A) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}-\frac {a (a+i b)^3 (-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}\right )}{a}\right )\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {2 a A (a+b \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (-\frac {2 a (5 a B+8 A b) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (\frac {2 \left (15 a^2 A-35 a b B-23 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {15 \left (\frac {a (a-i b)^3 (B+i A) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}}-\frac {a (a+i b)^3 (-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}\right )}{a}\right )\right )\) |
(-2*a*A*(a + b*Tan[c + d*x])^(3/2))/(5*d*Tan[c + d*x]^(5/2)) + ((-2*a*(8*A *b + 5*a*B)*Sqrt[a + b*Tan[c + d*x]])/(3*d*Tan[c + d*x]^(3/2)) + ((-15*(-( (a*(a + I*b)^3*(I*A - B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a - b]*d)) + (a*(a - I*b)^3*(I*A + B)*ArcTanh[ (Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a + b]*d)))/a + (2*(15*a^2*A - 23*A*b^2 - 35*a*b*B)*Sqrt[a + b*Tan[c + d*x]])/ (d*Sqrt[Tan[c + d*x]]))/3)/5
3.5.47.3.1 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_))/((e_.) + (f_.)*(x _)), x_] :> With[{q = Denominator[m]}, Simp[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] && L tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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] && (Gt Q[a, 0] || LtQ[b, 0])
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_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[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]
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[(A + I*B)/2 Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 - I*T an[e + f*x]), x], x] + Simp[(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]
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] :> Simp[(A*b^2 - a*(b*B - a*C))*(a + b*Tan[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] + Simp[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)*Ta n[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] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 1.02 (sec) , antiderivative size = 2652302, normalized size of antiderivative = 10738.06
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 18802 vs. \(2 (201) = 402\).
Time = 5.06 (sec) , antiderivative size = 18802, normalized size of antiderivative = 76.12 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{7/2}} \,d x \]