Integrand size = 35, antiderivative size = 296 \[ \int \frac {\cot ^{\frac {7}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {(i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {i a-b} d}-\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {i a+b} d}+\frac {2 \left (15 a^2 A-8 A b^2+10 a b B\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{15 a^3 d}+\frac {2 (4 A b-5 a B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 a^2 d}-\frac {2 A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 a d} \]
(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)*tan(d*x+c)^(1/2)/d/(I*a-b)^(1/2)-(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/(I*a+b)^(1/2)+2/15*(4*A*b-5*B*a)*cot(d*x+c)^(3/2)*(a+b*tan(d*x+c))^(1 /2)/a^2/d-2/5*A*cot(d*x+c)^(5/2)*(a+b*tan(d*x+c))^(1/2)/a/d+2/15*(15*A*a^2 -8*A*b^2+10*B*a*b)*cot(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2)/a^3/d
Time = 6.31 (sec) , antiderivative size = 444, normalized size of antiderivative = 1.50 \[ \int \frac {\cot ^{\frac {7}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (-\frac {(A-i B) \sqrt {a+b \tan (c+d x)}}{5 a \tan ^{\frac {5}{2}}(c+d x)}-\frac {(A+i B) \sqrt {a+b \tan (c+d x)}}{5 a \tan ^{\frac {5}{2}}(c+d x)}+\frac {(A+i B) \left (\frac {(5 i a+4 b) \sqrt {a+b \tan (c+d x)}}{a \tan ^{\frac {3}{2}}(c+d x)}-\frac {i \left (\frac {15 (-1)^{3/4} a^2 \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}+\frac {\left (15 i a^2+10 a b-8 i b^2\right ) \sqrt {a+b \tan (c+d x)}}{a \sqrt {\tan (c+d x)}}\right )}{a}\right )}{15 a}-\frac {(A-i B) \left (\frac {\left (5 i-\frac {4 b}{a}\right ) \sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {3}{2}}(c+d x)}-\frac {i \left (\frac {15 (-1)^{3/4} a^2 \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}+\frac {\left (10 a b-i \left (15 a^2-8 b^2\right )\right ) \sqrt {a+b \tan (c+d x)}}{a \sqrt {\tan (c+d x)}}\right )}{a}\right )}{15 a}\right )}{d} \]
(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(-1/5*((A - I*B)*Sqrt[a + b*Tan[c + d*x]])/(a*Tan[c + d*x]^(5/2)) - ((A + I*B)*Sqrt[a + b*Tan[c + d*x]])/(5*a *Tan[c + d*x]^(5/2)) + ((A + I*B)*((((5*I)*a + 4*b)*Sqrt[a + b*Tan[c + d*x ]])/(a*Tan[c + d*x]^(3/2)) - (I*((15*(-1)^(3/4)*a^2*ArcTan[((-1)^(1/4)*Sqr t[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[a + I*b] + (((15*I)*a^2 + 10*a*b - (8*I)*b^2)*Sqrt[a + b*Tan[c + d*x]])/(a*Sqrt[Tan[c + d*x]])))/a))/(15*a) - ((A - I*B)*(((5*I - (4*b)/a)*Sqrt[a + b*Tan[c + d *x]])/Tan[c + d*x]^(3/2) - (I*((15*(-1)^(3/4)*a^2*ArcTan[((-1)^(1/4)*Sqrt[ -a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[-a + I*b] + ((10*a*b - I*(15*a^2 - 8*b^2))*Sqrt[a + b*Tan[c + d*x]])/(a*Sqrt[Tan[c + d *x]])))/a))/(15*a)))/d
Time = 1.85 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.02, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.514, Rules used = {3042, 4729, 3042, 4092, 27, 3042, 4132, 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.
