\(\int \frac {A+B \tan (c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx\) [457]

Optimal result

Integrand size = 35, antiderivative size = 256 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\frac {(i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a-b} d}-\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a+b} d}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2 A-8 A b^2+10 a b B\right ) \sqrt {a+b \tan (c+d x)}}{15 a^3 d \sqrt {\tan (c+d x)}} \] Output:

(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)^(1/2)/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)^(1/2)/d-2/5*A*(a+b*tan(d*x+c))^(1/2)/a/d/tan(d*x+c)^(5/2) 
+2/15*(4*A*b-5*B*a)*(a+b*tan(d*x+c))^(1/2)/a^2/d/tan(d*x+c)^(3/2)+2/15*(15 
*A*a^2-8*A*b^2+10*B*a*b)*(a+b*tan(d*x+c))^(1/2)/a^3/d/tan(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 3.73 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.89 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\frac {-\frac {15 \sqrt [4]{-1} (A-i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}+\frac {15 \sqrt [4]{-1} (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}+\frac {2 \sqrt {a+b \tan (c+d x)} \left (-3 a^2 A-a (-4 A b+5 a B) \tan (c+d x)+\left (15 a^2 A-8 A b^2+10 a b B\right ) \tan ^2(c+d x)\right )}{a^3 \tan ^{\frac {5}{2}}(c+d x)}}{15 d} \] Input:

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

Output:

((-15*(-1)^(1/4)*(A - I*B)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + 
d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[-a + I*b] + (15*(-1)^(1/4)*(A + I*B 
)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + 
d*x]]])/Sqrt[a + I*b] + (2*Sqrt[a + b*Tan[c + d*x]]*(-3*a^2*A - a*(-4*A*b 
+ 5*a*B)*Tan[c + d*x] + (15*a^2*A - 8*A*b^2 + 10*a*b*B)*Tan[c + d*x]^2))/( 
a^3*Tan[c + d*x]^(5/2)))/(15*d)
 

Rubi [A] (verified)

Time = 1.70 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.10, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.457, Rules used = {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.

\(\displaystyle \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \tan (c+d x)}{\tan (c+d x)^{7/2} \sqrt {a+b \tan (c+d x)}}dx\)

\(\Big \downarrow \) 4092

\(\displaystyle -\frac {2 \int \frac {4 A b \tan ^2(c+d x)+5 a A \tan (c+d x)+4 A b-5 a B}{2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx}{5 a}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {4 A b \tan ^2(c+d x)+5 a A \tan (c+d x)+4 A b-5 a B}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx}{5 a}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \frac {4 A b \tan (c+d x)^2+5 a A \tan (c+d x)+4 A b-5 a B}{\tan (c+d x)^{5/2} \sqrt {a+b \tan (c+d x)}}dx}{5 a}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 4132

\(\displaystyle -\frac {-\frac {2 \int -\frac {15 A a^2+15 B \tan (c+d x) a^2+10 b B a-8 A b^2-2 b (4 A b-5 a B) \tan ^2(c+d x)}{2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx}{3 a}-\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {\int \frac {15 A a^2+15 B \tan (c+d x) a^2+10 b B a-8 A b^2-2 b (4 A b-5 a B) \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx}{3 a}-\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {\int \frac {15 A a^2+15 B \tan (c+d x) a^2+10 b B a-8 A b^2-2 b (4 A b-5 a B) \tan (c+d x)^2}{\tan (c+d x)^{3/2} \sqrt {a+b \tan (c+d x)}}dx}{3 a}-\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 4132

\(\displaystyle -\frac {\frac {-\frac {2 \int -\frac {15 \left (a^3 B-a^3 A \tan (c+d x)\right )}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {2 \left (15 a^2 A+10 a b B-8 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {\frac {15 \int \frac {a^3 B-a^3 A \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {2 \left (15 a^2 A+10 a b B-8 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {\frac {15 \int \frac {a^3 B-a^3 A \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {2 \left (15 a^2 A+10 a b B-8 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 4099

\(\displaystyle -\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {-\frac {2 \left (15 a^2 A+10 a b B-8 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}+\frac {15 \left (\frac {1}{2} a^3 (B+i A) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} a^3 (-B+i A) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )}{a}}{3 a}}{5 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {-\frac {2 \left (15 a^2 A+10 a b B-8 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}+\frac {15 \left (\frac {1}{2} a^3 (B+i A) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} a^3 (-B+i A) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )}{a}}{3 a}}{5 a}\)

\(\Big \downarrow \) 4098

\(\displaystyle -\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {-\frac {2 \left (15 a^2 A+10 a b B-8 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}+\frac {15 \left (\frac {a^3 (B+i A) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}-\frac {a^3 (-B+i A) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}\right )}{a}}{3 a}}{5 a}\)

\(\Big \downarrow \) 104

\(\displaystyle -\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {-\frac {2 \left (15 a^2 A+10 a b B-8 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}+\frac {15 \left (\frac {a^3 (B+i A) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}-\frac {a^3 (-B+i A) \int \frac {1}{\frac {(i a-b) \tan (c+d x)}{a+b \tan (c+d x)}+1}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}\right )}{a}}{3 a}}{5 a}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {-\frac {2 \left (15 a^2 A+10 a b B-8 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}+\frac {15 \left (\frac {a^3 (B+i A) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}-\frac {a^3 (-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}\right )}{a}}{3 a}}{5 a}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {-\frac {2 \left (15 a^2 A+10 a b B-8 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}+\frac {15 \left (\frac {a^3 (B+i A) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}}-\frac {a^3 (-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}\right )}{a}}{3 a}}{5 a}\)

Input:

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

Output:

(-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]*Sqrt[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)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 104

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]
 

rule 216

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])
 

rule 219

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])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4092

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])))
 

rule 4098

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]
 

rule 4099

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]
 

rule 4132

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])))
 

Maple [B] (warning: unable to verify)

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

Time = 2.08 (sec) , antiderivative size = 1892796, normalized size of antiderivative = 7393.73

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

Input:

int((A+B*tan(d*x+c))/tan(d*x+c)^(7/2)/(a+b*tan(d*x+c))^(1/2),x)
 

Output:

result too large to display
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 10125 vs. \(2 (210) = 420\).

Time = 2.62 (sec) , antiderivative size = 10125, normalized size of antiderivative = 39.55 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\text {Too large to display} \] Input:

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(7/2)/(a+b*tan(d*x+c))^(1/2),x, algo 
rithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

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

integrate((A+B*tan(d*x+c))/tan(d*x+c)**(7/2)/(a+b*tan(d*x+c))**(1/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(7/2)/(a+b*tan(d*x+c))^(1/2),x, algo 
rithm="maxima")
 

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(7/2)/(a+b*tan(d*x+c))^(1/2),x, algo 
rithm="giac")
 

Output:

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 Value
 

Mupad [F(-1)]

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

int((A + B*tan(c + d*x))/(tan(c + d*x)^(7/2)*(a + b*tan(c + d*x))^(1/2)),x 
)
 

Output:

int((A + B*tan(c + d*x))/(tan(c + d*x)^(7/2)*(a + b*tan(c + d*x))^(1/2)), 
x)
 

Reduce [F]

\[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}}{\tan \left (d x +c \right )^{4}}d x \] Input:

int((A+B*tan(d*x+c))/tan(d*x+c)^(7/2)/(a+b*tan(d*x+c))^(1/2),x)
 

Output:

int((sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*b + a))/tan(c + d*x)**4,x)