\(\displaystyle \int \frac {\cot ^{\frac {7}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cot (c+d x)^{7/2} (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}}dx\) |
\(\Big \downarrow \) 4729 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {A+B \tan (c+d x)}{\tan (c+d x)^{7/2} \sqrt {a+b \tan (c+d x)}}dx\) |
\(\Big \downarrow \) 4092 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2 \int \frac {4 A b \tan ^2(c+d x)+5 a A \tan (c+d x)+4 A b-5 a B}{2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx}{5 a}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {\int \frac {4 A b \tan ^2(c+d x)+5 a A \tan (c+d x)+4 A b-5 a B}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx}{5 a}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {\int \frac {4 A b \tan (c+d x)^2+5 a A \tan (c+d x)+4 A b-5 a B}{\tan (c+d x)^{5/2} \sqrt {a+b \tan (c+d x)}}dx}{5 a}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 4132 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {-\frac {2 \int -\frac {15 A a^2+15 B \tan (c+d x) a^2+10 b B a-8 A b^2-2 b (4 A b-5 a B) \tan ^2(c+d x)}{2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx}{3 a}-\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {\frac {\int \frac {15 A a^2+15 B \tan (c+d x) a^2+10 b B a-8 A b^2-2 b (4 A b-5 a B) \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx}{3 a}-\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {\frac {\int \frac {15 A a^2+15 B \tan (c+d x) a^2+10 b B a-8 A b^2-2 b (4 A b-5 a B) \tan (c+d x)^2}{\tan (c+d x)^{3/2} \sqrt {a+b \tan (c+d x)}}dx}{3 a}-\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 4132 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {\frac {-\frac {2 \int -\frac {15 \left (a^3 B-a^3 A \tan (c+d x)\right )}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {2 \left (15 a^2 A+10 a b B-8 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {\frac {\frac {15 \int \frac {a^3 B-a^3 A \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {2 \left (15 a^2 A+10 a b B-8 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {\frac {\frac {15 \int \frac {a^3 B-a^3 A \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {2 \left (15 a^2 A+10 a b B-8 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 4099 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {-\frac {2 \left (15 a^2 A+10 a b B-8 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}+\frac {15 \left (\frac {1}{2} a^3 (B+i A) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} a^3 (-B+i A) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )}{a}}{3 a}}{5 a}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {-\frac {2 \left (15 a^2 A+10 a b B-8 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}+\frac {15 \left (\frac {1}{2} a^3 (B+i A) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} a^3 (-B+i A) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )}{a}}{3 a}}{5 a}\right )\) |
\(\Big \downarrow \) 4098 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {-\frac {2 \left (15 a^2 A+10 a b B-8 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}+\frac {15 \left (\frac {a^3 (B+i A) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}-\frac {a^3 (-B+i A) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}\right )}{a}}{3 a}}{5 a}\right )\) |
\(\Big \downarrow \) 104 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {-\frac {2 \left (15 a^2 A+10 a b B-8 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}+\frac {15 \left (\frac {a^3 (B+i A) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}-\frac {a^3 (-B+i A) \int \frac {1}{\frac {(i a-b) \tan (c+d x)}{a+b \tan (c+d x)}+1}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}\right )}{a}}{3 a}}{5 a}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {-\frac {2 \left (15 a^2 A+10 a b B-8 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}+\frac {15 \left (\frac {a^3 (B+i A) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}-\frac {a^3 (-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}\right )}{a}}{3 a}}{5 a}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {-\frac {2 \left (15 a^2 A+10 a b B-8 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}+\frac {15 \left (\frac {a^3 (B+i A) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}}-\frac {a^3 (-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}\right )}{a}}{3 a}}{5 a}\right )\) |
Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((-2*A*Sqrt[a + b*Tan[c + d*x]])/(5* a*d*Tan[c + d*x]^(5/2)) - ((-2*(4*A*b - 5*a*B)*Sqrt[a + b*Tan[c + d*x]])/( 3*a*d*Tan[c + d*x]^(3/2)) + ((15*(-((a^3*(I*A - B)*ArcTan[(Sqrt[I*a - b]*S qrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a - b]*d)) + (a^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 - 8*A*b^2 + 10*a*b*B)*Sqrt[a + b* Tan[c + d*x]])/(a*d*Sqrt[Tan[c + d*x]]))/(3*a))/(5*a))
3.7.36.3.1 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*(A*b - a*B)*(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[b* B*(b*c*(m + 1) + a*d*(n + 1)) + A*(a*(b*c - a*d)*(m + 1) - b^2*d*(m + n + 2 )) - (A*b - a*B)*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b*d*(A*b - a*B)*(m + n + 2)*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] && LtQ[m, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) && !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 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^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*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])))
Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m Int[ActivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownTangentIntegrandQ[u, x]
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 2.07 (sec) , antiderivative size = 1890883, normalized size of antiderivative = 6388.12
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 10265 vs. \(2 (242) = 484\).
Time = 2.93 (sec) , antiderivative size = 10265, normalized size of antiderivative = 34.68 \[ \int \frac {\cot ^{\frac {7}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {\cot ^{\frac {7}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {\cot ^{\frac {7}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \cot \left (d x + c\right )^{\frac {7}{2}}}{\sqrt {b \tan \left (d x + c\right ) + a}} \,d x } \]
Exception generated. \[ \int \frac {\cot ^{\frac {7}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Valuesym2poly/r2sym(con st gen &
Timed out. \[ \int \frac {\cot ^{\frac {7}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\int \frac {{\mathrm {cot}\left (c+d\,x\right )}^{7/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}} \,d x \]